Positive solutions for some boundary value problems via a new fixed point theorem

Abstract

In this paper, we establish some new fixed point theorems of mixed monotone operator with a perturbation. Moreover, we prove the existence and the uniqueness of positive solutions of a second order Neumann boundary value problem, a second order Sturm Liouville boudary value problem and a nonlinear elliptic boundary value problem for the Lane–Emden–Fowler equation.

1 Introduction

Nonlinear boundary value problems arise in a variety of different areas of applied mathematics, physics, chemistry and biology. Therefore, these problems have been widely studied and recently, there have been many papers dealing with the existence and uniqueness of positive solutions by the use of techniques of nonlinear analysis as fixed point theorems in cones, Leray–Schauder theory, the coincidence degree theory of Mahwin, Krasnoselskii’s fixed point theorems... (see [2–4, 7, 9, 14]).

Mixed monotone operators were introduced by Guo and Lakshmikantham in [5] and they are an important class of operators that are used in nonlinear and integral equations. In [10], Sang investigated the existence and the uniqueness of positive solutions to operator equation

$$\begin{aligned} A(x, x)=x, \end{aligned}$$

where $A: P\times P\rightarrow P$ is a mixed monotone operator with certain concavity, where $P$ is a cone in a real Banach space by using the partial order theory and monotone iterative technique.

In this paper we study the existence and the uniqueness of positive fixed points for more general mixed montone operators with perturbation. i.e., we prove the existence and the uniqueness of positive solutions of the following operator equation in a real Banach space $E,$

$$\begin{aligned} A(x, x)+Bx=x, ~x\in E, \end{aligned}$$

where $A$ is a mixed monotone operator verifying a more general concave property, and $B$ is sublinear and without assuming operators to be continuous or compact.

Our results are used to prove positive solutions to a second order Neumann boundary value problem, a second order Sturm Liouville boundary value problem and a nonlinear elliptic boundary value problem for the Lane–Emden–Fowler equation.

2 Preliminaries

In this section, we present some definitions, that will be used in the proofs of our theorems. Let $E$ be a real Banach space. A nonempty convex closed set $P$ is called a cone if it satisfies the following conditions:

• 1. $x\in P, \lambda \ge 0$ implies $\lambda x\in P;$

• 2. $x\in P$ and $-x\in P$ imply $x=\theta ,$ where $\theta$ denotes the zero element of $E.$

Let $E$ be partially ordered by a cone $P$ of $E,$ i.e., $x\le y,$ if and only if $y- x\in P$ for any $x,y\in E.$ Recall that the cone $P$ is said to be solid if the interior $\mathring{P}$ is nonempty and we denote $x>>\theta$ if $x\in \mathring{P}.$ $P$ is said to be normal if there exists a positive constant $M$ such that $\theta \le x\le y$ implies that $\Vert x\Vert \le M\Vert y\Vert ,$ the smallest $M$ is called the normal constant of $P.$ An operator $A:P\times P\rightarrow E$ is said to be mixed monotone if $A(x,y)$ is nondecreasing in $x$ and nonincreasing in $y,$ that is, for any $x,y\in P,$

$$\begin{aligned} x_1 ,x_2 \in P, x_1\le x_2 \Longrightarrow A(x_1,y)\le A(x_2,y) \end{aligned}$$

and

$$\begin{aligned} y_1,y_2\in P, y_1\le y_2 \Longrightarrow A(x,y_1)\ge A(x,y_2). \end{aligned}$$

An element $x^*\in P$ is called a fixed point of $A$ if it satisfies $A(x^*, x^*)=x^*.$

An operator $B: E\rightarrow E$ is called a sublinear operator if $B(tx)\le tBx,$ for $x\in P$ and $t\ge 1.$

We always assume that $E$ is a real Banach space with a partial order induced by a cone $P$ of $E.$

3 Main results

Our first result is the following

Theorem 3.1

Let $E$ be a real Banach space and $P$ be a normal cone in $E.$ Let $u_0, v_0\in P$ with $u_0\le v_0$ and $A: P\times P\rightarrow P$ be a mixed monotone operator. Let $B: E\rightarrow E$ be sublinear and let $a, b$ be real numbers with $a<b$. Suppose that

• 1. there exists a surjection $\tau :(a, b)\rightarrow (0,1)$ and there exists a function $\varphi :(a, b]\times P\times P\rightarrow (0, 1]$ such that, for any $t\in (a, b],$ $\varphi (t, u, v)$ is nondecreasing in $u$ for fixed $v$ and non-increasing in $v$ for fixed $u;$

• 2. $\varphi (t, u, v)>\tau (t)$ for all $t\in (a, b)$ and $u, v\in P;$

• 3. $A(\tau (t)u, v)\ge \varphi (t, u, v)A(u, v)$ for all $t\in (a, b)$ and $u, v\in P;$

• 4. $(I-B)^{-1}: E\rightarrow E$ exists and is an increasing operator;

• 5. there exists a real positive number $r_0$ such that $u_0\ge r_0v_0,$ and $u_0$ and $v_0$ are such that

  $$\begin{aligned} u_0\le A(u_0, v_0)+Bu_0,~~A(v_0, u_0)+Bv_0\le v_0; \end{aligned}$$

• 6. for fixed $u\in [u_0, v_0],$ there exists $N>0$ such that

  $$\begin{aligned} A(u, v_1)-B(Nv_1)-A(u, v_2)+B(Nv_2)\ge -N (v_1-v_2),\quad \forall v_1\ge v_2,\\ v_1, v_2\in [u_0, v_0]. \end{aligned}$$

Then the operator equation $A(u, u)+Bu=u$ has exactly one fixed point $x^*$ in $[u_0, v_0].$

Proof 3.1

We divide the proof into three steps.

Step $1.$ We prove that for any fixed $u\in [u_0,v_0], (I-B)^{-1}A(u,.)$ has exactly one fixed point $T(u)\in [(I-B)^{-1}A(u_0,v_0), (I-B)^{-1}A(v_0,u_0)]$ such that $(I-B)^{-1}A(u,T(u)) =T(u).$ The proof of this step is inspired by [15].

We denote $C=(I-B)^{-1}A.$ By the fact that the operator $B$ is sublinear, we have $B\theta \ge \theta ,$ which together with hypothesis $(4)$ give

$$\begin{aligned} \theta \le (I-B)^{-1}\theta \le (I-B)^{-1}x,~x\in P. \end{aligned}$$

Consequently $C:P\times P\rightarrow P.$ By $(6)$ and $(4),$ for fixed $u\in [u_0, v_0],$ there exists $N > 0$ such that $(I-B)^{-1}A(u,v) +Nv$ is increasing in $v\in [u_0, v_0].$ We let

$$\begin{aligned} D(u,v) = \frac{C(u,v) +Nv}{N +1}, \end{aligned}$$

then $D(u,v)=v$ is equivalent to $C(u,v) =v.$ Let $x_0 =C(u_0,v_0), y_0 =C(v_0,u_0)$ and $x_{n+1}=D(u,x_n), y_{n+1} =D(u,y_n),$ where $u_0\le u\le v_0.$ Using hypotheses $(4-5),$ we obtain that $x_0\le v_0, y_0\ge x_0$ and then

$$\begin{aligned} C(u,x_0)\ge C(u_0, ,v_0) =x_0, \end{aligned}$$

$$\begin{aligned} C(u,y_0)\le C(v_0,u_0) =y_0, \end{aligned}$$

which gives

$$\begin{aligned} x_1 =D(u,x_0) = \frac{C(u,x_0) +Nx_0}{N+1}\ge \frac{x_0+Nx_0}{N+1}=x_0, \end{aligned}$$

$$\begin{aligned} y_1 =D(u,y_0) = \frac{C(u,y_0) +Ny_0}{N+1}\le \frac{y_0+Ny_0}{N+1}=y_0. \end{aligned}$$

Since for fixed $u, D(u, v)$ is increasing in $v,$ and for fixed $v, D(u, v)$ is increasing in $u,$ then $x_0\le x_1 \le y_1 \le y_0.$ By induction, we have $x_n \le x_{n+1}\le y_{n+1} \le y_n.$ Thus we get that

$$\begin{aligned} \theta \!\le \! y_n - x_n \!=\! \frac{C(u,y_{n-1}) \!+\! Ny_{n-1}- C(u,x_{n-1})- Nx_{n-1}}{N + 1} \le \frac{N}{N\!+\!1}(y_{n-1} - x_{n-1}). \end{aligned}$$

Since $P$ is a normal cone and by induction, we get that

$$\begin{aligned} \Vert y_n - x_n\Vert \le C\left( \frac{N}{N+1}\right) ^n\Vert y_0-x_0\Vert , \end{aligned}$$

(1)

where $C$ is the normal constant of $P.$ Moreover (for any natural number $p$),

$$\begin{aligned} x_{n+p}- x_n \le y_{n+p} - x_n \le y_n - x_n, \end{aligned}$$

$$\begin{aligned} \Vert x_{n+p} - x_n\Vert \le C\Vert y_n - x_n\Vert \le C^2\left( \frac{N}{N+1}\right) ^n\Vert y_0 - x_0\Vert . \end{aligned}$$

This gives that $\{x_n\}$ is a Cauchy sequence. Noticing $E$ is complete, $\{x_n\}$ converges to some element, we denote it by $T(u).$ By Eq. (1), it follows that $\{y_n\}$ also converges to $T(u).$ Using that

$$\begin{aligned} x_n\le T(u)\le y_n \end{aligned}$$

and

$$\begin{aligned} x_{n+1}=D(u,x_n)\le D(u, T(u))\le D(u, y_n)= y_{n+1}, \end{aligned}$$

and the normality of the cone $P,$ we can conclude that $\{x_n\}, \{y_n\}$ also converge to $D(u,T(u)).$ Thus $D(u,T(u)) =T(u).$

If there exists an $x$ such that $D(u,x) = x,$ then we have for any integer $n, x_n \le x \le y_n,$ which implies by taking the limit that $x=T(u),$ i.e., $T(u)$ is the unique fixed point of $C(u,.)$ in $[C(u_0, v_0), C(v_0, u_0)].$

Step $2.$ We prove that $T$ is increasing in $u.$

If $u, v \in [u_0, v_0], u\le v,$ then we let $x_0 =\alpha _0=C(u_0, v_0)$ in step $1.$ Since $D(u, v)$ is increasing in both variables, we obtain

$$\begin{aligned} x_1=D(u, x_0)\le D(v, \alpha _0)=\alpha _1. \end{aligned}$$

By induction we know $x_n\le \alpha _n.$ Taking the limit, we have $T(u)\le T(v).$

Step $3.$ We prove that $T$ has a unique fixed point in $[u_0, v_0].$

Let $u_{n+1}=T(u_n), v_{n+1}=T(v_n),$ we have $u_0\le u_1, v_1\le v_0.$ Since $T$ is increasing, we get $u_n\le u_{n+1}, v_{n+1}\le v_n$ and $u_n\le v_n.$ Since $B$ is sublinear, we know that for any $x\in P$ and $r\in (0,1),$ we obtain

$$\begin{aligned} (I-B)(rx)\le r(I-B)x. \end{aligned}$$

Then

$$\begin{aligned} (I-B)(r(I-B)^{-1}x)\le r(I-B)(I-B)^{-1}x=rx. \end{aligned}$$

Therefore, we have

$$\begin{aligned} r(I-B)^{-1}x\le (I-B)^{-1}(rx). \end{aligned}$$

Suppose now that, for $n\ge 1,$ there exists $r_n\in (0,1)$ such that $u_n\ge r_nv_n.$ Then

$$\begin{aligned} \begin{aligned} u_{n+1}&=C(u_{n}, T(u_{n}))\ge C(r_nv_n, T(u_n))\\&=C(\tau (t_n) v_n, T(u_n))\ge (I-B)^{-1}\varphi (t_n, v_n, T(u_n)) A(v_n, T(u_n))\\&\ge \varphi (t_n, v_n, T(u_n)) C(v_n, T(v_n))\ge \varphi (t_n, v_n, T(u_n))v_{n+1}. \end{aligned} \end{aligned}$$

Set, for all $n\in \mathbb N,$

$$\begin{aligned} \alpha _n=\sup \{\alpha >0; u_n\ge \alpha v_n\}. \end{aligned}$$

It is clear that $\alpha _n\in (0, 1]$ and $u_n\ge \alpha _nv_n.$ Since $u_{n+1}\ge u_n\ge \alpha _n v_n\ge \alpha _nv_{n+1},$ then $\alpha _n\le \alpha _{n+1}.$ So $\{\alpha _n\}_n$ is an increasing sequence in $(0, 1].$ Thus, there exists $\alpha ^*\in (0, 1]$ such that $\alpha _n\rightarrow \alpha ^*$ as $n\mapsto +\infty .$ Then $\alpha ^*=1.$ Indeed, if $0<\alpha ^*<1,$ then there exists $t^*\in (a, b)$ such that $\tau (t^*)=\alpha ^*.$

Case $1.$ If there exists $N\in \mathbb N$ such that $\alpha _N=\alpha ^*.$ Therefore, $\alpha _n=\alpha ^*$ for all $n\ge N.$ So, for $n\ge N,$ we have

$$\begin{aligned} \begin{aligned} u_{n+1}&=C(u_{n}, T(u_{n}))\ge C(\alpha ^*v_n, T(v_n))\\&=C(\tau (t^*) v_n, T(v_n))\ge \varphi (t^*, v_n, T(v_n)) C(v_n, T(v_n))\\&\ge \varphi (t^*, v_n, T(v_n))v_{n+1}. \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \alpha _{n+1}=\alpha ^*\ge \varphi (t^*, v_n, T(v_n))>\tau (t^*)=\alpha ^*, \end{aligned}$$

which is a contradiction.

Case $2.$ If $\alpha _n<\alpha ^*$ for all $n\in \mathbb N.$ Then $0<\frac{\alpha _n}{\alpha ^*}<1.$ By assumption $(1),$ there exists $t_n\in (a, b)$ such that $\tau (t_n)=\frac{\alpha _n}{\alpha ^*}.$ We have

$$\begin{aligned} \begin{aligned} u_{n+1}&=C(u_{n}, T(u_{n}))\ge C(\alpha _n v_n, T(v_n))\\&\ge C\left( \frac{\alpha _n}{\alpha ^*}\alpha ^*v_n, T(v_n)\right) =C(\tau (t_n ) \alpha ^*v_n, T(v_n))\\&\ge \varphi (t_n , \alpha ^*v_n, T(v_n)) C(\alpha ^*v_n, T(v_n))\\&\ge \varphi (t_n , \alpha ^*v_n, T(v_n))\varphi (t^*, v_n, T(v_n))C(\alpha ^*v_n, T(v_n))\\&\ge \varphi (t_n , \alpha ^*v_n, T(v_n))\varphi (t^*, v_n, T(v_n))v_{n+1}. \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} \alpha _{n+1}&\ge \varphi (t_n , \alpha ^*v_n, T(v_n))\varphi (t^*, v_n, T(v_n))\\&\ge \tau (t_n)\varphi (t^*, v_n, T(v_n))\ge \frac{\alpha _n}{\alpha ^*}\varphi (t^*, v_n, T(v_n))\\&\ge \frac{\alpha _n}{\alpha ^*}\varphi (t^*, u_0, T(v_0)). \end{aligned} \end{aligned}$$

As $n\mapsto +\infty ,$ we obtain that

$$\begin{aligned} \alpha _*\ge \varphi (t^*, u_0, T(v_0)). \end{aligned}$$

Then

$$\begin{aligned} \alpha _*\ge \varphi (t^*, u_0, T(v_0))>\tau (t^*)=\alpha ^*. \end{aligned}$$

Which is a contradiction. Hence we have that $\alpha ^*=1.$ We have, for all $n,p\in \mathbb N,$

$$\begin{aligned} \theta \le v_n-u_n\le v_n-\alpha _nv_n=(1-r_n)v_n\le (1-r_n)v_0 \end{aligned}$$

and

$$\begin{aligned} \theta \le u_{n+p}-u_n\le v_{n+p}-u_n\le v_n-u_n \end{aligned}$$

$$\begin{aligned} \theta \le v_n-v_{n+p}\le v_{n}-u_{n+p}\le v_n-u_n. \end{aligned}$$

By the normality of $P,$ it is easy to see that $v_n-u_n\rightarrow 0$ as $n\mapsto +\infty$ and hence $\{u_n\}$ and $\{v_n\}$ are Cauchy sequences. Therefore, there exist $u^*,v^*$ such that $u_n\rightarrow u^*, v_n\rightarrow v^*$ as $n\mapsto +\infty$ and $u^*=v^*.$ Let us denote $x^*=u^*=v^*.$ Now, we prove that $x^*$ is a fixed point of $T.$ We have that

$$\begin{aligned} T(x^*)\ge T(u_n)=u_{n+1}\rightarrow x^*, \end{aligned}$$

and so $T(x^*)\ge x^*.$ On the other hand

$$\begin{aligned} T(x^*)\le T(v_n)=v_{n+1}\rightarrow x^*, \end{aligned}$$

which gives that $T(x^*)\le x^*$ and then $T(x^*)=x^*.$

As in the proof of Step $1,$ we can conclude that the fixed point of $T$ is unique. Finally, from Step $1$ we get that $x^*$ is the fixed point of $C$ in $[u_0, v_0]$ such that $T(x^*) = x^*$ and $C(x^*, x^*) = x^*.$ We can easily see that it is unique. Indeed, if $y^*$ satisfies $C(y^*, y^*) = y^*.$ By Step $1,$ $T(y^*)$ is the unique fixed point of $C(y^*, .),$ thus $T(y^*) = y^*.$ Since now the fixed point of $T$ is unique, we get $x^*= y^*.$ Which gives the result. $\square$

Remark 1

In Theorem 3.1, we don’t require $A$ and $B$ to be compact and continuous. Moreover, we don’t assume that $B$ is positive.

If we take $B=0,$ then we obtain the following Corollary which improves Theorem $3.1$ in [10].

Corollary 3.1

Let $E$ be a real Banach space and $P$ be a normal cone in $E.$ Let $u_0, v_0\in P$ with $u_0\le v_0$, $A: P\times P\rightarrow P$ be a mixed monotone operator and let $a, b$ be real numbers with $a<b$. Suppose that

• 1. there exists a surjection $\tau :(a, b)\rightarrow (0,1)$ and there exists a function $\varphi :(a, b]\times P\times P\rightarrow (0, 1]$ such that, for any $t\in (a, b],$ $\varphi (t, u, v)$ is nondecreasing in $u$ for fixed $v$ and non-increasing in $v$ for fixed $u;$

• 2. $\varphi (t, u, v)>\tau (t)$ for all $t\in (a, b)$ and $u, v\in P;$

• 3. $A(\tau (t)u, v)\ge \varphi (t, u, v)A(u, v)$ for all $t\in (a, b)$ and $u, v\in P;$

• 4. there exists a real positive number $r_0$ such that $u_0\ge r_0v_0,$ and $u_0$ and $v_0$ are such that

  $$\begin{aligned} u_0\le A(u_0, v_0),~~~~~~A(v_0, u_0)\le v_0; \end{aligned}$$

• 5. for fixed $u\in [u_0, v_0],$ there exists $N>0$ such that

  $$\begin{aligned} A(u, v_1)-A(u, v_2)\ge -N (v_1-v_2),~~~~~\forall v_1\ge v_2, v_1, v_2\in [u_0, v_0]. \end{aligned}$$

Then the operator equation $A(u, u)=u$ has exactly one fixed point $x^*$ in $[u_0, v_0].$

If $B$ is a linear operator, then by the proof of [6] in corollary $2.5,$ we have the following results.

Corollary 3.2

Let $E$ be a real Banach space and $P$ be a normal cone in $E.$ Let $u_0, v_0\in P$ with $u_0\le v_0$ and $A: P\times P\rightarrow P$ be a mixed monotone operator. Let $B: E\rightarrow E$ be linear and let $a, b$ be real numbers with $a< b$. Suppose that

• 1. there exists a surjection $\tau :(a, b)\rightarrow (0,1)$ and there exists a function $\varphi :(a, b)\times P\times P\rightarrow (0, 1)$ such that, for any $t\in (a, b),$ $\varphi (t, u, v)$ is nondecreasing in $u$ for fixed $v$ and non-increasing in $v$ for fixed $u;$

• 2. $\varphi (t, u, v)>\tau (t)$ for all $t\in (a, b)$ and $u, v\in P;$

• 3. $A(\tau (t)u, v)\ge \varphi (t, u, v)A(u, v)$ for all $t\in (a, b)$ and $u, v\in P;$

• 4. $\Vert B\Vert <1$ and there exists some number $c\ge 0$ such that $B+cI\ge 0;$

• 5. there exists a real positive number $r_0$ such that $u_0\ge r_0v_0,$ and $u_0$ and $v_0$ are such that

  $$\begin{aligned} u_0\le A(u_0, v_0)+Bu_0,~~A(v_0, u_0)+Bv_0\le v_0; \end{aligned}$$

• 6. for fixed $u\in [u_0, v_0],$ there exists $N>0$ such that

  $$\begin{aligned} A(u, v_1)\!-\!NBv_1\!-\!A(u, v_2)\!+\!NBv_2\!\ge \! -N (v_1\!-\!v_2),~~~~\forall v_1\!\ge \! v_2, v_1, v_2\!\in \! [u_0, v_0]. \end{aligned}$$

Then the operator equation $A(u, u)+Bu=u$ has exactly one fixed point $x^*$ in $[u_0, v_0].$

By the proof of [6] in corollary $2.6,$ we have the following results.

Corollary 3.3

Let $E$ be a real Banach space and $P$ be a normal and solid cone in $E.$ Let $u_0, v_0\in P$ with $u_0\le v_0$ and $A: P\times P\rightarrow P$ be a mixed monotone operator. Let $B: E\rightarrow E$ be linear (unbounded) and let $a, b$ be real numbers with $a<b$. Suppose that

• 1. there exists a surjection $\tau :(a, b)\rightarrow (0,1)$ and there exists a function $\varphi :(a, b)\times P\times P\rightarrow (0, 1)$ such that, for any $t\in (a, b),$ $\varphi (t, u, v)$ is nondecreasing in $u$ for fixed $v$ and non-increasing in $v$ for fixed $u;$

• 2. $\varphi (t, u, v)>\tau (t)$ for all $t\in (a, b)$ and $u, v\in P;$

• 3. $A(\tau (t)u, v)\ge \varphi (t, u, v)A(u, v)$ for all $t\in (a, b)$ and $u, v\in P;$

• 4. $1\in \rho (B),$ where $\rho (B)$ is the resolvent set of $B;$

• 5. there exist $u\in \mathring{P},$ where $\mathring{P}$ is the interior of $P$ and a number $c\ge 0$ such that $Bu\le \theta$ and $B+cI\ge 0;$

• 6. there exists a real positive number $r_0$ such that $u_0\ge r_0v_0,$ and $u_0$ and $v_0$ are such that

  $$\begin{aligned} u_0\le A(u_0, v_0)+Bu_0,~~A(v_0, u_0)+Bv_0\le v_0; \end{aligned}$$

• 7. for fixed $u\in [u_0, v_0],$ there exists $N>0$ such that

  $$\begin{aligned} A(u, v_1)\!-\!NBv_1\!-\!A(u, v_2)\!+\!NBv_2\!\ge \! -N (v_1\!-\!v_2),~~~~\forall v_1\!\ge \! v_2, v_1, v_2\!\in \! [u_0, v_0]. \end{aligned}$$

Then the operator equation $A(u, u)+Bu=u$ has exactly one fixed point $x^*$ in $[u_0, v_0].$

4 Applications

4.1 Positive solution of second order Neumann boundary value problem

In this section we consider the second order two point Neumann boundary value problem:

$$\begin{aligned} \left\{ \begin{array}{ll} - u^{\prime \prime }(t)+Mu(t)=f(t, u(t), u(t))+ G(t) u(t),~ t\in (0,1)\\ u^\prime (0)=u^{\prime }(1)=0,\end{array}\right. \end{aligned}$$

(2)

where $M>0, f:[0,1]\times [0, +\infty )\times [0, +\infty )\rightarrow [0, +\infty )$ and $G:[0,1]\rightarrow [0, +\infty )$ are given functions.

Let $C([0, 1])$ be the set of all real valued continuous functions $x(t)$ defined on $[0,1].$ Then $C([0, 1])$ is a Banach space with respect to the norm

$$\begin{aligned} \Vert x\Vert =\sup \{|x(t)|, t\in [0, 1]\}. \end{aligned}$$

Let

$$\begin{aligned} P=\{x\in C([0, 1]); x(t)\ge 0, \forall t\in [0, 1]\}. \end{aligned}$$

It is clear that $P$ is a normal cone in $C([0,1])$ and the normality constant is $1.$ Let $K(t, s)$ be the green function of the problem

$$\begin{aligned} \left\{ \begin{array}{ll} - u^{\prime \prime }(t)+Mu(t)=0,~ t\in (0,1)\\ u^\prime (0)=u^{\prime }(1)=0,\end{array}\right. \end{aligned}$$

then we have (see [11, 12])

$$\begin{aligned} K(t, s)=\left\{ \begin{array}{ll}\frac{\cosh (m(1-t))\cosh (ms)}{m\sinh m},~~~~~~0\le s\le t\le 1,\\ \frac{\cosh (m(1-s))\cosh (mt)}{m\sinh m},~~~~~~0\le t\le s\le 1, \end{array}\right. \end{aligned}$$

where $m=\sqrt{M}.$ Define the operator $A$ by

$$\begin{aligned} A(u, v)(t)=\int _0^1K(t, s)f(s, u(s), v(s))ds \end{aligned}$$

and the operator $B$ by

$$\begin{aligned} Bu (t)=\int _0^1K(t, s)G(s)u(s)ds. \end{aligned}$$

We assume now the following hypotheses

• 1. $f:[0,1]\times [0, +\infty )\times [0, +\infty )\rightarrow [0, +\infty )$ and $G:[0,1]\rightarrow [0, +\infty )$ are continuous.

• 2. For fixed $t\in [0,1]$ and $v\in P$ the function $u\mapsto f(t, u, v)$ is increasing and for fixed $t\in [0,1]$ and $u\in P$ the function $v\mapsto f(t, u, v)$ is decreasing.

• 3. there exists a surjection $\tau :(a, b)\rightarrow (0,1)$, ($a, b\in \mathbb R$ with $a<b$), and there exists a function $\varphi :(a, b]\times P\times P\rightarrow (0, 1]$ such that, for any $t\in (a, b],$ $\varphi (t, u, v)$ is nondecreasing in $u$ for fixed $v$ and non-increasing in $v$ for fixed $u, \varphi (t, u, v)>\tau (t)$ for all $t\in (a, b)$ and $u, v\in P$ and

  $$\begin{aligned} \int _0^1K(t, s)f(s,\tau (z) u(s), v(s))ds\ge \varphi (z, u, v)\int _0^1K(t, s)f(s, u(s), v(s))ds \end{aligned}$$

  for all $t, z\in (a, b)$ and $u, v\in P;$

• 4. $\int _0^1\int _0^1K(t, s)^2G(s)^2dsdt<1;$

• 5. there exist $u_0, v_0\in P$ such that $u_0\le v_0$ and there exists a real positive number $r_0$ such that $u_0\ge r_0v_0,$ and $u_0$ and $v_0$ are such that

  $$\begin{aligned} u_0\le \int _0^1K(t, s)(f(s, u_0(s), v_0(s))+G(s)u_0(s))ds, \end{aligned}$$
  $$\begin{aligned} \int _0^1K(t, s)(f(s, v_0(s), u_0(s))+G(s)v_0(s))ds\le v_0, \end{aligned}$$

  and for fixed $u\in [u_0, v_0],$ there exists $N>0$ such that

  $$\begin{aligned} A(u, v_1)\!-\!NBv_1\!-\!A(u, v_2)\!+\!NBv_2\!\ge \! -N (v_1\!-\!v_2),~~~~\forall v_1\!\ge \! v_2, v_1, v_2\!\in \! [u_0, v_0]. \end{aligned}$$

Theorem 4.1

Under assumptions $1-5$ the problem (2) has a unique positive solution in $[u_0, v_0].$

Proof 4.1

Under hypothesis $(1),$ the operator $A: P\times P\rightarrow P$ and the operator $B:C([0, 1])\rightarrow C([0, 1])$ are well defined. So, to find a solution of problem (2) is equivalent to find a fixed point of the operator equation $A(u, u)+Bu=u.$ It is easy to see that the operators $A$ and $B$ satisfy all the hypotheses of Corollary 3.2 and then the result follows. $\square$

Example 4.1

Consider the second order two point Neumann boundary value problem (2) with $M=(\log 2)^2, G(t)=10^{-2}\frac{\log 2\sinh (t\log 2)}{\cosh ^2(\log 2)}$ and $f(t, u(t), v(t))=u(t)^{\frac{1}{3}}+v(t)^{-\frac{1}{4}}.$ Let us define

$$\begin{aligned} A(u, v)(t)=\int _0^1K(t, s)\left( u(s)^{\frac{1}{3}}+v(s)^{-\frac{1}{4}}\right) ds \end{aligned}$$

and

$$\begin{aligned} Bu (t)=10^{-2}\int _0^1K(t, s)\frac{\log 2\sinh (s\log 2)}{\cosh ^2(\log 2)}u(s)ds. \end{aligned}$$

Now let us show that the operator $A$ and $B$ satisfies all the conditions of Corollary 3.2. Let we set $u_0=1$ and $v_0=10.$ It is easy to see that the operator $A: P\times P\rightarrow P$ is well defined and it is a mixed monotone operator. We have also that the operator $B: P\rightarrow P$ is well defined. We can easily show also that $u_0< v_0, u_0, v_0\in P$ and there exists a real number $0<r_0<1$ such that $u_0\ge r_0v_0.$ By a simple calculation we obtain that

$$\begin{aligned} \frac{1}{\log 2\sinh (\log 2)}\le K(t, s)\le \frac{\cosh ^2(\log 2)}{\log 2\sinh (\log 2)} . \end{aligned}$$

(3)

Let now $z\in (0, 1)$ and let $\tau :(0, 1)\rightarrow (0,1)$ such that $\tau (z)=z$ and let $u, v\in P.$ We have

$$\begin{aligned} \begin{aligned} A(\tau (z)u, v)(t)&=\int _0^1K(t, s)\left( (\tau (z)u(s))^{\frac{1}{3}}+(v(s))^{-\frac{1}{4}}\right) ds\\&\ge \int _0^1K(t, s)\left( (\tau (z)u(s))^{\frac{1}{3}}+\tau (z)^{\frac{1}{3}}(v(s))^{-\frac{1}{4}}\right) ds\\&=\tau (z)^{\frac{1}{3}}\int _0^1K(t, s)\left( (u(s))^{\frac{1}{3}}+(v(s))^{-\frac{1}{4}}\right) ds\\&=z^{\frac{1}{3}}A(u, v)(t)\\&\ge \sqrt{z}A(u, v)(t)=(\sqrt{z}-z+z)A(u, v)(t)\\&\ge \left( z+(\sqrt{z}-z)\left( \frac{u+1}{8(u+2)}+\frac{v+2}{8(v+1)}\right) \right) A(u, v)(t)\\ \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} A(\tau (z)u, v)(t)\ge \varphi (z, u, v)A(u, v)(t) \end{aligned}$$

where $\varphi (z, u, v)=z+(\sqrt{z}-z)(\frac{u+1}{8(u+2)}+\frac{v+2}{8(v+1)}).$ It is clear that $\varphi : (0,1]\times P\times P\rightarrow (0,1]$ is such that $\varphi (z, u, v)>z$ for all $z\in (0, 1)$ and that for fixed $z$ and $v,$ $\varphi$ is increasing in $u$ and for fixed $z$ and $u, \varphi$ is decreasing in $v.$ For the operator $B,$ we have

$$\begin{aligned} \begin{aligned} \Vert B\Vert&\le \sqrt{\int _0^1\int _0^1(K(t, s)G(s))^2dsdt}\\&\le 10^{-2}<1 \end{aligned} \end{aligned}$$

and $Bu(t)\ge 0, \forall t\in [0,1].$ Using (3), we obtain that

$$\begin{aligned} A(u_0, v_0)(t)+Bu_0(t)&= \int _0^1K(t, s)((u_0(s))^{\frac{1}{3}}+(v_0(s))^{-\frac{1}{4}}+G(s)u_0(s))ds\\&\ge \int _0^1K(t, s)((u_0(s))^{\frac{1}{3}}+(v_0(s))^{-\frac{1}{4}}ds\\&\ge \frac{1}{\log 2\sinh (\log 2)}\big (1+10^{-\frac{1}{4}}\big )\\&\approx 3\ge 1=u_0 \end{aligned}$$

and

$$\begin{aligned} A(v_0, u_0)(t)+Bv_0(t)&= \int _0^1K(t, s)((v_0(s))^{\frac{1}{3}}+(u_0(s))^{-\frac{1}{4}}+G(s)v_0(s))ds\\&\le \frac{\cosh ^2(\log 2)}{\log 2\sinh (\log 2)}\left( 10^{\frac{1}{3}}+1+10^{-1}\frac{\log 2\sinh (\log 2)}{\cosh ^2(\log 2)}\right) \\&\le \frac{\cosh ^2(\log 2)}{\log 2\sinh (\log 2)}(10^{\frac{1}{3}}+1)+10^{-1}\\&\approx 6,57\le 10=v_0. \end{aligned}$$

For fixed $u\in P, \exists N=\frac{4}{5}$ such that

$$\begin{aligned} A(u, v_1)-NBv_1-A(u, v_2)+NBv_2\ge -N(v_1-v_2), ~\forall v_1, v_2\in [1, 10], v_1\ge v_2. \end{aligned}$$

Therefore, by means of Corollary 3.2, it follows that the operator equation $A(u, u)+Bu=u$ has a unique fixed point in $[1, 10].$

4.2 Nonlinear elliptic boundary value problem’s for the Lane–Emden–Fowler equation

Let $\Omega$ be a bounded domain with smooth boundary in $\mathbb R^N (N\ge 1).$ Consider the following Dirichlet problem for the Lane–Emden–Fowler equation:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} -\Delta u=\frac{1}{4}\left( \frac{u+1}{u+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{u+2}{u+1}\right) ,&{}\text {in}~\Omega \\ u(x)>0,&{}x\in \Omega ,\\ u(x)=0,&{}x\in \partial \Omega . \end{array}\right. \end{aligned}$$

(4)

Let $v_0$ be the positive solution of the equation below

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} -\Delta u=1,&{}\text {in}~\Omega \\ u(x)>0,&{}x\in \Omega ,\\ u(x)=0,&{}x\in \partial \Omega . \end{array}\right. \end{aligned}$$

We get the following result:

Proposition 4.1

Equation (4) has a unique positive solution $u^*\in [u_0, v_0]=[\frac{1}{4}v_0, v_0].$

Proof 4.2

We set $C(\overline{\Omega }),$ the Banach space of continuous functions on $\overline{\Omega }$ with the norm $\Vert u\Vert =\max \{|u(x)|, x\in \overline{\Omega }\}$ and we set $P=\{u\in C(\overline{\Omega }), u(x)\ge 0~\text {in}~x\in \overline{\Omega }\},$ the standard cone. It is clear that $P$ is a normal cone in $C(\overline{\Omega })$ and the normality constant is $1.$ Consider the following linear elliptic boundary value problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} -\Delta w=\frac{1}{4}\left( \frac{u+1}{u+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{v+2}{v+1}\right) ,&{}\text {in}~\Omega \\ w(x)>0,&{}x\in \Omega ,\\ w(x)=0,&{}x\in \partial \Omega , \end{array}\right. \end{aligned}$$

(5)

where $u,v\in P.$ By the classical theory of linear elliptic equations (see [8]), the problem (5) admits a unique strong solution $A(u, v)\in W^{2,l}(\Omega )\cap W_0^{1,l}(\Omega )$ for some $l>N.$ Using the Sobolev imbedding theory, $A(u, v)\in C^{1, \beta }(\overline{\Omega })$ with $\beta =1-\frac{N}{l}.$ Now the operator $A:C(\overline{\Omega })\times C(\overline{\Omega })\rightarrow C(\overline{\Omega })$ is well defined as a solution of (5) for given $u, v\in P.$

$u\mapsto \frac{1}{4}\left( \frac{u+1}{u+2}\right) ^{\frac{1}{3}}$ is increasing in $u$ and $v\mapsto \frac{1}{4}\left( \frac{v+2}{v+1}\right)$ is decreasing in $v,$ and we have also

$$\begin{aligned}&0<\frac{1}{4}\left( \frac{v_0+1}{v_0+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{u_0+2}{u_0+1}\right) \le 1\\&\frac{1}{4}\left( \frac{u_0+1}{u_0+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{v_0+2}{v_0+1}\right) \ge \frac{1}{4} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} -\Delta A(tu, v)&=\frac{1}{4}\left( \frac{tu+1}{tu+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{v+2}{v+1}\right) \\&\ge \frac{1}{4}\left( \frac{tu+1}{u+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{v+2}{v+1}\right) \\&\ge \frac{t^{\frac{1}{3}}}{4}[\left( \frac{u+1}{u+2}\right) ^{\frac{1}{3}}+\left( \frac{v+2}{v+1}\right) ]\\&\ge \sqrt{t}[\frac{1}{4}\left( \frac{u+1}{u+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{v+2}{v+1}\right) ]\\&\ge \sqrt{t}(-\Delta A(u, v)), ~~ \forall t\in (0,1). \end{aligned} \end{aligned}$$

By applying the comparaison principle, we have that the operator $A$ has a mixed monotone property and that $A(tu, v)\!\ge \! \sqrt{t}A(u, v)\!\ge \! \left( t\!+\!(\sqrt{t}\!-\!t)\left( \frac{u\!+\!1}{8(u+2)}+\frac{v+2}{8(v+1)}\right) \right) A(u, v).$ Moreover

$$\begin{aligned} A(u_0, v_0)\ge u_0~\text {and}~A(v_0, u_0)\le v_0. \end{aligned}$$

We have also

$$\begin{aligned} -\Delta (A_v^\prime (u, v))=-\frac{1}{4(v+1)^2}\ge -\frac{1}{4}. \end{aligned}$$

By the comparaison principle, it follows that

$$\begin{aligned} A_v^\prime (u, v)\ge -\frac{1}{4}v_0\ge -N ~~ (\text {for some}~N>0), \end{aligned}$$

thus for fixed $u\in [u_0, v_0],$ we get

$$\begin{aligned} A(u, v_1)-A(u, v_2)\ge -N(v_1, v_2),~\forall v_1, v_2\in P, v_1\ge v_2. \end{aligned}$$

Thus, by Corollary 3.1, $A$ has a unique fixed point $u^*$ in $[u_0, v_0],$ which is the unique solution of (4). By the theory of the linear elliptic equation, the problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} -\Delta w=\frac{1}{4}\left( \frac{u^*+1}{u^*+2}\right) ^{\frac{1}{3}}+\frac{1}{4}\left( \frac{u^*+2}{u^*+1}\right) ,&{}\text {in}~\Omega \\ w(x)>0,&{}x\in \Omega ,\\ w(x)=0,&{}x\in \partial \Omega , \end{array}\right. \end{aligned}$$

admits a unique solution $\overline{u^*}\in W^{2,l}(\Omega )\cap W_0^{1,l}(\Omega )$ and hence $\overline{u^*}\in C^{1, \beta }(\overline{\Omega }).$ Recalling the uniqueness of the solution of (4), one easily deduce that $\overline{u^*}=u^*.$ Thus the problem (4) has a unique classical solution $u^*\in C^{1, \beta }(\overline{\Omega }).$ $\square$

4.3 Nonlinear second order Sturm Liouville boundary value problem

Consider the following singular boundary value problem:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} \frac{1}{p(t)}(p(t)u^\prime (t))^\prime +f(u(t),u(t))=0, &{}0<t<1,\\ u(0)-\lim _{t\rightarrow 0^+}p(t)u^\prime (t),&{} u(1)+\lim _{t\rightarrow 1^-}p(t)u^\prime (t),\end{array}\right. \end{aligned}$$

(6)

where $p(t)=\frac{1}{3}\root 3 \of {t}$ and $f:(u, v)\in P\times P\mapsto 1.3u^{\frac{1}{3}}+v^{-\frac{1}{4}}\in P.$ Here, we set $C([0, 1]),$ the Banach space of continuous functions on $[0, 1]$ with the norm $\Vert u\Vert =\max \{|u(x)|, x\in \overline{\Omega }\}$ and we set $P=\{u\in C([0, 1]), u(x)\ge 0~\text {in}~x\in [0,1]\},$ the standard cone. This type of problem is referred to as a model for the deformation of an elastic beam under a variety of boundary conditions. Now we denote by $G(t, s)$ the Greens functions for the following boundary value problem:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} \frac{1}{p(t)}(p(t)u^\prime (t))^\prime =0, &{}0<t<1,\\ \alpha u(0)-\beta \lim _{t\rightarrow 0^+}p(t)u^\prime (t),&{} \gamma u(1)+\delta \lim _{t\rightarrow 1^-}p(t)u^\prime (t),\end{array}\right. \end{aligned}$$

It is well known that $G(t, s)$ can be written by

$$\begin{aligned} G(t, s)=\frac{1}{\omega }\left\{ \begin{array}{ll} (\beta +\alpha B(0, s))(\delta +\gamma B(t, 1))~~~~~0\le s\le t\le 1,\\ (\beta +\alpha B(0, t))(\delta +\gamma B(s, 1))~~~~~0\le t\le s\le 1,\end{array}\right. \end{aligned}$$

where $B(t, s)=\int _t^s\frac{d\tau }{p(\tau )}$ and $\omega =\alpha \delta +\alpha \gamma B(0,1)+\beta \gamma >0.$ So, if $\alpha =\beta =\gamma =\delta =1$ and $p(t)=\frac{1}{3}\root 3 \of {t},$ then we obtain that

$$\begin{aligned} G(t, s)=\frac{1}{70}\left\{ \begin{array}{ll} (2 +9 s^{\frac{2}{3}})(11 - 9 t^{\frac{2}{3}})~~~~~0\le s\le t\le 1,\\ (2 +9 t^{\frac{2}{3}})(11 - 9 s^{\frac{2}{3}})~~~~~0\le t\le s\le 1.\end{array}\right. \end{aligned}$$

It is clear that, for all $t\in (0, 1),$

$$\begin{aligned} \frac{1}{70}\le \int _0^1G(t, s)p(s)ds\le 0.432 \end{aligned}$$

Let us define the operator

$$\begin{aligned} \begin{aligned} T:&P\times P\rightarrow C([0,1])\\&(u, v)\mapsto T(u, v) \end{aligned} \end{aligned}$$

with $T(u, v)(t)=\int _0^1G(t, s)p(s)f(u(s), v(s))ds .$ Obviously, $u$ is a solution of the boundary value problem (6) if and only if $u$ verifies $T(u, u)=u.$ Now, let us show that the operator $T$ satisfies all the conditions of Corollary 3.1. We set $u_0=\frac{1}{60}$ and $v_0=2.$ It is clear that $T: P\times P\rightarrow P$ is a mixed monotone operator and we have

$$\begin{aligned} T(u_0, v_0)=\int _0^1G(t, s)p(s)\big (1.3u_0^{\frac{1}{3}}+v_0^{-\frac{1}{4}}\big )ds\ge \frac{1}{60}=u_0 \end{aligned}$$

and

$$\begin{aligned} T(v_0, u_0)=\int _0^1G(t, s)p(s)\big (1.3v_0^{\frac{1}{3}}+u_0^{-\frac{1}{4}}\big )ds\le 2=v_0. \end{aligned}$$

Let $z\in (0, 1)$ and $\tau (z)=z,$ then we have

$$\begin{aligned} T(zu, v)(t)\ge \varphi (z, u, v)T(u, v)(t), \end{aligned}$$

where $\varphi (z, u, v)=z+(\sqrt{z}-z)\big (\frac{u+1}{8(u+2)}+\frac{v+2}{8(v+1)}\big ).$ Finally, for fixed $u\in P,$ we have

$$\begin{aligned} T(u, v_1)-T(u, v_2)\ge -20 (v_1-v_2),~~~\forall v_1\ge v_2, v_1, v_2\in \left[ \frac{1}{60}, 2\right] . \end{aligned}$$

Therefore, we see that the boundary value problem (6) has a unique solution in $\big [\frac{1}{60}, 2\big ].$