Existence and uniqueness of positive solutions to boundary value problem with increasing homeomorphism and positive homomorphism operator

Abstract

In this paper, we consider the following nonlinear boundary value problem: ${(\phi (u^{\prime }(t)))}^{\prime }+a(t)f(u(t))=0$, $0<t<1$, $u(0)={\sum }_{i=1}^{m-2}{\alpha }_{i}u({\xi }_i)$, $u^{\prime }(1)=0$, where $\phi :\mathbb{R}⟶\mathbb{R}$ is an increasing homeomorphism and positive homomorphism with $\phi (0)=0$. By using a fixed-point theorem on partially ordered sets, we obtain sufficient conditions for the existence and uniqueness of positive and nondecreasing solutions to the above boundary value problem.

MSC:34B18, 34B27.

1 Introduction

In this paper, we consider the existence and uniqueness of a positive and nondecreasing solution to the following boundary value problem:

$${\left(\phi (u^{\prime }(t))\right)}^{\prime }+a(t)f(u(t))=0,0<t<1,$$

(1.1)

$$u(0)=\sum _{i=1}^{m-2}{\alpha }_{i}u({\xi }_i),u^{\prime }(1)=0,$$

(1.2)

where $\phi :\mathbb{R}\to \mathbb{R}$ is an increasing homeomorphism and positive homomorphism with $\phi (0)=0$. Here ${\xi }_i\in (0,1)$ with $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_{m-2}<1$ and ${\alpha }_i$ satisfy ${\alpha }_i\in [0,+\mathrm{\infty })$, $0<{\sum }_{i=1}^{m-2}{\alpha }_i<1$.

A projection $\phi :\mathbb{R}\to \mathbb{R}$ is called an increasing homeomorphism and positive homomorphism, if the following conditions are satisfied:

• (1) $\phi (x)\le \phi (y)$, for all $x,y\in \mathbb{R}$ with $x\le y$;

• (2) φ is a continuous bijection and its inverse mapping is also continuous;

• (3) $\phi (xy)=\phi (x)\phi (y)$, for all $x,y\in {\mathbb{R}}_{+}$.

In the above definition, we can replace the condition (3) by the following stronger condition:

• (4) $\phi (xy)=\phi (x)\phi (y)$, for all $x,y\in \mathbb{R}$, where $\mathbb{R}=(-\mathrm{\infty },+\mathrm{\infty })$.

Remark 1.1 If conditions (1), (2), and (4) hold, then it implies that φ is homogeneous generating a p-Laplace operator, i.e. $\phi (x)={|x|}^{p-2}x$, for some $p>1$.

Recently, the existence and multiplicity of positive solutions for the p-Laplacian operator, i.e., $\phi (x)={|x|}^{p-2}x$, for some $p>1$, have received wide attention, see [1–3] and references therein. We know that the oddness of a p-Laplacian operator is key to the proof. However, in this paper we define a new operator, which improves and generates a p-Laplacian operator for some $p>1$, and φ is not necessarily odd. Moreover research of increasing homeomorphisms and positive homomorphism operators has proceeded very slowly, see [4, 5].

In [4], Liu and Zhang studied the existence of positive solutions of quasilinear differential equation

$$\begin{array}{c}{\left(\phi \left(x^{\prime }\right)\right)}^{\prime }+a(t)f(x(t))=0,0<t<1,\hfill \\ x(0)-\beta x^{\prime }(0)=0,x(1)+\delta x^{\prime }(1)=0,\hfill \end{array}$$

where $\phi :\mathbb{R}\to \mathbb{R}$ is an increasing homeomorphism and positive homomorphism and $\phi (0)=0$. They obtain the existence of one or two positive solutions by using a fixed-point index theorem in cones. But the uniqueness of the solution is not treated.

In [5], the authors showed that there exist countably many positive solutions by using the fixed-point index theory and a new fixed-point theorem in cones. They also assumed that the operator $\phi :\mathbb{R}\to \mathbb{R}$ is an increasing homeomorphism and a positive homomorphism, and $\phi (0)=0$.

In [6], the authors established the existence and uniqueness of a positive and nondecreasing solution to a singular boundary value problem of a class of nonlinear fractional differential equation. Their analysis relies on a fixed-point theorem in partially ordered sets. The existence of a fixed point in partially ordered sets has been considered recently in [6–10].

But whether or not we can obtain the existence and uniqueness of a positive and nondecreasing solution to the boundary value problem (1.1)-(1.2) still remains unknown. So, motivated by all the works above, we will prove the existence and uniqueness of a positive and nondecreasing solution for the boundary value problems (1.1)-(1.2) by using a fixed-point theorem on partially ordered sets.

2 Some definitions and fixed-point theorems

Definition 2.1 Let $(E,\parallel \cdot \parallel )$ be a real Banach space. A nonempty, closed, convex set $P\subset E$ is said to be a cone provided the following are satisfied:

• (a) if $y\in P$ and $\lambda \ge 0$, then $\lambda y\in P$;

• (b) if $y\in P$ and $-y\in P$, then $y=0$.

If $P\subset E$ is a cone, we denote the order induced by P on E by ≤, that is, $x\le y$ if and only if $y-x\in P$.

The following fixed-point theorems in partially ordered sets are fundamental and important to the proofs of our main results.

Theorem 2.1 ([7])

Let $(E,\le )$ be a partially ordered set and suppose that there exists a metric d in E such that $(E,d)$ is a complete metric space. Assume that E satisfies the following condition:

$$\mathit{\text{if }}\{x_n\}\mathit{\text{ is a nondecreasing sequence in E such that }}{x}_n\to x\mathit{\text{, then }}{x}_n\le x,\mathrm{\forall }n\in \mathbb{N}.$$

(2.1)

Let $T:E\to E$ be a nondecreasing mapping such that

$$d(Tx,Ty)\le d(x,y)-\psi (d(x,y)),\mathit{\text{for }}x\ge y,$$

where $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is a continuous and nondecreasing function such that ψ is positive in $(0,+\mathrm{\infty })$, $\psi (0)=0$ and ${lim}_{t\to \mathrm{\infty }}\psi (t)=\mathrm{\infty }$. If there exists $x_0\in E$ with $x_0\le T(x_0)$, then T has a fixed point.

If we consider that $(E,\le )$ satisfies the following condition:

$$\text{for }x,y\in E\text{ there exists }z\in E\text{ which is comparable to }x\text{ and }y,$$

(2.2)

then we have the following result.

Theorem 2.2 ([8])

Adding condition (2.2) to the hypotheses of Theorem  2.1, we obtain uniqueness of the fixed point.

3 Main results

The basic space used in this paper is $E=C[0,1]$. Then E is a real Banach space with the norm $\parallel u\parallel ={max}_{0\le t\le 1}|u(t)|$. Note that this space can be equipped with a partial order given by

$$x,y\in C[0,1],x\le y\iff x(t)\le y(t),t\in [0,1].$$

In [8] it is proved that $(C[0,1],\le )$ with the classic metric given by

$$d(x,y)=\underset{0\le t\le 1}{sup}\left\{|x(t)-y(t)|\right\}$$

satisfies condition (2.1) of Theorem 2.1. Moreover, for $x,y\in C[0,1]$ as the function $max\{x,y\}\in C[0,1]$, $(C[0,1],\le )$ satisfies condition (2.2).

The main result of this paper is the following.

Theorem 3.1 The boundary value problem (1.1)-(1.2) has a unique positive solution $u(t)$ which is strictly increasing if the following conditions are satisfied:

• (A) $a(t)$ is a nonnegative measurable function defined in $[0,1]$ and $a(t)$ does not identically vanish on any subinterval of $[0,1]$ and

  $$0<{\int }_0^{1}a(t)dt<+\mathrm{\infty };$$

(f₁) $f:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is continuous and nondecreasing respect to u and $f(u(t))\not\equiv 0$ for $t\in Z\subset [0,1]$ with $\mu (Z)>0$ (μ denotes the Lebesgue measure);

(f₂) there exists $1<\lambda +1<\frac{1-{\sum }_{i=1}^{m-2}{\alpha }_i}{{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )}$ such that for $u,v\in [0,+\mathrm{\infty })$ with $u\ge v$ and $t\in [0,1]$

$$\phi (ln(v+2))\le f(v)\le f(u)\le \phi (ln(u+2){(u-v+1)}^{\lambda }).$$

Proof Consider the cone

$$K=\{u\in C[0,1]:u\ge 0\}.$$

As K is a closed set of $C[0,1]$, K is a complete metric space with the distance given by $d(u,v)={sup}_{t\in [0,1]}|u(t)-v(t)|$.

Now, we consider the operator T defined by

$$Tu(t)={\int }_0^t{\phi }^{-1}({\int }_s^{1}a(\tau )f(u(\tau ))d\tau )ds+\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\int }_0^{{\xi }_i}{\phi }^{-1}({\int }_s^{1}a(\tau )f(u(\tau ))d\tau )ds}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}.$$

By conditions (A), (f₁), we have $T(K)\subset K$.

We now show that all the conditions of Theorem 2.1 and Theorem 2.2 are satisfied.

Firstly, by condition (f₁), for $u,v\in K$ and $u\ge v$, we have

$$\begin{array}{rcl}Tu(t)& =& {\int }_0^t{\phi }^{-1}({\int }_s^{1}a(\tau )f(u(\tau ))d\tau )ds+\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\int }_0^{{\xi }_i}{\phi }^{-1}({\int }_s^{1}a(\tau )f(u(\tau ))d\tau )ds}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}\\ \ge & {\int }_0^t{\phi }^{-1}({\int }_s^{1}a(\tau )f(v(\tau ))d\tau )ds+\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\int }_0^{{\xi }_i}{\phi }^{-1}({\int }_s^{1}a(\tau )f(v(\tau ))d\tau )ds}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}\\ =& Tv(t).\end{array}$$

This proves that T is a nondecreasing operator. On the other hand, for $u\ge v$ and by (f₂) we have

$$\begin{array}{c}d(Tu,Tv)\hfill \\ =\underset{0\le t\le 1}{sup}|(Tu)(t)-(Tv)(t)|=\underset{0\le t\le 1}{sup}((Tu)(t)-(Tv)(t))\hfill \\ \le \underset{0\le t\le 1}{sup}{\int }_0^t[{\phi }^{-1}({\int }_s^{1}a(\tau )f(u(\tau ))d\tau )-{\phi }^{-1}({\int }_s^{1}a(\tau )f(v(\tau ))d\tau )]ds\hfill \\ +\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\int }_0^{{\xi }_i}[{\phi }^{-1}({\int }_s^{1}a(\tau )f(u(\tau ))d\tau )-{\phi }^{-1}({\int }_s^{1}a(\tau )f(v(\tau ))d\tau )]ds}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}\hfill \\ \le {\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )(ln(u+2){(u-v+1)}^{\lambda }-ln(v+2))\hfill \\ +\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\xi }_i{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}(ln(u+2){(u-v+1)}^{\lambda }-ln(v+2))\hfill \\ \le [{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )+\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\xi }_i{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}](ln\frac{(u+2){(u-v+1)}^{\lambda }}{v+2})\hfill \\ \le (\lambda +1)ln(u-v+1)\frac{{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}.\hfill \end{array}$$

Since the function $h(x)=ln(x+1)$ is nondecreasing, and condition (f₂), then we have

$$\begin{array}{rcl}d(Tu,Tv)& \le & (\lambda +1)ln(\parallel u-v\parallel +1)\frac{{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}<ln(\parallel u-v\parallel +1)\\ =& \parallel u-v\parallel -(\parallel u-v\parallel -ln(\parallel u-v\parallel +1)).\end{array}$$

Let $\psi (x)=x-ln(x+1)$. Obviously $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is continuous, nondecreasing, positive in $(0,+\mathrm{\infty })$, $\psi (0)=0$, and ${lim}_{x\to +\mathrm{\infty }}\psi (x)=+\mathrm{\infty }$. Thus, for $u\ge v$, we have

$$d(Tu,Tv)\le d(u,v)-\psi (d(u,v)).$$

By conditions (A) and (f₁), we know that

$$\begin{array}{rcl}(T0)(t)& =& {\int }_0^t{\phi }^{-1}({\int }_s^{1}a(\tau )f(0)d\tau )ds+\frac{{\sum }_{i=1}^{m-2}{\alpha }_i{\int }_0^{{\xi }_i}{\phi }^{-1}({\int }_s^{1}a(\tau )f(0)d\tau )ds}{1-{\sum }_{i=1}^{m-2}{\alpha }_i}\\ \ge & 0.\end{array}$$

Therefore, by Theorem 2.1 we know that problem (1.1)-(1.2) has at least one nonnegative solution. As $(K,\le )$ satisfies condition (2.2), thus, Theorem 2.2 implies the uniqueness of the solution. By definition of T and conditions (A), (f₁), it is easy to prove that this solution $u(t)$ is strictly increasing. □

4 Example

Example 4.1 Consider the boundary value problem

$$\{\begin{array}{c}{(\phi (u^{\prime }(t)))}^{\prime }+\frac{1}{5}{t}^{4}f(u(t))=0,0<t<1,\hfill \\ u(0)=\frac{1}{4}u(\frac{1}{4})+\frac{1}{4}u(\frac{1}{2}),u^{\prime }(1)=0,\hfill \end{array}$$

(4.1)

where

$$\phi (u)=\{\begin{array}{cc}\frac{u^3}{1+u^2},\hfill & u\le 0,\hfill \\ u^2,\hfill & u>0,\hfill \end{array}$$

$a(t)=\frac{1}{5}{t}^4$ and $f(x)={[ln(x+2)]}^2$ for $x\in [0,+\mathrm{\infty })$.

Proof Note that f is a continuous function and $f(x)>0$. Moreover, f is nondecreasing with respect to x since $\frac{\partial f}{\partial x}=\frac{2}{x+2}ln(x+2)>0$. On the other hand, for $u\ge v$, we have

$$\begin{array}{rcl}\phi (ln(v+2))& =& {[ln(v+2)]}^2=f(v)\le f(u)={[ln(u+2)]}^2\\ \le & {(ln(u+2)(u-v+1))}^2\\ =& \phi (ln(u+2)(u-v+1)).\end{array}$$

In this case, $\lambda =1$ because $1<\lambda +1<\frac{1-{\sum }_{i=1}^{m-2}{\alpha }_i}{{\phi }^{-1}({\int }_0^{1}a(\tau )d\tau )}=\frac{5}{2}$. Thus Theorem 3.1 implies that the boundary value problem (4.1) has a unique positive solution which is strictly increasing. □