Existence and Nonexistence Results for a Fourth-Order Boundary Value Problem with Sign-Changing Green’s Function

Abstract

In this paper, we consider a fourth-order three-point boundary value problem. Despite the fact that the corresponding Green’s function changes its sign on the square of its definition, we obtain the existence of at least one positive and decreasing solution under some suitable conditions. The results are based on the classical Krasosel’skii’s fixed point theorem in cones. Then, we impose some sufficient conditions that allow us to deduce nonexistence results. In the end, some examples are given in order to illustrate our main results.

1. Introduction

Fourth-order ordinary differential equations arise from different areas of applied mathematics and physics. Some interesting particular examples and mathematical models coupled with the basic theory of these types of equations can be found in the classical monographs [1,2,3,4]. Recently, many authors studied the existence of single or multiple solutions to some fourth-order problems by using different techniques such as Leray–Schauder nonlinear alternative [5,6], fixed point index theory [7,8,9], monotone iterative technique [10,11] or the method of upper and lower solutions [12,13,14]. However, it is important to point out that the results in almost all of these works are based on the fundamental fact that the corresponding Green’s functions are strictly positive. There are only a few papers where this condition does not hold as the corresponding Green’s functions change their sign on the square of its definition.

For example, one of the first results in this direction was given by Ma in [15], where the author studied the existence of positive solutions for the second-order periodic boundary value problem:

$$\begin{array}{c}{u}^{″}\left(t\right)+a\left(t\right)u\left(t\right)=\lambda b\left(t\right)f\left(u\right),\mathrm{a}.\mathrm{e}.t\in \left[0,T\right],\\ u\left(0\right)=u\left(T\right),u^{\prime }\left(0\right)=u^{\prime }\left(T\right),\end{array}$$

where the positive part of Green’s function is greater than the negative part.

Recently, in [16], the authors studied the existence, nonexistence and multiplicity of solutions of the third-order nonlinear differential equation:

$$\begin{array}{c}{u}^{‴}\left(t\right)=-\lambda p\left(t\right)f\left(u\left(t\right)\right),\mathrm{a}.\mathrm{e}.t\in [0,1]\equiv I,\\ u\left(0\right)=0,u^{″}\left(\eta \right)=\alpha u^{\prime }\left(1\right),u^{\prime }\left(1\right)=\beta u\left(1\right),\end{array}$$

where $\lambda >0$ is a parameter, $0\le \alpha \le 1$, ${\displaystyle 0\le \beta <\frac{2}{2-\alpha }}$, ${\displaystyle \eta \in \left[0,\frac{1}{3}\right]}$, $f:[0,\infty )\to [0,\infty )$ is a continuous function and $p\in L^{\infty }\left(I\right)$ such that $p<0$ a.e. on $[0,\eta ]$ and $p>0$ a.e. on $[\eta ,1]$. The results in this paper were based on the fixed point index theory. The idea of this paper was to study a three-layer beam formed by parallel layers of different materials. For an equally loaded beam of this type, the deflection u is governed by an ordinary third-order linear differential equation. The beam has a vanishing deflection at the left end. The second boundary condition represents that the shearing force registered in a point $\eta$ reacting to the bending moment at the right end. The third boundary condition represents that the bending moment reacts to the deflection at the right end.

Motivated by the above mentioned works, in this paper, we study the following fourth-order problem:

$$u^{\left(4\right)}\left(t\right)=f(t,u\left(t\right)),t\in [0,1]\equiv I,$$

(1)

with

$$u^{\prime }\left(0\right)=u^{″}\left(0\right)=u\left(1\right)=0,u^{‴}\left(\eta \right)=\alpha u^{\prime }\left(1\right),0\le \alpha <2,{\displaystyle \frac{1}{2}}\le \eta <1.$$

(2)

Equation (1) models the stationary states of the deflections of an elastic beam of length 1. The boundary conditions (2) describe that the beam neither rotates nor has a nonzero bending moment at the left end, and at the right end, the beam has a vanishing deflection, and the shearing force reacts to the bending moment registered in a point $\eta$.

The paper is organized as follows: In Section 2, we obtain the exact expression of the Green’s function related to the linear problem and we prove some of its properties. Then, in Section 3, we impose some suitable conditions in order to obtain an existence of at least one positive solution, while in Section 4, we establish conditions under which there is no solution for the considered problem. Finally, in Section 5, we give some examples in order to illustrate our main results in this paper.

We point out that our results complement previous ones that were given recently in the literature for fourth-order equations with sign-changing Green’s functions [17,18].

2. Linear Problem

Now, for $y\in C\left(I\right)$, let us consider the following fourth-order linear problem:

$$u^{\left(4\right)}\left(t\right)=y\left(t\right),$$

(3)

coupled with condition (2).

Lemma 1. 

The Green’s function related to problem (3)-(2) has the following expression:

$$G\left(t,s\right)={\displaystyle \frac{1}{6}}\left\{\begin{array}{ccc}{\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)+2\left(1-t^3\right)}{2-\alpha }}+{\left(t-s\right)}^3,& & s\le min\left\{t,\eta \right\},\\ {\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)+2\left(1-t^3\right)}{2-\alpha }},& & t\le s\le \eta ,\\ {\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)}{2-\alpha }}+{\left(t-s\right)}^3,& & \eta \le s\le t,\\ {\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)}{2-\alpha }},& & s\ge max\left\{t,\eta \right\}.\end{array}\right.$$

(4)

Proof. 

Integrating (3) with respect to t on both sides gives us

$$\begin{array}{ccc}\hfill u^{‴}\left(t\right)& =& {\int }_0^{t}y\left(s\right)ds+A,\hfill \\ \hfill u^{″}\left(t\right)& =& {\int }_0^t\left(t-s\right)y\left(s\right)ds+At+B,\hfill \\ \hfill u^{\prime }\left(t\right)& =& {\int }_0^t{\displaystyle \frac{{\left(t-s\right)}^2}{2}}y\left(s\right)ds+A{\displaystyle \frac{t^2}{2}}+Bt+C,\hfill \\ \hfill u\left(t\right)& =& {\int }_0^t{\displaystyle \frac{{\left(t-s\right)}^3}{6}}y\left(s\right)ds+A{\displaystyle \frac{t^3}{6}}+B{\displaystyle \frac{t^2}{2}}+Ct+D.\hfill \end{array}$$

From conditions $u^{\prime }\left(0\right)=u^{″}\left(0\right)=0$, it follows that $B=C=0.$ Now, $u\left(1\right)=0$ is equivalent to

$$D={\int }_0^1{\displaystyle \frac{{\left(s-1\right)}^3}{6}}y\left(s\right)ds-{\displaystyle \frac{A}{6}}.$$

Finally, condition $u^{‴}\left(\eta \right)=\alpha u^{\prime }\left(1\right)$ implies that

$$A={\displaystyle \frac{\alpha }{2-\alpha }}{\int }_0^1{\left(1-s\right)}^{2}y\left(s\right)ds-{\displaystyle \frac{2}{2-\alpha }}{\int }_0^{\eta }y\left(s\right)ds.$$

Thus,

$$D={\int }_0^1{\displaystyle \frac{{\left(s-1\right)}^2}{6}}\left({\displaystyle \frac{2s-s\alpha -2}{2-\alpha }}\right)y\left(s\right)ds+{\displaystyle \frac{1}{3\left(2-\alpha \right)}}{\int }_0^{\eta }y\left(s\right)ds.$$

As a result,

$$\begin{array}{ccc}\hfill u\left(t\right)& =& {\int }_0^t{\displaystyle \frac{{\left(t-s\right)}^3}{6}}y\left(s\right)ds+{\int }_0^1{\displaystyle \frac{\alpha {\left(s-1\right)}^2{t}^3}{6\left(2-\alpha \right)}}y\left(s\right)ds-{\int }_0^{\eta }{\displaystyle \frac{t^3}{3\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ & & +{\int }_0^1{\displaystyle \frac{{\left(s-1\right)}^2}{6}}\left({\displaystyle \frac{2s-s\alpha -2}{2-\alpha }}\right)y\left(s\right)ds+{\displaystyle \frac{1}{3\left(2-\alpha \right)}}{\int }_0^{\eta }y\left(s\right)ds\hfill \\ & =& {\int }_0^1{\displaystyle \frac{{\left(s-1\right)}^2\left(2s-s\alpha -2+\alpha t^3\right)}{6\left(2-\alpha \right)}}y\left(s\right)ds+{\int }_0^t{\displaystyle \frac{{\left(t-s\right)}^3}{6}}y\left(s\right)ds\hfill \\ & & +{\int }_0^{\eta }{\displaystyle \frac{1-t^3}{3\left(2-\alpha \right)}}y\left(s\right)ds,\hfill \end{array}$$

which proves our claim. □

Lemma 2. 

The Green’s function defined in (4) has the following properties:

$$G\left(t,s\right)\le 0\mathit{for}s\ge \eta \mathit{and}G\left(t,s\right)\ge 0\mathit{for}s\le \eta .$$

(5)

Moreover,

$${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\ge 0\mathit{for}s\ge \eta \mathit{and}{\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\le 0\mathit{for}s\le \eta .$$

(6)

Proof. 

First, suppose that $1\ge s\ge max\left\{t,\eta \right\}$. Since $\alpha <2$ and ${\left(1-s\right)}^2\ge 0,$ then $G\left(t,s\right)\le 0$ is equivalent to

$$2\left(1-s\right)+\alpha \left(s-t^3\right)\ge 0,$$

which is true since $s\ge t\ge t^3.$

Next, assume that $\eta \le s\le t\le 1$. From $\alpha <2$, it follows that $G\left(t,s\right)\le 0$ is equivalent to

$${\left(1-s\right)}^2\left(\alpha s+2-2s-\alpha t^3\right)\ge {\left(t-s\right)}^3\left(2-\alpha \right).$$

Considering that ${\left(1-s\right)}^2\ge {\left(t-s\right)}^2,$ it is enough to show that

$$\alpha s+2-2s-\alpha t^3\ge \left(t-s\right)\left(2-\alpha \right),$$

which can be written as

$$2\left(1-t\right)+\alpha t\left(1-t^2\right)\ge 0$$

and the last one clearly holds.

Now, if $t\le s\le \eta \le 1$, considering that $\alpha <2,$ we can determine that $G\left(t,s\right)\ge 0$ is equivalent to

$$2\left(1-t^3\right)\ge {\left(1-s\right)}^2\left(\alpha s+2-2s-\alpha t^3\right).$$

Considering that $1-t\ge 1-s,$ it is enough to show that

$$2\left(1+t+t^2\right)\ge \left(1-s\right)\left(\alpha s+2-2s-\alpha t^3\right).$$

The last inequality can be written as

$$2t+2t^2+s\left(2-\alpha \right)+2s\left(1-s\right)+\alpha s^2+\alpha t^3\left(1-s\right)\ge 0,$$

which obviously holds.

Finally, if $s\le min\left\{t,\eta \right\}\le 1,$ since $\alpha <2,$ then $G\left(t,s\right)\ge 0$ is equivalent to

$$2\left(1-t^3\right)+{\left(t-s\right)}^3\left(2-\alpha \right)\ge {\left(1-s\right)}^2\left(\alpha s+2-2s-\alpha t^3\right),$$

which is the same as

$$-\alpha s{\left(t-1\right)}^2\left(2t-2s+1-st\right)+2\left(1-t^3\right)+2{\left(t-s\right)}^3-2{\left(1-s\right)}^3\ge 0.$$

Denote

$$\phi \left(\alpha \right):=-\alpha s{\left(1-t\right)}^2\left(2t-2s+1-st\right)+2\left(1-t^3\right)+2{\left(t-s\right)}^3-2{\left(1-s\right)}^3.$$

Since ${\phi }^{\prime }\left(\alpha \right)\le 0$ for all $0\le s\le min\left\{t,\eta \right\}\le 1,$ we have that $\phi$ is decreasing in $\alpha .$ Thus,

$$\underset{0\le \alpha <2}{min}\phi \left(\alpha \right)>\phi \left(2\right)=2s\left(1-t^3\right)\left(2-s\right)\ge 0.$$

As a result, we have that $G\left(t,s\right)\ge 0$ for all $0\le s\le min\left\{t,\eta \right\}\le 1.$

Moreover, one can compute that

$$\begin{array}{cc}\hfill {\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)& \\ & ={\displaystyle \frac{1}{2}}\left\{\begin{array}{ccc}{\displaystyle \frac{s}{2-\alpha }}\left(\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-4t+2s\right),& & s\le min\left\{t,\eta \right\},\\ {\displaystyle \frac{t^2}{2-\alpha }}\left(\alpha s\left(s-2\right)+\alpha -2\right),& & t\le s\le \eta ,\\ {\displaystyle \frac{1}{2-\alpha }}\left(2{\left(t-s\right)}^2+\alpha s\left(1-t\right)\left(t-s+t\left(1-s\right)\right)\right),& & \eta \le s\le t,\\ {\displaystyle \frac{\alpha }{2-\alpha }}{t}^2{\left(1-s\right)}^2,& & s\ge max\left\{t,\eta \right\}.\end{array}\right.\hfill \end{array}$$

If $0\le s\le min\left\{t,\eta \right\}\le 1,$ denote

$$\psi \left(\alpha \right):=\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-4t+2s.$$

Clearly, $\psi$ increases on $\alpha$, which lead us to

$$\underset{0\le \alpha <2}{max}\psi \left(\alpha \right)<\psi \left(2\right)=2t^2\left(s-2\right)\le 0.$$

In other cases, it is obvious that

$$\begin{array}{cc}\hfill {\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)& ={\displaystyle \frac{t^2}{2-\alpha }}\left(\alpha s\left(s-2\right)+\alpha -2\right)\le 0\mathrm{if}t\le s\le \eta ,\hfill \\ \hfill {\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)& ={\displaystyle \frac{1}{2-\alpha }}\left(2{\left(t-s\right)}^2+\alpha s\left(1-t\right)\left(t-s+t\left(1-s\right)\right)\right)\ge 0\mathrm{if}\eta \le s\le t,\hfill \end{array}$$

and

$${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)={\displaystyle \frac{\alpha }{2-\alpha }}{t}^2{\left(1-s\right)}^2\ge 0\mathrm{if}s\ge max\left\{t,\eta \right\}.$$

□

Finally, we have that if $s\le \eta ,$ then $G\left(t,s\right)\ge 0$ and ${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\le 0,$ which gives us that

$$\underset{t\in I}{min}G\left(t,s\right)=G\left(1,s\right)=0$$

and

$$\underset{t\in I}{max}G\left(t,s\right)=G\left(0,s\right)={\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2\right)+2}{2-\alpha }}.$$

Similarly, if $s\ge \eta ,$ then $G\left(t,s\right)\le 0$ and ${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\ge 0,$ which implies

$$\underset{t\in I}{min}G\left(t,s\right)=G\left(0,s\right)={\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2\right)}{2-\alpha }}\mathrm{and}\underset{t\in I}{max}G\left(t,s\right)=G\left(1,s\right)=0.$$

Let $C\left(I\right)$ be equipped with the standard maximum norm $∥u∥={max}_{t\in I}\left|u\left(t\right)\right|$. Then, $C\left(I\right)$ is a Banach space. Now, define the cone

$$K_0:=\left\{y\in C\left(I\right):y\left(t\right)\mathrm{is}\mathrm{decreasing}\mathrm{and}\mathrm{non-negative}\mathrm{on}I\right\}.$$

Lemma 3. 

Let $0\le \alpha <2$, $G(t,s)$ be the related Green’s function to the linear problem (3)-(2) and $y\in K_0$. Then, the unique solution of the linear boundary value problem (3)-(2) is such that

$$u\left(t\right)={\int }_0^{1}G\left(t,s\right)y\left(s\right)ds\in K_0.$$

Moreover, $u\in C^2\left(I\right)$ and u is concave for all $t\le \eta .$

Proof. 

First, we will show that $u^{\prime }\left(t\right)\le 0$ on $I.$ This, compared with the boundary condition $u\left(1\right)=0$ ensures us that $u\left(t\right)\ge 0$ on $I.$

If $0\le t\le \eta ,$ we have that

$$\begin{array}{ccc}\hfill u\left(t\right)& =& {\int }_0^t\left({\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)+2\left(1-t^3\right)}{6\left(2-\alpha \right)}}+{\displaystyle \frac{{\left(t-s\right)}^3}{6}}\right)y\left(s\right)ds\hfill \\ & & +{\int }_t^{\eta }{\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)+2\left(1-t^3\right)}{6\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ & & +{\int }_{\eta }^1{\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)}{6\left(2-\alpha \right)}}y\left(s\right)ds.\hfill \end{array}$$

Now, using the fact that ${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\le 0$ for $0\le s\le \eta$, ${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\ge 0$ for $\eta \le s\le 1$ and since

$$\underset{0\le s\le t}{min}y\left(s\right)=y\left(t\right)\ge y\left(\eta \right)=\underset{t\le s\le \eta }{min}y\left(s\right)=\underset{\eta \le s\le 1}{max}y\left(s\right),$$

we obtain

$$\begin{array}{cc}\hfill u^{\prime }\left(t\right)& ={\int }_0^t{\displaystyle \frac{s\left(\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-4t+2s\right)}{2\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ \hfill & +{\int }_t^{\eta }{\displaystyle \frac{t^2\left(\alpha s\left(s-2\right)+\alpha -2\right)}{2\left(2-\alpha \right)}}y\left(s\right)ds+{\int }_{\eta }^1{\displaystyle \frac{\alpha t^2{\left(1-s\right)}^2}{2\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ & \le \underset{0\le s\le t}{min}y\left(s\right){\int }_0^t{\displaystyle \frac{s\left(\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-2t+s\right)}{2\left(2-\alpha \right)}}ds\hfill \\ \hfill & +\underset{t\le s\le \eta }{min}y\left(s\right){\int }_t^{\eta }{\displaystyle \frac{t^2\left(\alpha s\left(s-2\right)+\alpha -2\right)}{2\left(2-\alpha \right)}}ds\hfill \\ & +\underset{\eta \le s\le 1}{max}y\left(s\right){\int }_{\eta }^1{\displaystyle \frac{\alpha t^2{\left(1-s\right)}^2}{2\left(2-\alpha \right)}}ds\hfill \\ \hfill & =y\left(\eta \right){\int }_0^t{\displaystyle \frac{s\left(\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-2t+s\right)}{2\left(2-\alpha \right)}}ds\hfill \\ \hfill & +y\left(\eta \right){\int }_t^{\eta }{\displaystyle \frac{t^2\left(\alpha s\left(s-2\right)+\alpha -2\right)}{2\left(2-\alpha \right)}}ds+y\left(\eta \right){\int }_{\eta }^1{\displaystyle \frac{\alpha t^2{\left(1-s\right)}^2}{2\left(2-\alpha \right)}}ds\hfill \\ \hfill & ={\displaystyle \frac{y\left(\eta \right)t^2}{6\left(2-\alpha \right)}}\left(4t+\alpha -6\eta -\alpha t\right)\le 0.\hfill \end{array}$$

The latter one follows from the fact that for $\eta >{\displaystyle \frac{1}{2}},$ we have

$$\alpha -\alpha t+4t-6\eta \le 2\left(1-2\eta \right)+2\left(t-\eta \right)\le 0.$$

Moreover, since ${\displaystyle \frac{{\partial }^2}{\partial t^2}}G\left(t,s\right)\le 0$ for $0\le s\le \eta$, ${\displaystyle \frac{{\partial }^2}{\partial t^2}}G\left(t,s\right)\ge 0$ for $\eta \le s\le 1$ and again using

$$\underset{0\le s\le t}{min}y\left(s\right)=y\left(t\right)\ge y\left(\eta \right)=\underset{t\le s\le \eta }{min}y\left(s\right)=\underset{\eta \le s\le 1}{max}y\left(s\right),$$

we deduce

$$\begin{array}{ccc}\hfill u^{″}\left(t\right)& =& {\int }_0^t{\displaystyle \frac{s\left(\alpha -2+\alpha t\left(s-2\right)\right)}{2-\alpha }}y\left(s\right)ds\hfill \\ & & +{\int }_t^{\eta }{\displaystyle \frac{t\left(\alpha -2+\alpha s\left(s-2\right)\right)}{2-\alpha }}y\left(s\right)ds+{\int }_{\eta }^1{\displaystyle \frac{\alpha t{\left(s-1\right)}^2}{2-\alpha }}y\left(s\right)ds\hfill \\ & \le & \underset{0\le s\le t}{min}y\left(s\right){\int }_0^t{\displaystyle \frac{s\left(\alpha -2+\alpha t\left(s-2\right)\right)}{2-\alpha }}ds\hfill \\ & & +\underset{t\le s\le \eta }{min}y\left(s\right){\int }_t^{\eta }{\displaystyle \frac{t\left(\alpha -2+\alpha s\left(s-2\right)\right)}{2-\alpha }}ds\hfill \\ & & +\underset{\eta \le s\le 1}{max}y\left(s\right){\int }_{\eta }^1{\displaystyle \frac{\alpha t{\left(s-1\right)}^2}{2-\alpha }}ds\hfill \\ & =& {\displaystyle \frac{y\left(\eta \right)t}{6\left(2-\alpha \right)}}\left(6t-3\alpha t+2\alpha -12\eta \right)\le 0.\hfill \end{array}$$

The last inequality follows from the fact that

$$3t\left(2-\alpha \right)+2\alpha -12\eta \le 3\eta \left(2-\alpha \right)+2\alpha -12\eta \le 2\alpha -{\displaystyle \frac{3}{2}}\left(\alpha +2\right)={\displaystyle \frac{1}{2}}\alpha -3<0.$$

If $t\ge \eta ,$ we have that

$$\begin{array}{ccc}\hfill u\left(t\right)& =& {\int }_0^{\eta }\left({\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)+2\left(1-t^3\right)}{6\left(2-\alpha \right)}}+{\displaystyle \frac{{\left(t-s\right)}^3}{6}}\right)y\left(s\right)ds\hfill \\ & & +{\int }_{\eta }^t\left({\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)}{6\left(2-\alpha \right)}}+{\displaystyle \frac{{\left(t-s\right)}^3}{6}}\right)y\left(s\right)ds\hfill \\ & & +{\int }_t^1{\displaystyle \frac{{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha t^3\right)}{6\left(2-\alpha \right)}}y\left(s\right)ds.\hfill \end{array}$$

Now, using the fact that ${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\le 0$ for $0\le s\le \eta$, ${\displaystyle \frac{\partial }{\partial t}}G\left(t,s\right)\ge 0$ for $\eta \le s\le 1$ and since

$$\underset{0\le s\le \eta }{min}y\left(s\right)=\underset{\eta \le s\le t}{max}y\left(s\right)=y\left(\eta \right)\ge y\left(t\right)=\underset{t\le s\le 1}{max}y\left(s\right),$$

we obtain

$$\begin{array}{ccc}\hfill u^{\prime }\left(t\right)& =& {\int }_0^{\eta }{\displaystyle \frac{s\left(\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-4t+2s\right)}{2\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ & & +{\int }_{\eta }^t{\displaystyle \frac{2{\left(t-s\right)}^2+\alpha s\left(1-t\right)\left(t-s+t\left(1-s\right)\right)}{2\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ & & +{\int }_t^1{\displaystyle \frac{\alpha t^2{\left(1-s\right)}^2}{2\left(2-\alpha \right)}}y\left(s\right)ds\hfill \\ & \le & \underset{0\le s\le \eta }{min}y\left(s\right){\int }_0^{\eta }{\displaystyle \frac{s\left(\alpha \left(1-t\right)\left(t-s+t\left(1-s\right)\right)-4t+2s\right)}{2\left(2-\alpha \right)}}ds\hfill \\ & & +\underset{\eta \le s\le t}{max}y\left(s\right){\int }_{\eta }^t{\displaystyle \frac{2{\left(t-s\right)}^2+\alpha s\left(1-t\right)\left(t-s+t\left(1-s\right)\right)}{2\left(2-\alpha \right)}}ds\hfill \\ & & +\underset{t\le s\le 1}{max}y\left(s\right){\int }_t^1{\displaystyle \frac{\alpha t^2{\left(1-s\right)}^2}{2\left(2-\alpha \right)}}ds\hfill \\ & =& {\displaystyle \frac{y\left(\eta \right)t^2}{6\left(2-\alpha \right)}}\left(2t+\alpha -6\eta -\alpha t\right)\le 0.\hfill \end{array}$$

The last inequality follows from the fact that

$$2t+\alpha -6\eta -\alpha t$$

is increasing on $\alpha ,$ which gives us that

$$2t+\alpha -6\eta -\alpha t<2-6\eta <0.$$

□

Remark 1. 

Take $\mu \in \left(0,{\mu }_1\right]$ with ${\mu }_1$ being the least positive root of $2x^3-6x^2+1=0$, which is approximately equal to 0.4421. Denote by

$${\mu }^{*}={\displaystyle \frac{\eta -\mu }{\eta }}.$$

Since $0<\mu \le {\displaystyle \frac{1}{2}}\le \eta \le 1$, it follows that ${\mu }^{*}>0$.

Lemma 4. 

Assume that $y\in K_0.$ Then, the unique solution of (3)-(2) satisfies

$$\underset{t\in \left[0,\mu \right]}{min}u\left(t\right)\ge {\mu }^{*}∥u∥,$$

with μ and ${\mu }^{*}$ defined in Remark 1.

Proof. 

From the previous lemma, we know that $u\left(t\right)$ is concave on $\left[0,\eta \right],$ which gives us the following:

$$u\left(t\right)\ge {\displaystyle \frac{\eta -t}{\eta }}u\left(0\right)+{\displaystyle \frac{t}{\eta }}u\left(\eta \right)\mathrm{for}t\in \left[0,\eta \right].$$

From $u\in K_0,$ it follows that $u\in K_0,$$u\left(0\right)=∥u∥.$ Furthermore, since $u\left(\eta \right)>0$ for $t\in \left[0,\eta \right]$, then

$$u\left(t\right)\ge {\displaystyle \frac{\eta -t}{\eta }}∥u∥\mathrm{for}t\in \left[0,\eta \right].$$

If in the above inequality we take $t\in \left[0,\mu \right]$, we obtain

$$\underset{t\in \left[0,\mu \right]}{min}u\left(t\right)=u\left(\mu \right)\ge {\displaystyle \frac{\eta -\mu }{\eta }}∥u∥.$$

□

3. Existence Result

In this section, we will establish our existence result in the spirit of [19], based on the following classical Krasosel’skii’s fixed point theorem in cones [20].

Theorem 1. 

Let X be a Banach space and $K\subset X$ be a cone. Assume that ${\Omega }_1$ and ${\Omega }_2$ are open bounded subsets of X with $0\in {\Omega }_1,{\overline{\Omega }}_1\subset {\Omega }_2.$ If $T:K\cap \left({\overline{\Omega }}_2\setminus {\Omega }_1\right)\to K$ is a completely continuous operator such that

(i) 

$∥Tu∥\le ∥u∥,$ $u\in K\cap \partial {\overline{\Omega }}_1$ and $∥Tu∥\ge ∥u∥,$ $u\in K\cap \partial {\overline{\Omega }}_2$,

or

(ii) 

$∥Tu∥\ge ∥u∥,$ $u\in K\cap \partial {\overline{\Omega }}_1$ and $∥Tu∥\le ∥u∥,$ $u\in K\cap \partial {\overline{\Omega }}_2$.

Then, T has a fixed point in $K\cap \left({\overline{\Omega }}_2\setminus {\Omega }_1\right)$.

Let us denote

$$A={\int }_0^{\eta }G\left(0,s\right)ds\mathrm{and}B={\int }_0^{\mu }G\left(\eta ,s\right)ds.$$

Then, it is obvious that $0<B<A$.

Using the results in Lemma 4, let us define

$$K=\left\{u\in K_0:\underset{t\in \left[0,\mu \right]}{min}u\left(t\right)\ge {\mu }^{*}∥u∥\right\}.$$

(7)

Then, it is easy to see that K is a cone in $C\left(I\right)$. Now, we define an operator T on K by

$$\left(Tu\right)\left(t\right)={\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds,t\in I.$$

(8)

Obviously, if u is a fixed point of T in K, then u is a non-negative and decreasing solution of the problem (1)-(2).

Moreover, suppose that $f:I\times [0,+\infty )\to [0,+\infty )$ is a continuous function, satisfying the following conditions:

(F1)

For each $u\in [0,+\infty )$, the mapping $t⟼f(t,u)$ decreases;

(F2)

For each $t\in I$, the mapping $u⟼f(t,u)$ increases.

Our main result of this paper is as follows:

Theorem 2. 

Assume that conditions (F1) and (F2) hold and there exist two positive constants r and R with $r\ne R$ such that

$$f\left(0,r\right)\le {\displaystyle \frac{r}{A}}\mathit{and}f\left(\mu ,{\mu }^{*}R\right)\ge {\displaystyle \frac{R}{B}},$$

(9)

where μ and ${\mu }^{*}$ are as in Remark 1. Then, the problem (1)-(2) has a positive and decreasing solution u satisfying

$$min\left\{r,R\right\}\le ∥u∥\le max\left\{r,R\right\}.$$

Moreover, the obtained solution $u\left(t\right)$ is concave on $[0,\eta ]$.

Proof. 

We will show that there exists a fixed point of T in K. First, using arguments similar to the ones given in Lemmas 3 and 4, it is trivial to show that $T:K\to K$ is completely continuous. Next, for any $u\in K$, we claim that

$${\int }_{\mu }^{\eta }G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds+{\int }_{\eta }^{1}G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds\ge 0.$$

(10)

Indeed, if $u\in K$, we have that $G(t,s)\ge 0$ for $0\le s\le \eta$ and $G(t,s)\le 0$ for $\eta \le s\le 1$. Then, for $\eta \in \left[{\displaystyle \frac{1}{2}},1\right),$ we obtain that

$$\begin{array}{cc}\hfill & {\int }_{\mu }^{\eta }G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds+{\int }_{\eta }^{1}G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds\hfill \\ \hfill & \ge f\left(\eta ,u\left(\eta \right)\right)\left({\int }_{\mu }^{\eta }G\left(\eta ,s\right)ds+{\int }_{\eta }^{1}G\left(\eta ,s\right)ds\right)\hfill \\ \hfill & \ge {\displaystyle \frac{f\left(\eta ,u\left(\eta \right)\right)}{6(2-\alpha )}}\hfill \\ & \times [{\int }_{\mu }^{\eta }\left({\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha {\eta }^3\right)+2\left(1-{\eta }^3\right)+\left(2-\alpha \right){\left(\eta -s\right)}^3\right)ds\hfill \\ & +{\int }_{\eta }^1{\left(1-s\right)}^2\left(2s-\alpha s-2+\alpha {\eta }^3\right)ds]\hfill \\ \hfill & \ge {\displaystyle \frac{\left(1-\eta \right)f\left(\eta ,u\left(\eta \right)\right)}{12\left(2-\alpha \right)}}g\left(\eta \right),\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill g\left(\eta \right)=18\eta -\alpha -36{\mu }^2+& 24{\mu }^3+18{\eta }^2+18{\eta }^3-\alpha \eta +6\alpha {\mu }^2-8\alpha {\mu }^3-\alpha {\eta }^2\hfill \\ & +3\alpha {\eta }^3-36{\mu }^2\eta +6\alpha {\mu }^2\eta +4\alpha {\mu }^3\eta -12\alpha {\mu }^2{\eta }^2+4\alpha {\mu }^3{\eta }^2-6.\hfill \end{array}$$

Note that

$$\begin{array}{cc}\hfill g^{\prime }\left(\eta \right)=36\left(\eta -{\mu }^2\right)+54{\eta }^2+6\alpha {\mu }^2+4\alpha {\mu }^3+& 9\alpha {\eta }^2\hfill \\ & +8\alpha {\mu }^3\eta +18-\alpha -2\alpha \eta -24\alpha {\mu }^2\eta >0,\hfill \end{array}$$

since $\eta \ge {\mu }^2$ and $\alpha +2\alpha \eta +24\alpha {\mu }^2\eta \le 2+4+12=18.$ Thus, $g\left(\eta \right)$ increases on $\eta$, and for any $0<\mu \le {\mu }_1$ and $0\le \alpha <2,$ we obtain that

$$g\left(\eta \right)>g\left({\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{39}{4}}-{\displaystyle \frac{11\alpha }{8}}+6\alpha {\mu }^2-54{\mu }^2+24{\mu }^3-5\alpha {\mu }^3=\varphi \left(\alpha \right).$$

Notice that

$${\varphi }^{\prime }\left(\alpha \right)=-{\displaystyle \frac{11}{8}}+6{\mu }^2-5{\mu }^3<0$$

which gives us that $\varphi \left(\alpha \right)$ is decreasing on $\alpha$. Therefore,

$$\varphi \left(\alpha \right)>\varphi \left(2\right)=7\left(1+2{\mu }^3-6{\mu }^2\right)\ge 0$$

for $0<\mu \le {\mu }_1$. Consequently, we have that

$$g\left(\eta \right)>\varphi \left(\alpha \right)>\varphi \left(2\right)=7\left(1+2{\mu }^3-6{\mu }^2\right)\ge 0$$

for $0<\mu \le {\mu }_1$, which proves our claim.

Now, without loss of generality, we assume that $r<R$. Let

$${\Omega }_1=\left\{u\in C\left(I\right):∥u∥<r\right\}\mathrm{and}{\Omega }_2=\left\{u\in C\left(I\right):∥u∥<R\right\}.$$

For any $u\in K\cap \partial {\Omega }_1$, we obtain $0\le u\left(s\right)\le r$ for $s\in I$, which, together with (9), implies that

$$\begin{array}{ccc}\hfill 0& \le & \left(Tu\right)\left(t\right)\le {\int }_0^{\eta }\underset{t\in I}{max}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds+{\int }_{\eta }^1\underset{t\in I}{max}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds\hfill \\ & =& {\int }_0^{\eta }G\left(0,s\right)f\left(s,u\left(s\right)\right)ds\le {\int }_0^{\eta }G\left(0,s\right)f\left(0,r\right)ds\le r=∥u∥.\hfill \end{array}$$

Thus, $∥Tu∥\le ∥u∥$ for $u\in K\cap \partial {\Omega }_1.$

For any $u\in K\cap \partial {\Omega }_2$, we obtain ${\mu }^{*}R\le u\left(s\right)\le R$ for $s\in \left[0,\mu \right]$, which, together with (9) and (10), implies that

$$\begin{array}{cc}\hfill \left(Tu\right)\left(\eta \right)& ={\int }_0^{1}G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds\hfill \\ \hfill & ={\int }_0^{\mu }G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds+{\int }_{\mu }^{\eta }G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds\hfill \\ & +{\int }_{\eta }^{1}G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds\hfill \\ \hfill & \ge {\int }_0^{\mu }G\left(\eta ,s\right)f\left(s,u\left(s\right)\right)ds\hfill \\ \hfill & \ge {\int }_0^{\mu }G\left(\eta ,s\right)f\left(\mu ,{\mu }^{*}R\right)ds\hfill \\ \hfill & \ge R=∥u∥.\hfill \end{array}$$

Thus, $∥Tu∥\ge ∥u∥$ for $u\in K\cap \partial {\Omega }_2.$

Therefore, it follows from Theorem 1 that the operator T has a fixed point $u\in K\cap \left(\overline{{\Omega }_2}\setminus {\Omega }_1\right)$, which is a positive and decreasing solution of the problem (1)-(2) with $r\le ∥u∥\le R$. Moreover, similar to the proof of Lemma 3, one can show that the obtained solution $u\left(t\right)$ is concave on $[0,\eta ]$. □

4. Nonexistence Results

In the following theorem, we give some conditions to ensure that the integral Equation (8) has no nontrivial solution in K.

Theorem 3. 

Assume that $f(t,u)<m^{*}u$ for every $0\le t\le 1$ and $u\ge 0$, where

$$m^{*}={\left[\underset{t\in I}{sup}{\int }_0^{1}G(t,s)ds\right]}^{-1}.$$

Then, problem (1)-(2) have no nontrivial solution in K, defined in (7).

Proof. 

Suppose, on the contrary, that there exists $u\in K$ such that $u=Tu$. Let $t_0\in I$ be such that $\parallel u\parallel =u\left(t_0\right)$. Then,

$$\begin{array}{cc}\hfill \parallel u\parallel & \le {\int }_0^1\left|G(t_0,s)\right|f(s,u\left(s\right))ds\hfill \\ & <m^{*}{\int }_0^1\left|G(t_0,s)\right|u\left(s\right)ds\hfill \\ & <m^{*}\parallel u\parallel {\int }_0^1\left|G(t_0,s)\right|ds\hfill \\ & <\parallel u\parallel ,\hfill \end{array}$$

which is a contradiction. □

Theorem 4. 

Let $[a,b]\subset I$, with $a>0$ and $b<1$ be given. Moreover, suppose that $f(t,u(t\left)\right)=r\left(t\right)g\left(u\right(t\left)\right)$, where $r\in L^{\infty }\left(I\right)$ is such that $r>0$ a.e. on $[0,\eta ]$ and $r<0$ a.e. on $[\eta ,1]$ and $g:[0,\infty )\to [0,\infty )$ is a continuous function. If $g\left(u\right)>m_{*}u$ for every $u\ge 0$, where

$$m_{*}={\left[\underset{t\in [a,b]}{inf}{\int }_b^{a}G(t,s)r\left(s\right)ds\right]}^{-1},$$

then, problem (1)-(2) have no nontrivial solution in K, defined in (7).

Proof. 

Suppose, on the contrary, that there exists $u\in K$ such that $u=Tu$. Let $t_0\in [a,b]$, such that $u\left(t_0\right)={min}_{t\in [a,b]}u\left(t\right)$. Then, for $t\in [a,b]$, the following inequalities hold:

$$\begin{array}{cc}\hfill u\left(t_0\right)& ={\int }_0^{1}G(t_0,s)r\left(s\right)g\left(u\left(s\right)\right)ds\hfill \\ & \ge {\int }_a^{b}G(t_0,s)r\left(s\right)g\left(u\left(s\right)\right)ds\hfill \\ & >m_{*}{\int }_a^{b}G(t_0,s)r\left(s\right)u\left(s\right)ds\hfill \\ & >m_{*}u\left(t_0\right)\underset{t\in [a,b]}{inf}{\int }_a^{b}G(t_0,s)r\left(s\right)ds\hfill \\ & \ge u\left(t_0\right),\hfill \end{array}$$

which is a contradiction. □

5. Examples

In this section, we provide two examples to illustrate the applicability of Theorems 2 and 3.

Example 1. 

Consider (1)-(2) with $f(t,u)={\displaystyle \frac{u^2}{t+1}}$, $\alpha =1$, $\eta ={\displaystyle \frac{1}{2}}$ and $\mu ={\displaystyle \frac{1}{4}}$. Clearly, ${\mu }^{*}={\displaystyle \frac{1}{5}}$, $A={\displaystyle \frac{11}{144}}$, and $B={\displaystyle \frac{49}{4608}}$. Further, if we choose $r\le {\displaystyle \frac{144}{11}}$ and $R\ge {\displaystyle \frac{23040}{49}}$, then all conditions of Theorem 2 hold, which shows that the considered problem has a positive and decreasing solution u, satisfying

$$r\le ∥u∥\le R.$$

Moreover, the obtained solution u is concave on $[0,{\displaystyle \frac{1}{2}}]$.

Example 2. 

Consider (1)-(2) with $f(t,u)=10tu$, $\alpha =1$ and $\eta ={\displaystyle \frac{1}{2}}$. Then,

$$\begin{array}{cc}\hfill {\int }_0^{1}G(t,s)ds& ={\int }_0^1{\displaystyle \frac{{\left(s-1\right)}^2\left(2s-s\alpha -2+\alpha t^3\right)}{6\left(2-\alpha \right)}}ds+{\int }_0^t{\displaystyle \frac{{\left(t-s\right)}^3}{6}}ds\hfill \\ \hfill & +{\int }_0^{\eta }{\displaystyle \frac{1-t^3}{3\left(2-\alpha \right)}}ds\hfill \\ & ={\displaystyle \frac{t^4}{24}}+{\displaystyle \frac{t^3(\alpha -6\eta )}{18(2-\alpha )}}+{\displaystyle \frac{1}{72(2-\alpha )}}\left(24\eta -\alpha -6\right)\hfill \end{array}$$

and

$$\underset{t\in I}{sup}{\int }_0^{1}G(t,s)ds=\underset{t\in [0,1]}{sup}\left[{\displaystyle \frac{t^4}{24}}-{\displaystyle \frac{t^3}{9}}+{\displaystyle \frac{5}{72}}\right]={\displaystyle \frac{5}{72}}.$$

Hence,

$$m^{*}={\left[\underset{t\in I}{sup}{\int }_0^{1}G(t,s)ds\right]}^{-1}={\displaystyle \frac{72}{5}}.$$

Clearly, $f(t,u)<m^{*}u$ for every $0\le t\le 1$ and $u\ge 0$. Thus, condition $\left(i\right)$ in Theorem 3 holds, which shows that the considered problem has no nontrivial solutions in K.