New results for positive solutions of singular fourth-order four-point p-Laplacian problem

Abstract

The existence and uniqueness of positive solutions are obtained for singular fourth-order four-point boundary value problem with p-Laplace operator $[\varphi_{p}(u''(t))]''=f(t,u(t))$, $0< t<1$, $u(0)=0$, $u(1)=au(\xi)$, $u''(0)=0$, $u''(1)=bu''(\eta)$, where $f(t,u)$ is singular at $t=0,1$ and $u=0$. A fixed point theorem for mappings that are decreasing with respect to a cone in a Banach space plays a key role in the proof.

1 Introduction

In this paper, we investigate the existence and uniqueness of positive solutions for the singular fourth-order differential equation involving the p-Laplace operator

$$ \bigl[\varphi_{p} \bigl(u''(t) \bigr) \bigr]''=f \bigl(t,u(t) \bigr), \quad t\in(0,1), $$

(1.1)

with the four-point boundary conditions

$$ u(0)=0,\qquad u(1)=au(\xi), \qquad u''(0)=0, \qquad u''(1)=bu''( \eta), $$

(1.2)

where $\varphi_{p}(t)=|t|^{p-2}t$, $p>1$, $0<\xi, \eta<1$, $0\leq a<1/\xi$, $0\leq b^{p-1}<1/\eta$, and $f(t,x)$ is singular at $t=0, 1$ and $x=0$. Here by a positive solution u of SBVP (1.1)-(1.2) we mean a solution $u\in C^{2}[0,1]$ with $\varphi_{p}(u'')\in C^{2}(0,1)\cap C[0,1]$ satisfying $u(t)>0$ on $(0,1)$.

It is well known that the bending of elastic beam can be described by some fourth-order boundary value problems. There are extensive studies on fourth-order boundary value problems with diverse boundary conditions by using different methods, for instance, [1–28] and the references therein.

Recently, in the case $0\leq a<1$, $0\leq b<1$, using the lower and upper solution method and the Schauder fixed-point theorem, Zhang and Liu [25] proved that the SBVP (1.1)-(1.2) has at least one positive solution under the following assumptions:

(H₁):

$f\in C((0,1)\times(0,\infty),[0,\infty))$, and $f(t,x)$ is nonincreasing in x;

(H₂):

For any constant $\lambda>0$, $0<\int_{0}^{1}H(s,s)f(s,\lambda s(1-s))\,\mathrm{d}s<\infty$;

(H₃):

There exist a continuous function $a(t)$ in $[0,1]$ and a fixed positive number k such that $a(t)\geq kt(1-t)$, $t \in [0,1]$, and

$$\begin{aligned}& \int_{0}^{1}G(t,r)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(r,s)f\bigl(s,a(s)\bigr)\,\mathrm{d}s \biggr)\,\mathrm{d}r:=b(t)\geq a(t), \quad t\in[0,1], \\& \int_{0}^{1}G(t,r)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(r,s)f\bigl(s,b(s)\bigr)\,\mathrm{d}s \biggr)\,\mathrm{d}r\geq a(t),\quad t\in[0,1], \end{aligned}$$

where $G(t,s)$, $H(t,s)$ will be given in Section 2.

The purpose of this paper is to improve the existence results of [25]. Using a fixed point theorem for mappings that are decreasing with respect to a cone in a Banach space, we obtain the existence and uniqueness of positive solutions of SBVP (1.1)-(1.2). We note that, in our proofs, we just assume that (H₁) and (H₂) of [25] with (H₃) of [25] removed. Our study is motivated by the papers [11, 29].

In addition, we note that we also obtained the uniqueness of a positive solution for SBVP (1.1)-(1.2).

The rest of the paper is organized as follows. The fixed point theorem of Gatica et al. [29] and some definitions and lemmas are given in Section 2. The main results on the existence of positive solutions for SBVP (1.1)-(1.2) are presented in Section 3.

2 Preliminary

Let B be a Banach space. A nonempty closed set $K\subset B$ is called a cone if the following conditions are satisfied:

• (i) $a u+b v \in K$ for all $u,v\in K$ and all $a, b \geq0$;

• (ii) $u, -u \in K$ imply $u=0$.

Given a cone K, a partial order ⪯ is induced on B as follows; $u\preceq v$ for $u,v\in B$ iff $v-u\in K$ (for clarity, we sometimes write $u \preceq v$ (w.r.t. K)). For $u,v\in B$ with $u\preceq v$, we denote by $\langle u,v\rangle$ the closed order interval between u and v, that is, $\langle u,v\rangle=\{w\in B: u\preceq w \preceq v\}$. A cone K is normal in B if there exists $\delta>0$ such that $\|e_{1}+e_{2}\|\geq\delta$ for all $e_{1}, e_{2} \in K$ with $\|e_{1}\|=\|e_{2}\|=1$.

Lemma 2.1

([11, 29])

Let B be a Banach space, K a normal cone in B, D a subset of K such that if $u,v\in D$ with $u\preceq v$, then $\langle u,v\rangle\subset D$, and let $T: D\rightarrow K$ be a continuous mapping that is compact on any closed order interval contained in D. Suppose that there exists $u_{0}\in D$ such that $T^{2}u_{0}$ is defined and $Tu_{0}$ and $T^{2}u_{0}$ are order-comparable to $u_{0}$. Then T has a fixed point in D, provided that either

• (I) $Tu_{0}\preceq u_{0}$ and $T^{2}u_{0}\preceq u_{0}$ or $u_{0}\preceq Tu_{0}$ and $u_{0}\preceq T^{2}u_{0}$, or

• (II) the complete sequence of iterates $\{T^{n}u_{0}\} _{n=0}^{\infty}$ is defined and there exists $v_{0}\in D$ such that $Tv_{0}\in D$ and $v_{0}\preceq T^{n}u_{0}$ for all $n\geq0$.

Let $G(t,s)$ denote the Green function for

$$-u''=0,\qquad u(0)=0,\qquad u(1)=\alpha u(\xi). $$

Then by [18] the Green function $G(t,s)$ can be expressed as follows:

$$G(t,s)= \left \{ \textstyle\begin{array}{@{}l} s\in[0,\xi]: \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{t}{1-a\xi}[(1-s)-a(\xi-s)], & t\leq s, \\ \frac{s}{1-a\xi}[(1-t)-a(\xi-t)], & s\leq t, \end{array}\displaystyle \right . \\ s\in[\xi,1]: \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{t}{1-a\xi}(1-s), & t\leq s, \\ \frac{1}{1-a\xi}[s(1-t)+a\xi(t-s)], & s\leq t. \end{array}\displaystyle \right . \end{array}\displaystyle \right . $$

Lemma 2.2

([25])

The Green function $G(t,s)$ has the following properties:

$$t(1-t)G(s,s)\leq G(t,s)\leq G(s,s) \quad\textit{for } (t,s)\in[0,1]\times[0,1]. $$

Let $B=C[0,1]$ denote the Banach space of continuous functions with norm

$$\|u\|=\sup_{t\in[0,1]}\bigl|u(t)\bigr|,\quad \forall u\in B, $$

and let $K=\{u\in B: u(t)\geq0 \mbox{ on } [0,1]\}$ be the cone of nonnegative functions in B. It is easy to see that K is a normal cone in B. Now we define the subset $D\subset K$ as

$$D:= \bigl\{ u\in K: \mbox{ there exists } \lambda(u)>0 \mbox{ such that } u(t) \geq \lambda t(1-t) \mbox{ on } [0,1] \bigr\} . $$

Moreover, define $T:D\rightarrow K$ by

$$(Tu) (t):= \int_{0}^{1}G(t,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau,s) f \bigl(s,u(s) \bigr)\, \mathrm{d}s \biggr) \,\mathrm{d}\tau,\quad \forall u\in D, $$

where

$$H(t,s)= \left \{ \textstyle\begin{array}{@{}l} s\in[0,\eta]: \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{t}{1-b^{p-1}\eta}[(1-s)-b^{p-1}(\eta-s)], & t\leq s, \\ \frac{s}{1-b^{p-1}\eta}[(1-t)-b^{p-1}(\eta-t)], & s\leq t, \end{array}\displaystyle \right . \\ s\in[\eta,1]: \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{t}{1-b^{p-1}\eta}(1-s), & t\leq s, \\ \frac{1}{1-b^{p-1}\eta}[s(1-t)+b^{p-1}\eta(t-s)], & s\leq t. \end{array}\displaystyle \right . \end{array}\displaystyle \right . $$

Then T is well defined. In fact, from Lemma 2.2 we have

$$t(1-t)H(s,s)\leq H(t,s)\leq H(s,s) \quad\mbox{for } (t,s)\in [0,1]\times[0,1]. $$

It can be easily verified that

$$\bigl|H_{t}'(t,s)\bigr|\leq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1+b}{1-b^{p-1}\eta}s , & 0 \leq s \leq t, \\ \frac{1+b}{1-b^{p-1}\eta}(1-s) , & 0 \leq t \leq s. \end{array}\displaystyle \right . $$

It follows from conditions (H₁) and (H₂) that, for each $u\in K$,

$$\begin{aligned}[b] \biggl\vert \int_{0}^{1}H_{\tau}'(\tau,s)f \bigl(s,u(s)\bigr)\,\mathrm{d}s\biggr\vert \leq{}& \int_{0}^{1}\bigl|H_{\tau}'( \tau,s)\bigr|f\bigl(s,\lambda s(1-s)\bigr)\,\mathrm{d}s \\ \leq{}& \frac{1+b}{1-b^{p-1}\eta} \biggl[ \int_{0}^{\tau }sf\bigl(s,\lambda s(1-s)\bigr)\, \mathrm{d}s\\ &{} + \int_{\tau}^{1}(1-s)f\bigl(s,\lambda s(1-s)\bigr)\, \mathrm{d}s \biggr]:=v(\tau). \end{aligned} $$

By the Fubini theorem and (H₂) it is easy to show that $v(\tau)\in L^{1}[0,1]$, and hence $\int_{0}^{1}H(\tau,s)f(s,u(s))\,\mathrm{d}s\in C[0,1]$. Thus, T is well defined, and also from Lemma 2.2 we have that, for all $u\in D$ and $t\in[0,1]$,

$$\begin{aligned} (Tu) (t) = & \int_{0}^{1}G(t,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau,s) f\bigl(s,u(s)\bigr)\, \mathrm{d}s \biggr) \,\mathrm{d}\tau \\ \leq& \int_{0}^{1}G(\tau,\tau)\varphi_{p}^{-1} \biggl( \int _{0}^{1}H(\tau,s) f\bigl(s,u(s)\bigr)\, \mathrm{d}s \biggr) \,\mathrm{d}\tau. \end{aligned}$$

Hence,

$$\|Tu\|\leq \int_{0}^{1}G(\tau,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau,s) f \bigl(s,u(s) \bigr)\, \mathrm{d}s \biggr)\,\mathrm{d}\tau. $$

On the other hand, by Lemma 2.2, for all $u\in D$ and $t\in [0,1]$, we have

$$\begin{aligned} (Tu) (t) = & \int_{0}^{1}G(t,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau,s) f\bigl(s,u(s)\bigr)\, \mathrm{d}s \biggr) \,\mathrm{d}\tau \\ \geq& t(1-t) \int_{0}^{1}G(\tau,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau,s) f\bigl(s,u(s)\bigr)\, \mathrm{d}s \biggr)\,\mathrm{d}\tau. \end{aligned}$$

Thus,

$$ (Tu) (t)\geq t(1-t)\|Tu\|,\quad t\in[0,1]. $$

(2.1)

This implies that $Tu\in D$, that is, $T:D\rightarrow D$, and hence, it can be verified that $u\in D$ is a positive solution of SBVP (1.1)-(1.2) iff $Tu=u$.

3 Main results

In this section, we first establish an existence theorem of positive solutions for SBVP (1.1)-(1.2) by applying Lemma 2.1.

Before proceeding with our existence result for SBVP (1.1)-(1.2), we will define a sequence of functions that are modifications of f and have none of the singularities of f at $u=0$. To this end, we define a sequence of functions $f_{n}: (0,1)\times[0,\infty)\rightarrow[0,\infty)$ by

$$f_{n}(t,u)=f \bigl(t,\max \bigl\{ u,t(1-t)/n \bigr\} \bigr). $$

Note that, for $n=1,2,\ldots$ , $f_{n}$ satisfies (H₁). Also, for $n=1,2,\ldots$ ,

$$\begin{aligned}& f_{n}(t,u)\leq f(t,u),\quad (t,u)\in(0,1)\times(0, \infty), \end{aligned}$$

(3.1)

$$\begin{aligned}& f_{n}(t,u)\leq f \bigl(t,t(1-t)/n \bigr),\quad (t,u)\in(0,1) \times[0, \infty). \end{aligned}$$

(3.2)

We now state and prove our existence result for SBVP (1.1)-(1.2).

Theorem 3.1

Assume that conditions (H₁) and (H₂) are satisfied. Then SBVP (1.1)-(1.2) has at least one positive solution $u^{*}\in D$.

Proof

We define the sequence of mappings $T_{n}: K\rightarrow K$ by

$$(T_{n}u) (t):= \int_{0}^{1}G(t,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau,s) f_{n} \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr) \,\mathrm{d}\tau,\quad \forall u\in K. $$

Then by (3.2) and condition (H₂), $T_{n}$ is well defined and bounded. We note that $T_{n}$ is a continuous mapping by Lebesgue’s dominated convergence theorem. Also, it is easy to show by (H₂) and the continuity of $G(t,s)$ that $\{(T_{n}u)(t):u\in K\}$ is equicontinuous, and hence $T_{n}$ is a compact mapping by the Arzelà-Ascoli theorem.

In addition, observe that for all n and $u\in K$, $T_{n}u$ satisfies the boundary conditions (1.2). Furthermore, for each n, since $T_{n}$ satisfies (H₁), it follows that $T_{n}$ is nonincreasing relative to the cone K. Also, it is clear that $0\preceq T_{n}(0)$ and $0\preceq T_{n}^{2}(0)$ for each n. Thus, by Lemma 2.1, for each n, there exists $u_{n}\in K$ such that $T_{n}u_{n}=u_{n}$. Hence, for each n, $u_{n}(t)$ satisfies the boundary conditions (1.2).

Now we claim that there exist $R>r>0$ such that

$$r\leq\|u_{n}\|\leq R \quad\mbox{for all } n . $$

Firstly, we shall prove the right-hand side inequality. Assume to the contrary that the inequality is false. Then by passing to a subsequence and relabeling, without loss of generality, we may assume that

$$\lim_{n\rightarrow\infty}\|u_{n}\|=\infty \quad\mbox{and}\quad \|u_{n}\|\leq\|u_{n+1}\| \quad\mbox{for all } n . $$

Similarly to the proof of (2.1), we can easily show that, for any $u\in K$,

$$ (T_{n}u) (t)\geq t(1-t)\|T_{n}u\|,\quad t \in[0,1]. $$

(3.3)

Since $T_{n}u_{n}=u_{n}$ for each n, it follows that

$$u_{n}(t)\geq t(1-t)\|u_{n}\|\geq t(1-t)\|u_{1}\|, \quad t\in[0,1]. $$

Then assumptions (H₁) and (H₂) and inequality (3.1) yield that, for any $0\leq t\leq1$ and n,

$$\begin{aligned} u_{n}(t) = & \int_{0}^{1}G(t,\tau)\varphi_{p}^{-1} \biggl( \int _{0}^{1}H(\tau,s) f_{n} \bigl(s,u_{n}(s)\bigr)\,\mathrm{d}s \biggr)\,\mathrm{d}\tau \\ \leq& \int_{0}^{1}G(\tau,\tau)\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(s,s) f\bigl(s,u_{n}(s)\bigr) \,\mathrm{d}s \biggr)\,\mathrm{d}\tau \\ \leq& \int_{0}^{1}G(\tau,\tau)\,\mathrm{d}\tau\cdot\varphi _{p}^{-1} \biggl( \int_{0}^{1}H(s,s) f\bigl(s,s(1-s)\|u_{1} \|\bigr)\,\mathrm{d}s \biggr):=M. \end{aligned}$$

Thus, $\|u_{n}\|\leq M$ for all n. This is a contradiction to $\lim_{n\rightarrow\infty}\|u_{n}\|=\infty$.

Next, we prove the left-hand side inequality. Assume to the contrary that the inequality is false. By passing to a subsequence and relabeling we may assume without loss of generality that $\lim_{n\rightarrow\infty}\|u_{n}\|=0$. Then we have

$$ \lim_{n\rightarrow\infty}u_{n}(t)=0 \quad \mbox{uniformly on } [0,1]. $$

(3.4)

Since $f(t,u)$ is singular at $u=0$, by the finite covering theorem there exists $\delta>0$ such that, for $t\in[\xi/2,\xi]$ and $0< x<\delta$, we have

$$f(t,x)> \frac{2}{\xi(1-\xi)} \biggl[ \int_{\xi/2}^{\xi}H(s,s)\,\mathrm{d}s \biggr]^{-1}. $$

By (3.4) there exists $n_{0}\geq1$ such that, for any $n\geq n_{0}$,

$$0< u_{n}(t)< \delta/2,\qquad 0< t(1-t)/n< \delta/2,\quad t\in [\xi/2,\xi]. $$

Hence, for any $n\geq n_{0}$, we have

$$\begin{aligned} u_{n}(\xi) = & \int_{0}^{1}G(\xi,\tau)\varphi_{p}^{-1} \biggl( \int _{0}^{1}H(\tau,s) f_{n} \bigl(s,u_{n}(s)\bigr)\,\mathrm{d}s \biggr)\,\mathrm{d}\tau \\ \geq& \int_{\xi/2}^{\xi}G(\xi,\tau)\varphi_{p}^{-1} \biggl( \int_{\xi/2}^{\xi}H(\tau,s) f_{n} \bigl(s,u_{n}(s)\bigr)\,\mathrm{d}s \biggr)\,\mathrm{d}\tau \\ \geq & \int_{\xi/2}^{\xi}G(\xi,\tau)\varphi_{p}^{-1} \biggl(\tau(1-\tau) \int_{\xi/2}^{\xi}H(s,s) f(s,\delta/2)\,\mathrm{d}s \biggr)\,\mathrm{d}\tau \\ \geq & \int_{\xi/2}^{\xi}G(\xi,\tau)\varphi_{p}^{-1} \biggl(\frac{\xi(1-\xi)}{2} \int_{\xi/2}^{\xi}H(s,s) f(s,\delta/2)\,\mathrm{d}s \biggr)\,\mathrm{d}\tau \\ \geq & \int_{\xi/2}^{\xi}G(\xi,\tau)\varphi _{p}^{-1}(1) \,\mathrm{d}\tau = \int_{\xi/2}^{\xi}G(\xi,\tau)\,\mathrm{d}\tau= \frac{3\xi^{2}(1-\xi )}{8(1-a\xi)}, \end{aligned}$$

which implies $\|u_{n}\|\geq\frac{3\xi^{2}(1-\xi)}{8(1-a\xi)}$. This is a contradiction to $\lim_{n\rightarrow\infty}\|u_{n}\|=0$.

In summary, we conclude our claim. Furthermore, from (3.3) we have

$$ rt(1-t) \preceq u_{n} \preceq R \quad(\mbox{w.r.t.}K), n=1,2,\ldots. $$

(3.5)

This implies that the sequence $\{u_{n}\}$ belongs to the closed order interval $\langle rt(1-t), R\rangle\subset D$. It is easy to see that the restriction of T to $\langle rt(1-t), R\rangle$ is a compact mapping. Hence, there exists a subsequence of $\{Tu_{n}\}$ that converges to some $u^{*}\in K$. Relabel the subsequence as the original sequence so that $\lim_{n\rightarrow \infty}\|Tu_{n}-u^{*}\|=0$.

Also, by (3.5) there exists $n_{0}$ such that, for all $n\geq n_{0}$,

$$t(1-t)/n\leq rt(1-t)\leq u_{n}(t) \quad\mbox{on } [0,1]. $$

For all $n\geq n_{0}$ and $t\in[0,1]$, we have

$$\begin{aligned} (Tu_{n}) (t)-u_{n}(t) = & (Tu_{n}) (t)-(T_{n}u_{n}) (t) \\ = & \int_{0}^{1}G(t,\tau) \biggl[\varphi_{p}^{-1} \biggl( \int _{0}^{1}H(\tau,s) f\bigl(s,u_{n}(s) \bigr)\,\mathrm{d}s \biggr) \\ &{} -\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(\tau ,s)f_{n} \bigl(s,u_{n}(s)\bigr)ds \biggr) \biggr]\,\mathrm{d}\tau \\ =& 0. \end{aligned}$$

This implies $\lim_{n\rightarrow\infty}\|Tu_{n}-u_{n}\|=0$. It follows, in turn, that $\lim_{n\rightarrow\infty}\|u_{n}-u^{*}\|=0$, and thus $u^{*}\in\langle rt(1-t), R\rangle\subset D$ and

$$u^{*}=\lim_{n\rightarrow\infty}Tu_{n}=T \Bigl(\lim _{n\rightarrow \infty}u_{n} \Bigr)=Tu^{*}. $$

In summary, we have $u^{*}\in D$ and $Tu^{*}=u^{*}$. This completes the proof of the theorem. □

Theorem 3.2

Assume that conditions (H₁) and (H₂) are satisfied. Then SBVP (1.1)-(1.2) has exactly one positive solution $u^{*}\in D$.

Proof

The existence of positive solution to SBVP (1.1)-(1.2) immediately follows from Theorem 3.1. Thus, we only need to show the uniqueness.

Suppose that $u_{1}(t)$ and $u_{2}(t)$ are two positive solutions of SBVP (1.1)-(1.2). Then, by (H₁) and (H₂), $u_{1}(t)$ and $u_{2}(t)$ are both the solutions of the following boundary value problem:

$$\begin{aligned}& u''(t)+\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f \bigl(s,u(s) \bigr)\, \mathrm{d}s \biggr)=0,\quad 0< t< 1, \\& u(0)=0,\qquad u(1)=au(\xi). \end{aligned}$$

Let $w(t)=u_{1}(t)-u_{2}(t)$ on $[0,1]$. Without loss of generality, we may assume that $w(1)\geq0$. Now we show that $w(t)\equiv0$ on $[0,1]$. There are two cases to consider.

Case 1. $w(1)>0$. In this case, we have $w(t)\geq0$ on $[0,1]$. Assume by contradiction that there exists $t_{0} \in(0,1)$ such that $w(t_{0})<0$. Since $w(0)=0$ and $w(1)>0$, there exist $t_{1}, t_{2}\in[0,1)$ with $t_{1}< t_{0}< t_{2}$ such that

$$w(t)< 0 \quad\mbox{on } (t_{1},t_{2}),\qquad w(t_{1})=w(t_{2})=0. $$

It follows that, for each $t\in(t_{1}, t_{2})$,

$$\begin{aligned} w''(t) = & u_{1}''(t)-u_{2}''(t) \\ = & -\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f\bigl(s,u_{1}(s)\bigr) \,\mathrm{d}s \biggr)+\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f\bigl(s,u_{2}(s)\bigr) \,\mathrm{d}s \biggr) \leq0. \end{aligned}$$

Hence, $w(t)\geq0$ on $[t_{1}, t_{2}]$, which is a contradiction to $w(t)<0$ on $(t_{1},t_{2})$. Therefore, $w(t)\geq0$ on $[0,1]$. Consequently, for each $t\in(0,1)$,

$$w''(t)= -\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f \bigl(s,u_{1}(s) \bigr) \,\mathrm{d}s \biggr)+\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f \bigl(s,u_{2}(s) \bigr) \,\mathrm{d}s \biggr)\geq0. $$

Thus, $w(t)$ is concave upward on $[0,1]$. Since $w(1)>0$ and $w(1)=\alpha w(\xi)$, we have $w(\xi)>0$, and hence since $0<\alpha<1/\xi$, we have

$$w(1)< \frac{1}{\xi}w(\xi), $$

which is a contradiction to the upward concavity of $w(t)$ on $[0,1]$.

Case 2. $w(1)=0$. In this case, we have $w(t)\equiv0$ on $[0,1]$. Assume to the contrary that the conclusion is false. Then, there exists $t_{0} \in(0,1)$ such that $w(t_{0})\neq0$. Without loss of generality, we may assume that $w(t_{0})>0$. Since $w(0)=w(1)=0$, there exist $t_{1}, t_{2}\in[0,1]$ with $t_{1}< t_{0}< t_{2}$ such that

$$w(t)>0 \quad\mbox{on } (t_{1},t_{2}),\qquad w(t_{1})=w(t_{2})=0. $$

It follows that, for each $t\in(t_{1}, t_{2})$,

$$w''(t)= -\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f \bigl(s,u_{1}(s) \bigr) \,\mathrm{d}s \biggr)+\varphi_{p}^{-1} \biggl( \int_{0}^{1}H(t,s) f \bigl(s,u_{2}(s) \bigr) \,\mathrm{d}s \biggr)\geq0. $$

Since $w(t_{1})=w(t_{2})=0$, we have that

$$w(t)\leq0 \quad\mbox{for } t\in(t_{1},t_{2}), $$

which is a contradiction to $w(t)>0$ on $(t_{1},t_{2})$.

In summary, $w(t)\equiv0$ on $[0,1]$. This completes the proof of the theorem. □