Positive solutions for a three-point boundary value problem at resonance

J. Math. Anal. Appl. 336 (2007) 556–568 www.elsevier.com/locate/jmaa

Positive solutions for a three-point boundary value problem at resonance Xiaoling Han Department of Mathematics, Northwest Normal University, Lanzhou 730070, People’s Republic of China Received 27 June 2005 Available online 3 March 2007 Submitted by A.C. Peterson

Abstract This paper deals with the second order three-point boundary value problem   x  (t) = f t, x(t) , t ∈ (0, 1), x  (0) = 0,

x(η) = x(1).

The existence and multiplicity of positive solutions are investigated by means of the fixed point theorem in cones. © 2007 Elsevier Inc. All rights reserved. Keywords: Three-point boundary value problem; Fixed point theorem; Positive solution; Resonance

1. Introduction Let f : [0, 1] × [0, ∞) → R be a continuous function, and let η ∈ (0, 1) be given. In this paper, we consider the three-point boundary value problem   (1.1) x  (t) = f t, x(t) , t ∈ (0, 1), x  (0) = 0,

x(η) = x(1).

(1.2)

The problem (1.1)–(1.2) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem E-mail address: [email protected]. 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.02.069

X. Han / J. Math. Anal. Appl. 336 (2007) 556–568

x  (t) = 0, x  (0) = 0,

557

t ∈ (0, 1), x(η) = x(1),

has x(t) = c, c ∈ R as a nontrivial solution. The study of multi-point boundary value problems was initiated by Il’in and Moiseev [1]. Motivated by the study of Il’in and Moiseev, Gupta [2] investigated three-point boundary value problems for nonlinear ordinary differential equations at non-resonance. Since then, more general nonlinear multi-point boundary value problems have been studied by several authors using the fixed point index theorem, Leray–Schauder continuation theorem, nonlinear alternative of Leray– Schauder, coincidence degree theory and fixed point theorem in cones. We refer the readers to [3–8] for some recent results in non-resonance cases and to [9–12] in resonance cases. However, as far as the positive solutions are concerned, most of the results are about non-resonance problems, for the existence of positive solutions of multi-point boundary value problems at resonance, the research proceeds very slowly, and to our knowledge, the existence of positive solutions for the problems (1.1)–(1.2) at resonance has not been studied. The purpose of this paper is to investigate the existence of positive solutions of the problem (1.1)–(1.2) at resonance. Define a linear operator L : D(L) ⊂ C 2 [0, 1] → C[0, 1] by setting    D(L) = u ∈ C 2 [0, 1]  u (0) = 0, u(η) = u(1) , and for x ∈ D(L), Lx := x  + β 2 x, where β is a suitable constant such that L is invertible. We make the following assumptions: (H1) β ∈ (0, π2 ) is a constant; (H2) f : [0, 1] × [0, ∞) → R is a continuous function and f (t, x)  −β 2 x. We will use the Banach space C[0, 1] with the norm x = maxt∈[0,1] |x(t)|, and the Banach space C 2 [0, 1] with the norm x2 = max{x, x  , x  }. Define g(t, x) := f (t, x) + β 2 x. Obviously, problem (1.1)–(1.2) is equivalent to the problem   x  (t) + β 2 x(t) = g t, x(t) , t ∈ (0, 1), x  (0) = 0,

x(η) = x(1).

(1.3) (1.4)

Condition (H1) guarantees that L is invertible, so we can rewrite (1.3)–(1.4) to an equivalent integral equation, which enable us to apply the following Guo–Krasnoselskii’s fixed point theorem [13] to prove the existence of positive solutions. Theorem A. (See [13].) Let X be a Banach space, and let K ⊂ X be a cone. Assume Ω1 , Ω2 are two open bounded subsets of X with 0 ∈ Ω1 , Ω 1 ⊂ Ω2 , and let A : K ∩ (Ω 2 \Ω1 ) → K be a completely continuous operator such that

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X. Han / J. Math. Anal. Appl. 336 (2007) 556–568

(i) Au  u, u ∈ K ∩ ∂Ω1 , and Au  u, u ∈ K ∩ ∂Ω2 ; or (ii) Au  u, u ∈ K ∩ ∂Ω1 , and Au  u, u ∈ K ∩ ∂Ω2 . Then A has a fixed point in K ∩ (Ω 2 \Ω1 ). 2. Preliminaries Consider the linear boundary value problem x  (t) + β 2 x(t) = h(t), 

x (0) = 0,

t ∈ (0, 1),

x(η) = x(1).

(2.1) (2.2)

Lemma 2.1. Let (H1) hold. Then for each h ∈ C[0, 1], (2.1)–(2.2) has a unique solution 1 x=

G(t, s)h(s) ds := T h,

(2.3)

0

where

⎧ ⎨1 sin β(t − s), 0  s  t  1, G(t, s) = β ⎩ 0, 0t s1 ⎧ β(2s − η − 1) ⎪ ⎪ cos , 0  s  η < 1, ⎨ cos βt 2 + sin β(1 − s) ⎪ β sin β(η+1) ⎪ , 0 < η  s  1. ⎩ 2 2 sin β(1−η) 2

(2.4)

Moreover (i) T : C[0, 1] → C[0, 1] is a completely continuous linear operator; (ii) for each h ∈ C[0, 1] with h(t)  0 on [0, 1], we have (T h)(t)  0 on [0, 1]. Proof. It is not difficult to verify that 1 x(t) =

G(t, s)h(s) ds 0

is a solution of the problem (2.1)–(2.2). Obviously T is a completely continuous linear operator. In addition, we know from (H1) and (2.4) that G(t, s)  0,

(t, s) ∈ [0, 1] × [0, 1],

which together with (2.3) implies that T h  0. This completes the proof of the lemma. 2 Set

   K0 = x ∈ C[0, 1]  x(t)  0, t ∈ [0, 1] ,

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namely, K0 is a positive cone in C[0, 1]. For every x ∈ K0 , denote F (x) = g(t, x), then F : K0 → K0 is continuous. Let A = T ◦ F, then A : K0 → K0 is completely continuous. Now, it is easy to verify the nonzero fixed points of the operator A are the positive solutions of the problem (1.3)–(1.4). Lemma 2.2. There exist M, M0 > 0 such that 0  G(t, s)  M, M0  G(t, s)  M,

(t, s) ∈ [0, 1] × [0, 1], (t, s) ∈ [0, 1] × [0, η].

Proof. By (2.4) and (H1), we have that G(t, s)  0 (t, s) ∈ [0, 1] × [0, 1]. On the other hand ⎧ ⎨1 sin β(t − s), 0  s  t  1, G(t, s) = β ⎩ 0, 0t s1 ⎧ β(2s − η − 1) ⎪ ⎪ , 0  s  η < 1, cos ⎨ cos βt 2 + sin β(1 − s) ⎪ β sin β(η+1) ⎪ , 0<ηs1 ⎩ 2 2 sin β(1−η) 2 ⎧ β(2s − η − 1) ⎪ ⎪ cos , 0  s  η < 1, 1 cos βt ⎨ 2  sin β + sin β(1 − s) ⎪ β β sin β(η+1) ⎪ , 0<ηs1 ⎩ 2 2 sin β(1−η) 2 ⎧ β(2s − η − 1) ⎪ ⎪ , 0  s  η < 1, cos ⎨ 1 1 2 < sin β + sin β(1 − s) ⎪ β β sin β(η+1) ⎪ , 0<ηs1 ⎩ 2 2 sin β(1−η) 2 cos β(1−η) 1 2 := M.  sin β + β(η+1) β β sin 2

(2.5)

Thus 0  G(t, s)  M,

(t, s) ∈ [0, 1] × [0, 1].

If (t, s) ∈ [0, 1] × [0, η], then G(t, s) 

1 β(η + 1) cos β cot := M0 . β 2

2

(2.6)

Lemma 2.3. There exists a continuous function Φ : [0, 1] → [0, ∞) and a constant c ∈ (0, 1] such that cΦ(s)  G(t, s)  Φ(s),

t, s ∈ [0, 1].

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Proof. Let φ(s) = 1 − s,

H (t, s) = μφ(s) − G(t, s).

Upper bounds. We only need to show that, if μ > 0 is sufficiently large, then H (t, s)st  0,

H (t, s)st  0,

(t, s) ∈ [0, 1] × [0, 1].

Case 1. s ∈ [0, η]. H (t, s)st = μ(1 − s) −  μ(1 − s) −

cos βt cos β(1+η−2s) 2 β sin β(η+1) 2 cos β(1−η) 2 β sin β(η+1) 2 cos β(1−η) 2

 μ(1 − η) −

β sin β(η+1) 2

,

so, for μ  μ1 :=

cos β(1−η) 2 β(1 − η) sin β(η+1) 2

,

we have H (t, s)st  0. On the other hand H (t, s)st = μ(1 − s) −

cos βt cos β(1+η−2s) 1 2 sin β(t − s) − β(η+1) β β sin 2

 μ(1 − s) −

cos βt cos β(1+η−2s) 1 2 sin β − β β sin β(η+1) 2

 μ(1 − s) −

cos β(1−η) 1 2 sin β − β(η+1) β β sin 2

 μ(1 − η) −

cos β(1−η) 1 2 sin β − , β β sin β(η+1) 2

so, for μ  μ2 :=

cos β(1−η) 1 2 , sin β + β(1 − η) sin β(η+1) 2

we have H (t, s)st  0. Case 2. s ∈ [η, 1]. H (t, s)st = μ(1 − s) −  μ(1 − s) −

cos βt sin β(1 − s) 2β sin β(η+1) sin β(1−η) 2 2 sin β(1 − s) 2β sin β(η+1) sin β(1−η) 2 2

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β(1 − s)

 μ(1 − s) −

β(η+1) 2

2β sin sin β(1−η) 2

1 , = (1 − s) μ − 2 sin β(η+1) sin β(1−η) 2 2 so, for μ  μ3 :=

1 2 sin

β(η+1) 2

sin β(1−η) 2

,

we have H (t, s)st  0. On the other hand cos βt sin β(1 − s) 1 sin β(t − s) − β 2β sin β(η+1) sin β(1−η) 2 2 cos βt sin β(1 − s) 1  μ(1 − s) − sin β(1 − s) − β 2β sin β(η+1) sin β(1−η)

H (t, s)st = μ(1 − s) −

2

2

cos βη sin β(1 − s) 1 > μ(1 − s) − sin β(1 − s) − β 2β sin β(η+1) sin β(1−η) 2 2 β(1 − s) cos βη 1  μ(1 − s) − · β(1 − s) − β 2β sin β(η+1) sin β(1−η) 2 2

cos βη = (1 − s) μ − 1 − , 2 sin β(η+1) sin β(1−η) 2 2 so, for μ  μ4 := 1 +

cos βη 2 sin

β(η+1) 2

sin β(1−η) 2

,

we have H (t, s)st  0. Let μ∗  max{μ1 , μ2 , μ3 , μ4 }, then H (t, s)  0 for (t, s) ∈ [0, 1] × [0, 1], and accordingly μ∗ φ(s)  G(t, s). Set Φ(s) := μ∗ φ(s), then G(t, s)  Φ(s),

t, s ∈ [0, 1].

Lower bounds. We only need to show that if μ > 0 is sufficiently small, then H (t, s)st  0, Case 1. s ∈ [0, η].

H (t, s)st  0,

(t, s) ∈ [0, 1] × [0, 1].

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X. Han / J. Math. Anal. Appl. 336 (2007) 556–568

H (t, s)st = μ(1 − s) −  μ(1 − s) − μ−

cos βt cos β(1+η−2s) 2 β sin β(η+1) 2 cos βη cos β(1+η) 2 β sin β(η+1) 2

cos βη cos β(1+η) 2 β sin β(η+1) 2

,

so, for μ  μ5 :=

cos βη cos β(1+η) 2 β sin β(η+1) 2

,

we have H (t, s)st  0. On the other hand H (t, s)st = μ(1 − s) −  μ(1 − s) −  μ(1 − s) −  μ − μ5 .

cos βt cos β(1+η−2s) 1 2 sin β(t − s) − β β sin β(η+1) 2 cos βt cos β(1+η−2s) 2 β sin β(η+1) 2 cos βη cos β(1+η) 2 β sin β(η+1) 2

Then for μ  μ5 we have H (t, s)st  0. Case 2. s ∈ [η, 1]. If s = 1, then G(t, s) = 0 and φ(s) = 0, the conclusion is true. If s ∈ [η, 1), then H (t, s)st = μ(1 − s) −

cos βt sin β(1 − s) 2β sin β(η+1) sin β(1−η) 2 2

.

Obviously there exists μ6 > 0 sufficiently small such that H (t, s)st  0. Now suppose that s  t, then cos βt sin β(1 − s) 1 sin β(t − s) − β 2β sin β(η+1) sin β(1−η) 2 2 cos βt sin β(1 − s)  μ(1 − s) − , 2β sin β(η+1) sin β(1−η) 2 2

H (t, s)st = μ(1 − s) −

by our analysis, if μ  μ6 , then H (t, s)st  0. Let 0 < μ0  min{μ5 , μ6 }, then for all (t, s) ∈ [0, 1] × [0, 1] we have H (t, s)  0, thus μ0 φ(s)  G(t, s),

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namely cΦ(s)  G(t, s),

t, s ∈ [0, 1],

/μ∗

where c := μ0 ∈ (0, 1]. This completes the proof of the lemma.

2

Define

   K = x ∈ C[0, 1]  x(t)  cx, t ∈ [0, 1] ,

obviously, K ⊂ K0 is a cone in C[0, 1]. Lemma 2.4. A = T ◦ F : K → K. Proof. Suppose x ∈ C[0, 1], x  0, by Lemma 2.3 we have   (Ax)(t) = T F (x) (t) =

1



 G(t, s)g s, x(s) ds  c

1

0

  Φ(s)g s, x(s) ds,

(2.7)

0

on the other hand 1 Ax = T ◦ F  = max

t∈[0,1]



 G(t, s)g s, x(s) ds 

0

1

  Φ(s)g s, x(s) ds,

(2.8)

0

for every t ∈ [0, 1], by (2.7) and (2.8) we have (Ax)(t)  cAx,

(2.9)

which means A(K) ⊂ K, so A is a completely continuous operator.

2

3. Existence results for (1.1)–(1.2) Let f (t, x) , x f (t, x) , f ∞ = lim sup max t∈[0,1] x x→+∞ f 0 = lim sup max x→0+

t∈[0,1]

f (t, x) , x f (t, x) f ∞ = lim inf min . x→+∞ t∈[0,1] x

f 0 = lim inf min

x→0+ t∈[0,1]

Theorem 3.1. Assume (H1) and (H2) hold. Then the problem (1.1)–(1.2) has at least one positive solution in the case (i) f 0 = −β 2 , f ∞ = ∞; or (ii) f 0 = ∞, f ∞ = −β 2 . Proof. For 0 < r < R < ∞, let    Ω1 = x ∈ C[0, 1]  x < r , then, 0 ∈ Ω1 , Ω 1 ⊂ Ω2 .

   Ω2 = x ∈ C[0, 1]  x < R ,

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Case 1. f 0 = −β 2 , f ∞ = ∞. First we show that for all r > 0 sufficiently small we have Ax  x ∀x ∈ K ∩ ∂Ω1 .

(3.1)

Since f 0 = −β 2 , there exists r > 0 such that   f (t, x)  m − β 2 x ∀0 < x  r, where m > 0 satisfies Mm  1.

(3.2)

For x ∈ K with x = r, we have g(s, x) = f (s, x) + β 2 x  mx,

s ∈ [0, 1].

Moreover 0 < cx  x(s)  x = r. So 1 (Ax)(t) = (T ◦ F )(t)  T ◦ F  = max

t∈[0,1]

  G(t, s)g s, x(s) ds

0

1 G(t, s) ds  Mmx  x,

 mx max

t∈[0,1] 0

accordingly Ax  x, (3.1) is true. Next we show that for R > 0 sufficiently large Ax  x ∀x ∈ K ∩ ∂Ω2 .

(3.3)

Since f ∞ = ∞, there exists R0 > 0 such that f (t, x)  ρx

∀x  R0 ,

where ρ > 0 satisfies   M0 cη ρ + β 2  1. Let R > max{R0 /c, r}, then x ∈ K with x = R implies x(s)  cx = cR > R0 , and       g s, x(s) = f s, x(s) + β 2 x(s)  ρx + β 2 x  c ρ + β 2 x. By Lemmas 2.2 and 2.3, we have

(3.4)

X. Han / J. Math. Anal. Appl. 336 (2007) 556–568

1 (Ax)(t) =

565

1

   G(t, s)g s, x(s) ds  c ρ + β 2 x 

G(t, s) ds

0

0

   c ρ + β 2 x

η

  G(t, s) ds  M0 cη ρ + β 2 x  x,

0

accordingly Ax  x, (3.3) is true. By the first part of Theorem A, it follows that A has a fixed point x0 in K ∩ (Ω 2 \Ω1 ), such x0 is a positive solution of (1.1)–(1.2). Case 2. f 0 = ∞, f ∞ = −β 2 . Since f 0 = ∞, there exists r0 > 0 such that f (t, x)  λx

∀0 < x  r0 ,

where λ > 0   M0 cη λ + β 2  1.

(3.5)

Thus, for 0 < r  r0 , if x ∈ K and x = r, then       g s, x(s) = f s, x(s) + β 2 x(s)  c λ + β 2 x. By using the method to get (3.3), we conclude 1 (Ax)(t) =

   G(t, s)g s, x(s) ds  c λ + β 2 x 

0

   c λ + β 2 x

1 G(t, s) ds 0

η

  G(t, s) ds  M0 cη λ + β 2 x  x.

0

So Ax  x ∀x ∈ K ∩ ∂Ω1 . Now, since f ∞ = −β 2 , there exists R0 > 0 such that   f (t, x)  γ − β 2 x ∀x  R0 , where γ > 0 satisfies Mγ  1. Let R > R1 := max{R0 /c, r0 }, then if x ∈ K with x = R, we have x(s)  cx = cR  R0 , accordingly g(s, x)  γ x  γ x. Using the same method to get (3.1), we follows that

(3.6)

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X. Han / J. Math. Anal. Appl. 336 (2007) 556–568

1 (Ax)(t) = (T ◦ F )(t)  T ◦ F  = max

t∈[0,1]

  G(t, s)g s, x(s) ds  Mγ x  x,

0

then for R > R1 , Ax  x ∀x ∈ K ∩ ∂Ω2 .

(3.7)

By (3.6), (3.7) and the second part of Theorem A, it follows that A has a fixed point x ∗ in K ∩ (Ω 2 \Ω1 ), such x ∗ is a positive solution of the problem (1.1)–(1.2). 2 To get the multiplicity results of positive solutions, we make the following assumptions: (H3) f 0 , f ∞ = ∞, and there exists ω > 0 such that      max f (t, x)  0  t  1, cω  x  ω < ε − β 2 ω,

(3.8)

where ε > 0 satisfies Mε  1. (H4) f 0 , f ∞ = −β 2 , and there exists ω > 0 such that      min f (t, x)  0  t  1, cω  x  ω > μ − cβ 2 ω,

(3.9)

where μ > 0 satisfies M0 μ  1. Theorem 3.2. Assume (H1), (H2) and (H3) hold. Then the problem (1.1)–(1.2) has at least two positive solutions. Proof. Let Ω1 , Ω2 be as in the proof of Theorem 3.1, let Ω3 = {x ∈ C[0, 1] | x < ω}, then for 0 < r < ω < R, we have Ω 1 ⊂ Ω3 , Ω 3 ⊂ Ω2 . Since f 0 = ∞,

f ∞ = ∞,

by the proof of Theorem 3.1, we can let r sufficiently small, and R sufficiently large, such that (3.6) and (3.3) hold. In the following, we show that Ax < x ∀x ∈ K ∩ ∂Ω3 .

(3.10)

For x ∈ K ∩ ∂Ω3 , s ∈ [0, 1], we have cω = cx  x(s)  x = ω, by (3.8), f (s, x) < (ε − β 2 )ω, then       g s, x(s) = f s, x(s) + β 2 x  f s, x(s) + β 2 ω < εω. So 1 (Ax)(t) = (T ◦ F )(t)  T ◦ F  = max

t∈[0,1] 0

  G(t, s)g s, x(s) ds < Mεx  x.

X. Han / J. Math. Anal. Appl. 336 (2007) 556–568

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Thus, Ax < x and (3.10) hold. This implies A satisfies condition (ii) of Theorem A in K ∩ (Ω 3 \Ω1 ), and A satisfies condition (i) of Theorem A in K ∩ (Ω 2 \Ω3 ). Then A has fixed / ∂Ω3 , so x1  < point x1 and x2 in K ∩ (Ω 3 \Ω1 ) and K ∩ (Ω 2 \Ω3 ). By (3.10) we know x1 , x2 ∈ ω < x2 . This completes the proof of the theorem. 2 Theorem 3.3. Assume (H1), (H2) and (H4) hold. Then the problem (1.1)–(1.2) has at least two positive solutions. Proof. Let 0 < r < ω < R, Ωi (i = 1, 2, 3) be as in the proof of Theorems 3.1 and 3.2. Since f 0 = f ∞ = −β 2 , by the proof of Theorem 3.1, we can take r sufficiently small, R sufficiently large, so that A satisfies (3.1) and (3.7). In the following, we show that Ax > x ∀x ∈ K ∩ ∂Ω3 .

(3.11)

For x ∈ K ∩ ∂Ω3 , s ∈ [0, 1], we have cω = cx  x(s)  x = ω, by (3.9), f (s, x(s)) > (μ − cβ 2 )ω, then       g s, x(s) = f s, x(s) + β 2 x  f s, x(s) + cωβ 2 > μω. So   min g s, x(s) > μx.

s∈[0,1]

Now we get from Lemmas 2.1 and 2.4 that 1 (Ax)(t) =



   G(t, s)g s, x(s) ds  min g s, x(s)

1 G(t, s) ds

s∈[0,1]

0

   min g s, x(s)

0

η G(t, s) ds > M0 μx  x,

s∈[0,1]

0

and then (3.11) holds. By Theorem A, A has two fixed points x1 and x2 with x1 ∈ K ∩ (Ω 3 \Ω1 ) and x2 ∈ K ∩ (Ω 2 \Ω3 ). By (3.11) we know x1 = x2 , so we get two different positive solutions of (1.1)–(1.2). 2 References [1] V. Il’in, E. Moiseev, Non-local boundary value problem of the first kind for a Sturm–Liouville operator in its differential and finite difference aspects, Differ. Equ. 23 (1987) 803–810. [2] C. Gupta, Solvability of a three-point nonlinear boundary value problem for a second order ordinary differential equation, J. Math. Anal. Appl. 168 (1992) 540–551. [3] G. Zhang, J. Sun, Positive solutions of m-point boundary value problems, J. Math. Anal. Appl. 291 (2004) 406–418. [4] Z. Wei, C. Pang, Positive solutions of some singular m-point boundary value problems at non-resonance, Appl. Math. Comput. 171 (2005) 433–449.

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