Nonlocal boundary value problems with resonant or non-resonant conditions

We study solvability of nonlocal boundary value problems for second-order differential equations with resonance or non-resonance. The method of proof relies on Schauder’s fixed point theorem. Some examples are presented to illustrate the main results.

MSC:34B05.

In this paper, we investigate the existence of solutions for the following boundary value problem:

where J=[0,1], for 1\le i\le k, , g:\mathbb{R}\to \mathbb{R}, f:J\times \mathbb{R}\to \mathbb{R}.

Nonlocal boundary value problems, studied by Il’in and Moiseev [1], have been addressed by many authors; see, for example, [2–9] and references therein. In the related literature, (1.1) is called resonance when {\sum }_{i=1}^k{a}_i=1, and non-resonance when {\sum }_{i=1}^k{a}_i\ne 1. For the boundary value problems at resonance, researchers usually use the continuity method or nonlinear alternative, which involves a complicated a priori estimate for the solution set; see [7, 10–12]. However, it is very difficult to obtain a related estimate for general differential equations. Here we list only a classical result about nonlocal boundary value problems at resonance of the form

where h:J\times \mathbb{R}\times \mathbb{R}\to \mathbb{R} is continuous, e:J\to \mathbb{R} is continuous and a_i>0 for 1\le i\le k, {\sum }_{i=1}^k{a}_i=1, .

Theorem 1.1 [13]

Suppose that there are two constants M,\delta >0 such that

(A1) x[h(t,x,0)+e(t)]>\delta for any |x|>M, t\in J;

(A2) there exist constants L_1,L_2:L_1>M, such that

(A3) for (t,x,p)\in J\times [-M,M]\times [L_2,L_1],

Then (1.2) has at least one solution.

For (1.1), condition (A2) in Theorem 1.1 implies that f(t,x)\equiv g(x) for (t,x)\in J\times [-M,M]. It follows that (1.1) has infinitely many solutions (x\equiv C\in [-M,M] is the solution of (1.1)). At this point, Theorem 1.1 has little significance for (1.1). Moreover, there are few papers considering multiple results at resonance. For the case with non-resonance, there is an extensive literature; see [14–17] and the references therein.

The main purpose of this article is to discuss the existence of solutions of equation (1.1) by means of Schauder’s fixed point theorem. We only need to consider the behavior of g and f on some closed sets. Consequently, information on the location of the solution is obtained and multiple results are obtained if g and f satisfy the given conditions on distinct regions. Our approach is valid for the cases at resonance or non-resonance. In addition, some of our conditions are easily certified (see Corollaries 3.1 and 3.2).

The paper is organized as follows. Section 2 introduces an important lemma. Section 3 is devoted to the existence results of (1.1). In Section 4 we extend some results of Section 3 to the general boundary conditions.

In this section, we consider the following boundary value problem for the linear differential equation:

where p>0, , for 1\le i\le k, and h\in C(J,\mathbb{R}).

Lemma 2.1

1. (1)

   Boundary value problem (2.1) has a unique solution x_h\in C^2(J,\mathbb{R}).

2. (2)

   If h\equiv C\in R on J and a=1, then x_h\equiv C/p on J.

3. (3)

   If h(t)\ge 0 for all t\in J, then x_h\ge 0 on J; if h(t)\le 0 for all t\in J, then x_h\le 0 on J.

4. (4)

   If |h(t)|\le C (C>0) on J, then |x_h|\le C/p on J.

5. (5)

   Define an operator A:C(J,\mathbb{R})\to C(J,\mathbb{R}) by A(h)=x_h, where \parallel h\parallel ={max}_{t\in J}|h(t)|; then A is completely continuous.

Proof (1) Any solution of the differential equation -x^{″}(t)+px(t)=h(t) can be written as

where c_1, c_2 are constants and \phi \in C^2(t,\mathbb{R}) is a particular solution of -x^{″}(t)+px(t)=h(t). From the boundary conditions, we obtain that

Since the above system has a unique solution (c_1,c_2), (2.1) has a unique solution x_h\in C^2(t,\mathbb{R}).

1. (2)

   The conclusion is obvious.

2. (3)

   Here we only prove the case of h\ge 0. We consider two cases.

Case 3.1 Assume that x_h(t)\le 0 for all t\in J. We show that x_h(t)\equiv 0, t\in J. From (2.1), we obtain that

which implies that x_h^{\prime } is nonincreasing on J. Noting x_h^{\prime }(0)=0, we obtain that x_h is nonincreasing. Thus x_h(1)={min}_{t\in J}{x}_h(t)\le 0.

Since x_h is continuous, by the intermediate value theorem, there exists \theta \in (0,1) such that a{x}_h(\theta )={\sum }_{i=1}^m{a}_i{x}_h({\eta }_i).

If a=1, one can obtain from the monotonicity of x_h that x_h(t)\equiv x_h(1) on [\theta ,1]. Hence,

which implies that x_h(1)=0. Hence, x_h(t)\equiv 0 on J.

If , then x_h(1)={\sum }_{i=1}^m{a}_i{x}_h({\eta }_i)=a{x}_h(\theta )\ge a{x}_h(1), which implies that x_h(1)=0. Hence, x_h(t)\equiv 0 on J.

Case 3.2 There exist t_1,t_2\in J such that and x_h(t_2)>0. We assume that . Otherwise, x_h(t)\le 0 for all t\in [t_1,1]. Similar to Case 3.1, one can show that x_h\equiv 0 for t\in [t_1,1], which is impossible.

If there is \epsilon >0 such that x_h(t)\le 0 for t\in (0,\epsilon ), it is easy to check that x_h(t)\le 0 for all t\in J, a contradiction. Since , there exists r\in (0,1) such that x_h(r)={min}_{t\in J}{x}_h(t). Noting that x_h^{″}(r)\ge 0 and , we obtain that

which is a contradiction.

From Cases 3.1 and 3.2, one can easily obtain that x_h\ge 0 for all t\in J.

1. (4)

   Since -C\le h\le C, using the conclusion of (3), we have

   x_{h-C}\le 0,x_{h+C}\ge 0,t\in J.

Noting that x_{h-C}=x_h-x_C, x_{h+C}=x_h-x_{-C}, we obtain that

If a=1, then x_C=C/p and x_{-C}=-C/p. Thus |x_h|\le C/p on J.

Assume that . There exist t_1,t_2\in [0,1) such that

Noting that x_C^{″}(t_1)\le 0 and x_{-C}^{″}(t_2)\ge 0, we have

and

From (2.3), (2.4) and (2.5), we obtain that |x_h|\le C/p on J.

1. (5)

   Let h_n\to h in C(J,\mathbb{R}). For any \epsilon >0, there exists N(\epsilon )>0 such that

Noting that A(h_n-h)=x_{h_n}-x_h, using the conclusion of (4), we obtain that

which implies that A is continuous. Let D be a bounded set in C(J,\mathbb{R}). Then there is M_1>0 such that \parallel h\parallel \le M_1 for all h\in D. From the conclusion of (4), we obtain that |A(h)|\le M_1/p, which implies that A(D) is a uniformly bounded set. Since A(h), h are bounded for h\in D and

there exists M_2>0 such that

which implies that A(D) is equicontinuous. It follows that A(D) is relatively compact in C(J,\mathbb{R}) and A is a completely continuous operator. The proof is complete. □

Remark 2.1 Let h\equiv 1 and

By a direct computation, one can obtain that and

The following well-known Schauder fixed point theorem is crucial in our arguments.

Lemma 2.2 [18]

Let X be a Banach space and D\subset X be closed and convex. Assume that T:D\to D is a completely continuous map; then T has a fixed point in D.

The following theorem is the main result of the paper.

Theorem 3.1 Assume that there exist constants M>m, p>0 such that g\in C([m,M],\mathbb{R}), f\in C(J\times [m,M],\mathbb{R}), pu+g(u) is nondecreasing in u\in [m,M] and {\alpha }_p^{h=1}m\le {\beta }_p^{h=1}M. Further suppose that

Then (1.1) has at least one solution x with m\le x\le M.

Proof From Lemma 2.1, if x is a solution of (1.1), x satisfies

where A\circ H is composition of A and H defined as (A\circ H)x=A(Hx), and the operator H is defined in C(J,\mathbb{R}) as

Note that

Obviously, a fixed point of A\circ H is a solution of (1.1). Set \mathrm{\Omega }=\{x\in C(J,\mathbb{R}):m\le x(t)\le M,t\in J\}. Since the function pu+g(u) is nondecreasing in u\in [m,M], we obtain that for x\in \mathrm{\Omega },

Using (3.1), we have

for any x\in \mathrm{\Omega }. Since {\beta }_p^{h=1}\le pA(1)\le {\alpha }_p^{h=1}, and A is one nondecreasing operator, we obtain that for x\in \mathrm{\Omega },

Hence, (A\circ H)(\mathrm{\Omega })\subset \mathrm{\Omega }.

Also, the fact that A is completely continuous and H is continuous gives that A\circ H:\mathrm{\Omega }\to \mathrm{\Omega } is a continuous, compact map. By Lemma 2.2, A\circ H has at least one fixed point in Ω. The proof is complete. □

Remark 3.1 In Theorem 3.1, the condition that pu+g(u) is nondecreasing in u\in [m,M] can been replaced by the weaker condition

Corollary 3.1 Assume that a=1 and the following condition holds:

(H₁) There exist constants such that g\in C^1([m,M],\mathbb{R}), f\in C(J\times [m,M],\mathbb{R}), and

Then (1.1) has at least one solution x with m\le x\le M.

Proof Since g\in C^1([m,M],\mathbb{R}), there exists p>0 such that the function pu+g(u) is nondecreasing in u\in [m,M]. When a=1, {\alpha }_p^{h=1}={\beta }_p^{h=1}=1. We directly apply Theorem 3.1 and this ends the proof. □

Corollary 3.2 Assume that and the following condition holds:

(H₂) There exists constant M>0 such that g\in C^1([0,M],\mathbb{R}), f\in C(J\times [0,M],\mathbb{R}), and

Then (1.1) has at least one solution x with 0\le x\le M.

Proof Since g\in C^1([0,M],\mathbb{R}), there exists p>0 such that the function pu+g(u) is nondecreasing in u\in [0,M]. Condition (3.1) is satisfied if (3.4) holds. The proof is complete. □

Example 3.1 Consider the differential equation

Let m_n=(2n+0.5)\pi, M_n=(2n+1.5)\pi, g(u)=sinu, f(t,u)=-t^2{e}^{-u} and n be a positive integer. For any t\in J and u\in [m_n,M_n],

Hence, by Corollary 3.1, (3.5) has a solution m_n\le x\le M_n. Since n is an arbitrary positive integer, (3.5) has infinitely many solutions.

Example 3.2 Consider the differential equation

Using Corollary 3.2, we obtain that (3.6) has at least a nonnegative solution \overline{x}. Moreover, it is not difficult to show that \overline{x}\in (0,5) for any t\in J.

In this section, we extend some results in the previous section to the following equation:

where U_1 and U_2 are linear operators defined as

where a_i,b_j,c_s,d_l\in \mathbb{R} and for 1\le i\le n_1, 1\le j\le n_2, 1\le s\le m_1, 1\le l\le m_2.

Consider the boundary value problem for the linear differential equation:

where p>0 is sufficiently large, h:J\to \mathbb{R}. We introduce the following assumptions.

(P₁) The condition h\in C(J,\mathbb{R}) implies that boundary value problem (4.2) has a unique solution x_h\in C^2(J,\mathbb{R}).

(P₂) The condition h\equiv C\in \mathbb{R} implies that x_h\equiv C/p on J.

(P₃) The condition h\in C(J,\mathbb{R}):h\ge 0 (t\in J) implies that x_h\ge 0 on J.

(P₄) The condition h\in C(J,\mathbb{R}):|h(t)|\le C (C>0) implies that |x_h|\le C/p on J.

We say that the boundary condition U_{1}x=U_{2}x=0 satisfies P₁₂₃ if (4.2) satisfies conditions (P₁), (P₂), (P₃), and P₁₃₄ if (4.2) satisfies conditions (P₁), (P₃), (P₄).

Theorem 4.1

1. (1)

   Assume that the boundary condition U_{1}x=U_{2}x=0 satisfies P₁₂₃ and (H₁) holds. Then (4.1) has at least one solution x with m\le x\le M.

2. (2)

   Assume that the boundary condition U_{1}x=U_{2}x=0 satisfies P₁₃₄ and (H₂) holds. Then (4.1) has at least one solution x with 0\le x\le M.

The proof of Theorem 4.1 is similar to that of Theorem 3.1 and we omit it.

Remark 4.1 The solution obtained in Corollary 3.2 or (2) of Theorem 4.1 may be trivial. Further suppose that

Then the solution obtained is nonnegative and nontrivial.

Remark 4.2 The boundary condition U_{1}x=U_{2}x=0 satisfies P₁₃₄ if it satisfies P₁₂₃.

Remark 4.3 Consider the two-point boundary conditions:

One can easily check that boundary conditions (4.3), (4.4) satisfy P₁₂₃, and conditions (4.5), (4.6), (4.7) satisfy P₁₃₄.

Next, we consider the boundary conditions

where α, β, λ, μ, b_i (1\le i\le k), c_j (1\le j\le n) are constants and \alpha ,\lambda \in (0,+\mathrm{\infty }), \beta ,\mu ,b_i,c_j\in [0,+\mathrm{\infty }), {\eta }_i,{\xi }_j\in (0,1).

Theorem 4.2 Set b={\sum }_{i=1}^k{b}_i, c={\sum }_{j=1}^n{c}_j.

1. (1)

   If b=\alpha, c=\lambda, then the boundary condition U_{1}x=U_{2}x=0 satisfies P₁₂₃.

2. (2)

   If b\in [0,\alpha ], c\in [0,\lambda ) or b\in [0,\alpha ), c\in [0,\lambda ], then the boundary condition U_{1}x=U_{2}x=0 satisfies P₁₃₄.

Proof Without loss of generality, we assume that m=n=1. For any sufficiently large p>0 and h\in C(J,\mathbb{R}), the linear differential equation

has a unique solution x_h\in C^2(J,\mathbb{R}).

Suppose that h\ge 0 on J; if x_h\ge 0 is not true, by the maximum principle, we get that or . If , then x_h^{\prime }(0)\ge 0. From the boundary conditions, we have

By the maximum principle, we obtain that x_h({\eta }_1)\ge 0, which is a contradiction. If , then x_h^{\prime }(1)\le 0. From the boundary conditions, we have

By the maximum principle, we obtain that x_h({\xi }_1)\ge 0, a contradiction.

If h\equiv C and b=\alpha, c=\lambda, then x_h\equiv C/p.

Now suppose that 0\le h\le C on J and h\not\equiv 0.

If there is \theta \in (0,1) such that x_h(\theta )={max}_{t\in J}x(t)>0, noting that x_h^{″}(\theta )\le 0, we have

If x_h(0)={max}_{t\in J}x(t)>0, from the boundary conditions, we obtain that b=\alpha, \beta x_h^{\prime }(0)=0, which implies that x_h({\eta }_1)=x_h(0)={max}_{t\in J}x(t). The case has been discussed. If x_h(1)={max}_{t\in J}x(t)>0, from the boundary conditions, we obtain that c=\lambda, \mu x_h^{\prime }(1)=0, which implies that x_h({\xi }_1)=x_h(1)={max}_{t\in J}x(t). The case has also been discussed. The proof is complete. □

Example 4.1 Consider the differential equation

where \lambda >0, are constants.

Let g(u)={sin}^{-1}u and f(t,u)=u^{\lambda }-tu. Set m_n=2n\pi +{(4n\pi )}^{-\lambda -1}, M_n=2n\pi +0.5\pi. If n is a sufficiently large, positive integer, then for any t\in J, u\in [m_n,M_n],

By Theorems 4.1 and 4.2, (4.10) has a solution m_n\le x\le M_n. Hence, (4.10) has infinitely many solutions.

Example 4.2 Consider the differential equation

where \mu >0, , 0\le \lambda \le 0.5 are constants.

In fact, g(u)=-u^{\mu }cosu and f(t)=-t^2. The boundary conditions in (4.11) satisfy P₁₃₄ for \lambda \in [0,0.5] and P₁₂₃ for \lambda =0.5.

1. (1)

   Equation (4.11) has a solution 0\le \tilde{x}\le 1 and \tilde{x}(t)>0, t\in (0,1] for all 0\le \lambda \le 0.5. Set M=1, then g(M)=-1\le f(t)\le g(0) for all t\in J. By Theorems 4.1 and 4.2, (4.11) has a solution 0\le \tilde{x}\le 1. Now we show that \tilde{x}(t)>0 for t\in (0,1]. Assume that there exists r\in (0,1) with \tilde{x}(r)=0. Since \tilde{x}(t)\ge 0, \tilde{x}(r) is minimum value and {\tilde{x}}^{\prime }(r)=0, {\tilde{x}}^{″}(r)\ge 0. On the other hand, , a contradiction. If \tilde{x}(1)=0, then \tilde{x}({\xi }_2)=0. This is impossible.

2. (2)

   Equation (4.11) has infinitely many solutions for \lambda =0.5. Set M_n=2n\pi +2\pi, m_n=2n\pi +1.5\pi, where n>0 is an integer. Since g(M_n)=-M_n^{\mu }\le f(t)\le g(m_n)=m_n^{\mu } for all t\in J, (4.11) has a solution m_n\le x\le M_n. Hence, (4.11) has infinitely many solutions.

The work is supported by the NNSF of China (11171085) and Hunan Provincial Natural Science Foundation of China.

Affiliations

1. Department of Mathematics, Hunan University of Science and Technology, Xiangtan, Hunan, 411201, P.R. China
   • Weibing Wang
2. Department of Mathematics, Hunan First Normal College, Changsha, Hunan, 410205, P.R. China
   • Xuxin Yang

Corresponding author

Correspondence to Weibing Wang.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Authors typed, read and approved the final draft.

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