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Hindawi Publishing CorporationInternational Journal of Differential EquationsVolume 2013 Article ID 297085 11 pageshttpdxdoiorg1011552013297085
Research ArticleOscillations of a Class of Forced Second-Order DifferentialEquations with Possible Discontinuous Coefficients
Siniša MiliIiT Mervan PašiT and Darko CubriniT
Department of Mathematics Faculty of Electrical Engeneering and Computing University of Zagreb 10000 Zagreb Croatia
Correspondence should be addressed to Mervan Pasic mervanpasicgmailcom
Received 29 January 2013 Accepted 16 April 2013
Academic Editor Ondrej Dosly
Copyright copy 2013 Sinisa Milicic et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We study the oscillation of all solutions of a general class of forced second-order differential equations where their second derivativeis not necessarily a continuous function and the coefficients of the main equation may be discontinuous Our main results are notincluded in the previously published known oscillation criteria of interval type Many examples and consequences are presentedillustrating the main results
1 Introduction
Let 1199050gt 0 and let 119860119862loc([1199050infin)R) denote the set of all real
functions absolutely continuous on every bounded interval[119886 119887] sub [119905
0infin) We study the oscillatory behaviour of all
solutions 119909 = 119909(119905) of the following class of forced second-order differential equations
(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
+ 119902 (119905) 119891 (119909 (119905)) = 119890 (119905)
ae in [1199050infin)
119909 119903Φ (119909 1199091015840) isin 119860119862loc ([1199050infin) R)
(1)
where the functions Φ R2 rarr R Φ = Φ(119906 V) 119891
R rarr R and 119891 = 119891(119906) satisfy some general conditionsgiven in Section 2 A continuous function 119909 = 119909(119905) is saidto be oscillatory if there is a sequence 119905
119899isin [1199050infin) such that
119909(119905119899) = 0 for all 119899 isin N and 119905
119899rarr infin as 119899 rarr infin A differential
equation is oscillatory if all its solutions are oscillatoryThe forcing term 119890(119905) is a sign-changing function (possi-
bly discontinuous) This can be formulated by the followinghypothesis for every 119879 ge 119905
0there exist two intervals (119886
1 1198871)
and (1198862 1198872) 119879 le 119886
1lt 1198871le 1198862lt 1198872 such that
119890 (119905) ge 0 119905 isin (1198861 1198871)
119890 (119905) le 0 119905 isin (1198862 1198872)
(2)
The coefficient 119902(119905) may be a discontinuous function on[1199050infin) and the case 119909 notin 119862
2((1199050infin)R) occurs in our
main results and examples too Two important classes offunctions Φ(119906 V) are included in the differential operator(119903(119905)Φ(119909 119909
1015840))1015840 as
Φ (119906 V) = 120601 (119906) V Φ (119906 V) =120601 (119906) V
radic1 + V2 (119906 V) isin R
2 (3)
The first one is the classic second-order differential operatorwhich is linear in 119909
1015840 and the second one is the so-calledone-dimensional mean curvature differential operator seeExamples 1 and 2
Depending on 119902(119905) we propose the following four simplemodels for (1)
(i) 119902(119905) is strictly positive and continuous on [1199050infin) as
11990910158401015840+ 41198982119891 (119909) = ℎ (sin (119898119905)) ae in [119905
0infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862
2((1199050infin) R)
(4)
(ii) 119902(119905) is nonnegative and continuous on [1199050infin) as
11990910158401015840+ 11989821205872[cos (119898119905)]+119891 (119909) = ℎ (sin (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(5)
2 International Journal of Differential Equations
5 10 15
051
minus1
minus05
Figure 1 Function 119909(119905) = |sin(119905)| sin(119905) is a solution of (4)
5 10 15
051
minus1
minus05
Figure 21199091015840(119905) = 2|sin(119905)| cos(119905) and hence11990910158401015840(119905) is not a continuousfunction
(iii) 119902(119905) is nonnegative and discontinuous on [1199050infin) as
11990910158401015840+ 41198982[sign (cos (119898119905))]+119891 (119909) = ℎ (sin (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(6)
(iv) 119902(119905) is sign changing and discontinuous on [1199050infin) as
11990910158401015840+ 41198982 sign (cos (119898119905)) 119891 (119909) = ℎ (sin (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(7)
where 119898 isin N and ℎ(119904) is an arbitrary function such thatℎ(119904) 119904 gt 0 for all 119904 = 0 for instance ℎ(119904) = 119904 or ℎ(119904) = sign(119904)According to Corollaries 7 and 10 we will show that (4)ndash(7) are oscillatory provided the function 119891 = 119891(119906) satisfies119891(119906)119906 ge 119870 ge 1 for all 119906 = 0 see Examples 8ndash13 It isinteresting that in particular for 119891(119906) = 119906 and ℎ(119904) =
21198982 sign(119904) (4) allows an explicit oscillatory solution 119909(119905) =
|sin(119898119905)| sin(119898119905) as shown in Figures 1 and 2Moreover as a consequence of Corollary 7 one can
show that all solutions of (4) are oscillatory for details seeExample 8 The main goal of this paper is to give somesufficient conditions on functions Φ(119906 V) 119891(119906) and thecoefficients 119903(119905) 119902(119905) and 119890(119905) such that (1) is oscillatory seeTheorems 3 and 4 It will also cover the model equations (4)ndash(7) as well as some other examples presented in Section 2
To the best of our knowledge it seems that there are onlyfew papers which study the oscillation of the second-orderdifferential equations with nonsmooth (local integrable)coefficients see [1ndash3]More precisely in [1] the author studiedthe interval oscillation criteria for the following second-orderhalf-linear differential equation
(119903 (119905)100381610038161003816100381610038161199091015840(119905)10038161003816100381610038161003816
120590minus1
1199091015840(119905))
1015840
+ 119902 (119905) |119909 (119905)|120590minus1
119909 (119905) = 0
ae in (0infin)
119909 11990310038161003816100381610038161003816119909101584010038161003816100381610038161003816
120590minus1
1199091015840isin 119860119862loc ((0infin) R)
(8)
where 120590 gt 1 and 1119903 119902 isin 119871 loc((0infin)R) such thatintinfin
0119903minus1120590
(119905)119889119903 = infin See also [2] but with the solutionspace1198621((0infin)R) instead of119860119862loc((0infin)R) that is 119909 and119903|1199091015840|120590minus1
1199091015840isin 1198621((0infin)R)
In [3] the authors consider the following second-orderdifferential equation
(119903 (119905) 1199091015840(119905))1015840
+ 119876 (119905 119909 (119905) 1199091015840(119905)) = 0 119905 ge 119905
0 (9)
where 119903(119905) gt 0 ae in [1199050infin) 1119903 isin 119871 loc([1199050infin)R) and
119876(119905 119910 119911) is locally integrable function in 119905 and continuous in(119910 119911) Equation (9) allows the forcing term 119890(119905) in the nextsense as follows
119910119876 (119905 119910 119911) ge 119902 (119905) 119910119891 (119910) minus 119890 (119905) 119910
forall (119905 119910 119911) isin [1199050infin) timesR
2
(10)
where 119890 = 119890(119905) satisfies (2) but the functions 119902 = 119902(119905) and119891 = 119891(119910) are smooth enough in their variables that is 119902 isin
119862([1199050infin)R) and 119891 isin 119862
1(RR)
On certain oscillation criteria for various classes offorced second-order differential equations with continuouscoefficients we refer the reader to [4ndash13] Our methodmodifies a recently used one in [14 15] and it containsthe classic Riccati transformation of the main equation ablow-up argument and pointwise comparison principle Thecomparison principle applies to all sub- and supersolutions ofa class of the generalized Riccati differential equations withnonlinear terms that are supposed to be locally integrablein the first variable and locally Lipschitz continuous in thesecond variable
2 Hypotheses Results and Consequences
First of all the functionΦ(119906 V)which appears in the second-order differential operator of (1) satisfies
|119906|120574minus2
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) forall119906 V isin R (11)
where 120574 ge 2 and 119892 R+rarr R+is a locally Lipschitz function
1198920 R+rarr R+satisfying
119892 (119888119904) ge 1198881205741198920 (119904) forall119888 gt 0 119904 gt 0
1198920(119904) + 119872
0ge 1199042 for some 119872
0ge 0 and all 119904 isin R
+
(12)
In most cases 119892(119904) = 1198920(119904)1198921(119904) 119904 gt 0 where 119892
0(119904) = 119904
120574120574 ge 2 and 119892
1(119904) is an arbitrary function satisfying 119892
1(119904) ge 1
Thus for such 119892(119904) with 1198921(119904) equiv 1 condition (11) became
|119906|120574minus2
VΦ (119906 V) ge |Φ (119906 V)|120574
forall119906 V isin R (13)
It is not difficult to check that if 1198920isin 1198621(R+) or 119892
0(119904) is a
convex function then it is locally Lipschitz onR+too see for
instance [16 Theorem 133]Two essential classes of the second-order differential
operators (119903(119905)Φ(119909 1199091015840))1015840 satisfy condition (13) as is shown inthe next examples
International Journal of Differential Equations 3
Example 1 We consider the second-order differential opera-tor which is linear in 1199091015840 as follows
(119903 (119905) Φ (119909 1199091015840))1015840
= 120572(120601 (119909) 1199091015840)1015840
(14)
where 120572 gt 0 and 0 le 120601(119906) le 1 for all 119906 isin R Obviously thefunctionΦ(119906 V) = 120601(119906)V satisfies condition (13) in particularfor 120574 = 2 Two usual choices for 120601(119906) are 120601(119906) = |sin 119906| and120601(119906) = |119906|(1 + |119906|)
Example 2 We consider a quasilinear differential operator(the so-called one-dimensional prescribed mean curvatureoperator) as follows
(119903 (119905) Φ (119909 1199091015840))1015840
= 120572(120601 (119909)1199091015840
radic1 + 11990910158402)
1015840
(15)
where 120572 gt 0 and 0 le 120601120574minus1(119906) le |119906|
120574minus2 for all 119906 isin R It is notdifficult to check that condition (13) is satisfied in particularfor Φ(119906 V) = 120601(119906)V(1 + V2)12 and for any 120574 gt 1 For 120601(119906)we can take the same choice as in the previous example
Next we suppose the existence of a constant119870 such that
119891 (119906)
119906ge 119870 gt 0 forall119906 = 0 (16)
In order to simplify our consideration here inmany exampleswe often use 119891(119906) = 119870119906
Condition (2) means that there exists a sequence of pairsof intervals 119869
1119895= [1198861119895 1198871119895] and 119869
2119895= [1198862119895 1198872119895] 119895 isin N
contained in (1199050infin) such that the sequences (119886
1119895)119895ge1
(1198871119895)119895ge1
(1198862119895)119895ge1
and (1198872119895)119895ge1
are increasing 1198861119895lt 1198871119895le 1198862119895lt 1198872119895for
each 119895 and
119890 (119905) ge 0 on 1198691119895 119890 (119905) le 0 on 119869
2119895
for each 119895 isin N
lim119895rarrinfin
1198861119895= infin
(17)
On the intervals 1198691119895and 1198692119895 the coefficient 119903(119905) satisfies
119903 (119905) gt 0 on 119869119894119895 1199031minus120574
isin 1198711(119869119894119895)
forall119894 isin 1 2 119895 isin N(18)
Let there be a real function 119862 = 119862(119905) 119862 isin
1198711
loc((1199050infin)R) and let there exist a sequence of positive realnumbers (120582
119895)119895isinN such that
119862 (119905) ge 0 on 119869119894119895 119888119894119895= int
119869119894119895
119862 (120591) 119889120591 gt 0
forall119894 isin 1 2 119895 isin N
1
119888119894119895
119862 (119905) le1
120587min(120582
119895119903 (119905))1minus120574
119870
1198720+ 1
120582119895119902 (119905)
forall119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(19)
where 120574 1198720 and 119870 are constants defined in (11) (12) and
(16) respectivelyThe proof of the following main result will be presented
in Section 4
Theorem 3 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)
satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0
and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (19) is
fulfilled then (1) is oscillatory
Condition (19) can be replaced by an equivalent onewhich has a more practical value and takes a simpler formsince we do not need a sequence of auxiliary parameters(120582119895)119895isinN let there be a real function 119862 = 119862(119905) 119862 isin
1198711
loc((1199050infin)R) such that
119862 (119905) ge 0 on 119869119894119895 119888119894119895= int
119868119894119895
119862 (120591) 119889120591 gt 0 forall119894 isin 1 2
119895 isin N
sup119905isin119869119894119895
[119862 (119905)
119902 (119905)] sup119905isin119869119894119895
[119903 (119905) 119862(119905)1(120574minus1)
] le119870
1198720+ 1
(
119888119894119895
120587)
120574(120574minus1)
forall119894 isin 1 2 119895 isin N
(20)
where 120574 1198720 and 119870 are constants defined in (11) (12) and
(16) respectively Since we will show that (19) and (20)are equivalent see page 8 the next oscillation criterionimmediately follows fromTheorem 3
Theorem 4 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)
satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0
and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (20) is
fulfilled then (1) is oscillatory
Remark 5 Assuming that 119871 = lim119895rarrinfin
1198861119895
lt infin wecan ensure the oscillation in the point 119871 Note that 119871 =
lim119895rarrinfin
1198862119895
since 1198861119895
lt 1198862119895
lt 1198861119895+1
Thus we can generatea one-sided (right) limit
Now we consider some consequences of Theorem 4which depend on the qualitative properties of the coefficient119902(119905)
Substituting 119862(119905) equiv 1 in (20) Theorem 4 implies thefollowing result involving lower bounds on the lengths ofintervals |119869
119894119895| = 119887119894119895minus 119886119894119895
Corollary 6 (119902(119905) is positive) Let the functionsΦ(119906 V) 119891(119906)119890(119905) and 119903(119905) satisfy (11) (12) (16) (17) and (18) respectivelyLet inf
119905isin119869119894119895119902(119905) gt 0 for each 119894 isin 1 2 119895 isin N If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
119870inf119905isin119869119894119895
119902 (119905))
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(21)
then (1) is oscillatory
4 International Journal of Differential Equations
Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903
0 1199020satisfying
0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902
0gt 0 forall119894 isin 1 2 119895 isin N (22)
If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N (23)
then (1) is oscillatory where 1205741198720 and119870 are constants defined
in (11) (12) and (16) respectively
Example 8 (oscillation of (4)) We know that 119909(119905) =
|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892
0(119904) = 119904
2
and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is
satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982
it is clear that conditions (18) and (22) are also satisfied inparticular for 119903
0= 1 and 119902
0= 4119898
2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887
1119895= (2119895 + 1)120587119898 = 119886
2119895and 1198872119895= (2119895 + 2)120587119898
Moreover10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816= 119887119894minus 119886119894=120587
119898gt 0 119894 isin 1 2 (24)
Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=120587
119898ge
120587
radic41198982
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(25)
Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory
Example 9 We consider the following class of equations
11990910158401015840minus 2
11987810158401015840(119905)
119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840
2
(119905) ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862
2((1199050infin) R)
(26)
where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905
119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905
119899rarr infin
there is a 1205910isin R such that 119905
119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and
119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given
oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904
0gt 0 such that
minus211987810158401015840(119905)
119878 (119905)ge 1199040 119905 isin (119905
119899 119905119899+1) 120591
0ge
120587
radic1199040
(27)
then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872
0= 0 119903
0= 1 119902
0= 1199040 and
119870 ge 1 that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1
minus 119905119895
10038161003816100381610038161003816ge 1205910ge
120587
radic1199040
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(28)
Hence Corollary 7 proves this result
As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0
Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869
119894119895
119894 isin 1 2 119895 isin N such that
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)
If
119902120574
119894119895
119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)
119870)
120574minus1
119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N
(30)
then (1) is oscillatory
Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582
119895such that
120582119895ge120587 (1198720+ 1)
119870
1
119902119894119895
1
119902119894119895
119902 (119905) le1
120587(120582119895119903 (119905))1minus120574
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(31)
The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869
119894119895 119902 = 0 equivalent to
120587 (1198720+ 1)
119870
1
119902119894119895
le 120582119895le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (32)
that is to
120587 (1198720+ 1)
119870
1
119902119894119895
le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (33)
This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second
inequality in (31) is trivially satisfied
International Journal of Differential Equations 5
The following three examples are simple consequences ofCorollary 10
Example 11 (oscillation of (5)) Equation (5) as well asequation
11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(34)
are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =
11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are
fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2
119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and
1198861119895=2119895120587
119898 119887
1119895=(2119895 + 12) 120587
119898
1198862119895=(2119895 + 32) 120587
119898 119887
2119895=(2119895 + 2) 120587
119898
(35)
Moreover
119902119894119895= int
119869119894119895
119902 (120591) 119889120591
= 11989821205872int
119887119894119895
119886119894119895
[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N
(36)
and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582
119895= 1(119898120587)
we have
120582119895=
1
119898120587ge1
119870
1
119898120587=120587 (1198720+ 1)
119870
1
1198981205872
=120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894119895
119902 (119905) =1
119898120587211989821205872[cos (119898119905)]+ le 119898
=1
120587
1
(1119898120587)=1
120587
1
(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(37)
Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too
In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905
0infin)
Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge
1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898
2 sign(cos(119898119905)) on 119869119894119895 120582119895=
1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872
0= 0
Indeed
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 = 41198982int
119869119894119895
[sign (cos (119898120591))]+119889120591
= 41198982 120587
2119898= 2119898120587 119894 isin 1 2 119895 isin N
120582119895=
1
2119898=
120587
2119898120587= 120587 (119872
0+ 1)
1
119902119894119895
ge120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894
119902 (119905) =1
211989812058741198982[sign (cos (119898119905))]+
le2119898
120587=1
120587
1
radic1 (41198982)
=1
120587(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(38)
which shows that (30) is satisfied
At the end of this section we consider the case when 119902(119905)changes sign on [119905
0infin) but with the help of Corollary 10 since
119902(119905) is strictly positive on all intervals 119869119894119895
Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge
1 then model equation (7) is oscillatory In fact let 119869119894119895be as
in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869
1119895and 119890(119905) le 0 on 119869
2119895 On
the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869
119894119895and
so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence
the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example
3 Proofs of Theorems 3 and 4
Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879
0lt 119879lowast and functions
119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation
1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879
0 119879lowast) (39)
which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+
119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) which satisfy respectively
1199081015840le 119866 (119905 119908) + 119887 (119905) 119908
1015840ge 119866 (119905 119908) + 119887 (119905)
ae in [1198790 119879lowast)
(40)
We are interested in studying the following property
119908 (1198790) le 119908 (119879
0) implies 119908 (119905) le 119908 (119905)
forall119905 isin [T0 119879lowast)
(41)
In this way we introduce the following definition
6 International Journal of Differential Equations
Definition 14 We say that comparison principle holds for (39)on an interval [119879
0 119879lowast) 119879 le 119879
0lt 119879lowast if property (41)
is fulfilled for all sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) of (39) on [119879
0 119879lowast)
Now we are able to state the next result
Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be
two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879
0 119879lowast119872 such that
119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871
1(1198790 119879lowast)
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816le 119871 (119905)
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]
(42)
Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879
0lt 119879lowast
Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader
Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally
Lipschitz function on R+ Let 119870 be an arbitrary real number
and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0
on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879
0 119879lowast) then comparison principle
(41) holds for the Riccati differential equation
1199081015840=
1
(120582119903 (119905))120574minus1
1198920 (|119908|) + 119870120582119902 (119905) ae in [119879
0 119879lowast)
(43)
Proof It is clear that (43) is a particular case of (39) inparticular for
119866 (119905 119906) =1
(120582119903 (119905))120574minus1
1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)
Since1198920is a locally Lipschitz function onR
+ for every119872 gt 0
there is an 1198710gt 0 depending on119872 such that
10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710
10038161003816100381610038161199041minus 1199042
1003816100381610038161003816 forall119904
1 1199041isin [minus119872119872]
(45)
Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816=
1
(120582119903 (119905))120574minus1
10038161003816100381610038161198920(10038161003816100381610038161199061
1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062
1003816100381610038161003816)1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
1003816100381610038161003816
10038161003816100381610038161199061
1003816100381610038161003816minus10038161003816100381610038161199062
1003816100381610038161003816
1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
(46)
Thus according to assumption 119903minus1+120574
isin 1198711(1198790 119879lowast) the
required condition (42) is fulfilled in particular for 119871(119905) =
1198710(120582119903(119905))
minus120574+1 and so Lemma 15 proves this corollary
Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886
1 1198871) or (119886
2 1198872) In the second one we
construct two subsolutions 1199081(119905) and 119908
2(119905) of (43) which
blow up on intervals (1198861 119879lowast
1) sube (119886
1 1198871) and (119886
2 119879lowast
2) sube (119886
2 1198872)
respectively
Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886
1 1198871] cup [119886
2 1198872] Let 119909 = 119909(119905) be a
nonoscillatory solution of (1) and let for some 119879 ge 1199050and
120582 gt 0 the function 119908 = 119908(119905) be defined by
119908 (119905) = minus
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
119909 (119905) 119905 ge 119879 (47)
Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)
where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0
Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905
0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well
defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where
119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0 and since
120582 gt 0 we have
minus120582119890 (119905)
119909 (119905)ge 0 in 119869 (49)
Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from
119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
119909 (119905)
+
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
1199092(119905)
1199091015840(119905)
=120582119903 (119905)
|119909 (119905)|120574Φ(119909 (119905) 119909
1015840(119905)) 1199091015840(119905) |119909 (119905)|
120574minus2
+ 120582119902 (119905)119891 (119909 (119905))
119909 (119905)minus 120582
119890 (119905)
119909 (119905)
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
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Stochastic AnalysisInternational Journal of
2 International Journal of Differential Equations
5 10 15
051
minus1
minus05
Figure 1 Function 119909(119905) = |sin(119905)| sin(119905) is a solution of (4)
5 10 15
051
minus1
minus05
Figure 21199091015840(119905) = 2|sin(119905)| cos(119905) and hence11990910158401015840(119905) is not a continuousfunction
(iii) 119902(119905) is nonnegative and discontinuous on [1199050infin) as
11990910158401015840+ 41198982[sign (cos (119898119905))]+119891 (119909) = ℎ (sin (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(6)
(iv) 119902(119905) is sign changing and discontinuous on [1199050infin) as
11990910158401015840+ 41198982 sign (cos (119898119905)) 119891 (119909) = ℎ (sin (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(7)
where 119898 isin N and ℎ(119904) is an arbitrary function such thatℎ(119904) 119904 gt 0 for all 119904 = 0 for instance ℎ(119904) = 119904 or ℎ(119904) = sign(119904)According to Corollaries 7 and 10 we will show that (4)ndash(7) are oscillatory provided the function 119891 = 119891(119906) satisfies119891(119906)119906 ge 119870 ge 1 for all 119906 = 0 see Examples 8ndash13 It isinteresting that in particular for 119891(119906) = 119906 and ℎ(119904) =
21198982 sign(119904) (4) allows an explicit oscillatory solution 119909(119905) =
|sin(119898119905)| sin(119898119905) as shown in Figures 1 and 2Moreover as a consequence of Corollary 7 one can
show that all solutions of (4) are oscillatory for details seeExample 8 The main goal of this paper is to give somesufficient conditions on functions Φ(119906 V) 119891(119906) and thecoefficients 119903(119905) 119902(119905) and 119890(119905) such that (1) is oscillatory seeTheorems 3 and 4 It will also cover the model equations (4)ndash(7) as well as some other examples presented in Section 2
To the best of our knowledge it seems that there are onlyfew papers which study the oscillation of the second-orderdifferential equations with nonsmooth (local integrable)coefficients see [1ndash3]More precisely in [1] the author studiedthe interval oscillation criteria for the following second-orderhalf-linear differential equation
(119903 (119905)100381610038161003816100381610038161199091015840(119905)10038161003816100381610038161003816
120590minus1
1199091015840(119905))
1015840
+ 119902 (119905) |119909 (119905)|120590minus1
119909 (119905) = 0
ae in (0infin)
119909 11990310038161003816100381610038161003816119909101584010038161003816100381610038161003816
120590minus1
1199091015840isin 119860119862loc ((0infin) R)
(8)
where 120590 gt 1 and 1119903 119902 isin 119871 loc((0infin)R) such thatintinfin
0119903minus1120590
(119905)119889119903 = infin See also [2] but with the solutionspace1198621((0infin)R) instead of119860119862loc((0infin)R) that is 119909 and119903|1199091015840|120590minus1
1199091015840isin 1198621((0infin)R)
In [3] the authors consider the following second-orderdifferential equation
(119903 (119905) 1199091015840(119905))1015840
+ 119876 (119905 119909 (119905) 1199091015840(119905)) = 0 119905 ge 119905
0 (9)
where 119903(119905) gt 0 ae in [1199050infin) 1119903 isin 119871 loc([1199050infin)R) and
119876(119905 119910 119911) is locally integrable function in 119905 and continuous in(119910 119911) Equation (9) allows the forcing term 119890(119905) in the nextsense as follows
119910119876 (119905 119910 119911) ge 119902 (119905) 119910119891 (119910) minus 119890 (119905) 119910
forall (119905 119910 119911) isin [1199050infin) timesR
2
(10)
where 119890 = 119890(119905) satisfies (2) but the functions 119902 = 119902(119905) and119891 = 119891(119910) are smooth enough in their variables that is 119902 isin
119862([1199050infin)R) and 119891 isin 119862
1(RR)
On certain oscillation criteria for various classes offorced second-order differential equations with continuouscoefficients we refer the reader to [4ndash13] Our methodmodifies a recently used one in [14 15] and it containsthe classic Riccati transformation of the main equation ablow-up argument and pointwise comparison principle Thecomparison principle applies to all sub- and supersolutions ofa class of the generalized Riccati differential equations withnonlinear terms that are supposed to be locally integrablein the first variable and locally Lipschitz continuous in thesecond variable
2 Hypotheses Results and Consequences
First of all the functionΦ(119906 V)which appears in the second-order differential operator of (1) satisfies
|119906|120574minus2
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) forall119906 V isin R (11)
where 120574 ge 2 and 119892 R+rarr R+is a locally Lipschitz function
1198920 R+rarr R+satisfying
119892 (119888119904) ge 1198881205741198920 (119904) forall119888 gt 0 119904 gt 0
1198920(119904) + 119872
0ge 1199042 for some 119872
0ge 0 and all 119904 isin R
+
(12)
In most cases 119892(119904) = 1198920(119904)1198921(119904) 119904 gt 0 where 119892
0(119904) = 119904
120574120574 ge 2 and 119892
1(119904) is an arbitrary function satisfying 119892
1(119904) ge 1
Thus for such 119892(119904) with 1198921(119904) equiv 1 condition (11) became
|119906|120574minus2
VΦ (119906 V) ge |Φ (119906 V)|120574
forall119906 V isin R (13)
It is not difficult to check that if 1198920isin 1198621(R+) or 119892
0(119904) is a
convex function then it is locally Lipschitz onR+too see for
instance [16 Theorem 133]Two essential classes of the second-order differential
operators (119903(119905)Φ(119909 1199091015840))1015840 satisfy condition (13) as is shown inthe next examples
International Journal of Differential Equations 3
Example 1 We consider the second-order differential opera-tor which is linear in 1199091015840 as follows
(119903 (119905) Φ (119909 1199091015840))1015840
= 120572(120601 (119909) 1199091015840)1015840
(14)
where 120572 gt 0 and 0 le 120601(119906) le 1 for all 119906 isin R Obviously thefunctionΦ(119906 V) = 120601(119906)V satisfies condition (13) in particularfor 120574 = 2 Two usual choices for 120601(119906) are 120601(119906) = |sin 119906| and120601(119906) = |119906|(1 + |119906|)
Example 2 We consider a quasilinear differential operator(the so-called one-dimensional prescribed mean curvatureoperator) as follows
(119903 (119905) Φ (119909 1199091015840))1015840
= 120572(120601 (119909)1199091015840
radic1 + 11990910158402)
1015840
(15)
where 120572 gt 0 and 0 le 120601120574minus1(119906) le |119906|
120574minus2 for all 119906 isin R It is notdifficult to check that condition (13) is satisfied in particularfor Φ(119906 V) = 120601(119906)V(1 + V2)12 and for any 120574 gt 1 For 120601(119906)we can take the same choice as in the previous example
Next we suppose the existence of a constant119870 such that
119891 (119906)
119906ge 119870 gt 0 forall119906 = 0 (16)
In order to simplify our consideration here inmany exampleswe often use 119891(119906) = 119870119906
Condition (2) means that there exists a sequence of pairsof intervals 119869
1119895= [1198861119895 1198871119895] and 119869
2119895= [1198862119895 1198872119895] 119895 isin N
contained in (1199050infin) such that the sequences (119886
1119895)119895ge1
(1198871119895)119895ge1
(1198862119895)119895ge1
and (1198872119895)119895ge1
are increasing 1198861119895lt 1198871119895le 1198862119895lt 1198872119895for
each 119895 and
119890 (119905) ge 0 on 1198691119895 119890 (119905) le 0 on 119869
2119895
for each 119895 isin N
lim119895rarrinfin
1198861119895= infin
(17)
On the intervals 1198691119895and 1198692119895 the coefficient 119903(119905) satisfies
119903 (119905) gt 0 on 119869119894119895 1199031minus120574
isin 1198711(119869119894119895)
forall119894 isin 1 2 119895 isin N(18)
Let there be a real function 119862 = 119862(119905) 119862 isin
1198711
loc((1199050infin)R) and let there exist a sequence of positive realnumbers (120582
119895)119895isinN such that
119862 (119905) ge 0 on 119869119894119895 119888119894119895= int
119869119894119895
119862 (120591) 119889120591 gt 0
forall119894 isin 1 2 119895 isin N
1
119888119894119895
119862 (119905) le1
120587min(120582
119895119903 (119905))1minus120574
119870
1198720+ 1
120582119895119902 (119905)
forall119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(19)
where 120574 1198720 and 119870 are constants defined in (11) (12) and
(16) respectivelyThe proof of the following main result will be presented
in Section 4
Theorem 3 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)
satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0
and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (19) is
fulfilled then (1) is oscillatory
Condition (19) can be replaced by an equivalent onewhich has a more practical value and takes a simpler formsince we do not need a sequence of auxiliary parameters(120582119895)119895isinN let there be a real function 119862 = 119862(119905) 119862 isin
1198711
loc((1199050infin)R) such that
119862 (119905) ge 0 on 119869119894119895 119888119894119895= int
119868119894119895
119862 (120591) 119889120591 gt 0 forall119894 isin 1 2
119895 isin N
sup119905isin119869119894119895
[119862 (119905)
119902 (119905)] sup119905isin119869119894119895
[119903 (119905) 119862(119905)1(120574minus1)
] le119870
1198720+ 1
(
119888119894119895
120587)
120574(120574minus1)
forall119894 isin 1 2 119895 isin N
(20)
where 120574 1198720 and 119870 are constants defined in (11) (12) and
(16) respectively Since we will show that (19) and (20)are equivalent see page 8 the next oscillation criterionimmediately follows fromTheorem 3
Theorem 4 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)
satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0
and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (20) is
fulfilled then (1) is oscillatory
Remark 5 Assuming that 119871 = lim119895rarrinfin
1198861119895
lt infin wecan ensure the oscillation in the point 119871 Note that 119871 =
lim119895rarrinfin
1198862119895
since 1198861119895
lt 1198862119895
lt 1198861119895+1
Thus we can generatea one-sided (right) limit
Now we consider some consequences of Theorem 4which depend on the qualitative properties of the coefficient119902(119905)
Substituting 119862(119905) equiv 1 in (20) Theorem 4 implies thefollowing result involving lower bounds on the lengths ofintervals |119869
119894119895| = 119887119894119895minus 119886119894119895
Corollary 6 (119902(119905) is positive) Let the functionsΦ(119906 V) 119891(119906)119890(119905) and 119903(119905) satisfy (11) (12) (16) (17) and (18) respectivelyLet inf
119905isin119869119894119895119902(119905) gt 0 for each 119894 isin 1 2 119895 isin N If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
119870inf119905isin119869119894119895
119902 (119905))
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(21)
then (1) is oscillatory
4 International Journal of Differential Equations
Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903
0 1199020satisfying
0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902
0gt 0 forall119894 isin 1 2 119895 isin N (22)
If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N (23)
then (1) is oscillatory where 1205741198720 and119870 are constants defined
in (11) (12) and (16) respectively
Example 8 (oscillation of (4)) We know that 119909(119905) =
|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892
0(119904) = 119904
2
and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is
satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982
it is clear that conditions (18) and (22) are also satisfied inparticular for 119903
0= 1 and 119902
0= 4119898
2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887
1119895= (2119895 + 1)120587119898 = 119886
2119895and 1198872119895= (2119895 + 2)120587119898
Moreover10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816= 119887119894minus 119886119894=120587
119898gt 0 119894 isin 1 2 (24)
Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=120587
119898ge
120587
radic41198982
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(25)
Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory
Example 9 We consider the following class of equations
11990910158401015840minus 2
11987810158401015840(119905)
119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840
2
(119905) ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862
2((1199050infin) R)
(26)
where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905
119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905
119899rarr infin
there is a 1205910isin R such that 119905
119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and
119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given
oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904
0gt 0 such that
minus211987810158401015840(119905)
119878 (119905)ge 1199040 119905 isin (119905
119899 119905119899+1) 120591
0ge
120587
radic1199040
(27)
then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872
0= 0 119903
0= 1 119902
0= 1199040 and
119870 ge 1 that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1
minus 119905119895
10038161003816100381610038161003816ge 1205910ge
120587
radic1199040
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(28)
Hence Corollary 7 proves this result
As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0
Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869
119894119895
119894 isin 1 2 119895 isin N such that
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)
If
119902120574
119894119895
119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)
119870)
120574minus1
119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N
(30)
then (1) is oscillatory
Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582
119895such that
120582119895ge120587 (1198720+ 1)
119870
1
119902119894119895
1
119902119894119895
119902 (119905) le1
120587(120582119895119903 (119905))1minus120574
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(31)
The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869
119894119895 119902 = 0 equivalent to
120587 (1198720+ 1)
119870
1
119902119894119895
le 120582119895le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (32)
that is to
120587 (1198720+ 1)
119870
1
119902119894119895
le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (33)
This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second
inequality in (31) is trivially satisfied
International Journal of Differential Equations 5
The following three examples are simple consequences ofCorollary 10
Example 11 (oscillation of (5)) Equation (5) as well asequation
11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(34)
are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =
11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are
fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2
119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and
1198861119895=2119895120587
119898 119887
1119895=(2119895 + 12) 120587
119898
1198862119895=(2119895 + 32) 120587
119898 119887
2119895=(2119895 + 2) 120587
119898
(35)
Moreover
119902119894119895= int
119869119894119895
119902 (120591) 119889120591
= 11989821205872int
119887119894119895
119886119894119895
[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N
(36)
and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582
119895= 1(119898120587)
we have
120582119895=
1
119898120587ge1
119870
1
119898120587=120587 (1198720+ 1)
119870
1
1198981205872
=120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894119895
119902 (119905) =1
119898120587211989821205872[cos (119898119905)]+ le 119898
=1
120587
1
(1119898120587)=1
120587
1
(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(37)
Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too
In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905
0infin)
Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge
1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898
2 sign(cos(119898119905)) on 119869119894119895 120582119895=
1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872
0= 0
Indeed
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 = 41198982int
119869119894119895
[sign (cos (119898120591))]+119889120591
= 41198982 120587
2119898= 2119898120587 119894 isin 1 2 119895 isin N
120582119895=
1
2119898=
120587
2119898120587= 120587 (119872
0+ 1)
1
119902119894119895
ge120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894
119902 (119905) =1
211989812058741198982[sign (cos (119898119905))]+
le2119898
120587=1
120587
1
radic1 (41198982)
=1
120587(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(38)
which shows that (30) is satisfied
At the end of this section we consider the case when 119902(119905)changes sign on [119905
0infin) but with the help of Corollary 10 since
119902(119905) is strictly positive on all intervals 119869119894119895
Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge
1 then model equation (7) is oscillatory In fact let 119869119894119895be as
in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869
1119895and 119890(119905) le 0 on 119869
2119895 On
the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869
119894119895and
so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence
the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example
3 Proofs of Theorems 3 and 4
Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879
0lt 119879lowast and functions
119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation
1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879
0 119879lowast) (39)
which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+
119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) which satisfy respectively
1199081015840le 119866 (119905 119908) + 119887 (119905) 119908
1015840ge 119866 (119905 119908) + 119887 (119905)
ae in [1198790 119879lowast)
(40)
We are interested in studying the following property
119908 (1198790) le 119908 (119879
0) implies 119908 (119905) le 119908 (119905)
forall119905 isin [T0 119879lowast)
(41)
In this way we introduce the following definition
6 International Journal of Differential Equations
Definition 14 We say that comparison principle holds for (39)on an interval [119879
0 119879lowast) 119879 le 119879
0lt 119879lowast if property (41)
is fulfilled for all sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) of (39) on [119879
0 119879lowast)
Now we are able to state the next result
Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be
two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879
0 119879lowast119872 such that
119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871
1(1198790 119879lowast)
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816le 119871 (119905)
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]
(42)
Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879
0lt 119879lowast
Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader
Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally
Lipschitz function on R+ Let 119870 be an arbitrary real number
and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0
on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879
0 119879lowast) then comparison principle
(41) holds for the Riccati differential equation
1199081015840=
1
(120582119903 (119905))120574minus1
1198920 (|119908|) + 119870120582119902 (119905) ae in [119879
0 119879lowast)
(43)
Proof It is clear that (43) is a particular case of (39) inparticular for
119866 (119905 119906) =1
(120582119903 (119905))120574minus1
1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)
Since1198920is a locally Lipschitz function onR
+ for every119872 gt 0
there is an 1198710gt 0 depending on119872 such that
10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710
10038161003816100381610038161199041minus 1199042
1003816100381610038161003816 forall119904
1 1199041isin [minus119872119872]
(45)
Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816=
1
(120582119903 (119905))120574minus1
10038161003816100381610038161198920(10038161003816100381610038161199061
1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062
1003816100381610038161003816)1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
1003816100381610038161003816
10038161003816100381610038161199061
1003816100381610038161003816minus10038161003816100381610038161199062
1003816100381610038161003816
1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
(46)
Thus according to assumption 119903minus1+120574
isin 1198711(1198790 119879lowast) the
required condition (42) is fulfilled in particular for 119871(119905) =
1198710(120582119903(119905))
minus120574+1 and so Lemma 15 proves this corollary
Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886
1 1198871) or (119886
2 1198872) In the second one we
construct two subsolutions 1199081(119905) and 119908
2(119905) of (43) which
blow up on intervals (1198861 119879lowast
1) sube (119886
1 1198871) and (119886
2 119879lowast
2) sube (119886
2 1198872)
respectively
Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886
1 1198871] cup [119886
2 1198872] Let 119909 = 119909(119905) be a
nonoscillatory solution of (1) and let for some 119879 ge 1199050and
120582 gt 0 the function 119908 = 119908(119905) be defined by
119908 (119905) = minus
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
119909 (119905) 119905 ge 119879 (47)
Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)
where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0
Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905
0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well
defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where
119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0 and since
120582 gt 0 we have
minus120582119890 (119905)
119909 (119905)ge 0 in 119869 (49)
Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from
119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
119909 (119905)
+
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
1199092(119905)
1199091015840(119905)
=120582119903 (119905)
|119909 (119905)|120574Φ(119909 (119905) 119909
1015840(119905)) 1199091015840(119905) |119909 (119905)|
120574minus2
+ 120582119902 (119905)119891 (119909 (119905))
119909 (119905)minus 120582
119890 (119905)
119909 (119905)
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 3
Example 1 We consider the second-order differential opera-tor which is linear in 1199091015840 as follows
(119903 (119905) Φ (119909 1199091015840))1015840
= 120572(120601 (119909) 1199091015840)1015840
(14)
where 120572 gt 0 and 0 le 120601(119906) le 1 for all 119906 isin R Obviously thefunctionΦ(119906 V) = 120601(119906)V satisfies condition (13) in particularfor 120574 = 2 Two usual choices for 120601(119906) are 120601(119906) = |sin 119906| and120601(119906) = |119906|(1 + |119906|)
Example 2 We consider a quasilinear differential operator(the so-called one-dimensional prescribed mean curvatureoperator) as follows
(119903 (119905) Φ (119909 1199091015840))1015840
= 120572(120601 (119909)1199091015840
radic1 + 11990910158402)
1015840
(15)
where 120572 gt 0 and 0 le 120601120574minus1(119906) le |119906|
120574minus2 for all 119906 isin R It is notdifficult to check that condition (13) is satisfied in particularfor Φ(119906 V) = 120601(119906)V(1 + V2)12 and for any 120574 gt 1 For 120601(119906)we can take the same choice as in the previous example
Next we suppose the existence of a constant119870 such that
119891 (119906)
119906ge 119870 gt 0 forall119906 = 0 (16)
In order to simplify our consideration here inmany exampleswe often use 119891(119906) = 119870119906
Condition (2) means that there exists a sequence of pairsof intervals 119869
1119895= [1198861119895 1198871119895] and 119869
2119895= [1198862119895 1198872119895] 119895 isin N
contained in (1199050infin) such that the sequences (119886
1119895)119895ge1
(1198871119895)119895ge1
(1198862119895)119895ge1
and (1198872119895)119895ge1
are increasing 1198861119895lt 1198871119895le 1198862119895lt 1198872119895for
each 119895 and
119890 (119905) ge 0 on 1198691119895 119890 (119905) le 0 on 119869
2119895
for each 119895 isin N
lim119895rarrinfin
1198861119895= infin
(17)
On the intervals 1198691119895and 1198692119895 the coefficient 119903(119905) satisfies
119903 (119905) gt 0 on 119869119894119895 1199031minus120574
isin 1198711(119869119894119895)
forall119894 isin 1 2 119895 isin N(18)
Let there be a real function 119862 = 119862(119905) 119862 isin
1198711
loc((1199050infin)R) and let there exist a sequence of positive realnumbers (120582
119895)119895isinN such that
119862 (119905) ge 0 on 119869119894119895 119888119894119895= int
119869119894119895
119862 (120591) 119889120591 gt 0
forall119894 isin 1 2 119895 isin N
1
119888119894119895
119862 (119905) le1
120587min(120582
119895119903 (119905))1minus120574
119870
1198720+ 1
120582119895119902 (119905)
forall119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(19)
where 120574 1198720 and 119870 are constants defined in (11) (12) and
(16) respectivelyThe proof of the following main result will be presented
in Section 4
Theorem 3 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)
satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0
and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (19) is
fulfilled then (1) is oscillatory
Condition (19) can be replaced by an equivalent onewhich has a more practical value and takes a simpler formsince we do not need a sequence of auxiliary parameters(120582119895)119895isinN let there be a real function 119862 = 119862(119905) 119862 isin
1198711
loc((1199050infin)R) such that
119862 (119905) ge 0 on 119869119894119895 119888119894119895= int
119868119894119895
119862 (120591) 119889120591 gt 0 forall119894 isin 1 2
119895 isin N
sup119905isin119869119894119895
[119862 (119905)
119902 (119905)] sup119905isin119869119894119895
[119903 (119905) 119862(119905)1(120574minus1)
] le119870
1198720+ 1
(
119888119894119895
120587)
120574(120574minus1)
forall119894 isin 1 2 119895 isin N
(20)
where 120574 1198720 and 119870 are constants defined in (11) (12) and
(16) respectively Since we will show that (19) and (20)are equivalent see page 8 the next oscillation criterionimmediately follows fromTheorem 3
Theorem 4 Let the functions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905)
satisfy (11) (12) (16) (17) and (18) respectively Let 119902(119905) ge 0
and 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2 119895 isin N If (20) is
fulfilled then (1) is oscillatory
Remark 5 Assuming that 119871 = lim119895rarrinfin
1198861119895
lt infin wecan ensure the oscillation in the point 119871 Note that 119871 =
lim119895rarrinfin
1198862119895
since 1198861119895
lt 1198862119895
lt 1198861119895+1
Thus we can generatea one-sided (right) limit
Now we consider some consequences of Theorem 4which depend on the qualitative properties of the coefficient119902(119905)
Substituting 119862(119905) equiv 1 in (20) Theorem 4 implies thefollowing result involving lower bounds on the lengths ofintervals |119869
119894119895| = 119887119894119895minus 119886119894119895
Corollary 6 (119902(119905) is positive) Let the functionsΦ(119906 V) 119891(119906)119890(119905) and 119903(119905) satisfy (11) (12) (16) (17) and (18) respectivelyLet inf
119905isin119869119894119895119902(119905) gt 0 for each 119894 isin 1 2 119895 isin N If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
119870inf119905isin119869119894119895
119902 (119905))
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(21)
then (1) is oscillatory
4 International Journal of Differential Equations
Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903
0 1199020satisfying
0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902
0gt 0 forall119894 isin 1 2 119895 isin N (22)
If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N (23)
then (1) is oscillatory where 1205741198720 and119870 are constants defined
in (11) (12) and (16) respectively
Example 8 (oscillation of (4)) We know that 119909(119905) =
|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892
0(119904) = 119904
2
and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is
satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982
it is clear that conditions (18) and (22) are also satisfied inparticular for 119903
0= 1 and 119902
0= 4119898
2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887
1119895= (2119895 + 1)120587119898 = 119886
2119895and 1198872119895= (2119895 + 2)120587119898
Moreover10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816= 119887119894minus 119886119894=120587
119898gt 0 119894 isin 1 2 (24)
Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=120587
119898ge
120587
radic41198982
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(25)
Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory
Example 9 We consider the following class of equations
11990910158401015840minus 2
11987810158401015840(119905)
119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840
2
(119905) ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862
2((1199050infin) R)
(26)
where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905
119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905
119899rarr infin
there is a 1205910isin R such that 119905
119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and
119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given
oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904
0gt 0 such that
minus211987810158401015840(119905)
119878 (119905)ge 1199040 119905 isin (119905
119899 119905119899+1) 120591
0ge
120587
radic1199040
(27)
then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872
0= 0 119903
0= 1 119902
0= 1199040 and
119870 ge 1 that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1
minus 119905119895
10038161003816100381610038161003816ge 1205910ge
120587
radic1199040
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(28)
Hence Corollary 7 proves this result
As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0
Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869
119894119895
119894 isin 1 2 119895 isin N such that
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)
If
119902120574
119894119895
119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)
119870)
120574minus1
119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N
(30)
then (1) is oscillatory
Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582
119895such that
120582119895ge120587 (1198720+ 1)
119870
1
119902119894119895
1
119902119894119895
119902 (119905) le1
120587(120582119895119903 (119905))1minus120574
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(31)
The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869
119894119895 119902 = 0 equivalent to
120587 (1198720+ 1)
119870
1
119902119894119895
le 120582119895le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (32)
that is to
120587 (1198720+ 1)
119870
1
119902119894119895
le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (33)
This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second
inequality in (31) is trivially satisfied
International Journal of Differential Equations 5
The following three examples are simple consequences ofCorollary 10
Example 11 (oscillation of (5)) Equation (5) as well asequation
11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(34)
are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =
11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are
fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2
119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and
1198861119895=2119895120587
119898 119887
1119895=(2119895 + 12) 120587
119898
1198862119895=(2119895 + 32) 120587
119898 119887
2119895=(2119895 + 2) 120587
119898
(35)
Moreover
119902119894119895= int
119869119894119895
119902 (120591) 119889120591
= 11989821205872int
119887119894119895
119886119894119895
[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N
(36)
and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582
119895= 1(119898120587)
we have
120582119895=
1
119898120587ge1
119870
1
119898120587=120587 (1198720+ 1)
119870
1
1198981205872
=120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894119895
119902 (119905) =1
119898120587211989821205872[cos (119898119905)]+ le 119898
=1
120587
1
(1119898120587)=1
120587
1
(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(37)
Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too
In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905
0infin)
Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge
1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898
2 sign(cos(119898119905)) on 119869119894119895 120582119895=
1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872
0= 0
Indeed
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 = 41198982int
119869119894119895
[sign (cos (119898120591))]+119889120591
= 41198982 120587
2119898= 2119898120587 119894 isin 1 2 119895 isin N
120582119895=
1
2119898=
120587
2119898120587= 120587 (119872
0+ 1)
1
119902119894119895
ge120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894
119902 (119905) =1
211989812058741198982[sign (cos (119898119905))]+
le2119898
120587=1
120587
1
radic1 (41198982)
=1
120587(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(38)
which shows that (30) is satisfied
At the end of this section we consider the case when 119902(119905)changes sign on [119905
0infin) but with the help of Corollary 10 since
119902(119905) is strictly positive on all intervals 119869119894119895
Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge
1 then model equation (7) is oscillatory In fact let 119869119894119895be as
in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869
1119895and 119890(119905) le 0 on 119869
2119895 On
the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869
119894119895and
so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence
the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example
3 Proofs of Theorems 3 and 4
Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879
0lt 119879lowast and functions
119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation
1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879
0 119879lowast) (39)
which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+
119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) which satisfy respectively
1199081015840le 119866 (119905 119908) + 119887 (119905) 119908
1015840ge 119866 (119905 119908) + 119887 (119905)
ae in [1198790 119879lowast)
(40)
We are interested in studying the following property
119908 (1198790) le 119908 (119879
0) implies 119908 (119905) le 119908 (119905)
forall119905 isin [T0 119879lowast)
(41)
In this way we introduce the following definition
6 International Journal of Differential Equations
Definition 14 We say that comparison principle holds for (39)on an interval [119879
0 119879lowast) 119879 le 119879
0lt 119879lowast if property (41)
is fulfilled for all sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) of (39) on [119879
0 119879lowast)
Now we are able to state the next result
Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be
two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879
0 119879lowast119872 such that
119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871
1(1198790 119879lowast)
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816le 119871 (119905)
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]
(42)
Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879
0lt 119879lowast
Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader
Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally
Lipschitz function on R+ Let 119870 be an arbitrary real number
and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0
on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879
0 119879lowast) then comparison principle
(41) holds for the Riccati differential equation
1199081015840=
1
(120582119903 (119905))120574minus1
1198920 (|119908|) + 119870120582119902 (119905) ae in [119879
0 119879lowast)
(43)
Proof It is clear that (43) is a particular case of (39) inparticular for
119866 (119905 119906) =1
(120582119903 (119905))120574minus1
1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)
Since1198920is a locally Lipschitz function onR
+ for every119872 gt 0
there is an 1198710gt 0 depending on119872 such that
10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710
10038161003816100381610038161199041minus 1199042
1003816100381610038161003816 forall119904
1 1199041isin [minus119872119872]
(45)
Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816=
1
(120582119903 (119905))120574minus1
10038161003816100381610038161198920(10038161003816100381610038161199061
1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062
1003816100381610038161003816)1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
1003816100381610038161003816
10038161003816100381610038161199061
1003816100381610038161003816minus10038161003816100381610038161199062
1003816100381610038161003816
1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
(46)
Thus according to assumption 119903minus1+120574
isin 1198711(1198790 119879lowast) the
required condition (42) is fulfilled in particular for 119871(119905) =
1198710(120582119903(119905))
minus120574+1 and so Lemma 15 proves this corollary
Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886
1 1198871) or (119886
2 1198872) In the second one we
construct two subsolutions 1199081(119905) and 119908
2(119905) of (43) which
blow up on intervals (1198861 119879lowast
1) sube (119886
1 1198871) and (119886
2 119879lowast
2) sube (119886
2 1198872)
respectively
Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886
1 1198871] cup [119886
2 1198872] Let 119909 = 119909(119905) be a
nonoscillatory solution of (1) and let for some 119879 ge 1199050and
120582 gt 0 the function 119908 = 119908(119905) be defined by
119908 (119905) = minus
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
119909 (119905) 119905 ge 119879 (47)
Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)
where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0
Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905
0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well
defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where
119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0 and since
120582 gt 0 we have
minus120582119890 (119905)
119909 (119905)ge 0 in 119869 (49)
Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from
119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
119909 (119905)
+
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
1199092(119905)
1199091015840(119905)
=120582119903 (119905)
|119909 (119905)|120574Φ(119909 (119905) 119909
1015840(119905)) 1199091015840(119905) |119909 (119905)|
120574minus2
+ 120582119902 (119905)119891 (119909 (119905))
119909 (119905)minus 120582
119890 (119905)
119909 (119905)
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
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4 International Journal of Differential Equations
Corollary 7 (119902(119905) is bounded from below by a positiveconstant) Let the functionsΦ(119906 V)119891(119906) 119890(119905) and 119903(119905) satisfy(11) (12) (16) (17) and (18) respectively Let there be twoconstants 119903
0 1199020satisfying
0 lt 119903 (119905) le 1199030 119902 (119905) ge 119902
0gt 0 forall119894 isin 1 2 119895 isin N (22)
If
10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816ge 120587(
(1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N (23)
then (1) is oscillatory where 1205741198720 and119870 are constants defined
in (11) (12) and (16) respectively
Example 8 (oscillation of (4)) We know that 119909(119905) =
|sin(119898119905)| sin(119898119905) is an oscillatory solution of (4) Howeveraccording to Corollary 7 we can show that all solutions of (4)are oscillatory Indeed since Φ(119906 V) equiv V the conditions (11)and (12) are satisfied especially for 120574 = 2 119892(119904) = 119892
0(119904) = 119904
2
and 1198720= 0 Next 119891(119906) equiv 119906 implies that condition (16) is
satisfied especially for 119870 = 1 Since 119903(119905) equiv 1 and 119902(119905) equiv 41198982
it is clear that conditions (18) and (22) are also satisfied inparticular for 119903
0= 1 and 119902
0= 4119898
2 Moreover since 119890(119905) =ℎ(sin(119898119905)) and ℎ(119904)119904 gt 0 119904 = 0 we have that (17) is fulfilled for1198861119895= 2119895120587119898 119887
1119895= (2119895 + 1)120587119898 = 119886
2119895and 1198872119895= (2119895 + 2)120587119898
Moreover10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816= 119887119894minus 119886119894=120587
119898gt 0 119894 isin 1 2 (24)
Hence we conclude that the required condition (23) isfulfilled that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=120587
119898ge
120587
radic41198982
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(25)
Thus all conditions of Corollary 7 are satisfied and hence (4)is oscillatory
Example 9 We consider the following class of equations
11990910158401015840minus 2
11987810158401015840(119905)
119878 (119905)119909 = 2 sign (119878 (119905)) 1198781015840
2
(119905) ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R) 119909 notin 119862
2((1199050infin) R)
(26)
where 119878 = 119878(119905) 119878 isin 1198622(R) is an oscillatory function such thatthe zeros 119905
119899of the function sign(119878(119905))11987810158402(119905) satisfy 119905
119899rarr infin
there is a 1205910isin R such that 119905
119899+1minus 119905119899ge 1205910gt 0 for all 119899 isin N and
119878(119905) = 0 on (119905119899 119905119899+1) This equation allows an explicitly given
oscillatory solution 119909(119905) = |119878(119905)|119878(119905) Moreover if there is aconstant 119904
0gt 0 such that
minus211987810158401015840(119905)
119878 (119905)ge 1199040 119905 isin (119905
119899 119905119899+1) 120591
0ge
120587
radic1199040
(27)
then by Corollary 7 we conclude that (26) is oscillatoryIndeed conditions (11) (12) and (16) are satisfied by the samereasons as in Example 8 Condition (18) is satisfied becauseof 119903(119905) equiv 1 Also from (27) it follows that (22) and (23) arefulfilled in particular for 120574 = 2119872
0= 0 119903
0= 1 119902
0= 1199040 and
119870 ge 1 that is10038161003816100381610038161003816119869119894119895
10038161003816100381610038161003816=10038161003816100381610038161003816119905119895+1
minus 119905119895
10038161003816100381610038161003816ge 1205910ge
120587
radic1199040
= 120587((1198720+ 1) 1199030
1199020
)
(120574minus1)120574
ge 120587((1198720+ 1) 1199030
1198701199020
)
(120574minus1)120574
(28)
Hence Corollary 7 proves this result
As the second consequence ofTheorem3 is unlike the firstone we consider the case when the coefficient 119902(119905) is not astrictly positive function Here by 119902 = 0 we denote the setof all 119905 isin R such that 119902(119905) = 0
Corollary 10 (119902(119905) is nonnegative but not equiv 0) Let thefunctions Φ(119906 V) 119891(119906) 119890(119905) and 119903(119905) satisfy (11) (12) (16)(17) and (18) respectively Let 119902(119905) ge 0 on each interval 119869
119894119895
119894 isin 1 2 119895 isin N such that
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 gt 0 119894 isin 1 2 119895 isin N (29)
If
119902120574
119894119895
119902 (119905)ge 120587120574((1198720+ 1) 119903 (119905)
119870)
120574minus1
119905 isin 119869119894119895 119902 = 0 119894 isin 1 2 119895 isin N
(30)
then (1) is oscillatory
Proof It suffices to show that (30) is equivalent to theexistence of a real number 120582
119895such that
120582119895ge120587 (1198720+ 1)
119870
1
119902119894119895
1
119902119894119895
119902 (119905) le1
120587(120582119895119903 (119905))1minus120574
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(31)
The claim will then follow fromTheorem 3 Inequality (31) isfor any 119905 isin 119869
119894119895 119902 = 0 equivalent to
120587 (1198720+ 1)
119870
1
119902119894119895
le 120582119895le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (32)
that is to
120587 (1198720+ 1)
119870
1
119902119894119895
le (
119902119894119895
120587119902 (119905))
1(120574minus1)1
119903 (119905) (33)
This inequality is easily seen to be equivalent to (30) for any119905 isin 119869119894119895 119902 = 0 Note that if 119905 isin 119902 = 0 then the second
inequality in (31) is trivially satisfied
International Journal of Differential Equations 5
The following three examples are simple consequences ofCorollary 10
Example 11 (oscillation of (5)) Equation (5) as well asequation
11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(34)
are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =
11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are
fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2
119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and
1198861119895=2119895120587
119898 119887
1119895=(2119895 + 12) 120587
119898
1198862119895=(2119895 + 32) 120587
119898 119887
2119895=(2119895 + 2) 120587
119898
(35)
Moreover
119902119894119895= int
119869119894119895
119902 (120591) 119889120591
= 11989821205872int
119887119894119895
119886119894119895
[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N
(36)
and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582
119895= 1(119898120587)
we have
120582119895=
1
119898120587ge1
119870
1
119898120587=120587 (1198720+ 1)
119870
1
1198981205872
=120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894119895
119902 (119905) =1
119898120587211989821205872[cos (119898119905)]+ le 119898
=1
120587
1
(1119898120587)=1
120587
1
(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(37)
Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too
In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905
0infin)
Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge
1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898
2 sign(cos(119898119905)) on 119869119894119895 120582119895=
1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872
0= 0
Indeed
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 = 41198982int
119869119894119895
[sign (cos (119898120591))]+119889120591
= 41198982 120587
2119898= 2119898120587 119894 isin 1 2 119895 isin N
120582119895=
1
2119898=
120587
2119898120587= 120587 (119872
0+ 1)
1
119902119894119895
ge120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894
119902 (119905) =1
211989812058741198982[sign (cos (119898119905))]+
le2119898
120587=1
120587
1
radic1 (41198982)
=1
120587(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(38)
which shows that (30) is satisfied
At the end of this section we consider the case when 119902(119905)changes sign on [119905
0infin) but with the help of Corollary 10 since
119902(119905) is strictly positive on all intervals 119869119894119895
Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge
1 then model equation (7) is oscillatory In fact let 119869119894119895be as
in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869
1119895and 119890(119905) le 0 on 119869
2119895 On
the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869
119894119895and
so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence
the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example
3 Proofs of Theorems 3 and 4
Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879
0lt 119879lowast and functions
119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation
1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879
0 119879lowast) (39)
which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+
119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) which satisfy respectively
1199081015840le 119866 (119905 119908) + 119887 (119905) 119908
1015840ge 119866 (119905 119908) + 119887 (119905)
ae in [1198790 119879lowast)
(40)
We are interested in studying the following property
119908 (1198790) le 119908 (119879
0) implies 119908 (119905) le 119908 (119905)
forall119905 isin [T0 119879lowast)
(41)
In this way we introduce the following definition
6 International Journal of Differential Equations
Definition 14 We say that comparison principle holds for (39)on an interval [119879
0 119879lowast) 119879 le 119879
0lt 119879lowast if property (41)
is fulfilled for all sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) of (39) on [119879
0 119879lowast)
Now we are able to state the next result
Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be
two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879
0 119879lowast119872 such that
119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871
1(1198790 119879lowast)
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816le 119871 (119905)
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]
(42)
Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879
0lt 119879lowast
Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader
Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally
Lipschitz function on R+ Let 119870 be an arbitrary real number
and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0
on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879
0 119879lowast) then comparison principle
(41) holds for the Riccati differential equation
1199081015840=
1
(120582119903 (119905))120574minus1
1198920 (|119908|) + 119870120582119902 (119905) ae in [119879
0 119879lowast)
(43)
Proof It is clear that (43) is a particular case of (39) inparticular for
119866 (119905 119906) =1
(120582119903 (119905))120574minus1
1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)
Since1198920is a locally Lipschitz function onR
+ for every119872 gt 0
there is an 1198710gt 0 depending on119872 such that
10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710
10038161003816100381610038161199041minus 1199042
1003816100381610038161003816 forall119904
1 1199041isin [minus119872119872]
(45)
Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816=
1
(120582119903 (119905))120574minus1
10038161003816100381610038161198920(10038161003816100381610038161199061
1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062
1003816100381610038161003816)1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
1003816100381610038161003816
10038161003816100381610038161199061
1003816100381610038161003816minus10038161003816100381610038161199062
1003816100381610038161003816
1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
(46)
Thus according to assumption 119903minus1+120574
isin 1198711(1198790 119879lowast) the
required condition (42) is fulfilled in particular for 119871(119905) =
1198710(120582119903(119905))
minus120574+1 and so Lemma 15 proves this corollary
Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886
1 1198871) or (119886
2 1198872) In the second one we
construct two subsolutions 1199081(119905) and 119908
2(119905) of (43) which
blow up on intervals (1198861 119879lowast
1) sube (119886
1 1198871) and (119886
2 119879lowast
2) sube (119886
2 1198872)
respectively
Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886
1 1198871] cup [119886
2 1198872] Let 119909 = 119909(119905) be a
nonoscillatory solution of (1) and let for some 119879 ge 1199050and
120582 gt 0 the function 119908 = 119908(119905) be defined by
119908 (119905) = minus
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
119909 (119905) 119905 ge 119879 (47)
Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)
where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0
Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905
0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well
defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where
119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0 and since
120582 gt 0 we have
minus120582119890 (119905)
119909 (119905)ge 0 in 119869 (49)
Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from
119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
119909 (119905)
+
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
1199092(119905)
1199091015840(119905)
=120582119903 (119905)
|119909 (119905)|120574Φ(119909 (119905) 119909
1015840(119905)) 1199091015840(119905) |119909 (119905)|
120574minus2
+ 120582119902 (119905)119891 (119909 (119905))
119909 (119905)minus 120582
119890 (119905)
119909 (119905)
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 5
The following three examples are simple consequences ofCorollary 10
Example 11 (oscillation of (5)) Equation (5) as well asequation
11990910158401015840+ 11989821205872[sin (119898119905)]+119891 (119909) = ℎ (cos (119898119905))
ae in [1199050infin)
119909 1199091015840isin 119860119862loc ([1199050infin) R)
(34)
are oscillatory where 119891(119906) satisfies (16) with 119870 ge 1 andℎ(119904)119904 gt 0 119904 = 0 Indeed in (5) we have 119903(119905) equiv 1 119902(119905) =
11989821205872[cos(119898119905)] and 119890(119905) = ℎ(sin(119898119905)) and so (17) and (18) are
fulfilled and 119902(119905) ge 0 119902(119905) equiv 0 on each interval 119869119894119895 119894 isin 1 2
119895 isin N where 1198691119895= [1198861119895 1198871119895] 1198692119895= [1198862119895 1198872119895] and
1198861119895=2119895120587
119898 119887
1119895=(2119895 + 12) 120587
119898
1198862119895=(2119895 + 32) 120587
119898 119887
2119895=(2119895 + 2) 120587
119898
(35)
Moreover
119902119894119895= int
119869119894119895
119902 (120591) 119889120591
= 11989821205872int
119887119894119895
119886119894119895
[cos (119898120591)]+119889120591 = 1198981205872 119894 isin 1 2 119895 isin N
(36)
and since 120574 = 2 119903(119905) equiv 1119870 ge 1 and1198720= 0 for 120582
119895= 1(119898120587)
we have
120582119895=
1
119898120587ge1
119870
1
119898120587=120587 (1198720+ 1)
119870
1
1198981205872
=120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894119895
119902 (119905) =1
119898120587211989821205872[cos (119898119905)]+ le 119898
=1
120587
1
(1119898120587)=1
120587
1
(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(37)
Thus the required condition (30) is fulfilled in this case andhence we may apply Corollary 10 to (5) to verify that thisequation is oscillatory Analogously we can show that (34)is oscillatory too
In the next example the coefficient 119902(119905) is a discontinuousfunction on [119905
0infin)
Example 12 (oscillation of (6)) If 119891(119906) satisfies (16) with119870 ge
1 then (6) is oscillatory since the required condition (30) issatisfied especially for 119902(119905) = 4119898
2 sign(cos(119898119905)) on 119869119894119895 120582119895=
1(2119898) and 119869119894119895as in (35) 120574 = 2 119903(119905) equiv 1119870 ge 1 and119872
0= 0
Indeed
119902119894119895= int
119869119894119895
119902 (120591) 119889120591 = 41198982int
119869119894119895
[sign (cos (119898120591))]+119889120591
= 41198982 120587
2119898= 2119898120587 119894 isin 1 2 119895 isin N
120582119895=
1
2119898=
120587
2119898120587= 120587 (119872
0+ 1)
1
119902119894119895
ge120587 (1198720+ 1)
119870
1
119902119894119895
119894 isin 1 2 119895 isin N
1
119902119894
119902 (119905) =1
211989812058741198982[sign (cos (119898119905))]+
le2119898
120587=1
120587
1
radic1 (41198982)
=1
120587(120582119895119903 (119905))120574minus1
119905 isin 119869119894119895 119894 isin 1 2 119895 isin N
(38)
which shows that (30) is satisfied
At the end of this section we consider the case when 119902(119905)changes sign on [119905
0infin) but with the help of Corollary 10 since
119902(119905) is strictly positive on all intervals 119869119894119895
Example 13 (oscillation of (7)) If 119891(119906) satisfies (16) with119870 ge
1 then model equation (7) is oscillatory In fact let 119869119894119895be as
in (35) If ℎ(119904)119904 gt 0 119904 = 0 it is clear that 119890(119905) = ℎ(sin(119898119905))satisfies (2) that is 119890(119905) ge 0 on 119869
1119895and 119890(119905) le 0 on 119869
2119895 On
the other hand we have 119902(119905) = 41198982 sign(cos(119898119905)) on 119869
119894119895and
so 119902(119905) ge 0 119902(119905) equiv 0 on 119869119894119895for all 119894 isin 1 2 119895 isin N Hence
the proof of the fact that all assumptions of Corollary 7 arefulfilled is the same as in the preceding example
3 Proofs of Theorems 3 and 4
Let 119860119862loc([119886 119887)R) denote the set of all real functions whichare absolutely continuous on every interval [119886 119887 minus 120576] where120576 isin (0 119887 minus 119886) For arbitrary numbers 119879
0lt 119879lowast and functions
119866 = 119866(119905 119906) and 119887 = 119887(119905) we consider the ordinary differentialequation
1199081015840= 119866 (119905 119908) + 119887 (119905) ae in [119879
0 119879lowast) (39)
which generalizes the classic Riccati equation 1199081015840 = 119886(119905)1199082+
119887(119905) where 119886(119905) is an arbitrary function We associate to(39) the corresponding sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) which satisfy respectively
1199081015840le 119866 (119905 119908) + 119887 (119905) 119908
1015840ge 119866 (119905 119908) + 119887 (119905)
ae in [1198790 119879lowast)
(40)
We are interested in studying the following property
119908 (1198790) le 119908 (119879
0) implies 119908 (119905) le 119908 (119905)
forall119905 isin [T0 119879lowast)
(41)
In this way we introduce the following definition
6 International Journal of Differential Equations
Definition 14 We say that comparison principle holds for (39)on an interval [119879
0 119879lowast) 119879 le 119879
0lt 119879lowast if property (41)
is fulfilled for all sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) of (39) on [119879
0 119879lowast)
Now we are able to state the next result
Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be
two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879
0 119879lowast119872 such that
119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871
1(1198790 119879lowast)
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816le 119871 (119905)
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]
(42)
Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879
0lt 119879lowast
Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader
Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally
Lipschitz function on R+ Let 119870 be an arbitrary real number
and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0
on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879
0 119879lowast) then comparison principle
(41) holds for the Riccati differential equation
1199081015840=
1
(120582119903 (119905))120574minus1
1198920 (|119908|) + 119870120582119902 (119905) ae in [119879
0 119879lowast)
(43)
Proof It is clear that (43) is a particular case of (39) inparticular for
119866 (119905 119906) =1
(120582119903 (119905))120574minus1
1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)
Since1198920is a locally Lipschitz function onR
+ for every119872 gt 0
there is an 1198710gt 0 depending on119872 such that
10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710
10038161003816100381610038161199041minus 1199042
1003816100381610038161003816 forall119904
1 1199041isin [minus119872119872]
(45)
Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816=
1
(120582119903 (119905))120574minus1
10038161003816100381610038161198920(10038161003816100381610038161199061
1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062
1003816100381610038161003816)1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
1003816100381610038161003816
10038161003816100381610038161199061
1003816100381610038161003816minus10038161003816100381610038161199062
1003816100381610038161003816
1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
(46)
Thus according to assumption 119903minus1+120574
isin 1198711(1198790 119879lowast) the
required condition (42) is fulfilled in particular for 119871(119905) =
1198710(120582119903(119905))
minus120574+1 and so Lemma 15 proves this corollary
Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886
1 1198871) or (119886
2 1198872) In the second one we
construct two subsolutions 1199081(119905) and 119908
2(119905) of (43) which
blow up on intervals (1198861 119879lowast
1) sube (119886
1 1198871) and (119886
2 119879lowast
2) sube (119886
2 1198872)
respectively
Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886
1 1198871] cup [119886
2 1198872] Let 119909 = 119909(119905) be a
nonoscillatory solution of (1) and let for some 119879 ge 1199050and
120582 gt 0 the function 119908 = 119908(119905) be defined by
119908 (119905) = minus
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
119909 (119905) 119905 ge 119879 (47)
Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)
where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0
Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905
0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well
defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where
119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0 and since
120582 gt 0 we have
minus120582119890 (119905)
119909 (119905)ge 0 in 119869 (49)
Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from
119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
119909 (119905)
+
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
1199092(119905)
1199091015840(119905)
=120582119903 (119905)
|119909 (119905)|120574Φ(119909 (119905) 119909
1015840(119905)) 1199091015840(119905) |119909 (119905)|
120574minus2
+ 120582119902 (119905)119891 (119909 (119905))
119909 (119905)minus 120582
119890 (119905)
119909 (119905)
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 International Journal of Differential Equations
Definition 14 We say that comparison principle holds for (39)on an interval [119879
0 119879lowast) 119879 le 119879
0lt 119879lowast if property (41)
is fulfilled for all sub- and supersolutions 119908119908 isin
119860119862loc([1198790 119879lowast)R) of (39) on [119879
0 119879lowast)
Now we are able to state the next result
Lemma 15 Let 119879 le 1198790lt 119879lowast Let 119866 = 119866(119905 119906) and 119887 = 119887(119905) be
two arbitrary function For any119872 gt 0 let there be a function119871 = 119871(119905) gt 0 depending on 119879
0 119879lowast119872 such that
119871 (119905) gt 0 on [1198790 119879lowast) 119871 isin 119871
1(1198790 119879lowast)
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816le 119871 (119905)
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
forall119905 isin [1198790 119879lowast) 1199061 1199062isin [minus119872119872]
(42)
Then comparison principle (41) holds for (39) on an interval[1198790 119879lowast) 119879 le 119879
0lt 119879lowast
Proof It is enough to use the idea of the proof of [14 Lemma41] Hence this proof is left to the reader
Corollary 16 Let 120582 gt 0 and let 1198920 R+rarr R+be a locally
Lipschitz function on R+ Let 119870 be an arbitrary real number
and 119902(119905) an arbitrary function Let 119879 le 1198790lt 119879lowast If 119903(119905) gt 0
on [1198790 119879lowast) and 119903minus1+120574 isin 1198711(119879
0 119879lowast) then comparison principle
(41) holds for the Riccati differential equation
1199081015840=
1
(120582119903 (119905))120574minus1
1198920 (|119908|) + 119870120582119902 (119905) ae in [119879
0 119879lowast)
(43)
Proof It is clear that (43) is a particular case of (39) inparticular for
119866 (119905 119906) =1
(120582119903 (119905))120574minus1
1198920 (|119906|) 119887 (119905) = 119870120582119902 (119905) (44)
Since1198920is a locally Lipschitz function onR
+ for every119872 gt 0
there is an 1198710gt 0 depending on119872 such that
10038161003816100381610038161198920(1199041) minus 1198920(1199042)1003816100381610038161003816le 1198710
10038161003816100381610038161199041minus 1199042
1003816100381610038161003816 forall119904
1 1199041isin [minus119872119872]
(45)
Hence for any119872 gt 0 and all 1199061 1199062isin [minus119872119872] we obtain
1003816100381610038161003816119866 (119905 119906
1) minus 119866 (119905 119906
2)1003816100381610038161003816=
1
(120582119903 (119905))120574minus1
10038161003816100381610038161198920(10038161003816100381610038161199061
1003816100381610038161003816) minus 1198920(10038161003816100381610038161199062
1003816100381610038161003816)1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
1003816100381610038161003816
10038161003816100381610038161199061
1003816100381610038161003816minus10038161003816100381610038161199062
1003816100381610038161003816
1003816100381610038161003816
le1198710
(120582119903 (119905))120574minus1
10038161003816100381610038161199061minus 1199062
1003816100381610038161003816
(46)
Thus according to assumption 119903minus1+120574
isin 1198711(1198790 119879lowast) the
required condition (42) is fulfilled in particular for 119871(119905) =
1198710(120582119903(119905))
minus120574+1 and so Lemma 15 proves this corollary
Before we give the proof ofTheorem 3 we state and provethe next two propositions In the first one by a nonoscillatorysolution 119909(119905) of the main equation (1) we get the existenceof a supersolution 119908(119905) of the Riccati differential equation(43) on the interval (119886
1 1198871) or (119886
2 1198872) In the second one we
construct two subsolutions 1199081(119905) and 119908
2(119905) of (43) which
blow up on intervals (1198861 119879lowast
1) sube (119886
1 1198871) and (119886
2 119879lowast
2) sube (119886
2 1198872)
respectively
Proposition 17 Let Φ(119906 V) and 119891(119906) satisfy (11) (12) and(16) respectively and let 119890(119905) satisfy (2) Let 119903(119905) gt 0 119902(119905) ge0 and 119902(119905) equiv 0 on [119886
1 1198871] cup [119886
2 1198872] Let 119909 = 119909(119905) be a
nonoscillatory solution of (1) and let for some 119879 ge 1199050and
120582 gt 0 the function 119908 = 119908(119905) be defined by
119908 (119905) = minus
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
119909 (119905) 119905 ge 119879 (47)
Then 119908 isin 119860119862loc([119879infin)R) and 119908(119905) satisfies the inequality
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (48)
where 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0
Proof Since 119909 = 119909(119905) is a nonoscillatory solution of (1) thereis a 119879 ge 119905
0such that 119909(119905) = 0 on [119879infin) Hence 119908(119905) is well
defined by (47)Because of (2) we have 119890(119905)119909(119905) le 0 for all 119905 isin 119869 where
119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 = (119886
2 1198872) if 119909(119905) gt 0 and since
120582 gt 0 we have
minus120582119890 (119905)
119909 (119905)ge 0 in 119869 (49)
Next since 119909 and 119903(119905)Φ(119909(119905) 1199091015840(119905)) are from
119860119862loc([1199050infin)R) we can take the first derivative of 119908(119905)for almost everywhere in 119869 which together with 119903(119905) gt 0119902(119905) ge 0 and 119902(119905) equiv 0 on 119869 gives
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
119909 (119905)
+
120582119903 (119905)Φ (119909 (119905) 1199091015840(119905))
1199092(119905)
1199091015840(119905)
=120582119903 (119905)
|119909 (119905)|120574Φ(119909 (119905) 119909
1015840(119905)) 1199091015840(119905) |119909 (119905)|
120574minus2
+ 120582119902 (119905)119891 (119909 (119905))
119909 (119905)minus 120582
119890 (119905)
119909 (119905)
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
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Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 7
ge120582119903 (119905)
|119909 (119905)|120574119892 (10038161003816100381610038161003816Φ (119909 (119905) 119909
1015840(119905))
10038161003816100381610038161003816) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
=120582119903 (119905)
|119909 (119905)|120574119892(
|119909 (119905)|
120582119903 (119905)|119908 (119905)|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)
ge120582119903 (119905)
|119909 (119905)|120574
|119909 (119905)|120574
(120582119903 (119905))1205741198920 (|119908 (119905)|)
+ 119870120582119902 (119905) minus 120582119890 (119905)
119909 (119905)ae in 119869
(50)
that is
1199081015840ge (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) minus 120582
119890 (119905)
119909 (119905)ae in 119869
(51)
Hence (49) and (51) prove the desired inequality (48)
Proposition 18 Let (19) be satisfied Let 1198771and 119877
2be two
arbitrary real numbers Then there are two points 119879lowast1isin (1198861 1198871)
and 119879lowast
2isin (1198862 1198872) and two continuous functions 119908
1(119905) and
1199082(119905) such that
119908119894(119886119894) le 119877119894 lim119905rarr119879
lowast
119894
119908119894(119905) = infin 119894 isin 1 2
1199081015840
119894le (120582119903 (119905))
1minus1205741198920(1003816100381610038161003816119908119894
1003816100381610038161003816) + 119870120582119902 (119905)
for ae 119905 isin (119886119894 119879lowast
119894) 119894 isin 1 2
(52)
Proof Since the function 119910(119904) = tan(119904) is a bijection fromthe interval (minus1205872 1205872) intoR we observe that there are two1199041 1199042isin (minus1205872 1205872) such that
tan (119904119894) le 119877119894 119894 isin 1 2 (53)
Next we define two functions 1198811(119905) and 119881
2(119905) by
119881119894 (119905) = 119904119894 +
120587
119888119894
int
119905
119886119894
119862 (120591) 119889120591 119905 isin [119886119894 119887119894] 119894 isin 1 2 (54)
where the real numbers 1198881 1198882and the function 119862(119905) are
defined in (19) From (19) and (54) one can immediatelyconclude that
119881119894(119886119894) = 119904119894lt120587
2 119881
119894(119887119894) = 119904119894+ 120587 gt
120587
2 119894 isin 1 2 (55)
Since 119862 isin 1198711
loc((1199050infin)R) it follows that 119862 isin 1198711((1199050infin)R)
which implies that 119881119894isin 119860119862([119886
119894 119887119894]R) 119894 isin 1 2 Therefore
exploiting the fact that in (55) we have 119881119894isin 119862([119886
119894 119887119894]R) we
obtain the existence of two points 119879lowast1isin (1198861 1198871) and 119879
lowast
2isin
(1198862 1198872) such that
119881119894(119879lowast
119894)=
120587
2 minus
120587
2lt119881119894(119905)lt
120587
2
forall119905 isin [119886119894 119879lowast
119894) 119894 isin 1 2
(56)
Now we are able to define the following two functions 1205961(119905)
and 1205962(119905) by
120596119894(119905) = tan (119881i (119905)) 119905 isin [119886
119894 119879lowast
119894) 119894 isin 1 2 (57)
That are well defined because of (56) Moreover from (53)(54) and (56) we obtain
120596119894(119886119894) = tan (119904
119894) le 119877119894
lim119905rarr119879
lowast
119898
120596119894(119905) = tan (119881
119894(119879lowast
119894)) = tan(120587
2) = infin
(58)
Also since119881119894isin 119860119862([119886
119894 119887119894]R) we can take the first derivative
of 120596119894(119905) and hence from (19) and (57) we obtain
1199081015840
119894(119905) =
120587
119888119894
119862 (119905)1
cos2 (119881119894 (119905))
=120587
119888119894
119862 (119905) (1003816100381610038161003816120596119894(119905)1003816100381610038161003816
2+ 1)
le120587
119888119894
119862 (119905) [1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119872
0+ 1]
le120587
119888119894
119862 (119905) 1198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) +
120587
119888119894
119862 (119905) (1198720+ 1)
le (120582119903 (119905))1minus1205741198920(1003816100381610038161003816120596119894(119905)1003816100381610038161003816) + 119870120582119902 (119905) ae in (119886
119894 119879lowast
119894)
(59)
which together with (58) proves this proposition
Now we are able to present the proof of Theorem 3 basedon Lemma 15 and Propositions 17 and 18
Proof of Theorem 3 Assuming the contrary then there is anonoscillatory solution 119909(119905) and 119879 ge 119905
0such that 119909(119905) = 0 for
all 119905 ge 119879 119909(119905) and 119903(119905)Φ(119909 1199091015840) are from 119860119862loc([1199050infin)R)In order to simplify notation let every 119895 isin N be fixed andomitted in the notation For instance instead of (119886
1119895 1198871119895) and
(1198862119895 1198872119895) we write (119886
1 1198871) and (119886
2 1198872) respectively and so on
On one hand we observe by Proposition 17 that thefunction 119908(119905) defined by (47) is a supersolution of thefollowing Riccati differential equation
1199081015840= (120582119903 (119905))
1minus1205741198920(|119908|) + 119870120582119902 (119905) ae in 119869 (60)
where 119908 isin 119860119862loc(119869R) and 119869 = (1198861 1198871) if 119909(119905) lt 0 and 119869 =
(1198862 1198872) if 119909(119905) gt 0 see the proof of Proposition 17 Let for
instance 119909(119905) lt 0 and thus 119869 = (1198861 1198871)
On the other hand by Proposition 18 there are a number119879lowast
1isin (1198861 1198871) and subsolutions 119908
1(119905) 1199081isin 119860119862([119886
1 119879lowast
1)R) of
the Riccati differential equation (60) such that
1199081(1198861) le 119908 (119886
1) lim
119905rarr119879lowast
1
1199081(119905) = infin (61)
(in the case when 119869 = (1198862 1198872) then we work with119879lowast
2isin (1198862 1198872)
and the other subsolution 1199082(119905) 119908
2isin 119860119862([119886
2 119879lowast
2)R))
Hence by Corollary 16 and (61) we conclude that1199081(119905) le 119908(119905)
for all 119905 isin [1198861 119879lowast
1) and therefore
infin = lim119905rarr119879
lowast
1
1199081(119905) le lim119905rarr119879
lowast
1
119908 (119905) (62)
which contradicts 119908 isin 119860119862loc([119879infin)R) Thus the assump-tion that 119909(119905) is nonoscillatory is not possible and hence everysolution of (1) is oscillatory
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
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Discrete Dynamics in Nature and Society
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 International Journal of Differential Equations
Proof of Theorem 4 There is only one difference betweenTheorems 3 and 4 and it is the difference between conditions(19) and (20) Hence Theorem 4 immediately follows fromTheorem 3 provided that we prove the equivalence between(19) and (20)
First of all we need the following two lemmas
Lemma 19 Let 119886 119887 and 119888 be positive real numbers and 120574 gt 1Then the existence of a positive real number 120582 such that
119888 le min 119886
120582120574minus1
119887120582 (63)
is equivalent to
119888 le 119886112057411988711205741015840
(64)
Here 1205741015840 = 120574(120574 minus 1) is the conjugate exponent of 120574
Proof Let us define 119892(120582) = min119886120582120574minus1 119887120582 Since 120582 997891rarr
119886120582120574minus1 is decreasing and 120582 997891rarr 119887120582 is increasing there exists
a unique point of maximum of 119892 It is achieved at 120582 = 1205820
which is a solution of 119886120582120574minus10
= 1198871205820 hence 120582
0= (119886119887)
1120574If there is a positive real number 120582 such that 119888 le 119892(120582)
then 119888 le 119892(1205820) = 119887120582
0= 119887(119886119887)
1120574= 119886112057411988711205741015840
Converselyif 119888 le 119886
112057411988711205741015840
then 119888 le 119892(1205820) and the equivalence in the
lemma is proved
The preceding lemma is a special case of the followingmore general statement
Lemma 20 Assume that 119886(119905) 119887(119905) and 119888(119905) are positive realfunctions defined on a subset 119869 sube R Then the existence of apositive real number 120582 such that for all 119905 isin 119869
119888 (119905) le min 119886 (119905)120582120574minus1
119887 (119905) 120582 (65)
is equivalent to
sup119905isin119869
119888 (119905)
119887 (119905)le (inf119905isin119869
119886 (119905)
119888 (119905))
1(120574minus1)
(66)
Proof The condition that there exists 120582 gt 0 such that (65)is true for all 119905 isin 119869 is equivalent with the following twoinequalities which have to hold simultaneously
119888 (119905) le119886 (119905)
120582120574minus1
119888 (119905) le 119887 (119905) 120582 (67)
that is with
119888 (119905)
119887 (119905)le 120582 le (
119886 (119905)
119888 (119905))
1(120574minus1)
forall119905 isin 119869 (68)
Taking the supremum of the left-hand side and the infimumof the right-hand side we obtain (66) Conversely if (66)is satisfied then since the left-hand side in (66) cannotbe equal to zero while the right-hand side is less than infinthen inequality (66) defines a nonempty closed interval in R
(possibly reducing to just one point) in which we can chooseany positive 120582 Hence (67) is satisfied for all 119905 isin 119869 andtherefore (65) as well
Remark 21 If 119886 = 119886(119905) 119887 = 119887(119905) and 119888 = 119888(119905) appearingin Lemma 20 are constant functions then condition (66) isequivalent to 119888119887 = (119886119888)1(120574minus1) that is to (64)
Proof of Equivalence of (19) and (20) Wewill use Lemma 20Let us fix 119869
119894119895 where 119894 isin 1 2 and 119895 isin N We see that that the
inequality appearing in the second line of condition (19) is ofthe form (65) where
119888 (119905) =1
119888119894119895
119862 (119905) 119886 (119905) =1
120587119903(119905)1minus120574
119887 (119905) =119870119902 (119905)
120587 (1198720+ 1)
(69)
Condition (66) is equivalent to
sup119905isin119869119894119895
119862 (119905)
119902 (119905)le
1198701198881205741015840
119894119895
1205871205741015840
(1198720+ 1)
inf119905isin119869119894119895
1
119903 (119905) 119862(119905)120574minus1
=
1198701198881205741015840
119894119895
(1198720+ 1) sup
119905isin119869119894119895119903 (119905) 119862(119905)
120574minus1(
119888119894119895
120587)
1205741015840
(70)
This inequality is equivalent to the corresponding one in (19)and the claim follows from Lemma 19
Proof of Corollary 6 Substituting inf119905isin119869119894119895
119902(119905) instead of 119902(119905)and sup
119905isin119869119894119895119903(119905) instead of 119903(119905) in the inequality appearing in
the second line of (20) after a short computation we obtain astronger inequality than (20) as
1
119888119894119895
sup119905isin119869119894119895
119862 (119905)
le1
120587(
119870inf119905isin119869119894119895
119902 (119905)
(1198720+ 1) sup
119905isin119869119894119895119903 (119905)
)
(120574minus1)120574
forall119894 isin 1 2 119895 isin N
(71)
Substituting119862(119905) equiv 1we obtain that the left-hand side of (71)is equal to |119869
119894119895|minus1 and the resulting inequality is equivalent to
(21) Since (21) implies (20) the claim is proved
Remark 22 Here we show that the choice of 119862(119905) equiv 1 inthe proof of Corollary 6 is the best possible To prove thisnote that on the left-hand side of (71) we have the expressiondepending on an auxiliary function 119862(119905) and this functionis absent on the right-hand side Therefore it has sense totry to find a function 119862(119905) 119905 isin 119869
119894119895 such that the value of
119876(119862) = (1119888119894119895)sup119905isin119869119894119895
119862(119905) is minimal (note that 119888119894119895depends
on the function 119862 = 119862(119905) as well) It is easy to see that theminimum is achieved for 119862(119905) equiv 1 (or any positive constant)Indeed since 119876(119862) = 119876(120583119888) for any 120583 gt 0 by taking 120583 =
(sup119905isin119869119894119895
119862(119905))minus1 it suffices to assume that sup
119905isin119869119894119895119862(119905) = 1
The value of 119876(119862) is minimal if 119888119894119895= int119869119894119895
119862(120591) 119889120591 is maximalpossible and since 0 le 119862(119905) le 1 it is clear that the maximumof 119888119894119895= 119888119894119895(119862) is achieved when 119862(119905) equiv 1
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 9
Proof of Corollary 7 It is clear that the condition (23) implies(21) Furthermore since the lengths |119869
119894119895| are uniformly
bounded from below by a positive constant see (23) thenobviously 119886
1119895rarr infin as 119895 rarr infin Thus the claim follows
immediately from Corollary 6
Proof of Corollary 10 Let 119862(119905) = 119902(119905) Then the requiredcondition (19) is clearly satisfied because of assumption(30) Hence this corollary immediately follows from Theo-rem 3
4 An Extension of Condition (12) tothe Case of 120574 gt 1
In this section we consider the oscillation of (1) in the casewhen 120574 isin (1 2) is allowed in the assumption (12) In this waythe assumption (11) is slightly modified by a real number 119901 gt1 such that
|119906|(119901minus1)120574minus119901
VΦ (119906 V) ge 119892 (|Φ (119906 V)|) (72)
where (12) is supposed that is 119892(119888119904) ge 1198881205741198920(119904) for every 119888 119904 gt
0 and for a locally Lipschitz function 1198920 [0infin) rarr [0infin)
for which there exists a constant1198720such that 119892
0(119904)+119872
0ge 119904119901
for all 119904 isin [0infin)For the function 119891 we assume that there exists a constant
119870 gt 0 such that
119891 (119906) sign (119906) ge 119870|119906|119901minus1 for every 119906 isin R (73)
We will need the following two lemmas
Lemma 23 Let 119886 lt 119887 and let 119891 [119886 119887] rarr [0infin) be a meas-urable function Then
int
119887
119886
119891 (119905) 119889119905 ge 1 (74)
if and only if there exists a function 119892 isin 1198711([119886 119887] [0infin))
1198921= 1 such that
119891 (119905) ge 119892 (119905) forall119905 isin [119886 119887] (75)
Proof We assume (74) If int119887119886119891(119905)119889119905 is finite we may choose
119892(119909) = 119891(119909)1198911
For int119887119886119891(119905)119889119905 = infin we may define for every natural
number 119899 a function
119891119899(119905) =
119891 (119905) 119891 (119905) le 119899
119899 119891 (119905) gt 119899(76)
Since 119891119899(119905) le 119899 we have 119891
119899isin 1198711([119886 119887]R) Obviously
lim119899int119887
119886119891119899(119905)119889119905 is also infiniteThis implies that there exists an
1198990such that int119887
1198861198911198990(119905)119889119905 ge 1 We define 119892(119905) = 119891
1198990(119905)119891
11989901
Now we assume (75) Integrating over the whole intervalwe get
int
119887
119886
119891 (119905) 119889119905 ge int
119887
119886
119892 (119905) 119889119905 = 1 (77)
Lemma 24 Let119901 gt 1There exists a unique function 119910 = 119910(119904)that satisfies the following Cauchy problem
1199101015840(119904) = 1 +
1003816100381610038161003816119910 (119904)
1003816100381610038161003816
119901 119904 isin R (78)
119910 (0) = 0 (79)
Furthermore there exists a finite real number 119878lowast such that
lim119904rarrplusmn119878
lowast119910 (119904) = plusmninfin (80)
and 119910 isin 1198621((minus119878lowast 119878lowast)R) is injective and monotonous We
have 119878lowast = (120587119901)(1(sin(120587119901))) and 119910 is odd
We will denote by tan119901the solution of (78)-(79)
Proof Obviously every solution to (78) is injective andmonotonous on a connected set since 1199101015840 = 1 + |119910|119901 ge 1 gt 0
Let 119891(119909) = 1(1 + |119909|119901) Obviously 1 ge 119891(119909) gt 0 for all
119909 isin R and 119862119901gt |1198911015840(119909)| gt 0 for some constants 119862
119901isin R
We may assume 119862119901gt 1 Hence 1199111015840 = 119891(119905) 119911(0) = 0 has (by
[17Theorem 31]) a locally unique solution on (minus1119862119901 1119862119901)
By the same argument 119911 can be extended to the whole of Rinterval by interval of type ((119899 minus 1)119862
119901 (119899 + 1)119862
119901) for all
119899 isin ZWe know from [18] that the for the integral of the left-
hand side we have
int
infin
0
119891 (119909) 119889119909 = int
infin
0
119889119909
1 + 119909119901=120587
119901
1
sin (120587119901)= 119878lowast (81)
so lim119905rarrplusmninfin
119911(119905) = plusmn119878lowast
Since 1199111015840(119905) = 0 119911 is injective monotonous and of class1198621(R (minus119878lowast 119878lowast)) and so we define 119910(119904) = 119911
minus1(119904) It imme-
diately follows that 119910 isin 1198621((minus119878lowast 119878lowast)R) and that 119910 is
injective and monotonous Also by continuity of 119911 we havelim119904rarrplusmn119878
lowast119910(119904) = lim119908rarrplusmninfin
119910(119911(119908)) = lim119908rarrplusmninfin
119908 = plusmninfinDifferentiating it also follows that 1199101015840 = 1 + |119910|
119901 and bydefinition of 119910 119910(0) = 119911minus1(0) = 0 So 119910 is a solution of (78)-(79) Since any other such solution must have 119911 as its inverse119910 is unique on its domain
Proposition 25 Let 119886 lt 119887 and 120574 gt 119901 gt 1 Assume (72) and(73) and let 119902(119905) ge 0 and 119890(119905) le 0 (or 119890(119905) ge 0 ) for all 119905 isin [119886 119887]If there exists a constant 120582 gt 0 such that
119901
2120587sin(120587
119901)int
119887
119886
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
119889119905 ge 1 (82)
then for any solution 119909(119905) of (1) with 119909(119886) ge 0 (or 119909(119886) le 0)there exists a number 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Proof We assume that 119909(119905) = 0 for all 119905 isin [119886 119887] Then we maydefine the function
120596 (119905) = minus
120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
sign (119909 (119905)) |119909 (119905)|119901minus1 (83)
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 International Journal of Differential Equations
This function is well defined on [119886 119887] and its derivative is
1199081015840(119905) = minus
120582(119903 (119905) Φ (119909 (119905) 1199091015840(119905)))1015840
sign (119909 (119905)) |119909 (119905)|119901minus1
+
(119901 minus 1) 120582119903 (119905) Φ (119909 (119905) 1199091015840(119905))
|119909 (119905)|119901
1199091015840(119905)
(84)
Using the fact that 119909(119905) is a solution of (1) and omitting 119905 from119909 and 1199091015840 for clarity we get
1199081015840(119905) =
120582 (119902 (119905) 119891 (119909) minus 119890 (119905))
sign (119909) |119909|119901minus1+
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
119909119901
1199091015840
=
(119901 minus 1) 120582119903 (119905) Φ (119909 1199091015840)
|119909|119901
1199091015840+ 120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
Φ(119909 1199091015840) 1199091015840|119909|(119901minus1)120574minus119901
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(72)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892 (10038161003816100381610038161003816Φ (119909 119909
1015840)10038161003816100381610038161003816)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
=(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
119892(|119909|119901minus1
120582119903 (119905)|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
(12)
ge(119901 minus 1) 120582119903 (119905)
|119909|120574(119901minus1)
(|119909|119901minus1
120582119903 (119905))
120574
1198920 (|120596 (119905)|)
+ 120582119902 (119905)119891 (119909)
sign (119909) |119909|119901minus1minus 120582
119890 (119905)
sign (119909) |119909|119901minus1
=119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|)+120582119902 (119905)
119891 (119909)
sign (119909) |119909|119901minus1
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(73)
ge119901 minus 1
(120582119903 (119905))120574minus1
1198920 (|120596 (119905)|) + 120582119902 (119905) 119870
minus 120582119890 (119905)
sign (119909) |119909|119901minus1
(85)
Since 119890(119905)(sign(119909)|119909|119901minus1) le 0 we have
1199081015840(119905) ge
119901 minus 1
(120582119903 (119905))120574minus1
1198920(|120596 (119905)|) + 120582119870119902 (119905) (86)
We apply Lemma 23 to (82) to obtain a function119862(119905) suchthat
int
119887
119886
119862 (119905) 119889119905 = 1 (87)
and by (75)
0 le 119862 (119905) le119901
2120587sin(120587
119901)min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
(88)
Let 1199040isin (minus(120587119901)(1 sin(120587119901)) (120587119901)(1 sin(120587119901))) be such
that tan119901(1199040) le 120596(119886) We define the function
119881 (119905) = 1199040+2120587
119901
1
sin (120587119901)int
119905
119886
119862 (120591) 119889120591 (89)
Note that (minus(120587119901)(1 sin(120587119901))) lt 1199040
= 119881(119886) lt
(120587119901)(1 sin(120587119901)) and 119881(119887) = 1199040+ (2120587119901)(1 sin(120587119901)) gt
(120587119901)(1 sin(120587119901)) Since119881 is continuous there exists a 119879lowast isin(119886 119887) such that 119881(119905) lt (120587119901)(1 sin(120587119901)) for 119905 isin [119886 119879lowast) and119881(119879lowast) = (120587119901)(1 sin(120587119901))
We define
120596 (119905) = tan119901(119881 (119905)) 119905 isin [119886 119879
lowast) (90)
and note that by Lemma 24 lim119905rarr119879
lowast120596(119905) = +infin wheretan119901(119904) = 119910(119904) and 119910(119904) is determined by Lemma 24The derivative of 120596 then satisfies
1205961015840(119905) = tan1015840
119901(119881 (119905)) 119881
1015840(119905)
= (1 + tan119901119901(119881 (119905))) 119881
1015840(119905)
=2120587
119901
1
sin (120587119901)(1 + 120596
119901(119905)) 119862 (119905)
le (1 +1198720+ 1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816))min
119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119870120582119902 (119905)
1 +1198720
le119901 minus 1
119903120574minus1
(119905) 120582120574minus1
1198920(1003816100381610038161003816120596 (119905)
1003816100381610038161003816) + 119870120582119902 (119905)
(91)
Thus we may apply the comparison principle ofLemma 15 to 120596(119886) le 120596(119886) and their respective differentialinequalities to conclude that 120596(119905) le 120596(119905) for all 119905 isin [119886 119887]But since 120596 rarr +infin as 119905 rarr 119879
lowastlt 119887 and 120596(119905) is well defined
on the whole segment [119886 119887] we arrive at a contradictionHence 119909(119905) gt 0 cannot hold for all 119905 isin [119886 119887] and since 119909(119905) iscontinuous there exists a 119905lowast isin [119886 119887] such that 119909(119905lowast) = 0
Theorem 26 Let 120574 gt 119901 gt 1 Assume (72) and (73) and let forany 119879 gt 119905
0there exist 119879 le 119886
1lt 1198871le 1198862lt 1198872such that 119892(119905) ge 0
on [1198861 1198871] cup [119886
2 1198872] and 119890(119905) le 0 on [119886
1 1198871] and 119890(119905) ge 0 on
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 11
[1198862 1198872] If for both 119894 isin 1 2 there exist constants 120582
119894gt 0 such
that
119901
2120587sin (120587119901)int
119887119894
119886119894
min119901 minus 1
119903120574minus1
(119905) 120582120574minus1
119894
119870120582119894119902 (119905)
1 +1198720
119889119905 ge 1
(92)
then (1) is oscillatory
Proof For (1) to be oscillatory it is sufficient that for every119899 isin N there exists a 119905
119899gt 119899 such that 119909(119905
119899) = 0
Let 119899 isin N and let 119899 lt 1198861lt 1198871le 1198862lt 1198872as assumed in the
theorem It is enough to show that every solution 119909(119905) of (1)has a zero on [119886
1 1198872] If 119909(119886
1) ge 0 we apply Proposition 25 to
the segment [1198861 1198871] to obtain the sought zero If 119909(119886
2) le 0 we
again apply Proposition 25 to the segment [1198862 1198872] Suppose
119909(1198861) lt 0 and 119909(119886
2) gt 0 Then since 119909 is continuous there
exists a number 1199051isin [1198861 1198862] such that 119909(119905
1) = 0
Acknowledgment
This work is supported by the Ministry of Science of theRepublic of Croatia under Grant no 036-0361621-1291
References
[1] Q Kong ldquoNonoscillation and oscillation of second order half-linear differential equationsrdquo Journal of Mathematical Analysisand Applications vol 332 no 1 pp 512ndash522 2007
[2] Q-R Wang and Q-G Yang ldquoInterval criteria for oscillationof second-order half-linear differential equationsrdquo Journal ofMathematical Analysis and Applications vol 291 no 1 pp 224ndash236 2004
[3] Q Long and Q-R Wang ldquoNew oscillation criteria of second-order nonlinear differential equationsrdquo Applied Mathematicsand Computation vol 212 no 2 pp 357ndash365 2009
[4] Z Zheng and F Meng ldquoOscillation criteria for forced second-order quasi-linear differential equationsrdquo Mathematical andComputer Modelling vol 45 no 1-2 pp 215ndash220 2007
[5] D Cakmak and A Tiryaki ldquoOscillation criteria for certainforced second-order nonlinear differential equationsrdquo AppliedMathematics Letters vol 17 no 3 pp 275ndash279 2004
[6] F Jiang and F Meng ldquoNew oscillation criteria for a class ofsecond-order nonlinear forced differential equationsrdquo Journalof Mathematical Analysis and Applications vol 336 no 2 pp1476ndash1485 2007
[7] Q Yang ldquoInterval oscillation criteria for a forced secondorder nonlinear ordinary differential equations with oscillatorypotentialrdquo Applied Mathematics and Computation vol 135 no1 pp 49ndash64 2003
[8] A Tiryaki and B Ayanlar ldquoOscillation theorems for certainnonlinear differential equations of second orderrdquo Computers ampMathematics with Applications vol 47 no 1 pp 149ndash159 2004
[9] J Jaros T Kusano and N Yoshida ldquoGeneralized Piconersquosformula and forced oscillations in quasilinear differential equa-tions of the second orderrdquoArchivumMathematicum vol 38 no1 pp 53ndash59 2002
[10] R P Agarwal S R Grace and D OrsquoRegan Oscillation Theoryfor Second Order Linear Half-Linear Superlinear and SublinearDynamic Equations Kluwer Academic Publishers DordrechtThe Netherlands 2002
[11] S Jing ldquoA new oscillation criterion for forced second-orderquasilinear differential equationsrdquoDiscrete Dynamics in Natureand Society vol 2011 Article ID 428976 8 pages 2011
[12] W-T Li and S S Cheng ldquoAn oscillation criterion for nonhomo-geneous half-linear differential equationsrdquoAppliedMathematicsLetters vol 15 no 3 pp 259ndash263 2002
[13] J S W Wong ldquoOscillation criteria for a forced second-orderlinear differential equationrdquo Journal of Mathematical Analysisand Applications vol 231 no 1 pp 235ndash240 1999
[14] M Pasic ldquoFite-Wintner-Leighton type oscillation criteria forsecond-order differential equations with nonlinear dampingrdquoAbstract and Applied Analysis vol 2013 Article ID 852180 10pages 2013
[15] M Pasic ldquoNew interval oscillation criteria for forced second-order differential equations with nonlinear dampingrdquo Interna-tional Journal of Mathematical Analysis vol 7 no 25 pp 1239ndash1255 2013
[16] L Gasinski and N S Papageorgiou Nonsmooth Critical PointTheory and Nonlinear Boundary Value Problems vol 8 of Seriesin Mathematical Analysis and Applications Edited by R PAgarwal and D OrsquoRegan Chapman amp HallCRC Boca RatonFla USA 2005
[17] S Lang Real and Functional Analysis vol 142 of GraduateTexts in Mathematics Springer-Verlag New York NY USA 3rdedition 1993
[18] I N Bronshtein K A Semendyayev G Musiol and HMuehligHandbook of Mathematics Springer Berlin Germany5th edition 2007
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of