Abstract
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 fourthorder semipositone multipoint boundary value problems. The interesting point is that the nonlinear term f not only involve with the firstorder and the secondorder 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.
Keywords:
semipositone; positive solutions; multipoint boundary value problems1 Introduction
In this article, we consider the existence of positive solutions to the following nonlinear fourthorder semipositone multipoint boundary value problems with derivatives
where f ∈ C((0, 1) × R×R×R, R) satisfies f(t, y_{1 }y_{2}, y_{3}) ≥ p(t), p ∈ L^{1 }((0,1), (0, +∞)). λ > 0, ξ_{i }∈ (0, 1) with 0 < ξ_{1 }< ξ_{2 }< ⋯ < ξ_{m2 }< 1, α_{i}, β_{i }∈ [0, +∞), i = 1, 2,... , m2, are given constants satisfying . 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 multipoint boundary value problems has been widely studied in recent years. For details, see [115] 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 multipoint boundary value problems
For further background information of multipoint 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 multipoint 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 [11,1316]. 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 signchanging. Firstly, we employ a exchange technique and construct an integral operator for the corresponding secondorder multipoint 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 fixedpoint 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.
2 Preliminaries and lemmas
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:
(H_{1}) For i ∈ {1,2, ⋯, m  2}, ξ_{i }∈ (0, 1), 0 < ξ_{1 }< ξ_{2 }< ⋯ < ξ_{m2 }< 1 and α_{i}, β_{i }∈ [0, +∞) satisfying , and .
(H_{2}) f ∈ C((0, 1) × R × R × R, R) and there exist functions p, q ∈ L^{1}((0, 1), (0, +∞)), g ∈ C(R × R × R, (0, +∞)) such that
(H_{3}) for t uniformly on [0,1].
Remark 2.1. From (H_{2}) we know that for given points t_{1}, t_{2},..., t_{m }on [0,1], the functions p, q = (0, 1)\{t_{i}, i = 1, 2,..., m} → (0, +∞) are continuous and integrable, that is . The condition (H_{2}) also implies that f may have finitely singularities at t_{1}, t_{2},..., t_{m }on [0,1].
Lemma 2.1. Suppose that (H_{1}) and (H_{2}) hold. Then the problem (1.1) has a positive solution if and only if the following nonlinear secondorder integrodifferential 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 secondorder integrodifferential equation multipoint boundary value problem (2.1).
Conversely, let x(t) be a positive solution of the problem (2.1), then is a positive solution of the problem (1.1). In fact, , 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 for x ∈ C[0, 1]. Let
Lemma 2.2. Suppose that (H_{1}) holds. In addition, assume that u(t) ∈ L^{1}(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 (H_{1}) 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 . Thus we know that (2.4) holds.
Again from x''(t) = u(t) ≤ 0, we see that x'(t) is nonincreasing on (0, 1). Combining the condition we have x' (0) ≤ 0 and for t ∈ (0, 1). Hence x(t) is nonincreasing on (0, 1). By making use of the concavity of x(t) on (0, 1) we get x = x(0) and . 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 (H_{1}) holds. In addition, assume that p ∈ L^{1}((0, 1), (0, +∞)).
Then the following boundary value problem
has a unique positive solution z satisfying z(t) ≥ 0, t ∈ [0,1], and
where
ω is given by (2.5).
Proof. From Lemma 2.2. we have z(t) ≥ 0 and t ∈ [0,1]. By making use of (2.3) we get
The proof is complete. □
Let
and
Lemma 2.4. Suppose that (H_{1}) and (H_{2}) hold. Then the following nonlinear secondorder integrodifferential 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 fourthorder multipoint 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 . Noticing that [x(u)  z(u)]^{* }≤ x(u) ≤ L and , by virtue of (H_{2}), 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 (H_{1})  (H_{3}) hold. Then T : K →K is a completely continuous operator.
Proof. It follows from Lemma 2.2. we see T (K) ⊂ K. Combining (H_{1}) 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 M_{0 }> 0 such that x ≤ M_{0 }for all x ∈ D. By virtue of (H_{2}) we obtain
which implies that T (K) is uniformly bounded.
On the other hand, for all x ∈ D, once again from (H_{2}) we have
here . So, for any 0 ≤ t_{1 }< t_{2 }≤ 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 AscoliArzela 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 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.
3 Main results
In this section, we shall apply Lemma 2.6. to establish the existence of at least one positive solutions of the problem (1.1).
Theorem 3.1. Suppose that (H_{1})(H_{3}) hold. Then there exists sufficiently small λ^{* }> 0 such that the problem (1.1) has at least one positive solution for any λ ∈ (0, λ*).
Proof. Let r > 1 and λ ∈ (0, λ*) be fixed, where
here . Choose Ω_{r }= {x ∈ C^{+ }[0,1]: x < r}. If there is a fixed point on ∂Ω_{r}, we complete the proof. Without loss of generality, we may assume that there is no fixed point on ∂Ω_{r}. Thus, for any x ∈ K ∩∂Ω_{r}, from (3.1) we get
So
It follows from Lemma 2.6. we know that
Let d be a real number such that
Choose R > max{r, λdω + 1} such that if , then
and
Let , where is the unique solution of the problem (2.13). Denote
From (2.14) we have
Thus
Combining (3.7) with (3.5) and by making use of Lemma 2.2. we know that
From (3.8) together with (3.4), we see that
Then
Thus, for any x ∈ K ∩ ∂Ω_{R}, it follows from (H_{3}) we know that
Therefore, in view of (2.18) and (3.9) together with (3.3) we get
which implies that Tx ≥  x  for all x ∈K ∩ ∂Ω_{R}. It follows from Lemma 2.6 that
Combining (3.2) with (3.10) and the additivity of fixed point index, we know that
As a result, there exists satisfying Tx^{* }= x^{* }and r ≤ x* ≤ R. From (3.1) we have
Let y(t) = x*(t)  z(t). Then y(t) is a positive solution of the problem (2.1). By virtue of Lemma 2.1. we know that is a positive solution of the fourthorder multipoint boundary value problem (1.1). This completes the proof. □
Remark 3.1. In the case, when f = f(t, u) and f has lower bound i. e. f(t, u) + M ≥ 0 for some M > 0, we can study the secondorder multipoint boundary value problem under suitable condition by making use of the similar method. In particular, if p(t) = M, the conclusion of Theorem 3.1. is still valid.
Remark 3.2. The constant λ in problem (1.1) is usually called the Thiele modulus, in applications, one is interested in showing the existence of positive solutions for semipositone problems for small enough λ > 0.
4 Examples
Example 4.1. Consider the following singular fourthorder semipositone boundary value problem:
Proof. Let
Then
where , , which implies that (H_{1})(H_{3}) hold. Since , then we know that
Take r = 2, then
It follows from Theorem 3.1 that the problem (4.1) has at least one positive solution for any . □
Example 4.2. Consider the following singular fourthorder semipositone boundary value problem:
Proof. Let
Then
where , , which implies that (H_{1})(H_{3}) hold. Since , then we know that
Take r = 3, then
It follows from Theorem 3.1 that the problem (4.2) has at least one positive solution for any . □
Competing interests
The author declares that she has no competing interests.
Acknowledgements
The author is very grateful to Editor of the Journal and the anonymous referees for their carefully reading of the first draft of the manuscript and making many valuable suggestions and comments. The author was supported financially by the Foundation of Shanghai Municipal Education Commission (Grant Nos. DZL803, 10YZ77, and DYL201105).
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