POSITIVE SOLUTIONS for THIRD-ORDER STURM-LIOUVILLE BOUNDARY-VALUE PROBLEMS with P-LAPLACIAN

Home , Tu (surname)

Electronic Journal of Diﬀerential Equations, Vol. 2009(2009), No. 154, pp. 1–9. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

POSITIVE SOLUTIONS FOR THIRD-ORDER STURM-LIOUVILLE BOUNDARY-VALUE PROBLEMS WITH p-LAPLACIAN

CHENGBO ZHAI, CHUNMEI GUO

Abstract. In this article, we consider the third-order Sturm-Liouville boundary value problem, with p-Laplacian, 00 0 (φp(u (t))) + f(t, u(t)) = 0, t ∈ (0, 1), αu(0) − βu0(0) = 0, γu(1) + δu0(1) = 0, u00(0) = 0, p−2 where φp(s) = |s| s, p > 1. By means of the Leggett-Williams ﬁxed- point theorems, we prove the existence of multiple positive solutions. As an application, we give an example that illustrates our result.

1. Introduction In this paper, we study the existence of multiple positive solutions for the following third-order Sturm-Liouville boundary value problem with p-Laplacian 00 0 (φp(u (t))) + f(t, u(t)) = 0, t ∈ (0, 1), (1.1) αu(0) − βu0(0) = 0, γu(1) + δu0(1) = 0, u00(0) = 0, (1.2) p−2 −1 1 1 where φp(s) = |s| s, p > 1, (φp) = φq, p + q = 1, α, β, γ, δ ≥ 0. During the past decades, wide attention has been paid to the study equations with p-Laplacian operator, which arises in the modelling of different physical and natural phenomena, non-Newtonian mechanics [3, 9], combustion theory [19], pop- ulation biology [17, 18], nonlinear flow laws [5, 13, 14], and system of Monge- Kantorovich partial differential equations [4]. There exist a very large number of papers devoted to the existence of solutions of the p-Laplacian operator. The second-order problem, 0 0 (φp(u (t))) + f(t, u(t)) = 0, t ∈ (0, 1), with various boundary conditions has been studied by many authors, see [8, 12, 15, 16, 20, 21, 22, 23] and the references therein. However, to the best of our knowledge, few papers can be found in the literature on the existence of multiple positive solutions for the third-order Sturm-Liouville boundary value problem (1.1), (1.2). The purpose here is to fill this gap in the literature. Motivated by the works

2000 Mathematics Subject Classiﬁcation. 34K10. Key words and phrases. Positive solution; Sturm-Liouville boundary value problem; p-Laplacian operator; concave functional; ﬁxed point. c 2009 Texas State University - San Marcos. Submitted September 1, 2008. Published November 28, 2009. 1 2 C. B. ZHAI,C. M. GUO EJDE-2009/154

[1] and [7], we shall establish the existence of at least two or at least three positive solutions to third-order Sturm-Liouville boundary value problem with p-Laplacian (1.1), (1.2) by using fixed point theorems in cones. By a positive solution of (1.1) and (1.2) we understand a function u(t) ∈ C2[0, 1] which is positive on 0 < t < 1 and satisfies the differential equation (1.1) and the boundary conditions (1.2). In this article, we use the following assumptions: 4δ+γ α+4β (A1) ρ := γβ + αγ + αδ > 0, 0 < σ := min 4(δ+γ) , 4(α+β) < 1. (A2) G(t, s) is the Green’s function of the differential equation u00(t) = 0, t ∈ (0, 1) with respect to the boundary value condition (1.2), i.e., ( 1 ρ (γ + δ − γt)(β + αs), 0 ≤ s ≤ t ≤ 1. G(t, s) = 1 ρ (β + αt)(γ + δ − γs), 0 ≤ t ≤ s ≤ 1. Evidently G(t, s) ≤ G(s, s), 0 ≤ t, s ≤ 1. (A3) f ∈ C([0, 1] × [0, ∞); [0, ∞)). For convenience, we denote ζ(a) = max{f(t, u) : 0 ≤ t ≤ 1, 0 ≤ u ≤ a}, 1 3 b ψ(b) = min{f(t, u): ≤ t ≤ , b ≤ u ≤ }. 4 4 σ2 Where σ is given as in (A1). Our main results are the following. Theorem 1.1. Assume (A1)–(A3), and that there exist constants 0 < a a, where t∈[ 4 , 4 ] Z 1 −1 6ρ m = G(s, s)ds = , 0 αγ + 3αδ + 3βγ + 6βδ 2 Z 3/4 1 −1 2 32ρ l = 1−q G( , s)ds = 1−q · . σ4 1/4 2 σ4 3αγ + 7αδ + 7βγ + 16βδ Theorem 1.2. Assume (A1)–(A3) and that there exist constants a, b, c such that 0 < a < b < σ2c implies ζ(a) < (ma)p−1, (1.5) ψ(b) ≥ (lb)p−1, (1.6) ζ(c) ≤ (mc)p−1. (1.7) Then the boundary value problem (1.1), (1.2) has at least three positive solutions u1, u2 and u3 with ku1k < a, min 1 3 u2(t) > b, ku3k > a and min 1 3 u3(t) < t∈[ 4 , 4 ] t∈[ 4 , 4 ] b, where σ is given as in (A1) and m, l are given as in Theorem 1.1. The proofs of theorems are based upon the Leggett-Williams fixed-point theorems [10]. These theorems have been useful technique for proving the existence of three or two solutions for boundary value problems of differential and difference equations, see [1, 2, 7]. EJDE-2009/154 POSITIVE SOLUTIONS 3

2. Preliminaries In this section we summarize some basic concepts and results which are taken from Guo and Lakshmikantham [6], and from Leggett and Williams [10]. Definition 2.1. Let E be a real Banach space and P be a nonempty, convex closed set in E. We say that P is a cone if it satisfies the following properties: (i) λu ∈ P for u ∈ P, λ ≥ 0; (ii) u, −u ∈ P implies u = θ(θ denotes the null element of E). If P ⊂ E is a cone, we denote the order induced by P on E by ≤. For u, v ∈ P , we write u ≤ v if v − u ∈ P . Definition 2.2. The map ϕ is said to be a nonnegative continuous concave functional on P provided that ϕ : P → [0, ∞) is continuous and ϕ(tx + (1 − t)y) ≥ tϕ(x) + (1 − t)ϕ(y) for all x, y ∈ P and 0 ≤ t ≤ 1. Definition 2.3. Let 0 < a < b be given and let ϕ be a nonnegative continuous concave functional on the cone P . Define the convex sets Pr, P¯r and P (ϕ, a, b) by Pr = {y ∈ P : kyk < r}, P¯r = {y ∈ P : kyk ≤ r}, P (ϕ, a, b) = {y ∈ P : a ≤ ϕ(y), kyk ≤ b}.

Theorem 2.4 (Leggett-Williams [10]). Let T : P¯c → P¯c be a completely continuous operator and let ϕ be a nonnegative continuous concave functional on P such that ϕ(y) ≤ kyk for all y ∈ P¯c. Suppose that there exist 0 < a b}= 6 ∅ and ϕ(T y) > b for y ∈ P (ϕ, b, d); (b’) kT yk < a for kyk ≤ a; (c’) ϕ(T y) > b for y ∈ P (ϕ, b, c) with kT yk > d.

Then T has at least three ﬁxed points y1, y2, y3 in P¯c satisfying ky1k < a, ϕ(y2) > b, ky3k > a and ϕ(y3) < b

Theorem 2.5 ([10]). Let T : P¯c → P be a completely continuous operator and let ϕ be a nonnegative continuous concave functional on P such that ϕ(y) ≤ kyk for all y ∈ P¯c. Suppose that there exist 0 < a b}= 6 ∅, and ϕ(T y) > b for y ∈ P (ϕ, b, c); (b”) kT yk < a for kyk ≤ a; b ¯ (c”) ϕ(T y) > c kT yk for y ∈ Pc with kT yk > c. Then T has at least two fixed points y1, y2 in P¯c satisfying ky1k < a, ky2k > a and ϕ(y2) < b. In the rest of this section we assume that (A1)-(A3) hold. Let E = C[0, 1] and C+[0, 1] = {x ∈ E|x(t) ≥ 0, t ∈ [0, 1]}. Define an operator T by Z 1 Z v + (T u)(t) = G(t, v)φq f(s, u(s))ds dv, ∀u ∈ C [0, 1]. 0 0 From (A2) and (A3), we can easily get (T u)(t) ≥ 0, t ∈ [0, 1] for u ∈ C+[0, 1]. Remark 2.6. Suppose that u ∈ C+[0, 1] satisfies of the operator equation, T u = u. We can obtain Z t Z v 0 γ u (t) = − (β + αv)φq f(s, u(s))ds dv ρ 0 0 α Z 1 Z v + (γ + δ − γv)φq f(s, u(s))ds dv, ρ t 0 4 C. B. ZHAI,C. M. GUO EJDE-2009/154

Z t 00 00 u (t) = (T u) (t) = −φq f(s, u(s))ds . 0 So we have Z t 00 φp(u (t)) = − f(s, u(s))ds, 0 00 0 and in consequence, (φp(u (t))) = −f(t, u(t)). Moreover, it is clear that Z 1 Z v 0 αu(0) − βu (0) = α G(0, v)φq f(s, u(s))ds dv 0 0 α Z 1 Z v − β · (γ + δ − γv)φq f(s, u(s))ds dv = 0; ρ 0 0

Z 1 Z v 0 γu(1) + δu (1) = γ G(1, v)φq f(s, u(s))ds dv 0 0 Z 1 Z v γ + − · δ (β + αv)φq f(s, u(s))ds dv = 0. ρ 0 0 Further, u00(0) = 0, that is to say, all the ﬁxed points of operator T are the solutions for the problem (1.1), (1.2). Lemma 2.7 ([11]). Suppose that G(t, s) is deﬁned as in (A2). Then G(t, s) ≤ 1 for t ∈ [0, 1], s ∈ [0, 1], G(s, s) G(t, s) 1 3 ≥ σ for t ∈ [ , ], s ∈ [0, 1]. G(s, s) 4 4 Lemma 2.8. The operator T : C+[0, 1] → C+[0, 1] is completely continuous; i.e., T is continuous and compact. Proof. Firstly, we show that T : C+[0, 1] → C+[0, 1] is continuous. From Remark + + + 2.6, we know that T : C [0, 1] → C [0, 1]. Suppose {un} ⊂ C [0, 1], un → u¯(n → + ∞). Thenu ¯ ∈ C [0, 1] and there exists a constant M0 > 0 such that kunk ≤ M0, ku¯k ≤ M0. Let M1 = max{f(t, u)|t ∈ [0, 1], u ∈ [0,M0]}. Then for t ∈ [0, 1] we have Z 1 Z v Z v |T un(t) − T u¯(t)| ≤ G(t, v) φq f(s, un(s))ds − φq f(s, u¯(s))ds dv 0 0 0 Z 1 Z v Z v ≤ G(v, v) φq f(s, un(s))ds − φq f(s, u¯(s))ds dv 0 0 0 Z 1 ≤ 2φq(M1)G(v, v)dv. 0 R v Note that f(t, u) is continuous. We know that φq( 0 f(s, u)ds) is continuous in u on [0, ∞). Then for for each ε > 0, there exists δ1 > 0, such that |u1 − u2| < δ1 and we Z v Z v ε φq f(s, u1(s))ds − φq f(s, u2(s))ds < . 0 0 G(v, v) EJDE-2009/154 POSITIVE SOLUTIONS 5

In view of un(s) → u¯(s), as n → ∞, there exists a natural number N > 0, for n > N with |un(s) − u¯(s)| < δ1, we have Z v Z v ε φq f(s, un(s))ds − φq f(s, u¯(s))ds < . 0 0 G(v, v) Thus for ε > 0, there exists N > 0, such that when n > N, Z v Z v G(v, v) φq f(s, un(s))ds − φq f(s, u¯(s))ds < ε, a.e. [0, 1]. 0 0 An application of Lebesgue’s dominated convergence theorem implies

|T un(t) − T u¯(t)| → 0(as n → ∞), t ∈ [0, 1]. So operator T : C+[0, 1] → C+[0, 1] is continuous. Next we prove that T is compact. Let Ω ⊂ C+[0, 1] be a bounded set. Then there exists R > 0 such that Ω ⊂ {u ∈ C+[0, 1]|kuk ≤ R}. Set M = max{f(t, u)|t ∈ [0, 1], u ∈ Ω}. For any u ∈ Ω, we have Z 1 Z v Z 1 |(T u)(t)| = G(t, v)φq f(s, u(s))ds dv ≤ G(v, v)φq(M)dv, 0 0 0 which implies that T (Ω) is uniformly bounded. Furthermore, for any u ∈ Ω and t ∈ [0, 1], we have γ Z t Z v 0 |(T u) (t)| = − (β + αv)φq f(s, u(s))ds dv ρ 0 0 α Z 1 Z v

+ (γ + δ − γv)φq f(s, u(s))ds dv ρ t 0 hγ Z t α Z 1 i ≤ φq(M) (β + αv)dv + (γ + δ − γv)dv ρ 0 ρ t = φq(M)t ≤ φq(M). 0 Hence k(T u) k ≤ φq(M). So we can easily prove that T (Ω) is equicontinuous. The Arzela-Ascoli Theorem guarantee that T (Ω) is relatively compact and therefore that T is compact.

3. Proofs of main results In this section, we prove the existence of multiplicity results. Let E = C[0, 1] be endowed with the maximum norm kyk = maxt∈[0,1] |y(t)|, and the ordering x ≤ y if x(t) ≤ y(t) for all t ∈ [0, 1]. Deﬁne the cone P ⊂ E by P = {u ∈ C+[0, 1] : min u(t) ≥ σkuk}, 1 3 t∈[ 4 , 4 ] where σ is given as in (A1). Next we show that T (P ) ⊂ P . For any u ∈ P and t ∈ [0, 1], from Lemma 2.7 we have Z 1 Z v Z 1 Z v T u(t) = G(t, v)φq f(s, u(s))ds dv ≤ G(v, v)φq f(s, u(s))ds dv. 0 0 0 0 Consequently, Z 1 Z v kT uk ≤ G(v, v)φq f(s, u(s))ds dv. 0 0 6 C. B. ZHAI,C. M. GUO EJDE-2009/154

1 3 Further, for u ∈ P and t ∈ [ 4 , 4 ], from Lemma 2.7 we obtain Z 1 Z v min T u(t) = min G(t, v)φq f(s, u(s))ds dv 1 3 1 3 t∈[ 4 , 4 ] t∈[ 4 , 4 ] 0 0 Z 1 Z v ≥ σ G(v, v)φq f(s, u(s))ds dv ≥ σkT uk. 0 0 From Lemma 2.8, we know that T : P → P is completely continuous. Let ϕ : P → [0, ∞) be the nonnegative continuous concave functional deﬁned by ϕ(u) = min u(t), u ∈ P. 1 3 t∈[ 4 , 4 ] Evidently, for each u ∈ P , we have ϕ(u) ≤ kuk. We are now in a position to proving the main results.

Proof of Theorem 1.1. It is easy to see that T : P¯b → P is completely continuous σ2 b b and 0 < a b. σ2 σ2 b b b So {u ∈ P (ϕ, b, σ2 ): ϕ(u) > b} 6= ∅. Hence, if u ∈ P (ϕ, b, σ2 ), then b ≤ u(t) ≤ σ2 1 3 1 3 for t ∈ [ 4 , 4 ]. Thus for t ∈ [ 4 , 4 ], from assumption (1.4), we have 1 3 f(t, u(t)) ≥ ψ(b) ≥ (lb)p−1, t ∈ [ , ]. 4 4 Hence 1 Z 1 1 Z v T u( ) = G( , v)φq f(s, u(s) ds)dv 2 0 2 0 Z 3/4 1 Z v ≥ G( , v)φq f(s, u(s))ds dv 1/4 2 0 Z 3/4 1 ≥ G( , v)lbvq−1dv 1/4 2 1 Z 3/4 1 2b b ≥ ( )q−1lb G( , v)dv = > . 4 1/4 2 σ σ Consequently, b b 1 3 min T u(t) ≥ σkT uk > σ × = b for b ≤ u(t) ≤ , t ∈ [ , ]. 1 3 2 t∈[ 4 , 4 ] σ σ 4 4 That is, b ϕ(T u) > b, ∀ u ∈ P ϕ, b, . σ2 Therefore, condition (a”) of Theorem 2.5 is satisﬁed. Now if u ∈ P¯a, then kuk ≤ a. By assumption (1.3), we have f(t, u(t)) ≤ ζ(a) < (ma)p−1, t ∈ [0, 1]. Consequently, Z 1 Z v kT uk = max |T u(t)| = max G(t, v)φq f(s, u(s))ds dv t∈[0,1] t∈[0,1] 0 0 Z 1 Z 1 < ma max G(t, v)dv ≤ ma G(v, v)dv = a. t∈[0,1] 0 0 EJDE-2009/154 POSITIVE SOLUTIONS 7

This shows that T : P¯a → Pa. That is, kT uk < a for u ∈ P¯a. This shows that condition (b”) of Theorem 2.5 is satisfied. Finally, we show that (c”) of Theorem ¯ b 2.5 also holds. Assume that u ∈ P b with kT uk > 2 , then by the definition of σ2 σ cone P , we have b ϕ(T u) = min T u(t) ≥ σkT uk > σ2kT uk = b/ kT uk. 1 3 2 t∈[ 4 , 4 ] σ So condition (c”) of Theorem 2.5 is satisfied. Thus using Theorem 2.5, T has at least two fixed points. That is to say, problem (1.1),(1.2) has at least two positive ¯ solutions u1, u2 in P b satisfying ku1k < a, mint∈[ 1 , 3 ] u2(t) a. σ2 4 4 Proof of Theorem 1.2. It follows from the conditions (1.5)-(1.7) in Theorem 1.2 b that a b}= 6 ∅, ϕ(T u) > b ∀u ∈ P ϕ, b, . σ2 σ2 b Moreover, for u ∈ P (ϕ, b, c) and kT uk > σ2 , we have b ϕ(T u) = min T u(t) ≥ σkT uk > > b. 1 3 t∈[ 4 , 4 ] σ So all the conditions of Theorem 2.4 are satisfied. Thus using Theorem 2.4, T has at least three fixed points. That is to say, the boundary value problem (1.1),(1.2) has at least three positive solutions u1, u2 u3 with ku1k < a, min 1 3 u2(t) > b, t∈[ 4 , 4 ] ku3k > a and min 1 3 u3(t) < b. t∈[ 4 , 4 ]

Corollary 3.1. Assume (A1)–(A3) and that there exist constants 0 < aj < bj < 2 σ aj+1 (j = 1, 2, . . . , n − 1), n ∈ N such that p−1 (B1) ζ(aj) < (maj) , 1 ≤ j ≤ n. p−1 (B2) ψ(bj) ≥ (lbj) , 1 ≤ j ≤ n − 1, where σ is given as in (A1) and m, l are given as in Theorem 1.1. Then the boundary value problem (1.1),(1.2) has at least 2n − 1 positive solutions. The proof of the above corollary is an immediate consequence of Theorem 1.2. Remark 3.2. When p = 2, problem (1.1), (1.2) is the usual form of third-order Sturm-Liouville boundary value problem u000(t) + f(t, u(t)) = 0, t ∈ (0, 1), αu(0) − βu0(0) = 0, γu(1) + δu0(1) = 0, u00(0) = 0. Using the same method, we can present some suﬃcient conditions that guarantee the existence of at least two or three positive solutions for the above boundary value problem. These results are also new and diﬀerent from previous results.

4. An example Now we consider an example to illustrate our results. Consider the third-order Sturm-Liouville boundary value problem, with p-Laplacian, 00 0 2 (φ3(u (t))) + [ϕ(t)h(u(t))] = 0, t ∈ (0, 1), (4.1) u(0) − u0(0) = 0, u(1) = 0, u00(0) = 0, (4.2) 8 C. B. ZHAI,C. M. GUO EJDE-2009/154 where ϕ(t) = 4t, t ∈ [0, 1] and  u/2, 0 ≤ u ≤ 3/512;   1021 3057 3 5  4 u − 2048 , 512 ≤ u ≤ 512 ;  5 5 h(u) = 1, 512 ≤ u ≤ 32 ;  8 231 5  u + , ≤ u ≤ 2;  59 236 32 5u/8, u ≥ 2. In this example, we note that p = 3, α = β = γ = 1, δ = 0. After a simple 1 1 2 calculation, we get q = 3/2, ρ = 2, σ = 4 < 1, G(s, s) = 2 (1 − s ) and 6ρ 2 32ρ 512 m = = 3, l = · = . αγ + 3αδ + 3βγ + 6βδ σ41−q 3αγ + 7αδ + 7βγ + 16βδ 5 3 5 2 We choose a = 512 , b = 512 , c = 2. Evidently, a < b < σ c and 3 (i) for t ∈ [0, 1], 0 ≤ u ≤ 512 , we have 1 3 2 f(t, u) = [ϕ(t)h(u)]2 ≤ 4 × × < (ma)2. 2 512 1 3 5 b 5 (ii) for t ∈ [ 4 , 4 ], 512 ≤ u ≤ σ2 = 32 , we have 1 2 f(t, u) = [ϕ(t)h(u)]2 ≥ 4 × × 1 = (lb)2. 4 (iii) for t ∈ [0, 1], 0 ≤ u ≤ 2, we have

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Chengbo Zhai School of Mathematical Sciences, Shanxi University, Taiyuan 030006, Shanxi, China E-mail address: [email protected]

Chunmei Guo School of Mathematical Sciences, Shanxi University, Taiyuan 030006, Shanxi, China E-mail address: [email protected]