Positive Solution of a Nonlinear Four-Point Boundary-Value Problem

Transcription

1 Nonlinear Analysis and Differential Equations, Vol. 5, 27, no. 8, HIKARI Ltd, Positive Solution of a Nonlinear Four-Point Boundary-Value Problem Weijiao Xu School of Science Tianjin University of Commerce Tianjin 334, P.R. China Huaying Wang College of Mathematics and Statistics Hebei University of Economics and Business Hebei 56, P.R. China Copyright c 27 Weijiao Xu and Huaying Wang. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract This paper studies equation ü(t) q(t)f(t) =, t (, ) with fourpoint boundary conditions u() =, u() = a u(ξ) u(η), where < ξ, η <, a <. The existence result of positive solution is obtained by applying the fixed point theorem in cones. The approaches developed here extend the ideas and techniques derived in recent literatures. Mathematics Subject Classification: 34B8 Keywords: Boundary-value problem, Positive solution, Fixed point Introduction It is well known, boundary value problems have been becoming an important aspect in differential equations, one can read [-8], etc. Many papers are Corresponding author

2 3 Weijiao Xu and Huaying Wang concerned about them, especially to the existence of solution (include multiple solution, positive solution, periodic solution, and extreme solution). A wide variety of approaches have been derived to this goal, such as nonlinear alternative of Leray-Schauder [,9], Mawhin s continuation theorem [,2], Krasnoselskii fixed point theorem [], upper and lower solution [,], monotone iterative method [3], etc. Among them, multiple-point boundary value problems have attracted much attention for their widely background in theory and practical application. The study of multi-point boundary value problems for linear second order ordinary differential equations was initiated by II in and Moiseev. Subsequently the more conclusions of nonlinear multi-point boundary value problems appeared. Ma [2] studied positive solutions of a nonlinear three-point boundary-value problem with the boundary conditions ü(t) a(t)f(t) =, t (, ) u() =, u() = αu(η), where < η <, and < α < η. Motivated by [2], in this paper, we consider the following problems with the boundary conditions where < ξ, η <. For convenience, we denote that ü(t) q(t)f(t) = () u() =, u() = a u(ξ) u(η), (2) = max{ξ, η}, a = a. From now on, we assume the following: (A ) f C([, ), [, )); (A 2 ) q C([, ), [, )), there exists x [, ] such that q(x ) >. Set f(u) f = lim u u, f f(u) = u lim u. The paper is organized as follows: In section 2, we show some lemmas which are necessary to the proof of our main result. In section 3, by applying the fixed point theorem in cones, we obtain the main results for BVPs (), (2).

3 Positive solution of a nonlinear four-point boundary-value problem 3 2 Preliminary Notes In this section, we establish some lemmas which are necessary to develop the main results in this paper. Lemm. Let a, then for y C[, ], the problem has a unique solution ü y(t) =, t (, ) (3) u() =, u() = a u(ξ) u(η) (4) t u(t) = (t s)y(s)ds (ξ s)y(s)ds a a η 2 (η s)y(s)ds ( s)y(s)ds. a a (5) Lemm.2 [] Assume that a i, i =, 2, 3..., m 2 have the same signal, if the function x satisfies: x() = m2 i= a i x(ξ i ), then exists η [ξ, ξ m2 ] such that x() = αx(η), where α = Lemm.3 Let a <. If for y C[, ] and y, then the problem (3) and (4) has the unique solution u satisfies u, t C[, ]. Proof. From the fact that ü = y(t) we know that the graph of u(t) is concave down on (, ) and by the unique solution we known the unique solution is positive when t =. If u() then the concavity of u and the boundary condition u() = imply that u(t) = t m2 y(t)ds < then u(t) is monotonous decrease function, so it has u, t [, ]. If u() <,then it has u(β) < and u() = au(β) > u(β), β (, ). This contradicts the monotonous decrease function of u. Lemm.4 Let a >, if y C[, ] and y(t), t (, ). then (3) and (4) has no positive solution. i= a i.

4 32 Weijiao Xu and Huaying Wang Proof. Assume that (3) and (4) has a positive solution u. If u() > then u(β) > and u() = au(β) > u(β) this contradicts the monotonous decrease function of u. If u() =, and u(τ) > for some τ (, ). Then u(β) = u() =, τ β. If τ (, β), then u(τ) > u(β) = u() this contradicts the definition of the concavity. If τ (β, ), then u(τ) > u(β) this contradicts the monotonous decrease function of u. If u() <, u() < it has no solution; u() <, u(), let u(t) =, β (, t ], u(t) = au(β) contradicts; β (t, ), u() = au(β), it exists t (, t ] such that u(t ) >. There is u(t ) u(β) β t = u(t ) u() a t β < u(t ) u() t β < u(t ) u(). t This contradicts the concavity of u, so u(t ) > is not true. Lemm.5 Let a <.If for y C[, ] and y,then the problem (3) and (4) has the unique solution u satisfies inf u(t) γ u where γ = t [β,] a( β) min{ aβ, a} and a = a. Proof. When < a < by the monotonous u(β) u(). Let If then and u(t) = u. t β <, min u(t) = u(), t [β,] u(t) u() u() u(β) ( ) u()( β a aβ ) = u(t) β a( β), If a( β) min u(t) t [β,] aβ u. β < t <,

5 Positive solution of a nonlinear four-point boundary-value problem 33 then by the monotonous min u(t) = u(), t [β,] u(β) u(t), 3 Main Results u() a u(t), u() au(t), min u(t) au(t) = a u. t [β,] Theorem 3. Assume (A ) and (A 2 ) hold. Then the problem () and (2) has at least one positive solution in the case (i)f = and f = (superlinear) or (ii)f = and f = (sublinear) where < a <. Proof. (i) Superlinear case. Suppose that f = and f =. We wish to show the existence of a positive solution of () and (2). Now () and (2) has a solution y = y(t) if and only if y solves the operator equation t y(t) = Denote (ξ s)q(s)f(y(s))ds a (η s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds a (t s)q(s)f(y(s))ds a = Ay(t). η K = {y y C[, ], y, min y(t) γ y }, t = max{ξ, η}. It is obvious that K is a cone in C[, ]. Moreover AK K. It is also easy to check that A : K K is completely continuous. (6)

6 34 Weijiao Xu and Huaying Wang Since f =, we may choose H > such that f(y) εy, for < y < H ε when ε > satisfies ( s)q(s)ds. Thus, if y K and a y = H, it has t Ay(t) = and we let (t s)q(s)f(y(s))ds a a a a ε a ε a η (η s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds ( s)q(s)εy(s)ds ( s)q(s)ds y ( s)q(s)dsh, (ξ s)q(s)f(y(s))ds a (7) Ω = {y C[, ] y < H }. (8) Then (7) shows that Ay y, for y K Ω. Since f =, there exists Ĥ2 > such that f(u) ρu, for u Ĥ2 where ρ > such that ργ a ( s)q(s)ds. (9) Ĥ 2 Let H 2 = max{2h, r } and Ω 2 = {y C[, ] y < H 2 }, the y K and y = H 2 implies min y(t) γ y H 2. <t< Ay() = ( s)q(s)f(y(s))ds a a η (η s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds (ξ s)q(s)f(y(s))ds a

7 Positive solution of a nonlinear four-point boundary-value problem 35 = = = = a ( s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds a ( s)q(s)f(y(s))ds a ( s)q(s)f(y(s))ds a ( s)q(s)f(y(s))ds a ( s)q(s)f(y(s))ds a a ( ( s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds) a ( q(s)f(y(s))ds sq(s)f(y(s))ds q(s)f(y(s))ds sq(s)f(y(s))ds) a ( sq(s)f(y(s))ds q(s)f(y(s))ds) (s )q(s)f(y(s))ds a ρ (s )q(s)y(s)ds. a Hence, for y K Ω 2, () Ay ργ a ( s)q(s)ds y y. Therefore, by the Fixed Point Theorem, it follows that A has a fixed point in K (Ω \ Ω ) such that H u H 2. This completes the superlinear part of the theorem. (ii) Sublinear case. Suppose next that f = and f = choose H 3 > such that f(y) My for < y < H 3 where Mγ a By using the method to get (), we can get that (s )q(s)y(s)ds. ()

8 36 Weijiao Xu and Huaying Wang Ay() = ( s)q(s)f(y(s))ds a a a M a Mγ a = H 3. η (η s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds (s )q(s)f(y(s))ds (s )q(s)y(s)ds (s )q(s)y(s)ds y Thus, we may let Ω 3 = {y C[, ] y < H 3 } such that Ay y, y K Ω 3. (ξ s)q(s)f(y(s))ds a (2) Now, since f =, exists H 4 > such that f(y) ϕy for y H 4 where ϕ > N satisfies ( s)q(s)y(s)ds. a We consider two cases: Case(i). Support f is bounded, say f(y) N for all y [, ) choose N H 4 = max{2h 3, (s)q(s)ds} such that, for y K with y = a H 4. We have t Ay(t) = (t s)q(s)f(y(s))ds (ξ s)q(s)f(y(s))ds a a a a N a N a H 4, η (η s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds (s )q(s)f(y(s))ds (s )q(s)y(s)ds (s )q(s)ds y (3)

9 Positive solution of a nonlinear four-point boundary-value problem 37 therefore Ay y. Case(ii). If f is unbounded, we know from (A ) that there is H 4 : H 4 > max{2h 3, γ H 4} such that f(y) f(h 4 ) for < y H 4 (we are able to do this since f is unbounded). Then for y K and y = H 4,we have t Ay(t) = (t s)q(s)f(y(s))ds a a a a a η (η s)q(s)f(y(s))ds ( s)q(s)f(y(s))ds (s )q(s)f(y(s))ds (s )q(s)f(h 4 )ds (s )q(s)ϕh 4 ds (ξ s)q(s)f(y(s))ds a H 4. (4) Therefore, in either case we may put Ω 4 = {y C[, ] y < H 4 } and for y K Ω 4, we may have Ay y. By the Fixed Point Theorem, it follows that the bounded valuable problem () and (2) has a positive solution. Therefore, we complete the proof of the theorem. Acknowledgements. We are very grateful to the anonymous referees for their valuable comments and suggestions, which led to an improvement of our original manuscript. This work was supported by National Natural Science Foundation of China (Grant No. 6267) and National Natural Science Foundation of China (Grant No. 6385). References [] W. Ge, Boundary Value Problems for Nonlinear Ordinary Differential Equations, Science Press, 27. [2] W. Ge and J. Ren, An extension of Mawhin s continuation theorem and its application to boundary value problems with a p-laplacian, Nonlinear Analysis: Theory, Methods and Applications, 58 (24),

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