Nontrivial Solutions for Boundary Value Problems of Nonlinear Differential Equation

Advances in Dynamical Systems and Applications ISSN 973-532, Volume 6, Number 2, pp. 24 254 (2 http://campus.mst.edu/adsa Nontrivial Solutions for Boundary Value Problems of Nonlinear Differential Equation Jia Mu, Yongxiang Li, and Pengyu Chen Northwest Normal University, Department of Mathematics Lanzhou, Gansu 73, People s Republic of China mujia88@63.com Abstract The nonlinear three-point boundary value problem { u (t = f(t, u(t, t I, βu( γu ( =, u( = αu(η, is discussed under some conditions concerning the first eigenvalue corresponding to a special linear operator, where I = [, ], η (,, α, β, γ [, with β + γ, f : [, ] (, + (, + is sign changing continuous function and may be unbounded from below. By applying the topological degree of a completely continuous field, we establish some new existence criteria of nontrivial solutions. At last, two examples are given to demonstrate the application of the main results. AMS Subject Classifications: 34B6, 47H. Keywords: Three-point boundary value problems, nontrivial solution, topological degree. Introduction Over the last thirty years, boundary value problems have attracted extensive attentions due to their wide range of applications in applied mathematics, physics, biology and engineering (see, for example, [6 8, 2 8] and references therein for more details. Most of them obtained the existence of positive solutions provided f : [, + [, + is nonnegative, continuous and superlinear or sublinear by employing the cone expansion or compression fixed point theorem, the method of upper and lower solutions, Schauder s fixed point theorem or the fixed point index. Recently, some authors have Received September 26, 2; Accepted December, 2 Communicated by Mehmet Ünal

242 J. Mu, Y. Li, P. Chen studied the existence of nontrivial solutions of boundary value problems when the nonlinear term is sign changing, such as [, 3, 5, 9, 2 23]. L. Liu, B. Liu, Y. Wu [] discussed the existence of nontrivial solutions of the problem (Lu(t + h(tf(t, u(t =, < t <, m 2 u( =, u( = a i u(ξ i, i= under the following conditions: (A There exist two nonnegative functions b(t, c(t C[, ] with c(t and one continuous even function B : (, + [, + such that f(t, u b(t c(tbu, for all t I, u R. Moreover, B is nondecreasing on [, + and satisfies Bu lim u + u = ; f(t, u (A 2 lim inf > λ u + u, uniformly on t I; (A 3 lim sup f(t, u u u < λ, uniformly on t I. In [], the result improved that in [3, 2]. Inspired by the above work, this paper is concerned with the nonlinear three-point boundary value problem { u (t = f(t, u(t, t I, βu( γu (. ( =, u( = αu(η, where I = [, ], η (,, α, β, γ [, with β +γ, f : [, ] (, + (, + is sign changing continuous function and may be unbounded from below. The new features of this paper mainly include the following aspects. Firstly, we give the Green function of (., and it is not necessarily symmetric. Thus our work can be applied to more general problems. Secondly, to cope with the difficulties caused by the asymmetry of Green function, a special linear operator is sought and a special cone is constructed for the study of the existence of nontrivial solutions. Thirdly, our conditions are weaker than the previous papers. Fourthly, two examples are constructed to demonstrate the application of the main results. Finally, the main technique used here is the topological degree theory. 2 Preliminaries In Banach space C(I in which the norm is defined by u = max t I u(t, we set P = {u C(I u(t, t I}, then P is a positive cone in C(I. The open ball in C(I is denoted by B r = {u C(I u < r, r > }.

Nontrivial Solutions for BVPs 243 For convenience, we give the following assumption: (H β + γ αβη αγ >. Then we set L(t, s = { (βt + γ( s +, t s, β + γ (βs + γ( t +, s t. (2. By (2., we easily obtain the following lemma. Lemma 2.. L(t, s possesses the following properties: (i L(t, s is continuous and L(t, s over I I; (ii L(t, s = L(s, t, L(t, s L(s, s for t, s I. Lemma 2.2. Assume that (H holds. Then for y C(I, the problem { u (t = y(t, t I, βu( γu ( =, u( = αu(η, (2.2 has a unique solution where u(t = G(t, s = L(t, s + G(t, sy(sds, (2.3 αl(η, s (βt + γ. (2.4 β + γ αβη αγ Proof. If G(t, s is the same as in (2.4, it is easy to check that the function defined by (2.3 is a solution of (2.2. Now we show that the function defined by (2.3 is a solution of (2.2 only if G(t, s is the same as in (2.4. By the method of variation of constants, we can obtain that any solution of (2.2 can be represented by In view of (2., we have t u(t = ( t + u(t = (βs + γy(sds β + γ + (βt + γ L(t, sy(sds + M(βt + γ. (2.5 t ( s + y(sds + M(βt + γ, (2.6 β + γ and u (t = y(t, (2.7

244 J. Mu, Y. Li, P. Chen Observe that we get u( = γ u ( = β Since u( = M(β + γ and u(η = boundary condition of (2.2, we get that so by (2.5 the proof is finished. ( s + y(sds + Mγ, ω ( s + y(sds + Mβ, ω βu( γu ( =. L(η, sy(sds + M(βη + γ, by the second M = α L(η, sy(sds β + γ αβη αγ, Lemma 2.3. Assume (H holds. Then G(t, s defined by (2.4 has the following properties: (i G(t, s is nonnegative and continuous on I I; (ii G(t, s DL(s, s for t, s I, where D = β + γ αβη + αβ β + γ αβη αγ. By the above discussion, we know that the problem (. is equivalent to the integral equation u(t = G(t, sf(s, u(sds := Au(t, t I. (2.8 And it is easy to see that u is a nontrivial solution of (. if and only if u C(I is a fixed point of A and u(t, t I. Define two linear operators K, T by (Ku(t = (T u(t = G(t, su(sds, t I. (2.9 G(s, tu(sds, t I. (2. By virtue of Krein Rutmann s theorem [4] and similar as the proof of [2, Lemma 5], we get the following lemma. Lemma 2.4. Suppose (H is satisfied. Then for the operator K, T defined by (2.9 and (2., respectively, we have (i K, T : C(I C(I are completely continuous positive linear operators;

Nontrivial Solutions for BVPs 245 (ii the spectral radius r(k, r(t, and K, T have positive eigenfunctions corresponding to their first eigenvalues λ = (r(k and λ = (r(t respectively. Let ϕ and ϕ 2 be the positive eigenfunctions of K and T respectively, i.e., λ Kϕ = ϕ, λ T ϕ 2 = ϕ 2. (2. Lemma 2.5. Assume that (H holds. Then there exist δ, δ 2 > such that δ G(t, s ϕ 2 (s δ 2 G(s, s, t, s I. Proof. If γ =, it follows from β + γ, β, γ [, + and (2.4 that β >, G(t, = G(t, = for t I. Then we obtain from (2. and (2. that ϕ 2 ( = ϕ 2 ( =, which implies that ϕ 2( >, ϕ 2( < (see [9]. Since we can define Φ(s = lim s + lim s Φ( = (β + γϕ 2( β ϕ 2 (s βs + γ = ϕ 2( β >, ϕ 2 (s s + = ϕ 2( >, (β + γϕ 2(s (βs + γ( s +, < s <,, Φ( = (β + γϕ 2(. (β + γ Then Φ(s is continuous on I and Φ(s >, s (,. So there exist ρ, δ 2 > such that ρ Φ(s δ 2, s (,, i.e., Then ρ β + γ (βs + γ( s + ϕ 2(s δ 2 (βs + γ( s +, s I. β + γ By Lemma 2.3 and (2.4, we obtain Set δ = ρ, we can find that D ρl(s, s ϕ 2 (s δ 2 L(s, s. D ρg(t, s ϕ 2(s δ 2 G(s, s, t, s I. δ G(t, s ϕ 2 (s δ 2 G(s, s, t, s I. In the case γ >, we can prove the same conclusions analogically.

246 J. Mu, Y. Li, P. Chen Lemma 2.6. K(P P, where P = { u P } ϕ 2 (tu(tdt λ δ u. Proof. For all u P, by (2. and Lemma 2.5, we have ( ϕ 2 (t(ku(tdt = ϕ 2 (t G(t, su(sds dt ( = G(t, sϕ 2 (tdt u(sds Thus = (T ϕ 2 (su(sds λ δ (Ku(t, t I. ϕ 2 (t(ku(tdt λ δ Ku, i.e., K(P P. Lemma 2.7. [2] Let E be a Banach space and Ω be a bounded open set in E. Suppose that A : Ω E is a completely continuous operator. If there exists u θ such that u Au µu for u Ω, µ, then the topological degree deg(i A, Ω, θ =. Lemma 2.8. [2] Let E be a Banach space and Ω be a bounded open set in E with θ Ω. Suppose that A : Ω E is a completely continuous operator. If Au µu for u Ω, µ, then the topological degree deg(i A, Ω, θ =. 3 Existence of Nontrivial Solutions Theorem 3.. Suppose (H holds. In addition, f satisfies the following conditions: (W There exist ε > and d C(I, such that f(t, u (λ + ε u d(t, t I, u R; (W 2 There exist < ε 2 < λ, (λ ε 2 λ < and e C(I such that f(t, u (λ ε 2 u e(t, t I, u R; (W 3 There exist < ε 3 < λ and r >, such that f(t, u(t (λ ε 3 u(t, t I, u R, u r, where λ, λ are defined by (2.. Then the problem (. has at least one nontrivial solution.

Nontrivial Solutions for BVPs 247 Proof. Since (λ ε 2 λ <, the operator I (λ ε 2 K has the bounded inverse operator (I (λ ε 2 K. We set R > max { r, Now we show that (I (λ ε 2 K ( δ ε ε 2 d(sϕ 2 (sds + δ } e(sϕ 2 (sds + Ke. (3. u Au µϕ, u B R, µ, (3.2 where ϕ is defined in (2.. In fact, if (3.2 is not true, then there exist u B R, µ satisfying u Au = µ ϕ. (3.3 Combining (W, (2. and (3.3 yields u (tϕ 2 (tdt So (Au (tϕ 2 (tdt ( G(t, s[(λ + ε u (s d(s]ds ϕ 2 (tdt ( = (λ + ε G(t, sϕ 2 (tdt u (sds ( G(t, sϕ 2 (tdt d(sds = (λ + ε λ u (sϕ 2 (sds ε u (sϕ 2 (sds λ d(sϕ 2 (sds. d(sϕ 2 (sds. (3.4 By (2., (3.3, (3.4, (W 2 and Lemma 2.6, we get u (λ ε 2 (Ku + (Ke P and u (λ ε 2 (Ku + (Ke [ λ δ u (t (λ ε 2 (Ku (t + (Ke(t ] ϕ 2 (tdt = δ ε 2 δ ε ε 2 u (sϕ 2 (sds + δ d(sϕ 2 (sds + δ e(sϕ 2 (sds e(sϕ 2 (sds. So R = u, which is less than or equal to the right-hand side of (3., hence contradicting (3.. Then (3.2 holds. By Lemma 2.7, deg(i A, B R, θ =. (3.5

248 J. Mu, Y. Li, P. Chen In the following, we prove that Au τu, u B r, τ. (3.6 Otherwise, there exist u 2 B r and τ such that Au 2 = τ u 2. We may suppose τ >. By (2.8,(W 3 and (2., then ( τ u 2 (t ϕ 2 (tdt G(t, s f(s, u2 (s ds ϕ 2 (tdt ( (λ ε 3 G(t, s u 2 (s ds ϕ 2 (tdt ( = (λ ε 3 G(t, sϕ 2 (tdt u 2 (s ds which implies = (λ ε 3 = (λ ε 3 = ( ε 3 λ (T ϕ 2 (s u 2 (s ds λ ϕ 2 (s u 2 (s ds u 2 (t ϕ 2 (tdt, u 2 (t ϕ 2 (tdt. (3.7 On the other hand, ϕ 2 (t > for all < t < by the maximum principle and u 2 (t attains zero on isolated point by Sturm s theorem. Hence u 2 (t ϕ 2 (tdt >, which contradicts (3.7. Thus (3.6 holds. By Lemma 2.8. By (3.5, (3.8 and the additivity of degree, we have deg(i A, B r, θ =. (3.8 deg(i A, B R \B r, θ = deg(i A, B R, θ deg(i A, B r, θ =. Thus A has at least one fixed point in B R \B r. This means that the problem (. has at least one nontrivial solution. Corollary 3.2. Assume that (H, (A, (A 2, (A 3 from Section hold. Then the problem (. has at least one nontrivial solution.

Nontrivial Solutions for BVPs 249 Proof. From (A, for < ε 2 < λ, (λ ε 2 λ <, there exists l < such that f(t, u (λ ε 2 u b(t, t I, u < l. (3.9 From (A 2, there exist ε > and a sufficiently large number l 2 > such that f(t, u (λ + ε u, t I, u > l 2. (3. Since f : [, ] (, + (, + is continuous and b(t C(I, then there exists a constant l 3 that f(t, u (λ + ε u l 3 (λ ε 2 u l 3, t I, u, f(t, u (λ ε 2 u l 3 (λ + ε u l 3, t I, u. Thus, (W and (W 2 hold. It is easy to see that (A 3 (W 3. So, by Theorem 3., the problem (. has at least one nontrivial solution. Remark 3.3. In our paper, the green function is asymmetric. In fact, our results can be applicable to the case that the green function is symmetric, and then K = T, λ = λ, ϕ = ϕ 2. Theorem 3.4. Suppose that (H holds, and f satisfies the following conditions: (W 4 There exist ε 4, r > such that f(t, u (λ + ε 4 u, t I, u C(I, u r; (W 5 There exist < ε 5 < λ, (λ ε 5 λ < and r > such that f(t, u (λ ε 5 u, t I, u C(I, u r; (W 6 There exist < ε 6 < λ and g C(I such that f(t, u(t (λ ε 6 u(t + g(t, t I, u R, where λ, λ are defined by (2.. Then the problem (. has at least one nontrivial solution. Proof. Now, we show that u Au µϕ, u B r, µ, (3. where ϕ is defined by (2.. Suppose on the contrary, then there are u B r, µ satisfying u Au = µ ϕ. (3.2

25 J. Mu, Y. Li, P. Chen Combining (W 4, (2. and (3.2 yields u (tϕ 2 (tdt (λ + ε 4 = (λ + ε 4 λ ( G(t, su (sds ϕ 2 (tdt u (sϕ 2 (sds. Thus u (tϕ 2 (tdt. (3.3 On the other hand, by (2., (3.2, (3.3, (W 5 and Lemma 2.6, we get u (λ ε 5 (Ku P and u (t (λ ε 5 (Ku (t λ δ = δ ε 5 u (sϕ 2 (sds. ( u (t (λ ε 5 (Ku (t ϕ 2 (tdt Then u, which contradicts u B r. Thus (3. holds. By Lemma 2.7, deg(i A, B r, θ =. (3.4 Let In the following, we prove that R > max { r, ε 6 g }. (3.5 Au τu, u B R, τ. (3.6 Otherwise, there is u 2 B R and τ such that Au 2 = τ u 2. We may suppose τ >. Then by (2. and (W 6, ( u 2 (t ϕ 2 (tdt (λ ε 6 G(t, s u(s ds ϕ 2 (tdt ( + G(t, s g(s ds ϕ 2 (tdt = ( ε 6 λ u 2 (s ϕ 2 (sds + λ g(s ϕ 2 (sds. So ( u 2 (s ε 6 g(s ϕ 2 (tdt. (3.7

Nontrivial Solutions for BVPs 25 Because ϕ 2 (t > for < t < by maximum principle, then u 2 (s ε 6 g(s for < s <. Thus, R = u 2 ε 6 g, which contradicts (3.5. So (3.6 holds. By Lemma 2.8, deg(i A, B R, θ =. (3.8 By (3.4, (3.8 and the additivity of degree, we have deg(i A, B R \B r, θ = deg(i A, B R, θ deg(i A, B r, θ =. Thus A has at least one fixed point in B R \B r. This means that the problem (. has at least one nontrivial solution. 4 Examples In this section, we give two examples to demonstrate the applications of our main results. Example 4.. Consider the boundary value problem { u (t = f(t, u(t = h u(t + u 2 (t, t I, u( u ( =, u( = u(η, (4. where η (,, < h < λ and λ is the first eigenvalue of T corresponding to (4.. The boundary value problem (4. can be regarded as a form of (., where β = γ = α =. Then (H holds. (i For u(t λ + h or u(t, f(t, u(t (λ + u(t; for < u(t < λ + h, f(t, u(t (λ + u(t (λ + h 2. So, f(t, u (λ + u (λ + h 2, t I, u C(I. This means that (W holds. (ii We select ε 2 that max{λ λ, } < ε 2 < λ, where λ is the first eigenvalue of K corresponding to BVP (4.. If λ ε 2 h >, then for u(t λ ε 2 h or u(t, f(t, u(t (λ ε 2 u(t; for < u(t < λ ε 2 h, f(t, u(t (λ ε 2 u(t (λ ε 2 h 2. If λ ε 2 h <, then for u(t or u(t λ ε 2 h, f(t, u(t (λ ε 2 u(t; for λ ε 2 h < u(t <, f(t, u(t (λ ε 2 u(t (λ ε 2 h 2. Thus f(t, u (λ ε 2 u (λ ε 2 h 2, t I, u C(I. This means that (W 2 holds. (iii We select ε 3 that < ε 3 < λ h, u(t λ ε 3 h, f(t, u(t (λ ε 3 u(t; for (λ ε 3 h u(t <, (λ ε 3 h u(t u 2 (t ( (λ ε 3 h u(t, and then (λ ε 3 u(t f(t, u(t (λ ε 3 u(t. So, f(t, u(t (λ ε 3 u(t t I, u C(I, u λ ε 3 h. Thus, (W 3 holds. By Theorem 3., BVP (4. has at least one nontrivial solution.

252 J. Mu, Y. Li, P. Chen Example 4.2. Consider the boundary value problem { u (t = f(t, u(t = h 2 u(t + u 2 3 (t, t I, u( u ( =, u( = u(η, (4.2 where < h 2 < λ and λ is the first eigenvalue of T corresponding to BVP (4.2. (i Similar as Example 4., we can find that (H holds. (ii For u(t (λ + h 2 3, f(t, u(t (λ + u(t. Thus f(t, u (λ + u, t I, u C(I, u (λ + h 2 3. Then (W 4 holds. (iii We select ε 5 that max{λ λ, } < ε 5 < λ. If λ ε 5 h 2 >, then for u(t (λ ε 5 h 2 3, f(t, u(t (λ ε 5 u(t. If λ ε 5 h 2 <, then for u(t (λ ε 5 h 2 3, f(t, u(t (λ ε 5 u(t. So f(t, u (λ ε 5 u, t I, u C(I, u (λ ε 5 h 2 3. Then (W 5 holds. (iv We select ε 6 that < ε 6 < λ h 2. Then for u(t (λ ε 6 h 2 3, f(t, u(t (λ ε 6 u(t; for < u(t < (λ ε 6 h 2 3, f(t, u(t (λ ε 6 u(t + (λ ε 6 h 2 2 ; for ((λ ε 6 h 2 3 u(t, f(t, u(t (λ ε 6 u(t + (λ ε 6 + h 2 2 ; for u(t < ( (λ ε 6 h 2 3, f(t, u(t (λ ε 6 u(t ; So, f(t, u(t (λ ε 6 u(t + (λ ε 6 + h 2 2 t I, u C(I. Thus, (W 6 holds. By Theorem 3.4, BVP (4.2 has at least one nontrivial solution. Acknowledgements This research was supported by NNSF of China (876, the NSF of Gansu Province (7RJZA3, and Project NWNU-KJCXGC-3-47. References [] Y. Cui, Nontrivial solutions of singular superlinear m point boundary value problems, Appl. Math. Comput. 87 (27 256 264. [2] K. Deimling, Nonlinear Functional Analysis, Springer Verlag, Berlin, 985. [3] G. Han, Y. Wu, Nontrivial solutions of singular two point boundary value problems with sign changing nonlinear terms, J. Math. Anal. Appl. 325 (27 327 338.

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