Fixed-point theorems for nonlinear operators with singular perturbations and applications

Abstract

In this paper, using fixed-point index theory and approximation techniques, we consider the existence and multiplicity of fixed points of some nonlinear operators with singular perturbation. As an application we consider the existence and multiplicity of positive solutions of singular systems of multi-point boundary value problems, which improve the results in the literature.

1 Introduction

In this paper we consider the problem

$$x=Ax+\lambda Bx,$$

where A is continuous and compact and B is a singular continuous and compact operator (defined in Section 2).

In the study of nonlinear phenomena many models give rise to singular boundary value problems (singular in the dependent variable) (see [1–3]). In [4], Taliaferro showed that the singular boundary value problem

$$\{\begin{array}{l}{y}^{″}+q(t)y^{-\alpha }=0,0<t<1,\\ y(0)=0=y(1),\end{array}$$

has a $C[0,1]\cap C^1(0,1)$ solution; here $\alpha >0$, $q\in C(0,1)$ with $q>0$ on $(0,1)$ and ${\int }_0^{1}t(1-t)q(t)dt<\mathrm{\infty }$. For more recent work we refer the reader to [5–14] and the references therein.

In this paper we consider abstract singular operators (defined in Section 2) and we consider the existence and multiplicity of fixed points of some nonlinear operators with singular perturbations. As an application we discuss the existence and multiplicity of positive solutions of singular systems of multi-point boundary value problems.

2 Fixed-point theorems

Let E be a Banach space, P a cone of E, $\mathrm{\Omega }\subseteq E$ bounded and open. The following theorems are needed in our paper.

Theorem 2.1 ([8])

Suppose $\theta \in \mathrm{\Omega }$, $A:P\cap \overline{\mathrm{\Omega }}\to P$ is continuous and compact and

$$Ax\ne \mu x,\mathrm{\forall }x\in P\cap \partial \mathrm{\Omega },\mu \ge 1.$$

Then

$$i(A,P\cap \mathrm{\Omega },P)=1.$$

Theorem 2.2 ([8])

Assume that $A:P\cap \overline{\mathrm{\Omega }}\to P$ is continuous and compact. If there exists a compact and continuous operator $K:P\cap \partial \mathrm{\Omega }\to P$ such that

• (1) ${inf}_{x\in P\cap \partial \mathrm{\Omega }}\parallel Kx\parallel >0$;

• (2) $x-Ax\ne \lambda Kx$, $\mathrm{\forall }x\in P\cap \partial \mathrm{\Omega }$, $\lambda \ge 0$,

then

$$i(A,P\cap \mathrm{\Omega },P)=0.$$

Now we give a new definition.

Definition 2.1 If $B:P-\{\theta \}\to P$ is continuous with

$$\underset{x\to \theta ,x\in (P-\{\theta \})}{lim}\parallel Bx\parallel =+\mathrm{\infty }$$

and $B(\{x\in P|r\le \parallel x\parallel \le R\})$ is relatively compact, for any $0<r<R<+\mathrm{\infty }$, then $B:P-\{\theta \}\to P$ is called a singular continuous and compact operator.

Remark Consider

$$\begin{array}{c}{x}^{″}(t)+a(t)x^{-\gamma }(t)=0,t\in (0,1),\hfill \\ x(0)=0,x(1)=0,\hfill \end{array}$$

where $1>\gamma >0$ and $a(t)\in C((0,1),(0,+\mathrm{\infty }))\cap L^1(0,1)$ or equivalently

$$x(t)={\int }_0^{1}G(t,s)a(s)x^{-\gamma }(s)ds,t\in [0,1],$$

where

$$G(t,s)=\{\begin{array}{ll}s(1-t),& 0\le s\le t\le 1,\\ t(1-s),& 0\le t\le s\le 1.\end{array}$$

Set

$$P:=\{x\in C[0,1]:x(t)\ge t(1-t)\parallel x\parallel \},$$

where $\parallel x\parallel ={max}_{t\in [0,1]}|x(t)|$. For $x\in P-\{\theta \}$, let

$$(Bx)(t):={\int }_0^{1}G(t,s)a(s)x^{-\gamma }(s)ds,t\in [0,1].$$

It is easy to see that $B:P-\{\theta \}\to P$ is a singular continuous and compact operator (see [7, 14]).

Theorem 2.3 Suppose that $\theta \in \mathrm{\Omega }$, $A:P\cap \overline{\mathrm{\Omega }}\to P$ is continuous and compact and $B:P-\{\theta \}\to P$ is singular continuous and compact. Assume that

$$Ax\ne \mu x,\mathrm{\forall }x\in P\cap \partial \mathrm{\Omega },\mu \ge 1.$$

(2.1)

Then there exists a ${\lambda }_{\ast }>0$ such that, for any $\lambda \in (0,{\lambda }_{\ast })$, there exist $x_{\lambda }\in P\cap \mathrm{\Omega }-\{\theta \}$ with

$$x_{\lambda }=A{x}_{\lambda }+\lambda B{x}_{\lambda }.$$

Proof Choose $x_0\in P-\{\theta \}$, and define

$$B_{n}x=B(x+\frac{1}{n}{x}_0),\mathrm{\forall }x\in P,n\in \mathbb{N}.$$

Set

$$\gamma :=\underset{(\mu ,x)\in [1,+\mathrm{\infty })\times P\cap \partial \mathrm{\Omega }}{inf}\parallel \mu x-Ax\parallel .$$

Now we claim that

$$\gamma >0.$$

(2.2)

If $\gamma =0$, there exists $\{({\mu }_n,x_n)\}\subseteq [1,+\mathrm{\infty })\times P\cap \partial \mathrm{\Omega }$ such that

$$\underset{n\to +\mathrm{\infty }}{lim}\parallel {\mu }_n{x}_n-A{x}_n\parallel =0.$$

(2.3)

First, we show $\{{\mu }_n\}$ is bounded.

To see this suppose $\{{\mu }_n\}$ is unbounded. Without loss of generality, we assume that ${lim}_{n\to +\mathrm{\infty }}{\mu }_n=+\mathrm{\infty }$. Then

$$0\leftarrow \parallel {\mu }_n{x}_n-A{x}_n\parallel \ge |{\mu }_n|\parallel x_n\parallel -\parallel A{x}_n\parallel \ge {\mu }_n\underset{x\in P\cap \partial \mathrm{\Omega }}{inf}\parallel x\parallel -\parallel A{x}_n\parallel \to +\mathrm{\infty },$$

and this is a contradiction.

Next, we show that there exists a $({\mu }_0,x_0)\in [1,+\mathrm{\infty })\times P\cap \partial \mathrm{\Omega }$ such that

$$\parallel {\mu }_0{x}_0-A{x}_0\parallel =0.$$

The boundedness of $\{{\mu }_n\}$ means that $\{{\mu }_n\}$ has a convergent subsequence. Without loss of generality, we assume that ${\mu }_n\to {\mu }_0\ge 1$. Since $\{x_n\}$ is bounded and A is continuous and compact, $\{A{x}_n\}$ has a convergent subsequence $\{A{x}_{n_i}\}$ with ${lim}_{n_i\to +\mathrm{\infty }}A{x}_{n_i}\to y_0$. From (2.3), we have

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel {\mu }_{n_i}{x}_{n_i}-A{x}_{n_i}\parallel =0,$$

which implies that

$${\mu }_{n_i}{x}_{n_i}\to y_0,n_i\to +\mathrm{\infty }.$$

Then

$$x_{n_i}\to \frac{1}{{\mu }_0}{y}_0,\text{as }{n}_i\to +\mathrm{\infty }.$$

Let $x_0=\frac{1}{{\mu }_0}{y}_0$. Clearly, $x_0\in P\cap \partial \mathrm{\Omega }$ and

$$\parallel {\mu }_0{x}_0-A{x}_0\parallel =\underset{n_i\to +\mathrm{\infty }}{lim}\parallel {\mu }_{n_i}{x}_{n_i}-A{x}_{n_i}\parallel =0,$$

which contradicts (2.1).

Let

$${\beta }_n=\underset{x\in P\cap \partial \mathrm{\Omega }}{sup}\parallel B_{n}x\parallel .$$

Now we claim that

$$\underset{n\in \mathbb{N}}{sup}{\beta }_n<+\mathrm{\infty }.$$

(2.4)

To see this suppose that

$$\underset{n\in \mathbb{N}}{sup}{\beta }_n=+\mathrm{\infty }.$$

Without loss of generality, assume that

$$\underset{n\to +\mathrm{\infty }}{lim}{\beta }_n=+\mathrm{\infty },$$

which implies that there exists a sequence $\{x_n\}\in P\cap \partial \mathrm{\Omega }$ such that

$$\underset{n\to +\mathrm{\infty }}{lim}\parallel B_n{x}_n\parallel =\underset{n\to +\mathrm{\infty }}{lim}\parallel B(x_n+\frac{1}{n}{x}_0)\parallel =+\mathrm{\infty }.$$

(2.5)

For all $x\in P\cap \partial \mathrm{\Omega }$, we have

$$\parallel x+\frac{1}{n}{x}_0\parallel \le \parallel x\parallel +\parallel x_0\parallel \le \underset{x\in P\cap \mathrm{\Omega }}{sup}\parallel x\parallel +\parallel x_0\parallel :=R<+\mathrm{\infty }.$$

Since, for any $r>0$, $B:P\cap \{x:r\le \parallel x\parallel \le R\}$ is relatively compact, (2.5) guarantees that there exists a subsequence $\{x_{n_i}\}\subseteq \{x_n\}$ such that

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}+\frac{1}{n_i}{x}_0\parallel =0.$$

Thus

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}\parallel =0,$$

which implies that $\theta \in P\cap \partial \mathrm{\Omega }$. This contradicts $\theta \in \mathrm{\Omega }$. Hence, (2.4) holds.

Set

$$\beta :=\underset{n\in \mathbb{N}}{sup}{\beta }_n<+\mathrm{\infty }$$

and

$${\lambda }_{\ast }:=\frac{\gamma }{\beta }>0.$$

For $0<\lambda <{\lambda }_{\ast }$, $x\in P\cap \partial \mathrm{\Omega }$, $\mu \ge 1$, we have

$$\parallel Ax+\lambda B_{n}x-\mu x\parallel \ge \parallel Ax-\mu x\parallel -\lambda \parallel B_{n}x\parallel \ge \gamma -\beta \lambda >0.$$

Theorem 2.1 guarantees that

$$i(A+\lambda B_n,P\cap \mathrm{\Omega },P)=1.$$

(2.6)

Note (2.6) guarantees that, for any $\lambda \in (0,{\lambda }_{\ast })$, there exists a $\{x_n\}\subseteq P\cap \mathrm{\Omega }$ such that

$$x_n=A{x}_n+\lambda B_n{x}_n,n\in \mathbb{N}.$$

(2.7)

Now we show that

$$\underset{n\in \mathbb{N}}{inf}\parallel x_n\parallel >0,$$

(2.8)

which implies that

$$\underset{n\in \mathbb{N}}{inf}\parallel x_n+\frac{1}{n}{x}_0\parallel >0.$$

To see this suppose that

$$\underset{n\in \mathbb{N}}{inf}\parallel x_n\parallel =0.$$

Then there exists a $\{x_{n_i}\}$ such that

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}\parallel =0,$$

(2.9)

and so

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}+\frac{1}{n_i}{x}_0\parallel =0.$$

Thus

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel B_{n_i}{x}_{n_i}\parallel =\underset{n_i\to +\mathrm{\infty }}{lim}\parallel B(x_{n_i}+\frac{1}{n_i}{x}_0)\parallel =+\mathrm{\infty }.$$

(2.10)

The compactness of A guarantees that $\{A{x}_{n_i}\}$ has a convergent subsequence. Without loss of generality, we assume that ${lim}_{n_i\to +\mathrm{\infty }}A{x}_{n_i}=y_0$. From (2.10), we have

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}\parallel =\underset{n_i\to +\mathrm{\infty }}{lim}\parallel A{x}_{n_i}+\lambda B_{n_i}{x}_{n_i}\parallel \ge \underset{n_i\to +\mathrm{\infty }}{lim}\lambda \parallel B_{n_i}{x}_{n_i}\parallel -\underset{n_i\to +\mathrm{\infty }}{lim}\parallel A{x}_{n_i}\parallel =+\mathrm{\infty },$$

which contradicts (2.9).

Now (2.8) guarantees that

$$0<\underset{n\in \mathbb{N}}{inf}\parallel x_n+\frac{1}{n}{x}_0\parallel \le \parallel x_n+\frac{1}{n}{x}_0\parallel \le \underset{x\in P\cap \mathrm{\Omega }}{sup}\parallel x\parallel +\parallel x_0\parallel <+\mathrm{\infty },n\in \mathbb{N}.$$

Then $\{A{x}_n+\lambda B_n{x}_n\}$ has a convergent subsequence. Without loss of generality, we assume that

$$A{x}_n+\lambda B_n{x}_n\to y_1,\text{as }n\to +\mathrm{\infty }.$$

Then

$$x_n\to y_1,\text{as }n\to +\mathrm{\infty }.$$

Now (2.8) guarantees that $y_1\ne \theta$. Letting $n\to +\mathrm{\infty }$ in (2.7), and we have

$$y_1=A{y}_1+\lambda B{y}_1$$

(2.11)

and $y_1\in P\cap \mathrm{\Omega }-\{\theta \}$. The proof is complete. □

Corollary 2.1 Suppose that $\theta \in \mathrm{\Omega }$, $A:P\cap \overline{\mathrm{\Omega }}\to P$ is continuous and compact and $B:P-\{\theta \}\to P$ is singular continuous and compact. Assume that

$$\parallel Ax\parallel <\parallel x\parallel ,\mathrm{\forall }x\in P\cap \partial \mathrm{\Omega }$$

(2.12)

or

$$Ax\ngeqq x,\mathrm{\forall }x\in P\cap \partial \mathrm{\Omega }.$$

(2.13)

Then there exists a ${\lambda }_{\ast }>0$ such that, for any $\lambda \in (0,{\lambda }_{\ast })$, there exist $x_{\lambda }\in P\cap \mathrm{\Omega }$ with

$$x_{\lambda }=A{x}_{\lambda }+\lambda B{x}_{\lambda }.$$

It is easy to see that (2.12) or (2.13) guarantees that (2.1) holds (see [8]).

Theorem 2.4 Suppose that ${\mathrm{\Omega }}_1$, ${\mathrm{\Omega }}_2$ are bounded open sets and $\theta \in {\mathrm{\Omega }}_1\subseteq {\mathrm{\Omega }}_2$, $A,K:P\cap {\overline{\mathrm{\Omega }}}_2\to P$ are continuous and compact and $B:P-\{\theta \}\to P$ is singular continuous and compact. Assume that

(C₁) $Ax\ne \mu x$, $\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_1$, $\mu \ge 1$;

(C₂) ${inf}_{x\in P\cap \partial {\mathrm{\Omega }}_2}\parallel Kx\parallel >0$;

(C₃) $x-Ax\ne \mu Kx$, $\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_2$, $\mu \ge 0$.

Then there exists a ${\lambda }_{\ast }>0$ such that, for any $\lambda \in (0,{\lambda }_{\ast })$, there exist $x_{\lambda ,1}\in (P\cap {\mathrm{\Omega }}_1-\{\theta \})$ and $x_{\lambda ,2}\in P\cap ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)$ with

$$x_{\lambda ,1}=A{x}_{\lambda ,1}+\lambda B{x}_{\lambda ,1},x_{\lambda ,2}=A{x}_{\lambda ,2}+\lambda B{x}_{\lambda ,2}.$$

Proof Choose $x_0\in P-\{\theta \}$, and define

$$B_{n}x=B(x+\frac{1}{n}{x}_0),\mathrm{\forall }x\in P,n\in \mathbb{N}.$$

Set

$${\gamma }_1:=\underset{(\mu ,x)\in [1,+\mathrm{\infty })\times P\cap \partial {\mathrm{\Omega }}_1}{inf}\parallel \mu x-Ax\parallel$$

and

$${\gamma }_2:=\underset{(\mu ,x)\in [0,+\mathrm{\infty })\times P\cap \partial {\mathrm{\Omega }}_2}{inf}\parallel x-Ax-\mu Kx\parallel .$$

We claim that

$${\gamma }_1>0,{\gamma }_2>0.$$

(2.14)

An argument similar to that in (2.2) shows that

$${\gamma }_1>0.$$

(2.15)

Now we show that

$${\gamma }_2>0.$$

(2.16)

To see this suppose that ${\gamma }_2=0$. Then there exists $\{({\mu }_n,x_n)\}\subseteq [0,+\mathrm{\infty })\times (P\cap \partial {\mathrm{\Omega }}_2)$ such that

$$\underset{n\to +\mathrm{\infty }}{lim}\parallel x_n-A{x}_n-{\mu }_{n}Kx\parallel =0.$$

(2.17)

Now since

$$\parallel x_n-A{x}_n-{\mu }_{n}Kx\parallel \ge {\mu }_n\underset{x\in \partial \cap {\mathrm{\Omega }}_2}{inf}\parallel Kx\parallel -\parallel x_n-A{x}_n\parallel ,$$

we have $\{{\mu }_n\}$ is bounded, which means that $\{{\mu }_n\}$ has a convergent subsequence. Without loss of generality, we assume that

$$\underset{n\to +\mathrm{\infty }}{lim}{\mu }_n={\mu }_0.$$

Since $\{x_n\}$ is bounded and A and K are compact, $\{A{x}_n\}$ and $\{K{x}_n\}$ have convergent subsequences $\{A{x}_{n_i}\}$ and $\{K{x}_{n_i}\}$ with ${lim}_{n_i\to +\mathrm{\infty }}A{x}_{n_i}=y_0$ and ${lim}_{n_i\to +\mathrm{\infty }}K{x}_{n_i}=z_0$. Now

$$\underset{n\to +\mathrm{\infty }}{lim}\parallel x_n-A{x}_n-{\mu }_{n}K{x}_n\parallel =0,$$

which implies that

$$\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}-y_0-{\mu }_0{z}_0\parallel =0.$$

Let $x_0=y_0+{\mu }_0{z}_0$. Clearly, $x_0\in P\cap \partial {\mathrm{\Omega }}_2$. Now

$$\parallel x_0-A{x}_0-{\mu }_{0}K{x}_0\parallel =\underset{n_i\to +\mathrm{\infty }}{lim}\parallel x_{n_i}-A{x}_{n_i}-{\mu }_{n_i}K{x}_{n_i}\parallel =0,$$

which contradicts condition (C₃).

Consequently, (2.16) is true, which together with (2.15) yields (2.14).

Let $\gamma =min\{{\gamma }_1,{\gamma }_2\}$. Obviously, $\gamma >0$.

Let

$${\beta }_{n,1}=\underset{x\in P\cap \partial {\mathrm{\Omega }}_1}{sup}\parallel B_{n}x\parallel ,{\beta }_{n,2}=\underset{x\in P\cap \partial {\mathrm{\Omega }}_2}{sup}\parallel B_{n}x\parallel .$$

An argument similar to that in (2.4) shows that

$$\underset{n\in \mathbb{N}}{sup}{\beta }_{n,1}<+\mathrm{\infty },\underset{n\in \mathbb{N}}{sup}{\beta }_{n,2}<+\mathrm{\infty }.$$

Let

$$\beta :=max\{\underset{n\in \mathbb{N}}{sup}{\beta }_{n,1},\underset{n\in \mathbb{N}}{sup}{\beta }_{n,1}\}<+\mathrm{\infty }$$

and

$${\lambda }_{\ast }:=\frac{\gamma }{\beta }>0.$$

For $0<\lambda <{\lambda }_{\ast }$, $x\in P\cap \partial {\mathrm{\Omega }}_1$, $\mu \ge 1$, we have

$$\parallel Ax+\lambda B_{n}x-\mu x\parallel \ge \parallel Ax-\mu x\parallel -\lambda \parallel B_{n}x\parallel \ge \gamma -\beta \lambda >0,$$

which guarantees that

$$i(A+\lambda B_n,P\cap {\mathrm{\Omega }}_1,P)=1,n\in \mathbb{N},$$

(2.18)

and for $x\in P\cap \partial {\mathrm{\Omega }}_2$, $\mu \ge 0$, we have

$$\parallel x-(Ax+\lambda B_{n}x)-\mu Kx\parallel \ge \parallel x-Ax-\mu Kx\parallel -\lambda \parallel B_{n}x\parallel \ge \gamma -\beta \lambda >0,$$

which guarantees that

$$i(A+\lambda B_n,P\cap {\mathrm{\Omega }}_2,P)=0.$$

Thus

$$i(A+\lambda B_n,P\cap ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1),P)=-1.$$

(2.19)

Now (2.18) and (2.19) guarantee that there exist $\{x_{n,1}\}\subseteq P\cap {\mathrm{\Omega }}_1$ and $\{x_{n,2}\}\subseteq P\cap ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)$ such that

$$x_{n,1}=A{x}_{n,1}+\lambda B_n{x}_{n,1},x_{n,2}=A{x}_{n,2}+\lambda B_n{x}_{n,2},\lambda \in (0,{\lambda }_{\ast }),n\in \mathbb{N}.$$

An argument similar to that in (2.11) shows that there exist $y_1\in P\cap {\mathrm{\Omega }}_1-\{\theta \}$ and $y_2\in P\cap ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)$ with

$$y_1=A{y}_1+\lambda B{y}_1,y_2=A{y}_2+\lambda B{y}_2,\lambda \in (0,{\lambda }_{\ast }).$$

The proof is complete. □

Corollary 2.2 Suppose that ${\mathrm{\Omega }}_1$, ${\mathrm{\Omega }}_2$ are bounded open sets and $\theta \in {\mathrm{\Omega }}_1\subseteq {\mathrm{\Omega }}_2$, $A:P\cap {\overline{\mathrm{\Omega }}}_2\to P$ is continuous and compact and $B:P-\{\theta \}\to P$ is singular continuous and compact. Assume that

(C₄)

$$\parallel Ax\parallel <\parallel x\parallel ,\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_1$$

or

$$Ax\ngeqq x,\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_1;$$

(C₅)

$$\parallel Ax\parallel >\parallel x\parallel ,\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_2$$

or $\mathrm{\exists }{u}_0\in P-\{\theta \}$ such that

$$x-Ax\ne \mu u_0,\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_2,\mu \ge 0$$

or

$$Ax\nleqq x,\mathrm{\forall }x\in P\cap \partial {\mathrm{\Omega }}_2.$$

Then there exists a ${\lambda }_{\ast }>0$ such that, for any $\lambda \in (0,{\lambda }_{\ast })$, there exist $x_{\lambda ,1}\in P\cap {\mathrm{\Omega }}_1-\{\theta \}$ and $x_{\lambda ,2}\in P\cap ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)$ with

$$x_{\lambda ,1}=A{x}_{\lambda ,1}+\lambda B{x}_{\lambda ,1}$$

and

$$x_{\lambda ,2}=A{x}_{\lambda ,2}+\lambda B{x}_{\lambda ,2}.$$

It is easy to see that (C₄) and (C₅) guarantee that (C₁)-(C₃) hold (see [8]).

3 Applications for singular systems of multi-point boundary value problems

In [9], Henderson and Luca considered the system of nonlinear second-order ordinary differential equations

$$\{\begin{array}{l}{u}^{″}(t)+f(t,v(t))=0,t\in (0,T),\\ v^{″}(t)+g(t,u(t))=0,t\in (0,T)\end{array}$$

(3.1)

with multi-point boundary conditions

$$\{\begin{array}{l}u(0)=0,u(T)={\sum }_{i=1}^{m-2}{b}_{i}u({\xi }_i),m\ge 3,\\ v(0)=0,v(T)={\sum }_{i=1}^{n-2}{c}_{i}v({\eta }_i),n\ge 3.\end{array}$$

(3.2)

The following conditions come from [9]:

(H1) $0<{\xi }_1<\cdots <{\xi }_{m-2}<T$, $0<{\eta }_1<\cdots <{\eta }_{n-2}<T$, $b_i>0$, $i=1,2,\dots ,m-2$, $c_i\ge 0$, $i=1,2,\dots ,n-2$, $d=T-{\sum }_{i=1}^{m-2}{b}_i{\xi }_i>0$, $e=T-{\sum }_{i=1}^{n-2}{c}_i{\eta }_i>0$, ${\sum }_{i=1}^{m-2}{b}_i{\xi }_i>0$, ${\sum }_{i=1}^{n-2}{c}_i{\eta }_i>0$,

(H2) we have the functions $f,g\in C([0,T]\times [0,+\mathrm{\infty }),[0,+\mathrm{\infty }))$ and $f(t,0)=0$, $g(t,0)=0$ for all $t\in [0,T]$,

(H3) there exists a positive constant $p\in (0,1]$ such that

• (1) $f_{\mathrm{\infty }}^i={lim}_{u\to +\mathrm{\infty }}inf{inf}_{t\in [0,T]}\frac{f(t,u)}{u^p}\in (0,+\mathrm{\infty }]$;

• (2) $g_{\mathrm{\infty }}^i={lim}_{u\to +\mathrm{\infty }}inf{inf}_{t\in [0,T]}\frac{g(t,u)}{u^{\frac{1}{p}}}=+\mathrm{\infty }$,

(H4) there exists a $r\in (0,+\mathrm{\infty })$ such that

• (1) $f_{\mathrm{\infty }}^s={lim}_{u\to +\mathrm{\infty }}sup{sup}_{t\in [0,T]}\frac{f(t,u)}{u^r}\in (0,+\mathrm{\infty }]$;

• (2) $g_{\mathrm{\infty }}^s={lim}_{u\to +\mathrm{\infty }}sup{sup}_{t\in [0,T]}\frac{g(t,u)}{u^{\frac{1}{r}}}=0$,

(H5)

• (1) $f_0^i={lim}_{u\to 0+}inf{inf}_{t\in [0,T]}\frac{f(t,u)}{u}\in (0,+\mathrm{\infty }]$;

• (2) $g_0^i={lim}_{u\to 0+}inf{inf}_{t\in [0,T]}\frac{g(t,u)}{u}=+\mathrm{\infty }$,

(H6) for each $t\in [0,T]$, $f(t,u)$ and $g(t,u)$ are nondecreasing with respect to u, and there exists a constant $N>0$ such that

$$f(t,m_0{\int }_0^{T}g(s,N)ds)<\frac{N}{m_0},\mathrm{\forall }t\in [0,T],$$

where $m_0=\frac{T^2}{4}max\{a_{1}T,{\tilde{a}}_1\}$ and $a_1$, ${\tilde{a}}_1$ are defined in [9].

Theorem 3.1 ([9])

Assume that (H1)-(H2) and (H4)-(H5) hold. Then Problem (3.1), (3.2) has at least one positive solution $(u(t),v(t))$, $t\in [0,T]$.

Theorem 3.2 ([9])

Assume that (H1)-(H3) and (H5)-(H6) hold. Then Problem (3.1), (3.2) has at least two positive solutions $(u_1(t),v_1(t))$, $(u_2(t),v_2(t))$, $t\in [0,T]$.

Here we consider

$$\{\begin{array}{l}{u}^{″}(t)+f(t,v(t))+\lambda u^{-\gamma }=0,t\in (0,T),\\ v^{″}(t)+g(t,u(t))=0,t\in (0,T)\end{array}$$

(3.3)

with multi-point boundary conditions (3.2), where $1>\gamma >0$.

Let $C[0,T]:=\{x:[0,T]\to R:x(t)\text{ is continuous on }[0,T]\}$ with norm

$$\parallel x\parallel =\underset{t\in [0,T]}{max}|x(t)|.$$

Obviously, $C[0,T]$ is a Banach space. Let

$$P:=\{x\in C[0,T]:x(t)\ge 0\text{ is concave and }\underset{t\in [{\theta }_0,T]}{inf}x(t)\ge \gamma \parallel x\parallel \},$$

where ${\theta }_0$ and γ are defined in Section 2 in [9].

For $u\in P$, define an operator

$$(Au)(t)={\int }_0^T{G}_1(t,s)f(s,{\int }_0^T{G}_2(s,\tau )g(\tau ,u(\tau ))d\tau )ds,t\in [0,T]$$

and for $u\in P-\{\theta \}$, define an operator

$$(Bu)(t)={\int }_0^T{G}_1(t,s)u^{-\gamma }(s)ds,t\in [0,T],$$

where $G_1(t,s)$ and $G_2(t,s)$ are defined in [9].

It is easy to see that $B:P-\{\theta \}\to P$ is a singular continuous and compact operator (see [9, 11]).

Theorem 3.3 Assume that (H1)-(H2) and (H4) hold. Then there exists a ${\lambda }^{\ast }>0$ such that Problem (3.3), (3.2) has at least one positive solution $(u(t),v(t))$, $t\in [0,T]$ for all $\lambda \in (0,{\lambda }^{\ast })$.

Proof Let $B_R$ be defined as that in Theorem 3.2 of [9]. From the proof in [9], it is easy to see that

$$\parallel Ax\parallel <\parallel x\parallel ,\mathrm{\forall }x\in P\cap \partial B_R.$$

Now Corollary 2.1 guarantees that there exists a ${\lambda }^{\ast }>0$ such that, for any $\lambda \in (0,{\lambda }^{\ast })$, there exist $u\in (P\cap B_R-\{\theta \})$ such that

$$u=Au+\lambda Bu.$$

Let

$$v(t)={\int }_0^T{G}_2(t,s)g(s,u(s))ds,t\in [0,T].$$

Then $(u(t),v(t))$ is a positive solution for (3.3), (3.2). The proof is complete. □

Theorem 3.4 Assume that (H1)-(H3) and (H6) hold. Then Problem (3.3), (3.2) has at least two positive solutions $(u_1(t),v_1(t))$, $(u_2(t),v_2(t))$, $t\in [0,T]$.

Proof Let $B_N$ and $B_L$ be defined as in Theorem 3.3 of [9]. From the proof in [9], it is easy to see that

$$\begin{array}{r}\parallel Ax\parallel <\parallel x\parallel ,\mathrm{\forall }x\in P\cap \partial B_N,\\ x\ne Ax+\lambda u_0,\mathrm{\forall }x\in P\cap \partial B_L,\lambda \ge 0,L>N.\end{array}$$

Now Corollary 2.2 guarantees that there exists a ${\lambda }^{\ast }>0$, for any $\lambda \in (0,{\lambda }^{\ast })$, such that there exist $u_1\in (P\cap B_N-\{\theta \})$ and $u_2\in P\cap (B_L-{\overline{B}}_N)$ with

$$u_1=A{u}_1+\lambda B{u}_1$$

and

$$u_2=A{u}_2+\lambda B{u}_2.$$

Let

$$v_1(t)={\int }_0^T{G}_2(t,s)g(s,u_1(s))ds,t\in [0,T]$$

and

$$v_2(t)={\int }_0^T{G}_2(t,s)g(s,u_2(s))ds,t\in [0,T].$$

Then $(u_1(t),v_1(t))$ and $(u_2(t),v_2(t))$ are two positive solutions for (3.3), (3.2). The proof is complete. □

Remark Note that f and g have no singularity at $u=0$ and $v=0$ in [9, 10], so Theorems 3.3 and 3.4 improve the results in [9, 10].