Research Article Global and Blow-Up Solutions for Nonlinear …downloads.hindawi.com/journals/ijde/2014/724837.pdf · 2019. 7. 31. · wave equation with nonlinear damping and source

Research ArticleGlobal and Blow-Up Solutions for Nonlinear HyperbolicEquations with Initial-Boundary Conditions

Uumllkuuml Dinlemez1 and Esra AktaG2

1 Department of Mathematics Faculty of Science Gazi University Teknikokullar Ankara Turkey2 Incirli Mahallesi Karaelmas Sokak Yunusemre Caddesi 5118 Incirli Ankara Turkey

Correspondence should be addressed to Ulku Dinlemez ulkugaziedutr

Received 24 December 2013 Revised 7 March 2014 Accepted 20 March 2014 Published 13 April 2014

Academic Editor D D Ganji

Copyright copy 2014 U Dinlemez and E Aktas This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider an initial-boundary value problem to a nonlinear string equations with linear damping term It is proved that undersuitable conditions the solution is global in time and the solution with a negative initial energy blows up in finite time

1 Introduction

We study the damped nonlinear string equation with sourceterm |119906|

120572

119906

119906119905119905+ 119906119905= (120590 (

100381610038161003816100381611990611990910038161003816100381610038162

) 119906119909)119909

+ |119906|120572

119906

(119909 119905) isin (0 1) times [0 119879]

(1)

where 1 lt 120572 120590(119904) is a smooth function for 0 le 119904 with theinitial conditions

119906 (119909 0) = 1199060(119909) 119906

119905(119909 0) = 119906

1(119909) 119909 isin [0 1] (2)

and boundary conditions

119906 (1 119905) = 0 119905 isin (0 119879)

120590 (1003816100381610038161003816119906119909 (0 119905)

10038161003816100381610038162

119906119909(0 119905)) minus 119906

119905(0 119905) = 2120601 (119905) 119905 isin [0 119879]

120590 (1003816100381610038161003816119906119909 (0 119905)

10038161003816100381610038162

119906119909(0 119905)) + 119906

119905(0 119905) = 2120595 (119905) 119905 isin [0 119879]

(3)

Theproblem (1)ndash(3) can be regarded asmodelling a nonlinearstring with vertical displacement function 119906(119909 119905) in R Andthis problem has nonlinear mechanical damping of the form|119906|120572

119906 The right end of the string makes it steady The input120601(119905) function and the output 120595(119905) function are applied on theleft

Wu and Li [1] studied the motion for a nonlinearbeammodel with nonlinear damping 119886|120601

119905|119898minus1

120601119905and external

forcing 119887|120601|119901minus1

120601 terms They showed that this model hasa unique global solution and blow-up solution under thesame conditions Levine et al [2] and Levine and Serrin [3]studied abstract version Georgiev and Todorova [4] studiednonlinear wave equations involving the nonlinear dampingterm |119906

119905|119898minus1

119906119905and source term of type |119906

119905|119901minus1

119906119905They proved

global existence theoremwith large initial data for 1 lt 119901 le 119898Hao and Li [5] studied the global solutions for a nonlinearstring with boundary input and output Dinlemez [6] provedthe global existence and uniqueness of weak solutions forthe initial-boundary value problem for a nonlinear waveequation with strong structural damping and nonlinearsource terms in R A lot of papers in connection with blow-up global solutions and existence of weak solutions werestudied in [7ndash15]

In this paper we first find energy equation for the problem(1)ndash(3) Then we prove the solutions of the problem (1)ndash(3)are global in time under some conditions on the function120590(119904) input 120601(119905) and the output 120595(119905) Finally we establish ablow-up result for solutionswith a negative initial energyOurapproach is similar to the one in [5]

2 Main Results

Nowwe give the following lemma for energy equation for theproblem (1)ndash(3)

Hindawi Publishing CorporationInternational Journal of Differential EquationsVolume 2014 Article ID 724837 5 pageshttpdxdoiorg1011552014724837

2 International Journal of Differential Equations

Lemma 1 Let 1 lt 120572 and 119906(119909 119905) be a solution of the problem(1)ndash(3) Then the energy equation of the problem (1)ndash(3) is

119864 (119905) =1

2

100381710038171003817100381711990611990510038171003817100381710038172

2minus

1

120572 + 2119906120572+2

120572+2+

1

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

(4)

119889

119889119905119864 (119905) = 120601

2

(119905) minus 1205952

(119905) minus1003817100381710038171003817119906119905

10038171003817100381710038172

2 (5)

Proof Multiplying (1) with 119906119905and integrating over (0 1) then

we get

119889

1198891199051

2

100381710038171003817100381711990611990510038171003817100381710038172

2minus

1

120572 + 2119906120572+2

120572+2

= int1

0

(120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 119906119909)119909

119906119905119889119909 minus

100381710038171003817100381711990611990510038171003817100381710038172

2

(6)

Applying integration by parts in the right hand side of (6)we find

int1

0

(120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 119906119909)119909

119906119905119889119909 = minus120590 (

1003816100381610038161003816119906119909 (0 119905)10038161003816100381610038162

) 119906119909(0 119905) 119906

119905(0 119905)

minus1

2

119889

119889119905int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

(7)

And using boundary conditions in equality (7) we obtain

119889

1198891199051

2

100381710038171003817100381711990611990510038171003817100381710038172

2minus

1

120572 + 2119906120572+2

120572+2+

1

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

= 1206012

(119905) minus 1205952

(119905) minus1003817100381710038171003817119906119905

10038171003817100381710038172

2

(8)

Hence the proof is completed

Next we give the following theorem for global solutionsin time

Theorem 2 Assume that 119906(119909 119905) is a solution of the problem(1)ndash(3) with 1 lt 120572 and

(i) 120590(119904) satisfies the following condition

|119904|120572

le 120590 (119904) for 119904 isin R+

cup 0 (9)

(ii) the input and the output functions satisfy

1206012

(119905) le 1205952

(119905) (10)

Then the solution 119906(119909 119905) is global in time

Proof Let

119866 (119905) = 119864 (119905) +2

120572 + 2119906120572+2

120572+2

=1

2

100381710038171003817100381711990611990510038171003817100381710038172

2+

1

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909 +1

120572 + 2119906120572+2

120572+2

(11)

Differentiating 119866(119905) with respect to 119905 and using (5) we get

119889

119889119905119866 (119905) = 120601

2

(119905) minus 1205952

(119905) minus1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 2int1

0

|119906|120572

119906119906119905119889119909 (12)

Using the Cauchy-Schwarz inequality in the last term of (12)we obtain

2int1

0

|119906|120572

119906119906119905119889119909 le 2int

1

0

|119906|120572+1 1003816100381610038161003816119906119905

1003816100381610038161003816 119889119909

le 1199062(120572+1)

2(120572+1)+1003817100381710038171003817119906119905

10038171003817100381710038172

2

(13)

and it follows from (12) (13) and (10) that we have

119889

119889119905119866 (119905) le 119906

2(120572+1)

2(120572+1)+1003817100381710038171003817119906119905

10038171003817100381710038172

2 (14)

By assumption (9) and integrating over (0 |119906119909|2

) and (0 1)respectively we yield

1

120572 + 1

100381710038171003817100381711990611990910038171003817100381710038172(120572+1)

2(120572+1)le int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909 (15)

Furthermore we have

|119906 (119909 119905)|2(120572+1)

=

100381610038161003816100381610038161003816100381610038161003816int1

119909

119906120585(120585 119905) 119889120585

100381610038161003816100381610038161003816100381610038161003816

2(120572+1)

le int1

119909

10038161003816100381610038161003816119906120585(120585 119905)

10038161003816100381610038161003816

2(120572+1)

119889120585

le int1

0

1003816100381610038161003816119906119909 (119909 119905)10038161003816100381610038162(120572+1)

119889119909 =1003817100381710038171003817119906119909 (119909 119905)

10038171003817100381710038172(120572+1)

2(120572+1)

(16)

and then

119906 (119909 119905)2(120572+1)

2(120572+1)le

1003817100381710038171003817119906119909 (119909 119905)10038171003817100381710038172(120572+1)

2(120572+1) (17)

Combining (11) (14) (15) and (17) we get

119889119866 (119905)

119889119905le

1

1205851

119866 (119905) (18)

where 1205851= min12 1(120572 + 1) Using Gronwallrsquos inequality

we have

119866 (119905) le 119866 (0) 119890(11205851)119905

(19)

Therefore together with the continuation principle and thedefinition of 119866(119905) we complete the proof of Theorem 2

Then we give the following theorem for the blow-upsolutions of the problem (1)ndash(3)

Theorem 3 Let 119906(119909 119905) be a solution of the problem (1)ndash(3)with 1 lt 120572 Assume that

(i) there exists 1 lt 120576 lt (120572 + 2)2 such that the function120590(119904) satisfies

120590 (119904) 119904 le120576

2int119904

0

120590 (120577) 119889120577 for 119904120598R+

cup 0 (20)

International Journal of Differential Equations 3

(ii) the initial values satisfy

119864 (0) le 0 0 lt int1

0

1199060(119909) 1199061(119909) 119889119909 (21)

(iii) the input and output functions satisfy

1206012

(119905) le 1205952

(119905)

(120595 (119905) + 120601 (119905)) (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0)) le 0

(22)

(iv) 119906(119909 119905) satisfies 1 le 119906

Then the solution 119906(119909 119905) blows up in finite time 119879max and

119879max le (120572 + 4

120572120578)119873minus120572(120572+4)

(0) (23)

where 120578 is some positive constant independent of the initialvalue 120572 and 119873(119905) are given by (25)

Proof We define

119872(119905) = minus119864 (119905) 120574 =120572

2 (120572 + 2) (24)

119873(119905) = 1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909 (25)

By virtue of (5) (21) (22) and (24) we get

119889119872 (119905)

119889119905=

100381710038171003817100381711990611990510038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905) ge 0 (26)

0 le 119872 (0) le 119872 (119905) for 0 le 119905 (27)

Taking a derivative of (25) and using (26) we have

119889119873 (119905)

119889119905

= (1 minus 120574)119872minus120574

(119905)1198721015840

(119905) + int1

0

1199062

119905119889119909 + int

1

0

119906119906119905119905119889119909

= (1 minus 120574)119872minus120574

(119905) (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905)) +1003817100381710038171003817119906119905

10038171003817100381710038172

2

+ int1

0

119906119906119905119905119889119909

(28)

Multiplying (1) by 119906 and integrating over the interval [0 1]and then using boundary conditions (3) we obtain

int1

0

119906119906119905119905119889119909

= 119906120572+2

120572+2minus int1

0

119906119906119905119889119909 minus (120595 (119905) + 120601 (119905))

times (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0))

minus int1

0

120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909119889119909

(29)

From the definition of119872(119905) we yield

0 = 120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

+120576

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

minus120576

120572 + 2119906120572+2

120572+2

(30)

Combining (29) and (30) in (28) we get

119889119873 (119905)

119889119905= (1 minus 120574)119872

minus120574

(119905) (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905)) +1003817100381710038171003817119906119905

10038171003817100381710038172

2

+ 119906120572+2

120572+2minus int1

0

119906119906119905119889119909 minus (120595 (119905) + 120601 (119905))

times (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0))

minus int1

0

120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909119889119909 + 120576119872 (119905) +

120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2

+120576

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

minus120576

120572 + 2119906120572+2

120572+2

(31)

Using (22) in (31) we obtain

(1 +120576

2)1003817100381710038171003817119906119905

10038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

2119906120572+2

120572+2

+ int1

0

(120576

2int|119906119909|2

0

120590 (120585) 119889120585 minus 120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909)119889119909

+ 120576119872 (119905) minus int1

0

119906119906119905119889119909 le

119889119873 (119905)

119889119905

(32)

Thanks to Youngrsquos inequality

119860119861 le120575119901

119901119860119901

+120575minus119902

119902119861119902

0 le 119860 119861 forall0 lt 1205751

119901+

1

119902= 1

(33)

for int10

119906119906119905119889119909 with 119901 = 119902 = 2 and 120574 = 2 and then we get

int1

0

119906119906119905119889119909 le int

1

0

10038161003816100381610038161199061199061199051003816100381610038161003816 119889119909 le

100381710038171003817100381711990611990510038171003817100381710038172

+1

41199062

(34)

From embedding for 119871119901(0 1) and using (iv) we have 1199062

2le

119906120572+2

120572+2and putting (34) in (32) we have

(1 +120576

2)1003817100381710038171003817119906119905

10038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

21199062

2

+ int1

0

(120576

2int|119906119909|2

0

120590 (120585) 119889120585 minus 120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909)119889119909 + 120576119872 (119905)

minus1003817100381710038171003817119906119905

10038171003817100381710038172

2minus

1

41199062

2le

119889119873 (119905)

119889119905

(35)


From (20) we get

120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

41199062

2le

119889119873 (119905)

119889119905

(36)

Choosing 120576 and 120581 = min1205762 (12minus120576(120572+2)) 14 we obtain

120581 119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2 le

119889119873 (119905)

119889119905 (37)

Thanks to (21) and (27) we yield

0 lt 119873 (0) le 119873 (119905) forall0 lt 119905 (38)

Now we estimate [119873(119905)]1(1minus120574) From Holderrsquos inequality

100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le 1199062

100381710038171003817100381711990611990510038171003817100381710038172 le 119906

120572+2

100381710038171003817100381711990611990510038171003817100381710038172 (39)

then using Youngrsquos inequality again we get100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le

1205752(1minus120574)

2 (1 minus 120574)

100381710038171003817100381711990611990510038171003817100381710038172(1minus120574)

2

+1 minus 2120574

2 (1 minus 120574)120575minus2(1minus120574)(1minus2120574)

1199062(1minus120574)(1minus2120574)

120572+2

(40)

where 0 lt 120575 and 1119901 + 1119902 = 1 with 119901 = 2(1 minus 120574) And so wehave

100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 21(1minus120574)

(1205752

(2 (1 minus 120574))1(1minus120574)

100381710038171003817100381711990611990510038171003817100381710038172

2

+(1 minus 2120574

2 (1 minus 120574))

1(1minus120574)

120575minus2(1minus2120574)

1199062(1minus2120574)

120572+2)

(41)

Choosing 120573 = max1205752(1 minus 120574)1(1minus120574) ((1 minus 2120574)(1 minus 120574))

1(1minus120574)

120575minus2(1minus2120574)

we obtain

100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 120573 (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2) (42)

Therefore we yield

(119873(119905))1(1minus120574)

= (1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909)

1(1minus120574)

le 21(1minus120574)

(119872(119905) +

100381610038161003816100381610038161003816100381610038161003816int1

0

119906(119909 119905)119906119905(119909 119905)119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

)

le 119862 (119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2)

(43)

where 119862 depends on 120575 and 120572 From (37) and (43) we have

120578(119873 (119905))1(1minus120574)

le119889119873 (119905)

119889119905 (44)

where 120578 = 120581119862 Integrating (44) over (0 119905) then we get

1

(119873 (0))minus120572(120572+4)

minus (120572 (120572 + 4)) 120578119905le (119873(119905))

120572(120572+4)

(45)

Hence119873(119905) blows up in finite time 119879max 119879max is given by theinequality as below

119879max le120572 + 4

120572120578(119873 (0))

minus120572(120572+4)

(46)

Consequently the solution blows up in finite time And theproof of Theorem 3 is now finished

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors would like to thank the referees for the carefulreading of this paper and for the valuable suggestions toimprove the presentation and style of the paper

References

[1] J-Q Wu and S-J Li ldquoGlobal solution and blow-up solutionfor a nonlinear damped beam with source termrdquo AppliedMathematics vol 25 no 4 pp 447ndash453 2010

[2] H A Levine P Pucci and J Serrin ldquoSome remarks on globalnonexistence for nonautonomous abstract evolution equationsrdquoContemporary Mathematics vol 208 pp 253ndash263 1997

[3] H A Levine and J Serrin ldquoGlobal nonexistence theorems forquasilinear evolution equations with dissipationrdquo Archive forRational Mechanics and Analysis vol 137 no 4 pp 341ndash3611997

[4] V Georgiev and G Todorova ldquoExistence of a solution of thewave equation with nonlinear damping and source termsrdquoJournal of Differential Equations vol 109 no 2 pp 295ndash3081994

[5] J Hao and S Li ldquoGlobal solutions and blow-up solutions fora nonlinear string with boundary input and outputrdquo NonlinearAnalysis Theory Methods and Applications vol 66 no 1 pp131ndash137 2007

[6] U Dinlemez ldquoGlobal existence uniqueness of weak solutionsand determining functionals for nonlinear wave equationsrdquoAdvances in Pure Mathematics vol 3 pp 451ndash457 2013

[7] Y Guo and M A Rammaha ldquoGlobal existence and decayof energy to systems of wave equations with damping andsupercritical sourcesrdquo Zeitschrift fur Angewandte Mathematikund Physik vol 64 no 3 pp 621ndash658 2013

[8] L BociuMRammaha andDToundykov ldquoOn awave equationwith supercritical interior and boundary sources and dampingtermsrdquo Mathematische Nachrichten vol 284 no 16 pp 2032ndash2064 2011


[9] C O Alves M M Cavalcanti V N Domingos CavalcantiM A Rammaha and D Toundykov ldquoOn existence uniformdecay rates and blow up for solutions of systems of nonlinearwave equations with damping and source termsrdquo Discrete andContinuous Dynamical Systems vol 2 no 3 pp 583ndash608 2009

[10] M M Cavalcanti V N Domingos Cavalcanti and I LasieckaldquoWell-posedness and optimal decay rates for the wave equationwith nonlinear boundary damping-source interactionrdquo Journalof Differential Equations vol 236 no 2 pp 407ndash459 2007

[11] C O Alves and M M Cavalcanti ldquoOn existence uniformdecay rates and blow up for solutions of the 2-D wave equationwith exponential sourcerdquo Calculus of Variations and PartialDifferential Equations vol 34 no 3 pp 377ndash411 2009

[12] M A Rammaha ldquoThe influence of damping and source termson solutions of nonlinear wave equationsrdquo Boletim da SociedadeParanaense de Matematica vol 25 no 1-2 pp 77ndash90 2007

[13] V Barbu I Lasiecka and M A Rammaha ldquoExistence anduniqueness of solutions to wave equations with nonlineardegenerate damping and source termsrdquo Control and Cybernet-ics vol 34 no 3 pp 665ndash687 2005

[14] M M Cavalcanti and V N Domingos Cavalcanti ldquoExistenceand asymptotic stability for evolution problems on manifoldswith damping and source termsrdquo Journal of MathematicalAnalysis and Applications vol 291 no 1 pp 109ndash127 2004

[15] M M Cavalcanti V N D Cavalcanti J S Prates Filho andJ A Soriano ldquoExistence and uniform decay of solutions of aparabolic-hyperbolic equation with nonlinear boundary damp-ing and boundary source termrdquo Communications in Analysisand Geometry vol 10 no 3 pp 451ndash466 2002

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of


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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of


OptimizationJournal of


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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Lemma 1 Let 1 lt 120572 and 119906(119909 119905) be a solution of the problem(1)ndash(3) Then the energy equation of the problem (1)ndash(3) is

119864 (119905) =1

2

100381710038171003817100381711990611990510038171003817100381710038172

2minus

1

120572 + 2119906120572+2

120572+2+

1

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

(4)

119889

119889119905119864 (119905) = 120601

2

(119905) minus 1205952

(119905) minus1003817100381710038171003817119906119905

10038171003817100381710038172

2 (5)

Proof Multiplying (1) with 119906119905and integrating over (0 1) then

we get

119889

1198891199051

2

100381710038171003817100381711990611990510038171003817100381710038172

2minus

1

120572 + 2119906120572+2

120572+2

= int1

0

(120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 119906119909)119909

119906119905119889119909 minus

100381710038171003817100381711990611990510038171003817100381710038172

2

(6)

Applying integration by parts in the right hand side of (6)we find

int1

0

(120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 119906119909)119909

119906119905119889119909 = minus120590 (

1003816100381610038161003816119906119909 (0 119905)10038161003816100381610038162

) 119906119909(0 119905) 119906

119905(0 119905)

minus1

2

119889

119889119905int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

(7)

And using boundary conditions in equality (7) we obtain

119889

1198891199051

2

100381710038171003817100381711990611990510038171003817100381710038172

2minus

1

120572 + 2119906120572+2

120572+2+

1

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

= 1206012

(119905) minus 1205952

(119905) minus1003817100381710038171003817119906119905

10038171003817100381710038172

2

(8)

Hence the proof is completed

Next we give the following theorem for global solutionsin time

Theorem 2 Assume that 119906(119909 119905) is a solution of the problem(1)ndash(3) with 1 lt 120572 and

(i) 120590(119904) satisfies the following condition

|119904|120572

le 120590 (119904) for 119904 isin R+

cup 0 (9)

(ii) the input and the output functions satisfy

1206012

(119905) le 1205952

(119905) (10)

Then the solution 119906(119909 119905) is global in time

Proof Let

119866 (119905) = 119864 (119905) +2

120572 + 2119906120572+2

120572+2

=1

2

100381710038171003817100381711990611990510038171003817100381710038172

2+

1

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909 +1

120572 + 2119906120572+2

120572+2

(11)

Differentiating 119866(119905) with respect to 119905 and using (5) we get

119889

119889119905119866 (119905) = 120601

2

(119905) minus 1205952

(119905) minus1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 2int1

0

|119906|120572

119906119906119905119889119909 (12)

Using the Cauchy-Schwarz inequality in the last term of (12)we obtain

2int1

0

|119906|120572

119906119906119905119889119909 le 2int

1

0

|119906|120572+1 1003816100381610038161003816119906119905

1003816100381610038161003816 119889119909

le 1199062(120572+1)

2(120572+1)+1003817100381710038171003817119906119905

10038171003817100381710038172

2

(13)

and it follows from (12) (13) and (10) that we have

119889

119889119905119866 (119905) le 119906

2(120572+1)

2(120572+1)+1003817100381710038171003817119906119905

10038171003817100381710038172

2 (14)

By assumption (9) and integrating over (0 |119906119909|2

) and (0 1)respectively we yield

1

120572 + 1

100381710038171003817100381711990611990910038171003817100381710038172(120572+1)

2(120572+1)le int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909 (15)

Furthermore we have

|119906 (119909 119905)|2(120572+1)

=

100381610038161003816100381610038161003816100381610038161003816int1

119909

119906120585(120585 119905) 119889120585

100381610038161003816100381610038161003816100381610038161003816

2(120572+1)

le int1

119909

10038161003816100381610038161003816119906120585(120585 119905)

10038161003816100381610038161003816

2(120572+1)

119889120585

le int1

0

1003816100381610038161003816119906119909 (119909 119905)10038161003816100381610038162(120572+1)

119889119909 =1003817100381710038171003817119906119909 (119909 119905)

10038171003817100381710038172(120572+1)

2(120572+1)

(16)

and then

119906 (119909 119905)2(120572+1)

2(120572+1)le

1003817100381710038171003817119906119909 (119909 119905)10038171003817100381710038172(120572+1)

2(120572+1) (17)

Combining (11) (14) (15) and (17) we get

119889119866 (119905)

119889119905le

1

1205851

119866 (119905) (18)

where 1205851= min12 1(120572 + 1) Using Gronwallrsquos inequality

we have

119866 (119905) le 119866 (0) 119890(11205851)119905

(19)

Therefore together with the continuation principle and thedefinition of 119866(119905) we complete the proof of Theorem 2

Then we give the following theorem for the blow-upsolutions of the problem (1)ndash(3)

Theorem 3 Let 119906(119909 119905) be a solution of the problem (1)ndash(3)with 1 lt 120572 Assume that

(i) there exists 1 lt 120576 lt (120572 + 2)2 such that the function120590(119904) satisfies

120590 (119904) 119904 le120576

2int119904

0

120590 (120577) 119889120577 for 119904120598R+

cup 0 (20)



119864 (0) le 0 0 lt int1

0

1199060(119909) 1199061(119909) 119889119909 (21)


1206012

(119905) le 1205952

(119905)

(120595 (119905) + 120601 (119905)) (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0)) le 0

(22)

(iv) 119906(119909 119905) satisfies 1 le 119906


119879max le (120572 + 4

120572120578)119873minus120572(120572+4)

(0) (23)


Proof We define

119872(119905) = minus119864 (119905) 120574 =120572

2 (120572 + 2) (24)

119873(119905) = 1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909 (25)


119889119872 (119905)

119889119905=

100381710038171003817100381711990611990510038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905) ge 0 (26)

0 le 119872 (0) le 119872 (119905) for 0 le 119905 (27)


119889119873 (119905)

119889119905

= (1 minus 120574)119872minus120574

(119905)1198721015840

(119905) + int1

0

1199062

119905119889119909 + int

1

0

119906119906119905119905119889119909

= (1 minus 120574)119872minus120574

(119905) (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905)) +1003817100381710038171003817119906119905

10038171003817100381710038172

2

+ int1

0

119906119906119905119905119889119909

(28)


int1

0

119906119906119905119905119889119909

= 119906120572+2

120572+2minus int1

0

119906119906119905119889119909 minus (120595 (119905) + 120601 (119905))

times (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0))

minus int1

0

120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909119889119909

(29)


0 = 120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

+120576

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

minus120576

120572 + 2119906120572+2

120572+2

(30)


119889119873 (119905)

119889119905= (1 minus 120574)119872

minus120574

(119905) (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905)) +1003817100381710038171003817119906119905

10038171003817100381710038172

2

+ 119906120572+2

120572+2minus int1

0

119906119906119905119889119909 minus (120595 (119905) + 120601 (119905))

times (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0))

minus int1

0

120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909119889119909 + 120576119872 (119905) +

120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2

+120576

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

minus120576

120572 + 2119906120572+2

120572+2

(31)


(1 +120576

2)1003817100381710038171003817119906119905

10038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

2119906120572+2

120572+2

+ int1

0

(120576

2int|119906119909|2

0

120590 (120585) 119889120585 minus 120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909)119889119909

+ 120576119872 (119905) minus int1

0

119906119906119905119889119909 le

119889119873 (119905)

119889119905

(32)


119860119861 le120575119901

119901119860119901

+120575minus119902

119902119861119902

0 le 119860 119861 forall0 lt 1205751

119901+

1

119902= 1

(33)

for int10


int1

0

119906119906119905119889119909 le int

1

0

10038161003816100381610038161199061199061199051003816100381610038161003816 119889119909 le

100381710038171003817100381711990611990510038171003817100381710038172

+1

41199062

(34)


2le

119906120572+2


(1 +120576

2)1003817100381710038171003817119906119905

10038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

21199062

2

+ int1

0

(120576

2int|119906119909|2

0

120590 (120585) 119889120585 minus 120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909)119889119909 + 120576119872 (119905)

minus1003817100381710038171003817119906119905

10038171003817100381710038172

2minus

1

41199062

2le

119889119873 (119905)

119889119905

(35)


From (20) we get

120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

41199062

2le

119889119873 (119905)

119889119905

(36)


120581 119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2 le

119889119873 (119905)

119889119905 (37)


0 lt 119873 (0) le 119873 (119905) forall0 lt 119905 (38)


100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le 1199062

100381710038171003817100381711990611990510038171003817100381710038172 le 119906

120572+2

100381710038171003817100381711990611990510038171003817100381710038172 (39)


0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le

1205752(1minus120574)

2 (1 minus 120574)

100381710038171003817100381711990611990510038171003817100381710038172(1minus120574)

2

+1 minus 2120574


1199062(1minus120574)(1minus2120574)

120572+2

(40)


100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 21(1minus120574)

(1205752

(2 (1 minus 120574))1(1minus120574)

100381710038171003817100381711990611990510038171003817100381710038172

2

+(1 minus 2120574

2 (1 minus 120574))

1(1minus120574)

120575minus2(1minus2120574)

1199062(1minus2120574)

120572+2)

(41)


1(1minus120574)

120575minus2(1minus2120574)

we obtain

100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 120573 (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2) (42)

Therefore we yield

(119873(119905))1(1minus120574)

= (1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909)

1(1minus120574)

le 21(1minus120574)

(119872(119905) +

100381610038161003816100381610038161003816100381610038161003816int1

0

119906(119909 119905)119906119905(119909 119905)119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

)

le 119862 (119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2)

(43)


120578(119873 (119905))1(1minus120574)

le119889119873 (119905)

119889119905 (44)


1

(119873 (0))minus120572(120572+4)

minus (120572 (120572 + 4)) 120578119905le (119873(119905))

120572(120572+4)

(45)


119879max le120572 + 4

120572120578(119873 (0))

minus120572(120572+4)

(46)




Acknowledgment


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119864 (0) le 0 0 lt int1

0

1199060(119909) 1199061(119909) 119889119909 (21)


1206012

(119905) le 1205952

(119905)

(120595 (119905) + 120601 (119905)) (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0)) le 0

(22)

(iv) 119906(119909 119905) satisfies 1 le 119906


119879max le (120572 + 4

120572120578)119873minus120572(120572+4)

(0) (23)


Proof We define

119872(119905) = minus119864 (119905) 120574 =120572

2 (120572 + 2) (24)

119873(119905) = 1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909 (25)


119889119872 (119905)

119889119905=

100381710038171003817100381711990611990510038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905) ge 0 (26)

0 le 119872 (0) le 119872 (119905) for 0 le 119905 (27)


119889119873 (119905)

119889119905

= (1 minus 120574)119872minus120574

(119905)1198721015840

(119905) + int1

0

1199062

119905119889119909 + int

1

0

119906119906119905119905119889119909

= (1 minus 120574)119872minus120574

(119905) (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905)) +1003817100381710038171003817119906119905

10038171003817100381710038172

2

+ int1

0

119906119906119905119905119889119909

(28)


int1

0

119906119906119905119905119889119909

= 119906120572+2

120572+2minus int1

0

119906119906119905119889119909 minus (120595 (119905) + 120601 (119905))

times (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0))

minus int1

0

120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909119889119909

(29)


0 = 120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

+120576

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

minus120576

120572 + 2119906120572+2

120572+2

(30)


119889119873 (119905)

119889119905= (1 minus 120574)119872

minus120574

(119905) (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 1205952

(119905) minus 1206012

(119905)) +1003817100381710038171003817119906119905

10038171003817100381710038172

2

+ 119906120572+2

120572+2minus int1

0

119906119906119905119889119909 minus (120595 (119905) + 120601 (119905))

times (int119905

0

(120595 (119904) minus 120601 (119904)) 119889119904 + 1199060(0))

minus int1

0

120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909119889119909 + 120576119872 (119905) +

120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2

+120576

2int1

0

int|119906119909|2

0

120590 (120585) 119889120585 119889119909

minus120576

120572 + 2119906120572+2

120572+2

(31)


(1 +120576

2)1003817100381710038171003817119906119905

10038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

2119906120572+2

120572+2

+ int1

0

(120576

2int|119906119909|2

0

120590 (120585) 119889120585 minus 120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909)119889119909

+ 120576119872 (119905) minus int1

0

119906119906119905119889119909 le

119889119873 (119905)

119889119905

(32)


119860119861 le120575119901

119901119860119901

+120575minus119902

119902119861119902

0 le 119860 119861 forall0 lt 1205751

119901+

1

119902= 1

(33)

for int10


int1

0

119906119906119905119889119909 le int

1

0

10038161003816100381610038161199061199061199051003816100381610038161003816 119889119909 le

100381710038171003817100381711990611990510038171003817100381710038172

+1

41199062

(34)


2le

119906120572+2


(1 +120576

2)1003817100381710038171003817119906119905

10038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

21199062

2

+ int1

0

(120576

2int|119906119909|2

0

120590 (120585) 119889120585 minus 120590 (1003816100381610038161003816119906119909

10038161003816100381610038162

) 1199062

119909)119889119909 + 120576119872 (119905)

minus1003817100381710038171003817119906119905

10038171003817100381710038172

2minus

1

41199062

2le

119889119873 (119905)

119889119905

(35)


From (20) we get

120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

41199062

2le

119889119873 (119905)

119889119905

(36)


120581 119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2 le

119889119873 (119905)

119889119905 (37)


0 lt 119873 (0) le 119873 (119905) forall0 lt 119905 (38)


100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le 1199062

100381710038171003817100381711990611990510038171003817100381710038172 le 119906

120572+2

100381710038171003817100381711990611990510038171003817100381710038172 (39)


0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le

1205752(1minus120574)

2 (1 minus 120574)

100381710038171003817100381711990611990510038171003817100381710038172(1minus120574)

2

+1 minus 2120574


1199062(1minus120574)(1minus2120574)

120572+2

(40)


100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 21(1minus120574)

(1205752

(2 (1 minus 120574))1(1minus120574)

100381710038171003817100381711990611990510038171003817100381710038172

2

+(1 minus 2120574

2 (1 minus 120574))

1(1minus120574)

120575minus2(1minus2120574)

1199062(1minus2120574)

120572+2)

(41)


1(1minus120574)

120575minus2(1minus2120574)

we obtain

100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 120573 (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2) (42)

Therefore we yield

(119873(119905))1(1minus120574)

= (1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909)

1(1minus120574)

le 21(1minus120574)

(119872(119905) +

100381610038161003816100381610038161003816100381610038161003816int1

0

119906(119909 119905)119906119905(119909 119905)119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

)

le 119862 (119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2)

(43)


120578(119873 (119905))1(1minus120574)

le119889119873 (119905)

119889119905 (44)


1

(119873 (0))minus120572(120572+4)

minus (120572 (120572 + 4)) 120578119905le (119873(119905))

120572(120572+4)

(45)


119879max le120572 + 4

120572120578(119873 (0))

minus120572(120572+4)

(46)




Acknowledgment


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










From (20) we get

120576119872 (119905) +120576

2

100381710038171003817100381711990611990510038171003817100381710038172

2+ (

1

2minus

120576

120572 + 2) 119906120572+2

120572+2+

1

41199062

2le

119889119873 (119905)

119889119905

(36)


120581 119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2 le

119889119873 (119905)

119889119905 (37)


0 lt 119873 (0) le 119873 (119905) forall0 lt 119905 (38)


100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le 1199062

100381710038171003817100381711990611990510038171003817100381710038172 le 119906

120572+2

100381710038171003817100381711990611990510038171003817100381710038172 (39)


0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816le

1205752(1minus120574)

2 (1 minus 120574)

100381710038171003817100381711990611990510038171003817100381710038172(1minus120574)

2

+1 minus 2120574


1199062(1minus120574)(1minus2120574)

120572+2

(40)


100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 21(1minus120574)

(1205752

(2 (1 minus 120574))1(1minus120574)

100381710038171003817100381711990611990510038171003817100381710038172

2

+(1 minus 2120574

2 (1 minus 120574))

1(1minus120574)

120575minus2(1minus2120574)

1199062(1minus2120574)

120572+2)

(41)


1(1minus120574)

120575minus2(1minus2120574)

we obtain

100381610038161003816100381610038161003816100381610038161003816int1

0

119906119906119905119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

le 120573 (1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2) (42)

Therefore we yield

(119873(119905))1(1minus120574)

= (1198721minus120574

(119905) + int1

0

119906 (119909 119905) 119906119905(119909 119905) 119889119909)

1(1minus120574)

le 21(1minus120574)

(119872(119905) +

100381610038161003816100381610038161003816100381610038161003816int1

0

119906(119909 119905)119906119905(119909 119905)119889119909

100381610038161003816100381610038161003816100381610038161003816

1(1minus120574)

)

le 119862 (119872 (119905) +1003817100381710038171003817119906119905

10038171003817100381710038172

2+ 119906120572+2

120572+2+ 1199062

2)

(43)


120578(119873 (119905))1(1minus120574)

le119889119873 (119905)

119889119905 (44)


1

(119873 (0))minus120572(120572+4)

minus (120572 (120572 + 4)) 120578119905le (119873(119905))

120572(120572+4)

(45)


119879max le120572 + 4

120572120578(119873 (0))

minus120572(120572+4)

(46)




Acknowledgment


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article Global and Blow-Up Solutions for Nonlinear …downloads.hindawi.com/journals/ijde/2014/724837.pdf · 2019. 7. 31. · wave equation with nonlinear damping and source