Cent. Eur. J. Math. • 9(5) • 2011 • 1156-1163DOI: 10.2478/s11533-011-0052-9

Central European Journal of Mathematics

Positivity conditions and bounds for Green’s functionsfor higher order two-point BVP

Research Article

Michael I. Gil’1∗

1 Department of Mathematics, Ben Gurion University of the Negev, P.O. Box 653, Beer-Sheva 84105, Israel

Received 30 November 2010; accepted 4 May 2011

Abstract: We consider a linear nonautonomous higher order ordinary differential equation and establish the positivity con-ditions and two-sided bounds for Green’s function for the two-point boundary value problem. Applications of theobtained results to nonlinear equations are also discussed.

MSC: 34CB05, 34CB27, 34B15

Keywords: Green’s function • Ordinary differential equation • Boundary value problem • Positivity • The Lidstone equation© Versita Sp. z o.o.

1. Introduction and preliminaries

The boundary value problem continues to attract attention of many specialists despite its long history. It is still oneof the most burning problems of the theory of ODEs, because of the absence of its complete solution. Classical resultson BVPs can be found, for instance, in the books [1, 17]. For the recent investigations of BVPs for linear and nonlinearequations see papers [4, 14–16, 20]. In particular, in the paper [14] the authors study a two-point boundary value problemfor the Riccati matrix differential equation of a certain form. They find conditions guaranteeing a unique solution to theconsidered problem. In the paper [10], for some matrix ordinary differential equations the authors give a matrix versionof Green’s functions. In the paper [20], a vector semilinear second-order two-point boundary value problem is studied.By using Krasnonsel’skii’s fixed point theorem of cone expansion-compressing type, some results on the existence of oneand two positive solutions are established. The paper [15] presents a theory of differential inequalities for two-pointboundary value problems associated with the system of the nth-order nonlinear differential equations. Moreover, theauthors establish the existence and uniqueness of solutions to three-point BVPs associated with the system of the nth-order nonlinear differential equations by using the idea of matching solutions. The authors of the paper [4] deal with the

∗ E-mail: gilmi@bezeqint.net

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M.I. Gil’

singular perturbation of the Dirichlet problem and Robin problem for a nonlinear systems of ODEs. Under appropriateconditions they prove the existence of solutions and give asymptotic solution estimates. Positivity of Green’s functions forthe Cauchy problem for higher order ODEs was investigated in the book [11] (equations with almost periodic coefficientson the whole line), and in the papers [7] (equations with alternate coefficients on the half-line), and [8] (equations withpositive coefficients on the half-line). Various existence results for positive solutions of nonlinear ODEs can be found inthe following well-known publications [2, 5, 6, 13, 18]. Our main object in this paper is the equation

n∑

k=0

an−k (x)Dku(x) = f(x), a0 ≡ 1, x ∈ (0, 1), D = − d2

dx2 , (1)

with real continuous coefficients ak (x), k = 1, . . . , n, and a given f . In addition, the boundary conditions

u(2k)(0) = u(2k)(1) = 0, k = 0, . . . , n− 1, (2)

are imposed. The equation of this type is well known under the name of Lidstone equation and has been widely studiedunder different kinds of conditions, in particular various positivity conditions, see [3, 9, 19]. In the paper [3], the equation

(−1)nu(2n)(x) = F(x, u(x), . . . , u(2n−1)(x)

), 0 < x < 1, (3)

with conditions (2) was considered. Here F is a real function. Under certain monotonicity conditions on nonlinearityof F an existence result for the considered problem is proved, which is based on the upper and lower solution methodcombined with the monotone iterative technique. The paper [9] also deals with problem (3)–(2), where the monotonemethod in the presence of upper and lower solutions is also developed. The paper [19] is concerned with the solutionsto a class of the 2nth-order Lidstone boundary value problems. Sufficient conditions for the existence and uniquenessof a solution are given. A monotone iteration is developed so that the iteration sequence converges monotonically toa maximal or a minimal solution. Certainly we could not survey the whole subject here and refer the reader to theabove listed publications and references given therein. In the present paper we establish the positivity conditions andtwo-sided bounds for Green’s function for the problem (1)–(2). Applications of the obtained results to nonlinear equationsare also discussed. We have restricted ourselves to the Dirichlet conditions only for simplicity; our reasonings beloware valid in the cases of the Neumann and mixed selfadjoint boundary conditions.

Let L2(0, 1) be the space of real functions defined on [0, 1] with the finite norm ‖f‖L2 =(∫ 1

0 f2(x)dx

)1/2and the unit

operator I. By L2+ we denote the cone of nonnegative functions from L2(0, 1).

We begin with the equationn∑

k=0

bn−kDku(x) = f(x), 0 < x < 1, b0 = 1, (4)

with constant real coefficients bk and an f ∈ L2(0, 1). It is assumed that all the roots rj , j = 1, . . . , n, of the polynomial

P(z) =n∑

k=0

bn−kzk ,

counted with their multiplicities, are real and satisfy

0 ≤ r1 ≤ . . . ≤ rn < π2. (5)

Recall that Green’s function for the problem (4)–(2) is the function G(x, s) defined for 0 ≤ x, s ≤ 1 and solving theproblem

n∑

k=0

bk∂2(n−k)G(x, s)∂x2(n−k) = δ(x − s), 0 < x, s < 1,

∂2kG(0, s)∂x2k = ∂2kG(1, s)

∂x2k = 0, k = 0, . . . , n− 1,

where δ(x − s) is the Dirac delta function.

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Positivity conditions and bounds for Green’s functions for higher order two-point BVP

Lemma 1.1.Let condition (5) hold. Then Green’s function for the problem (4)–(2) is nonnegative.

Proof. On the set

DomE ={u ∈ L2(0, 1) : u(k) ∈ L2(0, 1), k = 1, . . . , 2n; u(2j)(0) = u(2j)(1) = 0, j = 0, . . . , n− 1

}

define the operator

E0w(x) =n∑

k=0

bn−kDkw(x), w ∈ DomE.

For simplicity we put D − a2I = D − a2 for any number a. Clearly, E0 = P(D) = (D − r1) . . . (D − rn) and the inverseto E0 is

E−10 = (D − r1)−1 . . . (D − rn)−1. (6)

But for any real a, we have

(D − a2)−1 f(x) = −

(d2

dx2 + a2)−1

f(x) =∫ 1

0G2(a, x, s) f(s)ds, (7)

where G2(a, x, s) is Green’s function to the operator D − a2 defined on

DomD ={u ∈ L2(0, 1) : u(k) ∈ L2(0, 1), k = 1, 2; u(0) = u(1) = 0

}.

As it is well known,

G2(a, x, s) = 1a sina

{sin [a(1− s)] sin (ax) if 0 ≤ x ≤ s ≤ 1,sin (as) sin [a(1− x)] if 0 ≤ s ≤ x ≤ 1,

if a 6= 0, and G2(0, x, s) = G0(x, s), where

G0(x, s) ={x(1− s) if 0 ≤ x ≤ s ≤ 1,s(1− x) if 0 ≤ s ≤ x ≤ 1,

cf. [1, 17]. Clearly, G2(a, x, s) ≥ 0, provided 0 ≤ a < π. Hence due to (5), (D − rk )−1 =(D − (√rk )2)−1 ≥ 0. Thanks to

(6) we arrive at the required result.

Note that the positivity condition (5) was established in [19], but we prove it for the convenience of the reader. Inparticular, the relations (6) and (7) will be considerably used in the next sections.

2. The main result

Green’s function G(t, s) to the equation (1) is defined absolutely similarly as in the case of equation (4). To formulatethe result, introduce the notations

bk = sup0≤x≤1

ak (x), ck = sup0≤x≤1

(bk − ak (x)) = sup0≤x≤1

ak (x)− inf0≤x≤1

ak (x),

η =n∑

j=1

cjπ2(n−j) and ζ =n∏

k=1

(π2 − rk

)= P

(π2) .

Recall that rk , k = 1, . . . , n, are the roots of the polynomial P(z) defined in the previous section.

Now we are in a position to formulate our main result.

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M.I. Gil’

Theorem 2.1.Let the conditions (5) and

η < ζ (8)

hold. Then Green’s function to the problem (1)–(2) is nonnegative.

Proof. On DomE define the operator

Ew(x) =n∑

k=0

an−k (x)Dkw(x), w ∈ DomE, x ∈ (0, 1).

The equation (1) takes the formEu = f.

The function G is nonnegative if and only if, for any f ∈ L2+, a solution u ∈ DomE of problem (1)–(2) is nonnegative.

Put

W = E0 − E =n∑

k=1

[bk − ak (x)]Dn−k .

We haveEu = E0u−Wu = f.

Substitute u = E−10 y into the last equation. Then we obtain

y−WE−10 y = f.

Butn∑

k=0

bkDn−k =( n∑

k=0

bkD−k)Dn

and, according to (6),

WE−10 =

n∑

k=1

[bk − ak (x)]D−k

n∑

j=0

bjD−j

−1

.

Hence

WE−10 =

n∑

j=1

((bj − aj (x)

)D−j

n∏

k=1

(I − rkD−1)−1 . (9)

Take into account that the smallest eigenvalue of D is π2. Therefore, the spectral radius ρs(D−1) of D−1 satisfies theequality ρs(D−1) = 1/π2. But D is selfadjoint. Therefore,

∥∥D−1∥∥L2 = 1

π2 . (10)

Here and below ‖A‖L2 means the operator norm of a linear operator A in L2. Hence by (5),

(I − rkD−1)−1 =

∞∑

j=0

(rkD−1)j > 0.

Consequently, by (9) the operator WE−10 is positive. In addition,

∥∥∥(I − rkD−1)−1

∥∥∥L2≤ 1

1− rk /π2 . (11)

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Positivity conditions and bounds for Green’s functions for higher order two-point BVP

According to (9),

∥∥WE−10 f∥∥L2 ≤

∥∥∥∥∥∥

n∑

j=1

[bj − aj (x)]D−j( n∑

k=0

bkD−k)−1

f

∥∥∥∥∥∥L2

≤n∑

j=1

cjπ2j

n∏

k=1

11− rk /π2 ‖f‖L2 .

Thus ∥∥WE−10 f∥∥L2 ≤

ηζ ‖f‖L2 , f ≥ 0. (12)

Since WE−10 ≥ 0, its spectral radius ρs(WE−1

0 ) is less than η/ζ. If condition (8) is fulfilled, then

y =∞∑

k=0

(WE−1

0)k f. (13)

Consequently, y ≥ f , provided f ∈ L2+, and thanks to Lemma 1.1,

E−1f = u = E−10 y ≥ E−1

0 f ≥ 0, f ∈ L2+. (14)

This means that Green’s function is positive.

3. Bounds for Green’s function

Let C (0, 1) be the space of real continuous functions defined on [0, 1] with the sup norm ‖ · ‖C and C+ be the cone ofnonnegative functions from C (0, 1). Put

γ =(π2 − r1

) √r14√

3 sin√r1if r1 > 0, and γ = π2

4√

3if r1 = 0.

Theorem 3.1.Let conditions (5) and (8) hold. Then ∫ 1

0G(x, s)ds ≤ γ

ζ − η (15)

and

G(x, s) ≥∫ 1

0G0(x, s1)

∫ 1

0G0(s1, s2) . . .

∫ 1

0G0(sn−1, s)dsn−1 . . . ds2 ds1, 0 ≤ s, x ≤ 1. (16)

Proof. The inequality (12) yields∥∥(WE−1

0 )k∥∥L2 ≤ (η/ζ)k . Recall that ‖A‖L2 means the operator norm of a linear

operator A in L2. Therefore, by (13),

‖y‖L2 ≤∞∑

k=0

∥∥WE−10∥∥kL2 ‖f‖L2 ≤

∞∑

k=0

(ηζ

)k‖f‖L2 = ζ

ζ − η ‖f‖L2 . (17)

According to (10) and (11),

∥∥E−10∥∥L2 ≤

∥∥(D − r1)−1 . . . (D − rn)−1∥∥L2 ≤

n∏

k=1

1π2 − rk

= 1P(π2) = 1

ζ . (18)

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M.I. Gil’

Now (17) implies ‖u‖L2 =∥∥E−1

0 y∥∥L2 ≤ (ζ − η)−1‖f‖L2 . So

∥∥E−1∥∥L2 ≤

1ζ − η .

Since sinb ≤ b, b > 0, according to (7) we have

G2(a, x, s) ≤a

sinaG0(x, s) = asina

{x(1− s) if x ≤ s,(1− x)s if x ≥ s, 0 < a < π.

Clearly,

∫ 1

0G2

0 (x, s)ds = x2∫ 1

x(1− s)2 ds+ (1− x)2

∫ x

0s ds = 1

3[x2(1− x)3 + x3(1− x)2] = 1

3x2(1− x)2 ≤ 1

48 , 0 ≤ x ≤ 1.

But by the Schwarz inequality

∥∥∥(D − a2)−1 f

∥∥∥C≤[∫ 1

0G2

2 (a, x, s)ds]1/2

‖f‖L2 ≤a

4√

3 sina‖f‖L2 .

Put ‖A‖L2→C = supf∈L2 ‖Af‖C /‖f‖L2 . Then

∥∥(D − r1)−1∥∥L2→C ≤ γ

(π2 − r1

)−1 .

Thus, (18) yields the inequality

∥∥E−10∥∥L2→C =

∥∥∥∥∥

n∏

k=1

(D − rk )−1

∥∥∥∥∥L2→C

≤∥∥(D − r1)−1∥∥

L2→C

∥∥∥∥∥

n∏

k=2

(D − rk )−1

∥∥∥∥∥L2

≤ γζ .

Take into account that u = E−1f = E−10 y. Then by (17),

∥∥E−1f∥∥C =

∥∥E−10 y∥∥C ≤

∥∥E−10∥∥L2→C ‖y‖L2 ≤

‖f‖L2γζ − η ≤

‖f‖Cγζ − η .

Taking f ≡ 1, we arrive at (15). Furthermore, since the norm of rkD−1 in L2 is less than one, we have

(D − rk )−1 = D−1 (I − rkD−1)−1 = D−1∞∑

j=0

(rkD−1)j ≥ D−1, k = 1, . . . , n,

and thus E−10 ≥ D−n. So according to (14), E−1 ≥ D−n. This proves (16).

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Positivity conditions and bounds for Green’s functions for higher order two-point BVP

4. Nonlinear equations

Put C+(a) = {f ∈ C+ : f(x) ≤ a, 0 ≤ x ≤ 1} for a positive number a ≤ ∞. Let φ be a monotone mapping defined onC+(a): if f ≥ h for f, h ∈ C+(a), then φ(f) ≥ φ(h).

Consider the problemn∑

k=0

ak (x)y2(n−k)(x) = [φ(y)] (x), 0 < x < 1, (19)

y(2k)(0) = y(2k)(1), k = 0, . . . , n− 1. (20)

It is equivalent to the equation

y(x) =∫ 1

0G(x, s)[φ(y)](s)ds.

It is assumed that there are constants q0, l ≥ 0, such that

φ(f) ≤ q0f + l, f ∈ C+(a). (21)

Under the hypothesis of Theorem 2.1, by (15), for a f ∈ C+(a) we have

∫ 1

0G(x, s) [φ(f)] (s)ds ≤ q0a

∫ 1

0G(x, s)ds+ l

∫ 1

0G(x, s)ds ≤ γ(q0a+ l)

ζ − η .

So ifγq0 < ζ − η (22)

andγ (q0a+ l)ζ − η ≤ a, (23)

then

supx

∫ 1

0G(x, s) [φ(f)] (s)ds ≤ a

and by the well-known [12, Theorem 38.1], the considered problem has a solution from C+(a). We thus have proved

Theorem 4.1.Under the assumptions of Theorem 2.1, let φ be a monotone mapping satisfying (21). If, in addition, conditions (22) and(23) hold, then problem (19)–(20) has a solution y ∈ C+(a).

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