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Positive solutions to a semipositone superlinear elastic beam equation

  • A semipositone fourth-order two-point boundary value problem is considered. In mechanics, the problem describes the deflection of an elastic beam rigidly fastened on the left and simply supported on the right. Under some conditions concerning the first eigenvalue corresponding to the relevant linear operator, the existence of nontrivial solutions and positive solutions to this boundary value problem is obtained. The main results are obtained by using the topological method and the fixed point theory of superlinear operators.

    Citation: Haixia Lu, Li Sun. Positive solutions to a semipositone superlinear elastic beam equation[J]. AIMS Mathematics, 2021, 6(5): 4227-4237. doi: 10.3934/math.2021250

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  • A semipositone fourth-order two-point boundary value problem is considered. In mechanics, the problem describes the deflection of an elastic beam rigidly fastened on the left and simply supported on the right. Under some conditions concerning the first eigenvalue corresponding to the relevant linear operator, the existence of nontrivial solutions and positive solutions to this boundary value problem is obtained. The main results are obtained by using the topological method and the fixed point theory of superlinear operators.



    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

    $ \left\{ \begin{array}{cc} u^{(4)}(t) = \lambda f(t, u(t)), \ 0\leq t\leq 1, \\ u(0) = u^{'}(0) = u(1) = u^{''}(1) = 0, \end{array} \right.\ \ \ \ \ \ \ \ ({\rm P}) $

    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

    $ \begin{align} f(t, u)\geq 0, \ \forall u\geq 0, \end{align} $ (1)

    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.

    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

    $ Ax\neq \mu x, \ \forall x\in \partial\Omega, \mu\geq 1, $

    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

    $ u-Au\neq \mu u_{0}, \ \forall u\in \partial\Omega, \mu\geq 0, $

    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} $,

    $ {\rm deg}(I-A, T_{r}, \theta) = {\rm deg}(I-A_{\theta}^{'}, T_{r}, \theta) = (-1)^{\kappa}, $

    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

    $ \begin{align} G(t, s) = \left\{ \begin{array}{cc} \frac{1}{12}(1-t)s^{2}[3(1-s)-(1-t)^{2}(3-s)], \ \ 0\leq s\leq t\leq 1, \\ \frac{1}{12}(1-s)t^{2}[3(1-t)-(1-s)^{2}(3-t)], \ \ 0\leq t\leq s\leq 1. \end{array} \right. \end{align} $ (2)

    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} $),

    $ \begin{array}{llllll} G(t, s)&\geq\frac{1}{6}s^{2}(1-s)t^{2}(1-t)\\ &\geq\frac{1}{4}s^{2}(1-s)\frac{2}{3}t^{2}(1-t)\\ &\geq\frac{1}{4}s^{2}(1-s)\frac{3}{4}t^{3}(1-t)^{2}\\ &\geq\frac{3}{4}t^{3}(1-t)^{2}G(s, \tau)\\ & = \frac{3}{4}t^{3}(1-t)^{2}G(\tau, s), \end{array} $

    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

    $ u(t) = \lambda\int_{0}^{1}G(t, s)f(s, u(s))ds. $

    Let

    $ \begin{align} (Au)(t) = \int_{0}^{1}G(t, s)f(s, u(s))ds, \end{align} $ (3)
    $ \begin{align} (Bu)(t) = \int_{0}^{1}G(t, s)u(s)ds. \end{align} $ (4)

    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) $.

    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

    $ \begin{align} \liminf\limits_{u\rightarrow +\infty}\frac{f(t, u)}{u}\geq\alpha, \ \forall t\in[0, 1]. \end{align} $ (5)

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

    $ \begin{align} \limsup\limits_{u\rightarrow 0}\left|\frac{f(t, u)}{u}\right|\leq\beta, \ \forall t\in[0, 1]. \end{align} $ (6)

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

    $ \begin{align} f(t, u)\geq -b, \ \forall t\in[0, 1], u(t)\in \mathbb{R}. \end{align} $ (7)

    (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} $,

    $ \begin{align} u-\lambda Au\neq\mu u^{*}, \forall\mu\geq 0, u\in \partial T_{R}, \end{align} $ (8)

    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

    $ \begin{align} u_{0}-\lambda Au_{0} = \mu_{0} u^{*}. \end{align} $ (9)

    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

    $ \begin{align} f(t, u)\geq(\alpha-\varepsilon)u-l, \ \forall t\in [0, 1], u\geq 0, \end{align} $ (10)
    $ \begin{align} |f(t, u)|\leq(\beta+\varepsilon)|u|, \ \forall t\in [0, 1], |u|\leq r_{0}. \end{align} $ (11)

    It follows from (7) that

    $ \begin{align} f(t, u)\geq -b\geq(\alpha-\varepsilon)u-b, \forall t\in [0, 1], u\leq 0. \end{align} $ (12)

    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

    $ \begin{align} \lambda Au = \lambda BFu\geq\lambda\alpha_{1}Bu-u_{1}, \ \forall u\in E, \end{align} $ (13)

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

    $ \delta = \frac{2}{3}\lambda_{1}\int_{0}^{1}s^{2}(1-s)u^{*}(s)ds > 0. $

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

    $ \begin{array}{llllll} u^{*}(t)& = \lambda_{1}Bu^{*}(t) = \lambda_{1} \int_{0}^{1}G(t, s)u^{*}(s)ds\\ &\geq\lambda_{1} \int_{0}^{1}\frac{1}{6}s^{2}(1-s)t^{2}(1-t)u^{*}(s)ds\\ &\geq \frac{1}{6}\lambda_{1}t^{2}(1-t) \int_{0}^{1}s^{2}(1-s)u^{*}(s)ds\\ &\geq \frac{2}{3}\lambda_{1}G(t , s) \int_{0}^{1}s^{2}(1-s)u^{*}(s)ds\\ & = \delta G(t, s). \end{array} $

    Let

    $ P_{1} = \left\{u\in P\mid\int_{0}^{1}u^{*}(t)u(t)dt\geq\lambda_{1}^{-1}\delta\|u\|\right\}. $

    For any $ u\in P $, we have

    $ \begin{array}{llll} \int_{0}^{1}u^{*}(t)(\lambda Bu)(t)dt& = \int_{0}^{1}u^{*}(t)[\int_{0}^{1}G(t, s)\lambda u(s)ds]dt\\ & = \int_{0}^{1}\lambda u(s)[\int_{0}^{1}G(s, t)u^{*}(t)dt]ds\\ & = \int_{0}^{1}\lambda u(s)Bu^{*}(s)ds\\ &\geq\lambda_{1}^{-1}\delta \int_{0}^{1}G(t, s)\lambda u(s)ds\\ & = \lambda_{1}^{-1}\delta(\lambda Bu)(t).\\ \end{array} $

    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

    $ \begin{align} \int_{0}^{1}u^{*}(t)(u_{0}+\lambda B\omega)(t)dt \geq\lambda_{1}^{-1}\delta\|u_{0}+\lambda B\omega\|\geq\lambda_{1}^{-1}\delta\|u_{0}\|-\lambda_{1}^{-1}\delta\|\lambda B\omega\|. \end{align} $ (14)

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

    $ \begin{array}{lll} \int_{0}^{1}u^{*}(t)(\lambda Au_{0})(t)dt&\geq& \int_{0}^{1}u^{*}(t)\alpha_{1}(\lambda Bu_{0})(t)dt- \int_{0}^{1}u^{*}(t)u_{1}(t)dt\\ & = &\alpha_{1}\lambda r(B) \int_{0}^{1}u^{*}(t)u_{0}(t)dt- \int_{0}^{1}u^{*}(t)u_{1}(t)dt\\ & = & \int_{0}^{1}u^{*}(t)u_{0}(t)dt+\varepsilon_{0} \int_{0}^{1}u^{*}(t)(u_{0}+\lambda B\omega)(t)dt\\ && -\varepsilon_{0} \int_{0}^{1}u^{*}(t)(\lambda B\omega)(t)dt - \int_{0}^{1}u^{*}(t)u_{1}(t)dt.\\ \end{array} $

    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

    $ \begin{array}{lll} \int_{0}^{1}u^{*}(t)(\lambda Au_{0}-u_{0})(t)dt&\geq&\varepsilon_{0}[\lambda_{1}^{-1}\delta\|u_{0}\|-\lambda_{1}^{-1}\delta\|\lambda B\omega\|]\\ &&-\varepsilon_{0} \int_{0}^{1}u^{*}(t)(\lambda B\omega)(t)dt- \int_{0}^{1}u^{*}(t)u_{1}(t)dt\\ & > &\varepsilon_{0}\delta\lambda_{1}^{-1}R_{0}-\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\\ & = &0.\\ \end{array} $

    But we see from (9) that

    $ \int_{0}^{1}u^{*}(t)(u_{0}-\lambda Au_{0})(t)dt = \int_{0}^{1}u^{*}(t)\mu_{0}u^{*}(t)dt\geq 0, $

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

    $ \begin{align} {\rm deg}(I-\lambda A, T_{R}, \theta) = 0. \end{align} $ (15)

    Next we show that

    $ \begin{align} (\lambda A)u\neq \mu u, \ \forall u\in\partial T_{r}, \mu\geq 1, \end{align} $ (16)

    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

    $ \begin{align} \mu_{0}^{n}|u_{0}|\leq B_{1}^{n}|u_{0}|. \end{align} $ (17)

    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

    $ \|B_{1}^{n}\|\geq\frac{1}{\|u_{0}\|}\|B_{1}^{n}u_{0}\|\geq\frac{d}{\|u_{0}\|}\mu_{0}^{n}, \qquad n = 1, 2, \cdots, $

    which shows

    $ r(B_{1}) = \lim\limits_{n\rightarrow\infty}(\|B_{1}^{n}\|)^{1/n} \geq\lim\limits_{n\rightarrow\infty}\left(\frac{d}{\|u_{0}\|}\mu_{0}^{n}\right)^{1/n} = \mu_{0} > 1. $

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

    $ \begin{align} {\rm deg}(I-\lambda A, T_{r}, \theta) = 1. \end{align} $ (18)

    By (15) and (18), we have

    $ {\rm deg}(I-\lambda A, T_{R}\backslash\overline{T_{r}}, \theta) = {\rm deg}(I-\lambda A, T_{R}, \theta)-{\rm deg}(I-\lambda A, T_{r}, \theta) = -1. $

    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

    $ f(t, u)\geq-\frac{b^{*}}{K}, \ \forall t\in[0, 1], u\geq-\frac{\lambda_{1}b^{*}}{\beta}, $

    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

    $ f_{1}(t, u) = \left\{ \begin{array}{lll} f(t, u), & \forall u\geq-\frac{\lambda_{1}b^{*}}{\beta}, & t\in [0, 1], \ \ \ \ \\ f(t, -\frac{\lambda_{1}b^{*}}{\beta}), & \forall u < -\frac{\lambda_{1}b^{*}}{\beta}, & t\in [0, 1], \end{array} \right. $
    $ (A_{1}u)(t) = \int_{0}^{1}G(t, s)f_{1}(s, u(s))ds. $

    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

    $ v^{*}(t) = \lambda\int_{0}^{1}G(t, s)f_{1}(s, v^{*}(s))ds\geq-\lambda\frac{b^{*}}{K}\int_{0}^{1}G(t, s)ds\geq-\frac{\lambda_{1}b^{*}}{\beta}. $

    Thus

    $ v^{*}(t) = \lambda\int_{0}^{1}G(t, s)f_{1}(s, v^{*}(s))ds = \lambda\int_{0}^{1}G(t, s)f(s, v^{*}(s))ds = \lambda Av^{*}(t). $

    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

    $ \begin{align} \lim\limits_{u\rightarrow 0}\frac{f(t, u)}{u} = \rho. \end{align} $ (19)

    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}\} $,

    $ \begin{align} {\rm deg}(I-\lambda A, T_{r}, \theta) = {\rm deg}(I-\lambda A_{\theta}^{'}, T_{r}, \theta) = (-1)^{\kappa}\neq 0, \end{align} $ (20)

    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.

    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

    $ f(t, u)\geq -b_{1}, \forall t\in[0, 1], u\geq 0, $
    $ \begin{align} f(t, u)\geq \eta^{-1}Nb_{1}, \forall t\in[t_{1}, t_{2}], u\geq R_{1}, \end{align} $ (21)

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

    $ f_{2}(t, u) = \left\{ \begin{array}{ll} f(t, u), & u\geq 0, \\ f(t, -u), & u < 0. \end{array} \right. $

    Then

    $ \begin{align} f_{2}(t, u)\geq -b_{1}, \forall t\in[0, 1], u\in \mathbb{R}. \end{align} $ (22)

    Let

    $ (A_{2}u)(t) = \int_{0}^{1}G(t, s)f_{2}(s, u(s))ds. $

    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} $,

    $ \begin{align} {\rm deg}(I-\lambda A_{2}, T_{R}, \theta) = 0, \forall\lambda > 0. \end{align} $ (23)

    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

    $ \|\lambda A_{2}u\| = \max\limits_{0\leq t\leq 1}|\int_{0}^{1}\lambda G(t, s)f_{2}(s, u(s))ds| < \overline{\lambda}Mm = r = \|u\|. $

    Thus

    $ \begin{align} {\rm deg}(I-\lambda A_{2}, T_{r}, \theta) = 1, \ \forall \ 0 < \lambda < \overline{\lambda}. \end{align} $ (24)

    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

    $ \begin{align} \lim\limits_{\lambda\rightarrow 0^{+}, \ u_{\lambda} = \lambda A_{2}u_{\lambda}, \ \|u_{\lambda}\| > r}\|u_{\lambda}\| = +\infty. \end{align} $ (25)

    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

    $ \begin{align} u_{\lambda_{n}} = \lambda_{n} A_{2}u_{\lambda_{n}}. \end{align} $ (26)

    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

    $ \begin{align} R = R(\overline{\lambda}) = {\rm max}\{2\eta^{-1}R_{1}, \ 2\eta^{-1}\overline{\lambda}b_{1}M, \ 2\overline{\lambda}b_{1}M\}. \end{align} $ (27)

    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] $,

    $ \begin{align} \begin{array}{lll} u_{0}(\overline{t})& = \lambda_{0} \int_{0}^{1}G(\overline{t}, s)[f_{2}(s, u_{0}(s))+b_{1}]ds- \lambda_{0} \int_{0}^{1}b_{1}G(\overline{t}, s)ds\\ &\geq\lambda_{0}\eta \int_{0}^{1}G(\tau, s)f_{2}(s, u_{0}(s))ds-\overline{\lambda}b_{1}M\\ & = \eta u_{0}(\tau)-\overline{\lambda}b_{1}M.\\ \end{array} \end{align} $ (28)

    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

    $ \begin{array}{lll} u_{0}(\overline{t})&\geq\eta\|u_{0}\|-\overline{\lambda}b_{1}M\geq\eta R-\overline{\lambda}b_{1}M\\ & = \frac{\eta}{2} R+\frac{\eta}{2} R-\overline{\lambda}b_{1}M\geq\frac{\eta}{2} R\geq R_{1}.\\ \end{array} $

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

    $ \begin{align} f_{2}(s, u_{0}(s))\geq \eta^{-1}Nb_{1}, \forall s\in D_{0} = [t_{1}, t_{2}]. \end{align} $ (29)

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

    $ \begin{align} G(t, s) = G(s, t)\geq\eta G(\tau, t) = \eta G(t, \tau). \end{align} $ (30)

    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), $

    $ \begin{align} \frac{1}{N}G(t, s)f_{2}(s, u_{0}(s))\geq b_{1}G(t, \tau). \end{align} $ (31)

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

    $ \frac{1}{N}\int_{D_{0}}G(t, s)f_{2}(s, u_{0}(s))ds\geq\int_{D_{i}}b_{1}G(t, \tau)d\tau, \ i = 1, 2, \cdots, N. $

    Thus

    $ \begin{align} \begin{array}{lll} \int_{D_{0}}G(t, s)f_{2}(s, u_{0}(s))ds&\geq\sum\limits_{i = 1}^{N} \int_{D_{i}}b_{1}G(t, \tau)d\tau\\ &\geq \int_{D\backslash D_{0}}b_{1}G(t, \tau)d\tau\\ & = \int_{D\backslash D_{0}}b_{1}G(t, s)ds.\\ \end{array} \end{align} $ (32)

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

    $ u_{0}(t) = \lambda_{0}\int_{D_{0}}G(t, s)f_{2}(s, u_{0}(s))ds+\lambda_{0}\int_{D\backslash D_{0}}G(t, s)f_{2}(s, u_{0}(s))ds\geq 0. $

    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

    $ u_{\lambda}(t) = \lambda\int_{0}^{1}G(t, s)f_{2}(s, u_{\lambda}(s))ds = \lambda\int_{0}^{1}G(t, s)f(s, u_{\lambda}(s))ds = \lambda Au_{\lambda}(t). $

    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

    $ \left\{ \begin{array}{cc} u^{(4)}(t) = \lambda \left[(\sqrt{t}+1)u^{3}-\sqrt[3]{u}\right], \ 0\leq t\leq 1, \\ u(0) = u^{'}(0) = u(1) = u^{''}(1) = 0.\ \ \ \ \ \ \ \ \ \ \ \end{array} \right.\ \ \ \ \ \ \ \ ({\rm P_{1}}) $

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

    $ \lim\limits_{u\rightarrow +\infty}\frac{f(t, u)}{u} = \lim\limits_{u\rightarrow +\infty}\left((\sqrt{t}+1)u^{2}-\frac{1}{\sqrt[3]{u^{2}}}\right) = +\infty, $

    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.

    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).

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



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