Positive solutions to a semipositone superlinear elastic beam equation

1.   Introduction

The purpose of this paper is to investigate the existence of nontrivial solutions and positive solutions to the following nonlinear fourth-order two-point boundary value problem

where $ \lambda $ is a positive parameter, $ f:[0, 1]\times \mathbb{R}\rightarrow \mathbb{R} $ is continuous. One function $ u\in C[0, 1] $ is called a positive solution of problem (P) if $ u $ is a solution of (P) and $ u(t) > 0, 0 < t < 1. $

Fourth-order two-point boundary value problems appear in beam analysis (See ^{[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]}). The deflection of an elastic beam rigidly fastened on the left and simply supported on the right leads to problem (P). The existence and multiplicity of positive solutions for the elastic beam equations have been studied extensively when the nonlinear term satisfies

see for example ^{[1,2,3,6,7,8,9,11]} and references therein. Agarwal and Chow ^[1] investigated the problem (P) by using contraction mapping and iterative methods. Bai ^[3] applied upper and lower solution method and Yao ^[11] used Guo-Krasnosel'skii fixed point theorem of cone expansion-compression type. However there are only a few papers concerned with the non-positone or semipositone elastic beam equations. Yao ^[12] considered the existence of positive solutions of semipositone elastic beam equations by using a special cone and the fixed point theorem of cone expansion-compression type. In this paper, we assume that there exists a constant $ b > 0 $ such that $ f(t, u)\geq -b $ for all $ t\in[0, 1] $ and $ u(t)\in\mathbb{R} $, which implies that problem (P) is semipositone. We obtain the existence of nontrivial solutions and positive solutions to the semipositone boundary value problem (P) under some conditions concerning the first eigenvalue corresponding to the relevant linear operator by the topological method and the fixed point theory of superlinear operators.

2.   Preliminaries

In this section, we give some preliminaries required in our subsequent discussions.

Let $ E $ be an ordered Banach space in which the partial ordering $ \leq $ is induced by a cone $ P\subset E $, $ \theta $ the zero element of $ E $. For the concepts and properties about the cone we refer to ^[16,17,18].

Lemma 1. (see ^[19]). Let $ \Omega\subset E $ be a bounded open set, $ \theta\in \Omega $, and $ A: \overline{\Omega}\rightarrow E $ a completely continuous operator. If

then deg $ (I-A, \Omega, \theta) = 1. $

Lemma 2. (see ^[19]). Let $ \Omega\subset E $ be a bounded open set and $ A: \overline{\Omega}\rightarrow E $ a completely continuous operator. If there exists $ u_{0}\in E\backslash\{\theta\} $ such that

then deg$ (I-A, \Omega, \theta) = 0. $

The following is the famous Leray-Schauder theorem.

Lemma 3. (see ^[18]). Let $ \Omega\subset E $ be a bounded open set, $ \theta\in \Omega $ and $ A: \overline{\Omega}\rightarrow E $ a completely continuous operator with $ A\theta = \theta $. Suppose that the Frechet derivative $ A_{\theta}^{'} $ of $ A $ at $ \theta $ exists and 1 is not an eigenvalue of $ A_{\theta}^{'} $. Then there exists $ r_{0} > 0 $ such that for any $ 0 < r < r_{0} $,

where $ \kappa $ is the sum of algebraic multiplicities for all eigenvalues of $ A_{\theta}^{'} $ lying in the interval (0, 1) and $ T_{r} = \{x\in E\mid \|x\| < r\} $.

Let $ G(t, s) $ be the Green's function of homogeneous linear problem $ u^{(4)}(t) = 0, u(0) = u^{'}(0) = u(1) = u^{''}(1) = 0 $, which can be explicitly given by

By (2) and Yao ^[13] we have

(G$ _{1} $) $ G(t, s) = G(s, t), \ 0\leq t, s\leq 1; $

(G$ _{2} $) $ \frac{1}{6}s^{2}(1-s)t^{2}(1-t)\leq G(t, s)\leq\frac{1}{4}t^{2}(1-t), \ 0\leq t, s\leq 1; $

By (G$ _{1} $)(G$ _{2} $),

and we have

(G$ _{3} $) $ G(t, s)\geq\frac{3}{4}t^{3}(1-t)^{2}G(\tau, s), \ 0\leq t, s, \tau\leq 1. $

It is well known that the problem (P) is equivalent to the integral equation

Let

By the famous Krein-Rutmann theorem (See ^[16]) and similar to Lemma 3 in ^[20], we have

Lemma 4. Suppose that the linear operator $ B $ is defined by (4). Then the spectral radius $ r(B)\neq 0 $ and $ B $ has a positive eigenfunction corresponding to its first eigenvalue $ \lambda_{1} = r^{-1}(B) $.

3.   Nontrivial solutions

In the sequel we always take $ E = C[0, 1] $ with the norm $ \|u\| = \max_{0\leq t\leq1}|u(t)| $ and the cone $ P = \{u\in C[0, 1]\mid u(t)\geq 0, 0\leq t\leq 1\} $.

Now let us list the following conditions which will be used in this paper:

(H$ _{1} $) There exists a constant $ \alpha > 0 $ satisfying

(H$ _{2} $) There exists a constant $ \beta $ with $ 0 < \beta < \alpha $ satisfying

(H$ _{3} $) There exists a constant $ b > 0 $ such that

(H$ _{4} $) $ \lim\limits_{u\rightarrow +\infty} \frac{f(t, u)}{u} = +\infty $.

Theorem 5. Suppose that (H$ _{1} $)-(H$ _{3} $) hold. Then for any $ \lambda\in(\frac{\lambda_{1}}{\alpha}, \frac{\lambda_{1}}{\beta}) $, problem (P) has at least one nontrivial solution, where $ \lambda_{1} = r^{-1}(B) $ and $ B $ is given by (4).

Proof. Let $ (Fu)(t) = f(t, u(t)) $ for all $ u\in E $, then by (4), $ A = BF. $ Applying the Arzela-Ascoli theorem and a standard argument, we can prove that $ (\lambda A): E\rightarrow E $ is a completely continuous operator. It is known to all that the nonzero fixed points of the operator $ \lambda A $ are the nontrivial solutions of the boundary value problem (P).

Now we show that there exists $ R_{0} > 0 $, such that for any $ R > R_{0} $,

where $ u^{*} $ is the positive eigenfunction of $ B $ corresponding to its first eigenvalue $ \lambda_{1} = r^{-1}(B) $, $ T_{R} = \{u\in C[0, 1]\mid \|u\| < R\} $ is a bounded open subset of $ E $.

If (8) is not true, then there exist $ \mu _{0} > 0 $ (if $ \mu_{0} = 0 $, then Theorem 5 holds) and $ u_{0}\in \partial T_{R} $ such that

By (5) and (6), there exist $ l > 0, r_{0} > 0 $ and $ 0 < \varepsilon < {\rm min}\{\frac{\alpha\lambda-\lambda_{1}}{\lambda}, \frac{\lambda_{1}-\beta\lambda}{\lambda}\} $ such that

It follows from (7) that

Take $ \omega = {\rm max}\{b, l\} $, then by (10) and (12) we have $ f(t, u)\geq(\alpha-\varepsilon)u-\omega, \forall t\in [0, 1], u\in R. $ Let $ \alpha_{1} = \alpha-\varepsilon $, then

where $ u_{1} = \int_{0}^{1}G(t, s)\lambda\omega ds $. Take

By (4) and (G$ _{2} $) we have

Let

For any $ u\in P $, we have

And so $ \ \int_{0}^{1}u^{*}(t)(\lambda Bu)(t)dt\geq\lambda_{1}^{-1}\delta\|\lambda Bu\| $, i.e., $ \ (\lambda B)(P)\subset P_{1}. $ Since $ Fu_{0}+\omega\in P $, then $ (\lambda B)(Fu_{0}+\omega)\in P_{1} $ and $ \mu_{0}u^{*} = \mu_{0}\lambda_{1}Bu^{*}\in P_{1} $. By (9) we have $ u_{0}+\lambda B\omega = \lambda Au_{0}+\mu_{0}u^{*}+\lambda B\omega = \lambda B(Fu_{0}+\omega)+\mu_{0}u^{*}\in P_{1}. $ Thus

Take $ \varepsilon_{0} = \alpha_{1}\lambda r(B)-1 > 0 $. By (13) we have

Take $ R_{0} = \frac{\lambda_{1}}{\varepsilon_{0}\delta}[\varepsilon_{0}\delta\lambda_{1}^{-1}\|\lambda B\omega\| +\varepsilon_{0}\int_{0}^{1}u^{*}(t)(\lambda B\omega)(t)dt+\int_{0}^{1}u^{*}(t)u_{1}(t)dt]. $ For any $ \|u_{0}\| = R > R_{0} $, by (14) we have

But we see from (9) that

which is a contradiction. So (8) is true. By Lemma 2 we have

Next we show that

where $ 0 < r < {\rm min}\{r_{0}, R_{0}\} $. Assume on the contrary that there exist $ u_{0}\in\partial T_{r} $ and $ \mu_{0}\geq 1 $ such that $ (\lambda A)u_{0} = \mu_{0}u_{0} $. Since $ \lambda A $ has no fixed point on $ \partial T_{r} $, we have $ \mu_{0} > 1 $. Let $ B_{1} = \lambda(\beta+\varepsilon)B $, then $ r(B_{1}) < 1 $. By (11) we have $ |\lambda Au_{0}|\leq(\beta+\varepsilon)\lambda B|u_{0}| = B_{1}|u_{0}| $, then $ \mu_{0}|u_{0}|\leq B_{1}|u_{0}|, $ and therefore

Let $ D = \{v\mid v\geq|u_{0}|\} $. It follows from (17) that $ \{\mu_{0}^{-n}B_{1}^{n}|u_{0}|\mid n = 1, 2, \cdots\}\subset D. $ And $ u_{0}\in\partial T_{r} $ and $ \theta\in T_{r} $ imply that $ d = d(\theta, D) > 0 $. Then one can have that

which shows

This contradicts $ r(B_{1}) < 1. $ So (16) holds. By Lemma 1, we have

By (15) and (18), we have

Then $ \lambda A $ has at least one fixed point in $ T_{R}\backslash\overline{T_{r}} $. This means that problem (P) has at least one nontrivial solution.

Remark 6. For any $ \alpha > 0 $ and $ \beta = 0, $ the proof of Theorem 5 is valid. Then for any $ \lambda > 0 $, problem (P) has at least one nontrivial solution.

Corollary 7. Suppose that (H$ _{1} $) and (H$ _{2} $) hold. Assume there exists a constant $ b^{*} > 0 $ such that

where $ K = \max_{t\in[0, 1]}\int_{0}^{1}G(t, s)ds $. Then for any $ \lambda\in(\frac{\lambda_{1}}{\alpha}, \frac{\lambda_{1}}{\beta}) $ problem (P) has at least one nontrivial solution.

Proof. Let

Then all conditions of Theorem 5 hold for $ f_{1} $. By Theorem 5, $ \lambda A_{1} $ has at least one nonzero fixed point $ v^{*}(t) $, and

Thus

This indicates $ v^{*}(t) $ is a nontrivial solution of problem (P).

Theorem 8. Suppose that (H$ _{1} $) and (H$ _{3} $) hold. Let $ f(t, 0)\equiv 0, \forall t\in[0, 1] $ and

Then for any $ \lambda\in(\frac{\lambda_{1}}{\alpha}, +\infty) $ and $ \lambda\neq\frac{\lambda_{1}}{\rho} $, problem (P) has at least one nontrivial solution.

Proof. From the proof of Theorem 5, if (H$ _{1} $) and (H$ _{3} $) hold, then there exists $ R_{0} > 0 $ such that for any $ R > R_{0} $ and $ \lambda > \frac{\lambda_{1}}{\alpha} $ (15) holds.

Since $ f(t, 0)\equiv 0, \forall t\in[0, 1] $, then $ A\theta = \theta $. By (19) we have that the Frechet derivative $ A_{\theta}^{'} $ of $ A $ at $ \theta $ exists and $ (A_{\theta}^{'}u)(t) = \int_{0}^{1}G(t, s)\rho u(s)ds. $ Notice that $ \lambda\neq\frac{\lambda_{1}}{\rho} $, then 1 is not an eigenvalue of $ \lambda A_{\theta}^{'} $. By Lemma 3 there exists $ r_{0} > 0 $, for any $ 0 < r < {\rm min}\{r_{0}, R_{0}\} $,

where $ \kappa $ is the sum of algebraic multiplicities for all eigenvalues of $ \lambda A_{\theta}^{'} $ lying in the interval (0, 1).

By (15) and (20) $ \lambda A $ has at least one nonzero fixed point. Thus problem (P) has at least one nontrivial solution.

4.   Positive solutions

In many realistic problems, the positive solution is more significant. In this section we will study this question.

Theorem 9. Suppose that (H$ _{4} $) holds. Then there exists $ \lambda^{*} > 0 $ such that for any $ 0 < \lambda < \lambda^{*} $ problem (P) has at least one positive solution.

Proof. Let $ D = [0, 1], D_{0} = [t_{1}, t_{2}]\subset(0, 1)\subset D, \eta = \min_{t_{1}\leq t\leq t_{2}}\frac{3}{4}t^{3}(1-t)^{2} > 0 $. By (H$ _{4} $) there exist $ b_{1} > 0 $ and $ R_{1} > 0 $ such that

where $ N > \frac{1-(t_{2}-t_{1})}{t_{2}-t_{1}} $ is a natural number. Let

Then

Let

Obviously, $ A_{2}: E\rightarrow E $ is a completely continuous operator.

From Remark 6 and the proof of Theorem 5 there exists $ R_{0} > 0 $, for any $ R > R_{0} $,

Take $ 0 < r < R_{0} $. Let $ m = \max_{0\leq t\leq 1, |u|\leq r}|f_{2}(t, u)| $, $ M = \max_{0\leq s, t\leq 1}G(t, s) $, $ \overline{\lambda} = r(mM)^{-1} $. For any $ 0 < \lambda < \overline{\lambda}, u\in \partial T_{r} $, we have

Thus

From (23) and (24) we have that for any $ 0 < \lambda < \overline{\lambda} $, there exists $ u_{\lambda}\in C[0, 1] $ with $ \|u_{\lambda}\| > r $ such that $ u_{\lambda} = \lambda A_{2}u_{\lambda}. $ Now we show

In fact, if (25) doesn't hold, then there exist $ \lambda_{n} > 0, u_{\lambda_{n}}\in C[0, 1] $ such that $ \lambda_{n}\rightarrow 0, r < \|u_{\lambda_{n}}\| < c $ $ (c > 0 $ is a constant), and

Since $ A_{2} $ is completely continuous, then $ \{u_{\lambda_{n}}\} $ has a subsequence (assume without loss of generality that it is $ \{u_{\lambda_{n}}\} $) converging to $ u_{*}\in C[0, 1]. $ Let $ n\rightarrow+\infty $ in (26), we have $ u_{*} = \theta $, which is a contradiction of $ \|u_{\lambda_{n}}\| > r > 0 $. Then (25) holds.

Next we show that there exists $ R = R(\overline{\lambda}) > 0 $ such that if $ 0 < \lambda_{0}\leq\overline{\lambda}, \|u_{0}\|\geq R $ and $ u_{0} = \lambda_{0}Au_{0} $, then $ u_{0}(t)\geq 0. $ Take

Assume that $ 0 < \lambda_{0}\leq\overline{\lambda}, \|u_{0}\|\geq R $ and $ u_{0} = \lambda_{0}Au_{0}. $ Take any $ \overline{t}\in[t_{1}, t_{2}] $, by (G$ _{3} $) we have that for any $ \tau\in[0, 1] $,

On account of the continuity of $ u_{0} $, there exists $ t^{*}\in[0, 1] $ such that $ u_{0}(t^{*}) = \|u_{0}\|. $ Take $ \tau = t^{*} $ in (28), by (27) we have

Thus $ u_{0}(\overline{t})\geq R_{1}, $ for any $ \overline{t}\in[t_{1}, t_{2}] $. By (21) we have

It follows from (G$ _{1} $) and (G$ _{3} $) that for any $ s\in[t_{1}, t_{2}] $ and $ t, \tau\in[0, 1] $

Take $ D_{i}\subset D \ (i = 1, 2, \cdots, N) $ such that $ {\rm mes}D_{i} = {\rm mes}D_{0} $, $ \bigcup\limits_{i = 1}^{N}D_{i}\supset D\backslash D_{0} $. By (29) and (30) we have that for any $ t\in D, s\in D_{0}, \tau\in D_{i} \ (i = 1, 2, \cdots, N), $

Notice that $ {\rm mes}D_{i} = {\rm mes}D_{0}(i = 1, 2, \cdots, N), $ then

Thus

By (32) and (21) we have that for any $ t\in[0, 1] $,

For $ R $ in (27), by (25), there exists $ \lambda^{*} < \overline{\lambda} $ such that if $ 0 < \lambda\leq\lambda^{*}, \|u_{\lambda}\|\geq r $ and $ u_{\lambda} = \lambda A_{2}u_{\lambda}, $ then $ \|u_{\lambda}\|\geq R $, thus $ u_{\lambda}(t)\geq 0. $ By the definition of $ A_{2} $ and $ f_{2} $ we have

So $ u_{\lambda}(t) $ is a positive solution of problem (P).

Remark 10. In Theorem 9 we obtain the existence of positive solutions for the semipositone boundary value problem (P) without that assuming (1) holds.

Remark 11. Since we only study the existence of positive solutions for the boundary value problem (P), which is irrelevant to the value of $ f(t, u) $ when $ u\leq 0 $, we only suppose that $ f(t, u) $ is bounded below when $ u\geq 0 $. The nonlinear term $ f(t, u) $ may be unbounded from below when $ u\leq 0 $.

Example 12. Consider the fourth-order boundary value problem

In this example, $ f(t, u) = (\sqrt{t}+1)u^{3}-\sqrt[3]{u} $, then

which means that (H$ _{4}) $ holds. By Theorem 9 there exists $ \lambda^{*} > 0 $ such that for any $ 0 < \lambda < \lambda^{*} $ BVP $ ({\rm P_{1}}) $ has at least one positive solution.

Remark 13. In Example 12, the nonlinear term $ f $ doesn't satisfy (1) and (H$ _{3}) $, but the existence of positive solutions of BVP $ ({\rm P_{1}}) $ is obtained by using our result.

Acknowledgments

This work is supported by the Foundation items: National Natural Science Foundation of China (11501260), Natural Science Foundation of Jiangsu Higher Education Institutions(18KJB180027), and Postgraduate Research Innovation Program of Jiangsu Province(KYCX20_82).

Conflict of interest

The authors declare that there is no conflict of interests regarding the publication of this article.