Existence of positive solutions for fourth-order semipositone multi-point boundary value problems with a sign-changing nonlinear term

In this article, some new sufficient conditions are obtained by making use of fixed point index theory in cone and constructing some available integral operators together with approximating technique. They guarantee the existence of at least one positive solution for nonlinear fourth-order semipositone multi-point boundary value problems. The interesting point is that the nonlinear term f not only involve with the first-order and the second-order derivatives explicitly, but also may be allowed to change sign and may be singular at t = 0 and/or t = 1. Moreover, some stronger conditions that common nonlinear term f ≥ 0 will be modified. Finally, two examples are given to demonstrate the validity of our main results.

2000 Mathematics Subject Classification: 34B10; 34B18; 47N20.

In this article, we consider the existence of positive solutions to the following nonlinear fourth-order semipositone multi-point boundary value problems with derivatives

where f ∈ C((0, 1) × R×R×R, R) satisfies f(t, y₁ y₂, y₃) ≥ -p(t), p ∈ L¹ ((0,1), (0, +∞)). λ > 0, ξ_i ∈ (0, 1) with 0 < ξ₁ < ξ₂ < ⋯ < ξ_m-2 < 1, α_i, β_i ∈ [0, +∞), i = 1, 2,... , m-2, are given constants satisfying 0 < ∑ i = 1 m - 2 α i < 1 , 0 < ∑ i = 1 m - 2 β i < 1 . Here, by a positive solution of the problem (1.1) we mean a function y*(t) which is positive on (0, 1) and satisfies the problem (1.1).

The existence of positive solutions for multi-point boundary value problems has been widely studied in recent years. For details, see [1-15] and references therein. We note that the existence of n solutions and/or positive solutions to the following semipositone elastic beam equation boundary value problem

was obtained by Yao [13] in a Banach space setting. Gupta [3] proved the existence of positive solutions for more general multi-point boundary value problems

For further background information of multi-point boundary value problems we refer the reader to [11,12,16]. However, in previous work, the positivity which imposed on nonlinear term plays an important role for boundary value problems. Naturally, one is interested in establishing the existence of positive solutions for multi-point boundary value problems under the relaxed conditions.

Inspired and motivated greatly by the above mentioned works, the present work may be viewed as a direct attempt to extend the results of [3,13] to a broader class of nonlinear boundary value problems in a general Banach spaces. When the nonlinearity is negative, such kinds of the problems are called semipositone problems, which occur in chemical rector theory, combustion and management of natural resources, see [11,13-16]. To our best knowledge, few results were obtained for the problem (1.1).

The purpose of the article is to establish some new criteria for the existence of positive solutions to the problem (1.1). The nonlinear term f may take negative values and the nonlinearity may be sign-changing. Firstly, we employ a exchange technique and construct an integral operator for the corresponding second-order multi-point boundary value problem. Then we establish a special cone associated with concavity of functions. Finally, the existence of positive solutions for the problem (1.1) is obtained by applying fixed-point index theory. The common restriction on f ≥ 0 is modified.

The plan of the article is as follows. Section 2 contains a number of lemmas useful to the derivation of the main results. The proof of the main results will be stated in Section 3. A class of examples are given to show that our main result is applicable to many problems in Section 4.

In this section, we shall state some necessary definitions and preliminaries.

Definition 2.1. Let E be a real Banach space. A nonempty closed convex set K ⊂ E is called a cone if it satisfies the following two conditions:

(1) x ∈ K, λ > 0 implies λx ∈ K;

(2) x ∈ K, -x ∈ K implies x = 0.

Definition 2.2. An operator T is called completely continuous if it is continuous and maps bounded sets into precompact sets.

For convenience, we list the following assumptions:

(H₁) For i ∈ {1,2, ⋯, m - 2}, ξ_i ∈ (0, 1), 0 < ξ₁ < ξ₂ < ⋯ < ξ_m-2 < 1 and α_i, β_i ∈ [0, +∞) satisfying 0 < ∑ i = 1 m - 2 α i < 1 , 0 < ∑ i = 1 m - 2 β i < 1 and 0 < ∑ i = 1 m - 2 α i ξ i < 1 .

(H₂) f ∈ C((0, 1) × R × R × R, R) and there exist functions p, q ∈ L¹((0, 1), (0, +∞)), g ∈ C(R × R × R, (0, +∞)) such that

(H₃) lim ( | x 1 | + | x 2 | + | x 3 | ) → + ∞ f ( t , x 1 , x 2 , x 3 ) | x 1 | + | x 2 | + | x 3 | = + ∞ for t uniformly on [0,1].

Remark 2.1. From (H₂) we know that for given points t₁, t₂,..., t_m on [0,1], the functions p, q = (0, 1)\{t_i, i = 1, 2,..., m} → (0, +∞) are continuous and integrable, that is 0 < ∫ 0 1 ( p ( t ) + q ( t ) ) d t < + ∞ . The condition (H₂) also implies that f may have finitely singularities at t₁, t₂,..., t_m on [0,1].

Lemma 2.1. Suppose that (H₁) and (H₂) hold. Then the problem (1.1) has a positive solution if and only if the following nonlinear second-order integro-differential equation

has a positive solution.

Proof. Let y(t) be a positive solution of the problem (1.1) and let x(t) = y''(t). Then it follows from the problem (1.1) and combining with exchanging the integral sequence we know that

Thus x(t) = y''(t) is a positive solution of the second-order integro-differential equation multi-point boundary value problem (2.1).

Conversely, let x(t) be a positive solution of the problem (2.1), then y ( t ) = ∫ 0 t ( t - u ) x ( u ) d u is a positive solution of the problem (1.1). In fact, y ′ ( t ) = ∫ 0 t x ( u ) d u , y ″ ( t ) = x ( t ) , which implies that y(0) = 0, y'(0) = 0. The proof is complete.   □

Now, let X = C[0,1]. Then X is a real Banach space with norm ∥ x ∥ = max t ∈ [ 0 , 1 ] | x ( t ) | for x ∈ C[0, 1]. Let

Lemma 2.2. Suppose that (H₁) holds. In addition, assume that u(t) ∈ L¹(0, 1) and u(t) ≥ 0. Then the following problem

has a unique positive solution

satisfies x(t) ≥ 0, t ∈ [0,1] and

where

Proof. From (2.2), we have x'' (t) = -u(t), 0 < t < 1. For t ∈ [0,1], integrating from 0 to t we get

Thus

For t ∈ [0, 1], integrating (2.6) from t to 1 yields

which means that

From (2.9), we have

It follows from (2.9) and (2.10) that

Combining (2.11) with (H₁) we know that

From the fact that x'' (t) = -u(t) ≤ 0, we know that the graph of x(t) is concave on [0,1].

Thus

If x(1) ≥ 0, we know that x(t) ≥ 0 for all t ∈[0,1].

If x(1) < 0, from the concavity of x once again we know that

for i ∈ {1, 2, ..., m - 2}. This implies

which contracts with the fact 0 < ∑ i = 1 m - 2 α i ξ i < 1 . Thus we know that (2.4) holds.

Again from x''(t) = -u(t) ≤ 0, we see that x'(t) is non-increasing on (0, 1). Combining the condition 0 < ∑ i = 1 m - 2 β i < 1 we have x' (0) ≤ 0 and x ′ ( t ) = x ′ ( 0 ) - ∫ 0 t u ( s ) d s ≤ 0 for t ∈ (0, 1). Hence x(t) is non-increasing on (0, 1). By making use of the concavity of x(t) on (0, 1) we get ||x|| = x(0) and min t ∈ [ 0 , 1 ] x ( t ) = x ( 1 ) . Therefore, for all i = 1, 2, ⋯, m - 2, we obtain

which implies that

where ω is given by (2.5). This completes the proof.   □

Lemma 2.3. Suppose that (H₁) holds. In addition, assume that p ∈ L¹((0, 1), (0, +∞)).

Then the following boundary value problem

has a unique positive solution z satisfying z(t) ≥ 0, t ∈ [0,1], min t ∈ [ 0 , 1 ] z ( t ) ≥ ω ∥ z ∥ and

where

ω is given by (2.5).

Proof. From Lemma 2.2. we have z(t) ≥ 0 and min t ∈ [ 0 , 1 ] z ( t ) ≥ ω ∥ z ∥ , t ∈ [0,1]. By making use of (2.3) we get

The proof is complete.   □

Let

and

Lemma 2.4. Suppose that (H₁) and (H₂) hold. Then the following nonlinear second-order integro-differential equation boundary value problem

has a positive solution x(t) with x(t) ≥ z(t) for t ∈ [0, 1] if and only if y(t) = x(t) - z(t) is a nonnegative solution (positive on (0, 1)) of the problem (2.1).

Proof. Assume that y(t) = x(t) - z(t) is a nonnegative solution (positive on (0,1)) of the problem (2.1). Then we know that x(t) ≥ z(t) and

Noticing that z is a positive solution of the problem (2.13). Thus we get

Therefore x(t) is a positive solution of the problem (2.15) with x(t) ≥ z(t) for t ∈ [0,1].

Conversely, we assume that x(t) and z(t) are positive solutions of the problem (2.15) and the problem (2.13), respectively, and it implies that the boundary conditions of the problem (2.13) are also satisfied. Thus y(t) = x(t) - z(t) is a nonnegative solution (positive on (0, 1)) of the problem (2.1). The proof is complete.   □

Remark 2.2. Combining Lemma 2.4. with Lemma 2.1. we know that if the problem (2.15) has a positive solution, then the fourth-order multi-point boundary value problem (1.1) has a positive solution. So, we need only to study the problem (2.15).

Remark 2.3. For any fixed x ∈ C⁺[0,1], let L = max t ∈ [ 0 , 1 ] x ( t ) . Noticing that [x(u) - z(u)]^* ≤ x(u) ≤ L and ∫ 0 s ( s - u ) [ x ( u ) - z ( u ) ] * d u ≤ ∫ 0 1 L d u = L , by virtue of (H₂), we obtain

where

We introduce an integral mapping T : C⁺[0, 1] → C⁺[0, 1] defined by

Denote

where ω is given by the problem (2.5). It is obvious that K is a positive cone of C[0,1].

Lemma 2.5. Suppose that (H₁) - (H₃) hold. Then T : K →K is a completely continuous operator.

Proof. It follows from Lemma 2.2. we see T (K) ⊂ K. Combining (H₁) with (2.18) we know that T (K) is equicontinuous and uniformly bounded. In fact, let D ⊂ C⁺[0,1] be a bounded set. Then there exists M₀ > 0 such that ||x|| ≤ M₀ for all x ∈ D. By virtue of (H₂) we obtain

which implies that T (K) is uniformly bounded.

On the other hand, for all x ∈ D, once again from (H₂) we have

here M * = ∫ 0 1 [ p ( s ) + q ( s ) ] d s . So, for any 0 ≤ t₁ < t₂ ≤ 1, and for all x ∈ D, we get

By the absolutely continuous of integral, we know that T(K) is equicontinuous on [0,1]. Thus, an application of the Ascoli-Arzela theorem we know that T(K) is a relatively compact set.

Now we show that T is continuous. Let x_n → x^* (n → ∞), x_n, x^* ∈ C⁺[0,1]. It follows from the Lebesgue control convergence theorem that we obtain

Therefore T : K →K is a completely continuous operator. The proof is complete.   □

Lemma 2.6. [17] Let X = (X, ||·||) be a Banach space and K ⊂ X be a cone. For r > 0 define K_r = {u ∈ K : ||u|| < r}. Assume that T : K ¯ r → K is a completely continuous operator, such that Tu ≠ u for u ∈ ∂K_r = {u ∈ K : ||u|| = r}, and

(1) If ||Tu|| ≥ ||u|| for u ∈ ∂K_r, then i(T, K_r, K) = 0,

(2) If ||Tu|| ≤ ||u|| for u ∈ ∂K_r, then i(T, K_r, K) = 1.