Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 1, 1-9; http://www.math.u-szeged.hu/ejqtde/

Positive solutions of boundary value problems for nth order ordinary differential equations ^{1}

Moustafa El-Shahed

Department of Mathematics, College of Education Qassim University, P.O.Box 3771 Unizah, Qassim, Saudi Arabia Tel: 00966501219508 e.mail: elshahedm@yahoo.com

Abstract

In this paper, we investigate the problem of existence and nonexistence of positive solutions for the nonlinear boundary value problem:

u^{(n)}(t) + λa(t)f(u(t)) = 0, 0 < t < 1,

satisfying three kinds of different boundary value conditions. Our analysis relies on Krasnoselskii’s fixed point theorem of cone. An example is also given to illustrate the main results.

1. Introduction

There is currently a great deal of interest in positive solutions for several types of boundary value problems. A large part of the literature on positive solutions to boundary value problems seems to be traced back to Krasnoselskii’s work on nonlinear operator equations [6], especially the part dealing with the theory of cones in Banach spaces. In 1994, Erbe and Wang [3] applied Krasnoselskii’s work to eigenvalue problems to establish intervals of the parameter λ for which there is at least one positive solution. In 1995, Eloe and Henderson [1] obtained the solutions that are positive to a cone for the boundary value problem

u^{(n)}(t) + a(t)f(u) = 0, 0 < t < 1,
u^{(i)}(0) = u^{(n−2)}(1), 0 ≤ i ≤ n − 2.

1AMS subject classification. 34 B15, 34 B18.

Key words: Higher order differential equations, boundary value problems, Krasnoselskii’s fixed point theorem, Green’s function, positive solutions.

Since this pioneering works, a lot research has been done in this area [2, 3, 5, 7, 8, 9]. The purpose of this paper is to establish the existence of positive solutions to nonlinear nth order boundary value problems:

u^{(n)}(t) + λ a(t)f(u(t)) = 0, 0 < t < 1,  (1)
u(0) = u′′(0) = u′′′(0) = ........ = u^{(n−1)} = 0, u′(1) = 0,  (2)
u(0) = u′(0) = u′′(0) = ........ = u^{(n−2)} = 0, u′(1) = 0,  (3)
u(0) = u′(0) = u′′(0) = ........ = u^{(n−2)} = 0, u′′(1) = 0,  (4)

where λ is a positive parameter. Throughout the paper, we assume that C1: f : [0, ∞) → [0, ∞) is continuous C2: a : (0, 1) → [0, ∞) is continuous function such that ∫_{0}^{1} a(t) dt > 0.

2. Preliminaries

For the convenience of the reader, we present here some notations and lemmas that will be used in the proof our main results. Definition 1. Let E be a real Banach space. A nonempty closed convex set K ⊂ E is called cone of E if it satisfies the following conditions:

1. x ∈ K, σ ≥ 0 implies σx ∈ K;

2. x ∈ K, −x ∈ K implies x = 0. Definition 2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets. Lemma 1. Let E be a Banach space and K ⊂ E is a cone in E. Assume that Ω_{1} and Ω_{2} are open subsets of E with 0 ∈ Ω_{1} and \overline{Ω}_{1} ⊂ Ω_{2}. Let T : K ∩ (\overline{Ω}_{2} \ Ω_{1}) ⟶ K be completely continuous operator. In addition suppose either: H1 : ∥T u∥ ≤ ∥u∥, ∀u ∈ K ∩ ∂Ω_{1} and ∥T u∥ ≥ ∥u∥, ∀u ∈ K ∩ ∂Ω_{2} or H2 : ∥T u∥ ≤ ∥u∥, ∀u ∈ K ∩ ∂Ω_{2} and ∥T u∥ ≥ ∥u∥, ∀u ∈ K ∩ ∂Ω_{1}. holds. Then T has a fixed pint in K ∩ (\overline{Ω}_{2} \ Ω_{1}).

3. Green functions and their properties

Lemma 2. Let y ∈ C[0, 1], then the boundary value problem

u_{2}^{(n)}(t) + y(t) = 0, 0 < t < 1,  (5)
u_{2}(0) = u′′_{2}(0) = u′′′_{2}(0) = ...........u_{2}^{(n−1)}(0) = 0, u′_{2}(1) = 0,  (6)

has a unique solution

u_{2}(t) = ∫_{0}^{1} G_{2}(t, s)y(s)ds,

where

G_{2}(t, s) = \{ \begin{matrix}\frac{t(1−s)^{n−2}}{(n−2)!} − \frac{(t−s)^{n−1}}{(n−1)!} & if 0 ≤ s ≤ t ≤ 1; \\ \frac{t(1−s)^{n−2}}{(n−2)!} & if 0 ≤ t ≤ s ≤ 1. \end{matrix}

Proof. Applying the Laplace transform to Eq(5) we get

s^{n}u_{2}(s) − s^{n−2}u′_{2}(0) = −y(s),  (7)

where u_{2}(s)and y(s) is the Laplace transform of u_{2}(t) and y(t) respectively. The Laplace inversion of Eq (7) gives the final solution as:

u_{2}(t) = ∫_{0}^{1} \frac{t(1 − s)^{n−2}}{(n − 2)!} y(s)ds − ∫_{0}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds.  (8)

The proof is complete.

It is obvious that

G_{2}(t, s) ≥ 0 and G_{2}(1, s) ≥ G_{2}(t, s), 0 ≤ t, s ≤ 1.  (9)

Lemma 3. G_{2}(t, s) ≥ q_{2}(t)G_{2}(1, s) for 0 ≤ t, s ≤ 1, where q_{2}(t) = \frac{(n−2)t}{n−1}.

Proof. If t ≥ s, then

\frac{G(t, s)}{G(1, s)} = \frac{(n − 1)t(1 − s)^{n−2} − (t − s)^{n−1}}{(n − 1)(1 − s)^{n−2} − (1 − s)^{n−1}}

= \frac{(n − 1)t(1 − s)^{n−2} − (t − s)(t − s)^{n−2}}{(1 − s)^{n−2}(n − 2 + s)} ≥ \frac{(n − 1)t}{n − 1}.

If t ≤ s, then \frac{G_{2}(t,s)}{G_{2}(1,s)} = t ≥ \frac{(n−1)t}{n−1}. The proof is complete.

u_{3}^{(n)}(t) + y(t) = 0, 0 < t < 1,  (10)
u_{3}(0) = u′_{3}(0) = u′′_{3}(0) = ...... = u_{3}^{n−2}(0) = 0, u′_{3}(1) = 0,  (11)

has a unique solution

u_{3}(t) = ∫_{0}^{1} G_{3}(t, s)y(s)ds,

where

G_{3}(t, s) = \{ \begin{matrix}\frac{t^{n−1}(1−s)^{n−2}}{(n−1)!} − \frac{(t−s)^{n−1}}{(n−1)!} & if 0 ≤ s ≤ t ≤ 1; \\ \frac{t^{n−1}(1−s)^{n−2}}{(n−1)!} & if 0 ≤ t ≤ s ≤ 1. \end{matrix}

Proof. Applying the Laplace transform to Eq(10) we get

s^{n}u_{2}(s) − u_{2}^{n−1}(0) = −y(s),  (12)

where u_{3}(s) is the Laplace transform of u_{3}(t). The Laplace inversion of Eq (12) gives the final solution as:

u_{3}(t) = ∫_{0}^{1} \frac{t^{n−1}(1 − s)^{n−2}}{(n − 1)!} y(s)ds − ∫_{0}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds.  (13)

The proof is complete.

It is obvious that

G_{3}(t, s) ≥ 0 and G_{3}(1, s) ≥ G_{3}(t, s), 0 ≤ t, s ≤ 1.  (14)

Lemma 5. G_{3}(t, s) ≥ q_{3}(t)G_{3}(1, s) for 0 ≤ t, s ≤ 1, where q_{3}(t) = t^{n−1}.

Proof. If t ≥ s, then

\frac{G_{3}(t, s)}{G_{3}(1, s)} = \frac{t^{n−1}(1 − s)^{n−2} − (t − s)^{n−1}}{(1 − s)^{n−2} − (1 − s)^{n−1}}

= \frac{t(t − ts)^{n−2} − (t − s)(t − s)^{n−2}}{s(1 − s)^{n−2}} ≥ t^{n−2} ≥ t^{n−1}.

If t ≤ s, then \frac{G_{3}(t,s)}{G_{3}(1,s)} = t^{n−1}. The proof is complete.

u_{4}^{(n)}(t) + y(t) = 0, 0 < t < 1,  (15)
u_{4}(0) = u′_{4}(0) = u′′_{4}(0) = ........ = u_{4}^{n−2}(0) = 0, u′′_{4}(1) = 0,  (16)

has a unique solution

u_{4}(t) = ∫_{0}^{1} G_{4}(t, s)y(s)ds,

where

G_{4}(t, s) = \{ \begin{matrix}\frac{t^{n−1}(1−s)^{n−3}}{(n−1)!} − \frac{(t−s)^{n−1}}{(n−1)!} & if 0 ≤ s ≤ t ≤ 1; \\ \frac{t^{n−1}(1−s)^{n−3}}{(n−1)!} & if 0 ≤ t ≤ s ≤ 1.. \end{matrix}

The proof of Lemma 6 is very similar to that of Lemma 4 and therefore omitted. It is obvious that

G_{4}(t, s) ≥ 0 and G_{4}(1, s) ≥ G_{4}(t, s), 0 ≤ t, s ≤ 1.  (17)

Lemma 7. G_{4}(t, s) ≥ q_{4}(t)G_{4}(1, s) for 0 ≤ t, s ≤ 1, where q_{4}(t) = \frac{t^{n−1}}{2} .

Proof. If t ≥ s, then

\frac{G_{4}(t, s)}{G_{4}(1, s)} = \frac{t^{n−1}(1 − s)^{n−3} − (t − s)^{n−1}}{(1 − s)^{n−3} − (1 − s)^{n−1}}

≥ \frac{t^{2}(t − ts)^{n−3} − (t − s)^{2}(t − ts)^{n−3}}{(1 − s)^{n−3}(2s − s^{2})} ≥ \frac{t^{n−1}}{2} .

If t ≤ s, then \frac{G_{4}(t,s)}{G_{4}(1,s)} = t^{n−1} ≥ \frac{t^{n−1}}{2} . The proof is complete.

3. Main results

In this section, we will apply Krasnoselskii’s fixed point theorem to the eigenvalue problem (1), (i) (i=2,3,4). We note that u_{i}(t) is a solution of (1),(i) if and only if

u_{i}(t) = λ ∫_{0}^{1} G_{i}(t, s) a(s)f(u_{i}(s)) ds, 0 ≤ t ≤ 1.

For our constructions, we shall consider the Banach space X = C[0, 1] equipped with standard norm ∥u_{i}∥ = max_{0≤t≤1} |u_{i}(t)|, u_{i} ∈ X. We define a cone P by

P = {u_{i} ∈ X : u_{i}(t) ≥ q(t) ∥u_{i}∥, t ∈ [0, 1]} ,

It is easy to see that if u_{i} ∈ P , then ∥u_{i}∥ = u_{i}(1) . Define an integral operator by:

T u_{i}(t) = λ ∫_{0}^{1} G_{i}(t, s) a(s)f(u_{i}(s)) ds, 0 ≤ t ≤ 1, u_{i} ∈ P. (18)

Lemma 8. T (P) ⊂ P.

Proof. Notice from (9), (12) and (15) that, for u_{i} ∈ P, T u_{i}(t) ≥ 0 on [0, 1] and

T u_{i}(t) = λ ∫_{0}^{1} G_{i}(t, s)a(s)f(u_{i}(s)) ds

≥ λq_{i}(t) ∫_{0}^{1} G_{i}(1, s)a(s)f(u_{i}(s)) ds

≥ λq_{i}(t) max_{0≤t≤1} ∫_{0}^{1} G_{i}(t, s) a(s)f(u_{i}(s)) ds

= q(t)∥T u_{i}(t)∥, for all t, s ∈ [0, 1].

Thus, T (P) ⊂ P.

By standard argument, it is easy to see that T : P ⟶ P is a completely continuous operator. Following Sun and Wen [8] , we define some important constants:

A = ∫_{0}^{1} G_{i}(1, s) a(s)q_{i}(s) ds, B = ∫_{0}^{1} G_{i}(1, s) a(s) ds,

F_{0} = lim_{u_{i}→0^{+}} sup \frac{f(u_{i})}{u_{i}}, f_{0} = lim_{u_{i}→0^{+}} inf \frac{f(u_{i})}{u_{i}},

F_{∞} = lim_{u_{i}→+∞} sup \frac{f(u_{i})}{u_{i}}, f_{∞} = lim_{u_{i}→+∞} inf \frac{f(u_{i})}{u_{i}}.

Here we assume that \frac{1}{Af_{∞}} = 0 if f_{∞} → ∞ and \frac{1}{BF_{0}} = ∞ if F_{0} → 0 and \frac{1}{Af_{0}} = 0 if f_{0} → ∞ and \frac{1}{BF_{∞}} = ∞ if F_{∞} → 0.

Theorem 1. Suppose that Af_{∞} > BF_{0}, then for each λ ∈ (\frac{1}{Af_{∞}} , \frac{1}{BF_{0}} ), the problem (1) , (i)(i=2,3,4) has at least one positive solution.

Proof. We choose ϵ > 0 sufficiently small such that (F_{0} + ϵ)λ B ≤ 1. By definition of F_{0}, we can see that there exists an l_{1} > 0, such that f(u_{i}) ≤ (F_{0} + ϵ)u_{i} for 0 < u_{i} ≤ l_{1}. If u_{i} ∈ P with ∥u_{i}∥ = l_{1}, we have

∥T u_{i}(t)∥ = T u_{i}(1) = λ ∫_{0}^{1} G_{i}(1, s) a(s)f(u_{i}(s)) ds

≤ λ ∫_{0}^{1} G_{i}(1, s)a(s)(F_{0} + ϵ)u_{i}(s) ds

≤ λ (F_{0} + ϵ)∥u_{i}∥ ∫_{0}^{1} G_{i}(1, s)a(s) ds
≤ λB (F_{0} + ϵ)∥u_{i}∥ ≤ ∥u_{i}∥.

Then we have ∥T u_{i}∥ ≤ ∥u_{i}∥. Thus if we let Ω_{1} = {u_{i} ∈ X : ∥u_{i}∥ < l_{1}}, then ∥T u_{i}∥ ≤ ∥u_{i}∥ for u_{i} ∈ P ∩ ∂Ω_{1}. Following Yang [9], we choose δ > 0 and c ∈ (0, \frac{1}{4}), such that

λ ((f_{∞} − δ) ∫_{c}^{1} G_{i}(1, s)a(s)q(s) ds) ≥ 1.

There exists l_{3} > 0, such that f(u_{i}) ≥ (f_{∞} − δ) u_{i} for u_{i} > l_{3}. Let l_{2} = max\{\frac{1}{q_{i}(c)}, 2l_{1}\}. If u_{i} ∈ P with ∥u_{i}∥ = l_{2}, then we have

u_{i}(t) ≥ q_{i}(t)l_{2} ≥ q_{i}(c)l_{2} ≥ l_{3}.

Therefore, for each u_{i} ∈ P with ∥u_{i}∥ = l_{2},we have

∥T u_{i}(t)∥ = (T u_{i})(1) = λ ∫_{0}^{1} G_{i}(1, s) a(s)f(u_{i}(s)) ds

≥ λ ∫_{c}^{1} G_{i}(1, s)a(s)(f_{∞} − ϵ)u_{i}(s) ds

≥ λ (f_{∞} − ϵ)∥u_{i}∥ ∫_{c}^{1} G_{i}(1, s)a(s) q_{i}(s) ds ≥ ∥u_{i}∥.

Thus if we let Ω_{2} = \{u_{i} ∈ E : ∥u_{i}∥ < l_{2}\}, then Ω_{1} ⊂ \overline{Ω}_{2} and ∥T u_{i}∥ ≥ ∥u_{i}∥ for u_{i} ∈ P ∩ ∂Ω_{2}. Condition ( H1) of Krasnoselskii’s fixed point theorem is satisfied. So there exists a fixed point of T in P. This completes the proof.

Theorem 2. Suppose that A f_{0} > B F_{∞}, then for each λ ∈ (\frac{1}{Af_{0}} , \frac{1}{BF_{∞}} ) the problem (1), (i) (i = 2, 3, 4) has at least one positive solution.

The proof of Theorem 2 is very similar to that of Theorem 1 and therefore omitted.

Theorem 3. Suppose that λBf(u_{i}) < u_{i} for u_{i} ∈ (0, ∞). Then the problem (1), (i)(i=2,3,4) has no positive solution.

Proof. Following Sun and Wen [8], assume to the contrary that u_{i} is a positive solution of (1),(i). Then

u_{i}(1) = λ ∫_{0}^{1} G_{i}(1, s) a(s)f(u_{i}(s)) ds < \frac{1}{B} ∫_{0}^{1} G_{i}(1, s)a(s)u_{i}(s) ds

≤ \frac{1}{B} u_{i}(1) ∫_{0}^{1} G_{i}(1, s)a(s) ds = u_{i}(1).

This is a contradiction and completes the proof.

Theorem 4. Suppose that λAf(u_{i}) > u_{i} for u_{i} ∈ (0, ∞). Then the problem (1), (i)(i=2,3,4) has no positive solution.

The proof of Theorem 4 is very similar to that of Theorem 3 and therefore omitted. Example 1. Consider the equation

u_{1}^{5}(t) + λ(5t + 2) \frac{7u_{1}^{2} + u_{1}}{u_{1} + 1} (8 + sin u_{1}) = 0, 0 ≤ t ≤ 1,  (19)

u_{1}(0) = u′′_{1}(0) = u′′′_{1}(0) = u′′′′_{1}(0) = 0, u′_{1}(1) = 0,  (20)

Then F_{0} = f_{0} = 5, F_{∞} = 45, f_{∞} = 27 and 5u_{1} < f(u_{1}) < 45u_{1}. By direct calculations, we obtain that A = 0.0193452 and B = 0.101389. From theorem 1 we see that if λ ∈ (1.05495, 1.23288), then the problem (19)-(20) has a positive solution. From theorem 3 we have that if λ < 0.156556, then the problem (19)-(20) has a positive solution. By theorem 4 we have that if λ > 6.46154, then the problem (19)-(20) has a positive solution.

References

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(Received August 20, 2007)