Positive Solutions of Three-Point Nonlinear Second Order Boundary Value Problem

YOUSSEF N. RAFFOUL

Department of Mathematics University of Dayton, Dayton, OH 45469-2316 email:youssef.raffoul@notes.udayton.edu

Abstract

In this paper we apply a cone theoretic fixed point theorem and obtain conditions for the existence of positive solutions to the three-point nonlinear second order boundary value problem

u′′(t) + λa(t)f(u(t)) = 0, t ∈ (0, 1)
u(0) = 0, αu(η) = u(1),

where 0 < η < 1 and 0 < α < \frac{1}{η} .

AMS Subject Classifications: 34B20.

Keywords: Cone theory; Three-point; Nonlinear second order boundary value problem; Positive solutions.

1 Introduction

In this paper, we are concerned with determining values for λ so that the three-point nonlinear second order boundary value problem

u′′(t) + λ a(t)f(u(t)) = 0, t ∈ (0, 1)  (1.1)
u(0) = 0, αu(η) = u(1),  (1.2)

where 0 < η < 1,

(A1) the function f : [0, ∞) → [0, ∞) is continuous,

(A2) a : [0, 1] → [0, ∞) is continuous and does not vanish identically on any subinterval,

(L1) lim_{x→0} \frac{f(x)}{x} = ∞,

(L2) lim_{x→∞} \frac{f(x)}{x} = ∞,

(L3) lim_{x→0} \frac{f(x)}{x} = 0,

(L4) lim_{x→∞} \frac{f(x)}{x} = 0,

(L5) lim_{x→0} \frac{f(x)}{x} = l with 0 < l < ∞,

and

(L6) lim_{x→∞} \frac{f(x)}{x} = L with 0 < L < ∞

has positive solutions. In the case λ = 1, Ruyun Ma [11] showed the existence of positive solutions of (1.1)-(1.2) when f is superlinear (l = 0 and L = ∞), or f is sublinear (l = ∞ and L = 0). In this research it is not required that f be either sublinear or superlinear. As in [8] and [11], the arguments that we present here in obtaining the existence of a positive solution of (1.1)-(1.2), rely on the fact that solutions are concave downward. In arriving at our results, we make use of Krasnosel’skii fixed point theorem [10]. The existence of positive periodic solutions of nonlinear functional differential equations have been studied extensively in recent years. For some appropriate references we refer the reader to [1], [2], [3], [4], [5], [6], [8], [9], [12], [13], [14], [15], [16] and the references therein. In section 2, we state some known results and Krasnosel’skii fixed point theorem [10]. In section 3, we construct the cone of interest and present a lemma, four theorems and a corollary. In each of the theorems and the corollary, an open interval of eigenvalues is determined, which in return, imply the existence of a positive solution of (1.1)-(1.2) by appealing to Krasnosel’skii fixed point theorem. We say that u(t) is a solution of (1.1)-(1.2) if u(t) ∈ C[0, 1] and u(t) satisfies (1.1)-(1.2).

2 Preliminaries

Theorem 2.1 (Krasnosel’skii) Let B be a Banach space, and let P be a cone in B. Suppose Ω_{1} and Ω_{2} are bounded open subsets of B such that 0 ∈ Ω_{1} ⊂ \overline{Ω}_{1} ⊂ Ω_{2} and suppose that

T : P ∩ (\overline{Ω}_{2}\Ω_{1}) → P

is a completely continuous operator such that

(i) ∥T u∥ ≤ ∥u∥, u ∈ P ∩ ∂Ω_{1}, and ∥T u∥ ≥ ∥u∥, u ∈ P ∩ ∂Ω_{2}; or

(ii) ∥T u∥ ≥ ∥u∥, u ∈ P ∩ ∂Ω_{1}, and ∥T u∥ ≤ ∥u∥, u ∈ P ∩ ∂Ω_{2}.

Then T has a fixed point in P ∩ (\overline{Ω}_{2}\Ω_{1}).

In arriving at our results, we need to state four preliminary Lemmas. Consider the boundary value problem

u′′(t) + y(t) = 0, t ∈ (0, 1), (I)

u(0) = 0, αu(η) = u(1), (II)

Lemma 2.2 Let αη ≠ 1. Then, for y ∈ C[0, 1], the boundary value problem (I) − (II) has the unique solution

u(t) = λ[ − ∫_{0}^{t}(t − s)y(s)ds − \frac{αt}{1 − αη} ∫_{0}^{η}(η − s)y(s)ds
+\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)y(s)ds].  (2.1)

The proof of (2.1) follows along the lines of the proof that is given in [7] in the case λ = 1, and hence we omit it.

The proofs of the next three lemmas can be found in [11].

Lemma 2.3 Let 0 < α < \frac{1}{η} and assume (A1) and (A2) hold. Then, the unique solution of (I) − (II) is non-negative for all t ∈ (0, 1).

Lemma 2.4 Let αη > 1 and assume (A1) and (A2) hold. Then, (I) − (II) has no positive solution.

Lemma 2.5 Let 0 < α < \frac{1}{η} and assume (A1) and (A2) hold. Then, the unique solution of (I) − (II) satisfies

inf u(t)_{t∈[η,1]} ≥ γ||u||,

where γ = min\{αη, \frac{α(1−η)}{1−αη}, η\}.

The proofs of Lemmas 2.3, 2.4 and 2.5 depend on the fact that under conditions (A1) and (A2) the solution u(t) concave downward for t ∈ (0, 1).

3 Main Results

Assuming (A1) and (A2), it follows from Lemmas 2.3 and 2.4, that (1.1)-(1.2) has a non-negative solution if and only if α < \frac{1}{η}. Therefore, throughout this paper we assume that α < \frac{1}{η}. Let

B = C[0, 1], with ||y|| = sup_{t∈[0,1]}|y(t)|.

Define a cone, P, by

P = {y ∈ C[0, 1] : y(t) ≥ 0, t ∈ (0, 1) and min_{t∈[η,1]}y(t) ≥ γ∥y∥}.

Define an integral operator T : P → B

T u(t) = λ[ − ∫_{0}^{t}(t − s)a(s)f(u(s))ds − \frac{αt}{1 − αη} ∫_{0}^{η}(η − s)a(s)f(u(s))ds
+\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)a(s)f(u(s))ds].  (3.1)

By Lemma 2.2, (1.1)-(1.2) has a solution u = u(t) if and only if u solves the operator defined by (3.1). Note that, for 0 < α < 1/η, the first two terms on the right of (3.1) are less than or equal to zero. We seek a fixed point of T in the cone P.

For the sake of simplicity, we let

A = \frac{∫_{0}^{1}(1 − s)a(s)ds}{1 − αη},  (3.2)

and

B = \frac{η ∫_{η}^{1}(1 − s)a(s)ds}{1 − αη}.  (3.3)

Lemma 3.1 Assume that (A1) and (A2) hold. If T is given by (3.1), then T : P → P and is completely continuous.

Proof: Let ϕ, ψ ∈ C[0, 1]. In view of A1, given an ϵ > 0 there exists a δ > 0 such that for ||ϕ − ψ|| < δ we have

sup_{t∈[0,1]}|f(ϕ) − f(ψ)| < \frac{ϵ}{A[2 + α(1 − η)]}.

Using (3.1) we have for t ∈ (0, 1),

|(T ϕ)(t) − (T ψ)(t)| ≤ ∫_{0}^{1}(1 − s)a(s)|f(ϕ(s)) − f(ϕ(s))|ds
+ \frac{α}{1 − αη} ∫_{0}^{1}(1 − s)a(s)|f(ϕ(s)) − f(ϕ(s))|ds

+\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)|f(ϕ(s)) − f(ϕ(s))|ds

≤ [(1 − αη)A + αA + A]|f(ϕ(s)) − f(ϕ(s))|
≤ A[2 + α(1 − η)] sup_{t∈[0,1]}|f(ϕ) − f(ψ)| < ϵ.

Thus, T is continuous. Notice from Lemma 2.3 that, for u ∈ P, T u(t) ≥ 0 on [0, 1]. Also, by Lemma 2.5, T P ⊂ P. Thus, we have shown that T : P → P. Next, we show that f maps bonded sets into bounded sets. Let D be a positive constant and define the set

K = {x ∈ C[0, 1] : ||x|| ≤ D}.

Since A1 holds, for any x, y ∈ K, there exists a δ > 0 such that if ||x − y|| < δ, implies

|f(x) − f(y)| < 1.

We choose a positive integer N so that δ > \frac{D}{N} . For x(t) ∈ C[0, 1], define x_{j}(t) = \frac{jx(t)}{N}, for j = 0, 1, 2, ...., N. For x ∈ K,

||x_{j} − x_{j−1}|| = sup_{t∈[0,1]}|\frac{jx(t)}{N} − \frac{(j − 1)x(t)}{N}|
≤ \frac{||x||}{N} ≤ \frac{D}{N} < δ.

Thus, |f(x_{j}) − f(x_{j} − 1)| < 1. As a consequence, we have

f(x) − f(0) = ∑_{j=1}^{N} (f(x_{j}) − f(x_{j−1})),

which implies that

|f(x)| ≤ ∑_{j=1}^{N} |f(x_{j}) − f(x_{j−1})| + |f(0)|
< N + |f(0)|.

Thus, f maps bounded sets into bounded sets. It follows from the above inequality and (3.1), that

||(T x)(t)|| ≤ λ\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)a(s)|f(x(s))|ds
≤ \frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)(N + |f(0)|)
≤ A(N + |f(0)|).

Next, for t ∈ (0, 1), we have

(T x)′(t) = λ[ − ∫_{0}^{t} a(s)f(u(s))ds − \frac{α}{1 − αη} ∫_{0}^{η}(η − s)a(s)f(u(s))ds

+\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)f(u(s))ds].

Hence,

|(T x)′(t)| ≤ \frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)|f(x(s))|ds
≤ A(N + |f(0)|).

Thus, the set

{(T x) : x ∈ P, ||x|| ≤ D}

is a family of uniformly bounded and equicontinuous functions on the set t ∈ [0, 1]. By Ascoli-Arzela Theorem, the map T is completely continuous. This completes the proof.

Theorem 3.2 Assume that (A1), (A2), (L5) and (L6) hold. Then, for each λ satisfying

\frac{1}{γBL} < λ < \frac{1}{Al}  (3.4)

(1.1)-(1.2) has at least one positive solution.

Proof: We construct the sets Ω_{1} and Ω_{2} in order to apply Theorem 2.1. Let λ be given as in (3.4), and choose ϵ > 0 such that

\frac{1}{γB(L − ϵ)} ≤ λ ≤ \frac{1}{A(l + ϵ)}.

By condition (L5), there exists H_{1} > 0 such that f(y) ≤ (l + ϵ)y, for 0 < y ≤ H_{1}. So, choosing u ∈ P with ||u|| = H_{1}, we have

(T u)(t) ≤ λ\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)a(s)f(u(s))ds
≤ λ\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)a(s)(l + ϵ)u(s)ds
≤ λ\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)(l + ϵ)||u||ds
= λ\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)(l + ϵ)H_{1}ds
≤ λA(l + ϵ)∥u∥ ≤ ∥u∥.

Consequently, ||T u|| ≤ ||u||. So, if we set

Ω_{1} = {y ∈ P : ∥y∥ < H_{1}},

then

||T u|| ≤ ||u||, for u ∈ P ∩ ∂Ω_{1}.  (3.5)

Next we construct the set Ω_{2}. Considering (L6) there exists \overline{H_{2}} such that f(y) ≥ (L − ϵ)y, for all y ≥ \overline{H_{2}}. Let H_{2} = max\{2H_{1}, \frac{\overline{H_{2}}}{γ} \} and set

Ω_{2} = {y ∈ P : ∥y∥ < H_{2}}.

If u ∈ P with ||u|| = H_{2}, then

min_{t∈[η,1]}y(t) ≥ γ||y|| ≥ \overline{H_{2}}.

Thus, by a similar argument as in [11], we have

(T u)(η) ≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)f(u(s))ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)(L − ϵ)u(s)ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)(L − ϵ)γ||u||ds

= λ\frac{γη}{1 − αη} ∫_{η}^{1}(1 − s)a(s)(L − ϵ)H_{2}ds
≥ λBγ(L − ϵ)∥u∥
≥ ∥u∥.

Thus, ||T u∥ ≥ ||u||. Hence

||T u∥ ≥ ||u||, for u ∈ P ∩ ∂Ω_{2}.  (3.6)

Applying (i) of Theorem 2.1 to (3.5) and (3.6) yields that T has a fixed point u ∈ P ∩ (\overline{Ω}_{2}\Ω_{1}). The proof is complete.

Theorem 3.3 Assume that (A1), (A2), (L5) and (L6) hold. Then, for each λ satisfying

\frac{1}{γBl} < λ < \frac{1}{AL}  (3.7)

(1.1)-(1.2) has at least one positive solution.

Proof: We construct the sets Ω_{1} and Ω_{2} in order to apply Theorem 2.1. Let λ be given as in (3.7), and choose ϵ > 0 such that

\frac{1}{γB(l − ϵ)} ≤ λ ≤ \frac{1}{A(L + ϵ)}.

By condition (L5), there exists H_{1} > 0 such that f(y) ≤ (l − ϵ)y, for 0 < y ≤ H_{1}. So, choosing u ∈ P with ||u|| = H_{1}, we have

(T u)(η) ≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)f(u(s))ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)(l − ϵ)u(s)ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)(l − ϵ)γ||u||ds

= λ\frac{γη}{1 − αη} ∫_{η}^{1}(1 − s)a(s)(l − ϵ)H_{1}ds
≥ λBγ(l − ϵ)∥u∥
≥ ∥u∥.

Thus, ||T u∥ ≥ ||u||. So, if we let

Ω_{1} = {y ∈ P : ∥y∥ < H_{1}},

then

||T u|| ≥ ||u||, for u ∈ P ∩ ∂Ω_{1}.  (3.8)

Next we construct the set Ω_{2}. Considering (L6) there exists \overline{H_{2}} such that f(y) ≤ (L + ϵ)y, for all y ≥ \overline{H_{2}}. We consider two cases; f is bounded and f is unbounded. The case where f is bounded is straight forward. If f(y) is bounded by Q > 0, set

H_{2} = max{2H_{1}, λQA}.

Then if u ∈ P and ||u|| = H_{2}, we have

(T u)(t) ≤ λ\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)a(s)f(u(s))ds
≤ λ\frac{Q}{1 − αη} ∫_{0}^{1}(1 − s)a(s)ds
= λAQ
≤ H_{2}
= ∥u∥.

Consequently, ||T u|| ≤ ||u||. So, if we set

Ω_{2} = {y ∈ P : ∥y∥ < H_{2}},

then

||T u|| ≤ ||u||, for u ∈ P ∩ ∂Ω_{2}.  (3.9)

When f is unbounded, we let H_{2} > max\{2H_{1}, \overline{H_{2}}\} be such that f(y) ≤ f(H_{2}), for 0 < y ≤ H_{2}. For u ∈ P with ||u|| = H_{2},

(T u)(t) ≤ λ\frac{t}{1 − αη} ∫_{0}^{1}(1 − s)a(s)f(u(s))ds
≤ λ\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)f(H_{2})ds
≤ λ\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)(L + ϵ)H_{2}ds
= λ\frac{1}{1 − αη} ∫_{0}^{1}(1 − s)a(s)(L + ϵ)||u||ds
= λA(L + ϵ)||u||
≤ ∥u∥.

Consequently, ||T u|| ≤ ||u||. So, if we set

Ω_{2} = {y ∈ P : ∥y∥ < H_{2}},

then

||T u|| ≤ ||u||, for u ∈ P ∩ ∂Ω_{2}.  (3.10)

Applying (ii) of Theorem 2.1 to (3.8) and (3.9) yields that T has a fixed point u ∈ P ∩ (\overline{Ω}_{2}\Ω_{1}). Also, applying (ii) of Theorem 2.1 to (3.8) and (3.10) yields that T has a fixed point u ∈ P ∩ (\overline{Ω}_{2}\Ω_{1}). The proof is complete.

Theorem 3.4 Assume that (A1), (A2), (L1) and (L6) hold. Then, for each λ satisfying

0 < λ < \frac{1}{AL}  (3.11)

(1.1)-(1.2) has at least one positive solution.

Proof: Apply (L1) and choose H_{1} > 0 such that if 0 < y < H_{1}, then

f(y) ≥ \frac{y}{λγB} .

Define

Ω_{1} = {y ∈ P : ∥y∥ < H_{1}}.

If y ∈ P ∩ ∂Ω_{1}, then

(T u)(η) ≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)f(u(s))ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)\frac{u(s)}{λγB} ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)\frac{γ||u||}{λγB}ds
≥ ∥u∥.

In particular, ||T u∥ ≥ ||u||, for all u ∈ P ∩ ∂Ω_{1}. In order to construct Ω_{2}, we let λ be given as in (3.11), and choose ϵ > 0 such that

0 ≤ λ ≤ \frac{1}{A(L + ϵ)}.

The construction of Ω_{2} follows along the lines of the construction of Ω_{2} in Theorem 3.3, and hence we omit it. Thus, by (ii) of Theorem 2.1, (1.1)-(1.2) has at least one positive solution.

Theorem 3.5 Assume that (A1), (A2), (L2) and (L5) hold. Then, for each λ satisfying

0 < λ < \frac{1}{Al}  (3.12)

(1.1)-(1.2) has at least one positive solution.

Proof: Assume (L5) holds. Then, we may take the set Ω_{1} to be the one obtained for Theorem 3.1. That is,

Ω_{1} = {y ∈ P : ∥y∥ < H_{1}}.

Hence, we have

||T u|| ≤ ||u||, for u ∈ P ∩ ∂Ω_{1}.

Next, we assume (L2). Choose \overline{H_{2}} > 0 such that f(y) ≥ \frac{y}{λγB} , for y ≥ \overline{H_{2}}. Let H_{2} =

max\{2H_{1}, \frac{\overline{H_{2}}}{γ} \} and set

Ω_{2} = {y ∈ P : ∥y∥ < H_{2}}.

If u ∈ P with ||u|| = H_{2},

(T u)(η) ≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)f(u(s))ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)\frac{u(s)}{λγB} ds

≥ λ\frac{η}{1 − αη} ∫_{η}^{1}(1 − s)a(s)\frac{γ||u||}{λγB}ds
≥ ∥u∥.

Consequently,

||T u|| ≥ ||u||, for u ∈ P ∩ ∂Ω_{2}.

Applying (i) of Theorem 2.1 yields that T has a fixed point u ∈ P ∩ (\overline{Ω}_{2}\Ω_{1}). We state the next results as corollary, because by now, its proof can be easily obtained from the proofs of the previous results.

Corollary 3.6 Assume that (A1) and (A2) hold. Also, if either (L3) and (L6) hold, or, (L4) and (L5) hold, then (1.1)-(1.2) has at least one positive solution if λ satisfies either 1/(γBL) < λ, or, 1/(γBl) < λ.

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