ABSTRACT

Using the method of lower and upper solutions, we study the following singular nonlinear three-point boundary value problems: $\{\begin{array}{l}-x^{″}(t)+K(t)x^{-q}(t)=\lambda x^p(t),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ),\hfill \end{array}$ , where $K\in C[0,1]$ , $0<a<1$ , $0<\eta <1$ and $\lambda$ is a positive parameter and present the existence, uniqueness, and the dependency on parameters of the positive solutions under various assumptions. Our result improves those in the previous literatures.

Keywords:

Three-Point Boundary Value Problem, Positive Solution, Lower and Upper Solutions, Eigenvalue and Eigenfunction

1. Introduction and Main Results

In this paper, we consider the three-point boundary value problem

$\{\begin{array}{l}-x^{″}(t)+K(t)x^{-q}(t)=\lambda x^p(t),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ),\hfill \end{array}$ (1.1)

where $K\in C[0,1]$ , $0<a<1$ , $0<\eta <1$ , and $\lambda$ is a positive parameter.

The m-point boundary value problem for linear second-order ordinary differential equations was initiated by Ilin and Moiseev [1] [2]. Since then, there are many results on the existence of general nonlinear multi-point boundary value problems, see [3] [4] [5] [6] and their references. For examples, in [6], Rynne studied the $m$-point boundary value problem

$\{\begin{array}{l}-u^{″}=f(u),on(0,1),u\in R\times X,\hfill \\ u(0)=0,u(1)={\displaystyle \underset{i=1}{\overset{m-2}{\sum }}{\alpha }_i}u({\eta }_i),\hfill \end{array}$

where $m\ge 3$ , ${\eta }_i\in (0,1)$ , ${\alpha }_i>0$ with ${\displaystyle \underset{i=1}{\overset{m-2}{\sum }}{\alpha }_i}<1$ and presented the existence of the sign changing solutions by Rabinowitz bifurcation theorem. Especially, Rynne ([7]) discussed the three-point boundary value problem

$\{\begin{array}{l}-u^{″}=f(u)+h,on(0,1),\hfill \\ u(0)=0,u(1)=\alpha u(\eta ),\hfill \end{array}$

and showed the solvability and non-solvability results from either the half-eigenvalue or the Fucik spectrum approach. As we known, the method of upper and lower solutions is very important for the study of the boundary value problems, see [8]-[18]. Therefore, establishing the method of upper and lower solutions for three-point boundary value problems is necessary and important.

In [19], when f is nondecreasing on x, Du and Zhao got the methods of upper and lower solutions of

$\{\begin{array}{l}-x^{″}(t)=f(t,x(t)),t\in (0,1),\hfill \\ x(0)=ax(\eta ),x(1)=0,\hfill \end{array}$

and used iterative techniques to study the existence of positive solutions. And in [3] when f is decreasing on u, Du and Zhao considered the existence and uniqueness of positive solutions of the problem

$\{\begin{array}{l}-u^{″}(t)=f(t,u(t)),t\in (0,1),\hfill \\ u(0)={\displaystyle \underset{i=1}{\overset{m-2}{\sum }}{\alpha }_i}u({\eta }_i),u(1)=0\hfill \end{array}$

by constructing lower and upper solutions. Wei ([15]) constructed the method of upper and lower solutions for three-point boundary value problems and gave the sufficient and necessary conditions for the existence of positive solutions of the problem

$\{\begin{array}{l}-x^{″}(t)=f(t,x(t)),t\in (0,1),\hfill \\ x(0)=ax(\eta ),x(1)=0.\hfill \end{array}$

On the other hand, singular boundary problems arise in the contexts of chemical heterogeneous catalysts, non-Newtonian fluids and also the theory of heat conduction in electrically conducting materials, see [20]-[25] for a detailed discussion. An interesting result comes from [25], in which, using method of upper and lower solutions, Shi and Yao discussed the following problem

$\{\begin{array}{l}-\Delta u+K(x)u^{-q}=\lambda u^p,x\in \Omega ,\hfill \\ u(x)>0,\forall x\in \Omega ,\hfill \\ u{|}_{\partial \Omega }=0,\hfill \end{array}$

where $K\in C^{2,\beta }(\bar{\Omega })$ , $p,q\in (0,1)$ and $\lambda$ is a positive parameter. Under various appropriate assumptions on $K(x)$ , Shi and Yao obtained the existence and uniqueness of classical solutions.

Motivated by above works, under various appropriate assumptions on p, q and $K(t)$ , we will obtain the existence and uniqueness of positive solution of problem (1.1) for $\lambda$ in different circumstances. In our proof, the upper and lower solutions theorem (see [16]) plays an important role in the paper.

Define

$K^{*}=\underset{{}_{t\in [0,1]}}{\max }K(t),K_{*}=\underset{t\in [0,1]}{\min }K(t).$

The main results of this paper are stated in the following theorems.

Theorem 1.1. When $K_{*}>0$ ,

1) If $0<p,q<1$ , there exists $\bar{\lambda }>0$ such that the problem (1.1) has at least one C[0,1] positive solution $x_{\lambda }(t)$ for $\lambda >\bar{\lambda }$.

2) For $\lambda >\bar{\lambda }$ , (1.1) has a maximal solution ${\bar{x}}_{\lambda }(t)$ and ${\bar{x}}_{\lambda }(t)$ is increasing with respect to $\lambda$.

Theorem 1.2. When $K^{*}<0$ ,

1) If $0<p<1,0<q$ , (1.1) has at least one C[0,1] positive solution for all $\lambda >0$.

2) If $0<p,q<1$ , (1.1) has an unique $C^1[0,1]$ positive solution $x_{\lambda }(t)$ for all $\lambda >0$.

3) $x_{\lambda }(t)$ in (2) is increasing with respect to $\lambda$.

Theorem 1.3. When $K_{*}<0<K^{*}$ ,

1) If $0<p,q<1$ , there exists a ${\lambda }_{*}>0$ such that the problem (1.1) has at least one C[0,1] posit- ive solution $x_{\lambda }(t)$ for $\lambda >{\lambda }_{*}$.

2) For $\lambda >{\lambda }_{*}$ , $x_{\lambda }(t)$ in (1) is increasing with respect to $\lambda$.

Remark 1.1: Note $K(t)>0$ in Theorem (1.1). This is different from the conditions in [3] [15] [19] because $K(t)<0$ in these references.

Remark 1.2: The unique result in Theorem 1.2 is different from that in [3] because we remove the monotonicity of nonlinearity f in x.

Remark 1.3: Note $K(t)$ is sigh-changing in Theorem 1.3. This is different from the conditions in [3] [15] [19] because $K(t)<0$ in these references and is different from conditions in [1] [2] [4] [5] [6] [7] [26] because f is continuous at $x=0$ in these references.

This paper is organised as follows. Some preliminary lemmas are stated and proved in Section 2. And Section 3 is devoted to prove the results.

2. Preliminaries

In this section, we first consider the following problem

$\{\begin{array}{l}-x^{″}(t)=f(t,x(t),x^{\prime }(t)),t\in (0,1),\hfill \\ x(0)=0,x(\eta )=ax(1),\hfill \end{array}$ (2.1)

where $\eta \in (0,1)$ , $0<a<1$ and $f\in [0,1]\times ℝ\times ℝ$.

Let $C^1[0,1]=\{x:[0,1]\to ℝ|x(t)$ is differential continuous on $[0,1]\}$ with norm

$\left|\right|x\left|\right|=\max \left\{\right|x{|}_{\infty },\left|x^{\prime }{|}_{\infty }\right\}$ ,

where $|x^{\prime }{|}_{\infty }=\underset{t\in [0,1]}{\max }|x(t)|$. Obviously, $C^1[0,1]$ is a Banach space. Now we give the definitions of lower and upper solutions for problem (2.1).

Definition 2.1. A function $\alpha (t)$ is called a lower solution to the problem (2.1), if $\alpha (t)\in C[0,1]\cap C^2(0,1)$ and satisfies

$\{\begin{array}{l}-{\alpha }^{″}(t)\le f(t,\alpha (t),{\alpha }^{\prime }(t)),t\in (0,1),\hfill \\ \alpha (0)\le 0,\alpha (1)\le a\alpha (\eta ).\hfill \end{array}$ (2.2)

Upper solution is defined by reversing the above inequality signs in problem (2.2).

If there exists a lower solution $\alpha (t)$ and an upper solution $\beta (t)$ to problem (2.1) such that $\alpha (t)\le \beta (t)$ , then $(\alpha (t),\beta (t))$ is called a couple of upper and lower solutions of problem (2.1).

Set $D_{\alpha }^{\beta }=\{(t,x)\in (0,1)\times {ℝ}^{+},\alpha (t)\le x\le \beta (t),t\in (0,1)\}.$

We list a lemma for the eigenvalues and eigenfunctions for the following linear problem

$\{\begin{array}{l}-x^{″}(t)=\lambda x(t),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ).\hfill \end{array}$ (2.3)

Lemma 2.1. (see [6]) The spectrum $\sigma (L)$ of problem (2.3) consists of a strictly increasing sequence ofeigenvalues ${\lambda }_k>0$ , $k=1,2,\cdots$ , with eigenfuctions ${\varphi }_k=\sin ({\lambda }_k^{\frac{1}{2}}t)$. In addition,

1) $\underset{k\to +\infty }{\lim }{\lambda }_k=+\infty$ ;

2) ${\varphi }_k(t)$ has exact $k-1$ simple zeros in $(0,1)$ , $k=2,3,\cdots$ and ${\varphi }_1$ is strictly positive on $(0,1)$.

Lemma 2.2. Suppose that $h\in L^1(0,1)$. Then, for each $\lambda >0$ , the problem

$\{\begin{array}{l}-x^{″}(t)+\lambda x=h(t),t\in (0,1),\hfill \\ x(0)=0,x(\eta )=\alpha x(1)\hfill \end{array}$ (2.4)

has an unique solution in C[0,1].

Proof. Assume that $v_1(t)$ and $v_2(t)$ satisfies that

$\{\begin{array}{l}-x^{″}(t)+\lambda x=h(t),t\in (0,1),\hfill \\ x(0)=0,x^{\prime }(0)=1\hfill \end{array}$

and

$\{\begin{array}{l}-x^{″}(t)+\lambda x=h(t),t\in (0,1),\hfill \\ x(1)=0,x^{\prime }(1)=-1\hfill \end{array}$

respectively. Define

$G(t,s)=\frac{1}{\omega }\{\begin{array}{l}{v}_2(t)v_1(s),0\le s\le t\le 1,\hfill \\ v_1(t)v_1(s),0\le t\le s\le 1,\hfill \end{array}$

and

$x(t)={\displaystyle {\int }_0^{1}G}(t,s)h(s)ds+\frac{e_1(t)}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G}(\eta ,s)h(s)ds,s\in [0,1].$

Then

$\begin{array}{c}-x\text{'}\text{'}(t)+\lambda x(t)=-\frac{1}{\omega }[{\displaystyle {\int }_0^t{v}_2(t)v_1(s)h(s)ds+}{\displaystyle {\int }_t^1{v}_1(t)v_2(s)h(s)ds}]\text{'}\text{'}\\ -\frac{e_1\text{'}\text{'}(t)}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}+\lambda x(t)\\ =-\frac{1}{\omega }[v_2\text{'}(t)v_1(t)-v_1\text{'}(t)v_2(t)]h(t)\\ -\frac{1}{\omega }[\lambda {\displaystyle \int \begin{array}{c}t\\ 0\end{array}{v}_2\left(t\right)v_1\left(s\right)h(s)ds+\lambda {\displaystyle {\int }_t^1{v}_1\left(t\right)v_2\left(s\right)h(s)ds}}]\\ -\frac{\lambda e_1(t)}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}+\lambda x(t)\\ =h(t)-\lambda \frac{1}{\omega }[{\displaystyle {\int }_0^t{v}_2(t)v_1(s)h(s)ds}+{\displaystyle {\int }_t^1{v}_1(t)v_2(s)h(s)ds}]\\ -\lambda \frac{e_1(t)}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}+\lambda x(t)\\ =h(t),\begin{array}{cc}& \end{array}t\in (0,1)\end{array}$

and

$\begin{array}{c}x(1)-\alpha x(\eta )={\displaystyle {\int }_0^{1}G(1,s)h(s)ds}+\frac{e_1(1)}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}\\ -\alpha [{\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}+\frac{e_1(\eta )}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}]\\ =\frac{e_1(1)}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}\\ -\alpha [{\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}+\frac{e_1(\eta )}{e_1(1)-\alpha e_1(\eta )}\alpha {\displaystyle {\int }_0^{1}G(\eta ,s)h(s)ds}]\\ =0.\end{array}$

Hence, $x(t)$ is a C[0,1] solution to problem(2.4). Since $\lambda >0$ , Lemma 2.1 guarantees that problem (2.4) has an unique C[0,1] solution. The proof is complete. $\square$

Theorem 2.1. Let $\alpha$ and $\beta \in C([0,1])\cap C^1(0,1)$ be lower and upper solutions of (2.1) such that $\alpha \le \beta$. Let $\bar{\psi }\in L^1[0,1]$ and $\bar{\varphi }:{ℝ}^{+}\to {ℝ}_0^{+}$ be a continuous function that satisfies

${\displaystyle {\int }_0^{\infty }\frac{1}{\bar{\varphi }(s)}ds}=+\infty .$ (2.5)

Suppose $f:D_{\alpha }^{\beta }\times ℝ\to ℝ$ is an $L^1$ -Carathéodory-function such that

$|f(t,x,v)|\le \bar{\psi }(t)\bar{\varphi }(|v|),\forall (t,x)\in D_{\alpha }^{\beta },v\in ℝ.$ (2.6)

Then the problem (2.1) has at least one solution $x\in C^1[0,1]$ such that for all $t\in [0,1]$ ,

$\alpha (t)\le x(t)\le \beta (t).$

Proof. The proof proceeds in five steps.

Step 1. We consider a new modified problem. From (2.5), there is an $R>0$ be large enough so that

${\displaystyle {\int }_0^R\frac{1}{\bar{\varphi }(s)}ds}>\left|\right|\psi |{|}_1.$ (2.7)

And (2.6) guarantees that there is an $\bar{N}(t)$ with $\bar{N}\in L^1[0,1]$ such that

$|f(t,x,v)|\le \bar{N}(t),\forall (t,x)\in D_{\alpha }^{\beta },\left|v\right|\le R.$ (2.8)

Define then

$\chi (t,x)=\{\begin{array}{l}\alpha (t),ifx<\alpha (t),\hfill \\ x,if\alpha (t)\le x\le \beta (t),\hfill \\ \beta (t),ifx>\beta (t)\hfill \end{array}$ (2.9)

and

$g(t,x,v)=\max \left\{\min \right\{f(t,\chi (t,x),v),\bar{N}(t)\},-\bar{N}(t)\}.$ (2.10)

Choose a $\lambda >0$ and consider the new boundary value problem

$\{\begin{array}{l}-x^{″}(t)+\lambda x=g(t,x(t),x^{\prime }(t))+\lambda \chi (t,x(t)),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ),\hfill \end{array}$ (2.11)

where $0<a<1$ , $0<\eta <1$.

Step 2. We discuss the existence of a $C^1[0,1]$ solution of (2.11).

Now Lemma 2.2 guarantees that for each $h\in L^1[0,1]$ , the linear problem

$\{\begin{array}{l}-x^{″}(t)+\lambda x=h,t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta )\hfill \end{array}$

has an unique C[0,1] solution

$v(t)={\displaystyle {\int }_0^{1}G}(t,s)h(s)ds+\frac{e_1(t)}{e_1(1)-a{e}_1(\eta )}a{\displaystyle {\int }_0^{1}G}(\eta ,s)h(s)ds,s\in [0,1].$

For $x\in C^1[0,1]$ , define

$(Fx)(t)=g(t,x(t),x^{\prime }(t))+\lambda \chi (t,x(t)),t\in [0,1]$

and

$(Tx)(t)={\displaystyle {\int }_0^{1}G}(t,s)(Fx)(s)ds+\frac{e_1(t)}{e_1(1)-a{e}_1(\eta )}a{\displaystyle {\int }_0^{1}G}(\eta ,s)(Fx)(s)ds,s\in [0,1].$

From (2.9) and (2.10), we have

$|g(t,x(t),x^{\prime }(t))+\lambda \chi (t,x(t))|\le \bar{N}(t)+\lambda \max \left\{\underset{t\in [0,1]}{\sup }\right|\alpha (t)|,\underset{t\in [0,1]}{\sup }|\beta (t)\left|\right\},$ which implies that the functions belonging to $\{(Tx)(t):x\in C^1[0,1]\}$ and $\{(Tx{)}^{\prime }(t):x\in C^1[0,1]\}$ are bounded and equicontinuous. The Arzela-Ascoli Theorem guarantees that $T{C}^1[0,1]$ is relatively compact. The proof of the continuity of T is standard. Using the Schauder’s fixed point theorem, we assert that T has at least one fixed point $x\in C^1[0,1]$.

Step 3. The solution x of (2.11) is such that $\alpha (t)\le x(t)\le \beta (t)$.

We prove that $x(t)\le \beta (t)$ for $t\in [0,1]$ only. In fact, suppose that there exist a $t_0\in [0,1)$ such that $x(t_0)>\beta (t_0)$. Since $x(0)=0\le \beta (0)$ , $t_0>0$. Let $w(t)=x(t)-\beta (t)$ , $t\in [0,1]$. Then $w(0)\le 0$ and $w(t_0)>0$.

Let $t^{\ast }=\sup \{t\mid w(s)>0,s\in [t_0,t]\}$ , $t_{\ast }=\inf \{t\mid w(s)>0,s\in [t,t_0]\}.$

It is obvious that $w(t)>0$ for all $t\in (t_{*},t^{*})$ , $w(t_{*})=0$ and $w(t^{*})\ge 0$. If $w(t^{*})=0$ , then there exists a $t^{\prime }\in (t_{*},t^{*})$ such that $w(t^{\prime })=\underset{t\in [t_{*},t^{*}]}{\max }w(t)$. If $w(t^{*})>0$ , obviously $t^{*}=1$ and $w(1)=x(1)-\beta (1)>0$. Since

$w(\eta )=x(\eta )-\beta (\eta )=\frac{1}{a}(x(1)-\beta (1))=\frac{1}{a}w(1)>w(1)$ , there exists $t^{\prime }\in (t_{*},t^{*})$

such that $w(t^{\prime })=\underset{t\in [t_{*},t^{*}]}{\max }w(t)$ also. Hence, $w^{\prime }(t^{\prime })=0$ (i.e., ${\beta }^{\prime }(t^{\prime })=x^{\prime }(t^{\prime })$ ) and $-w^{\prime \prime }(t^{\prime })\ge 0$. On the other hand, since

$\begin{array}{c}-w^{\prime \prime }(t^{\prime })={\beta }^{\prime \prime }(t^{\prime })-x^{\prime \prime }(t^{\prime })\\ \le -f(t^{\prime },\beta (t^{\prime }),\beta (t^{\prime }))+g(t^{\prime },x(t^{\prime }),x^{\prime }(t))+\lambda \chi (t^{\prime },x(t^{\prime }))-\lambda x(t^{\prime })\\ =-f(t^{\prime },\beta (t^{\prime }),{\beta }^{\prime }(t^{\prime }))+\max \left\{\min \right\{f(t^{\prime },\beta (t^{\prime }),{\beta }^{\prime }(t)),\bar{N}(t)\},-\bar{N}(t)\}\\ +\lambda \beta (t^{\prime })-\lambda x(t^{\prime })\\ =-f(t^{\prime },\beta (t^{\prime }),{\beta }^{\prime }(t^{\prime }))+f(t^{\prime },\beta (t^{\prime }),{\beta }^{\prime }(t^{\prime }))+\lambda \beta (t^{\prime })-\lambda x(t^{\prime })\\ =\lambda (\beta (t^{\prime })-x(t^{\prime }))<0.\end{array}$

This is a contradiction.

A similar argument holds to prove $x(t)\le \beta (t)$ for all $t\in [0,1]$.

Hence, from (2.10), one know that x satisfies that

$\{\begin{array}{l}-x^{″}(t)=g(t,x(t),x^{\prime }(t))=\max \left\{\min \right\{f(t,x(t),x^{\prime }(t)),\bar{N}(t)\},t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ).\hfill \end{array}$ (2.12)

Step 4. The solution x of (2.11) is such that $|x^{\prime }{|}_{\infty }\le R$.

On the contrary, suppose that there is a $t^{\prime }\in (0,1)$ such that $|x^{\prime }(t^{\prime })|>R$. Without loss of generality, we assume that $x^{\prime }(t^{\prime })>R$. Since $x(0)=0$ and $x(1)=ax(\eta )$ with $0<a<1$ , there is a $t_0\in (0,1)$ such that $x^{\prime }(t_0)=0$. Without loss of generality, we assume that $x^{\prime }(t)>0$ for all $(t^{\prime },t_0)$. Observe that, for all $(t,x)\in D_{\alpha }^{\beta }$ , $v\in ℝ$ ,

$\max \left\{\min \right\{f(t,x,v),\bar{N}(t)\},-\bar{N}(t)\}\le \bar{\psi }(t)\bar{\varphi }(|v|).$

Then, from (2.12), one has

$\begin{array}{c}{\displaystyle {\int }_0^R\frac{1}{\bar{\varphi }(s)}ds}=\left|{\displaystyle {\int }_{x^{\prime }(t_0)}^{x^{\prime }(t^{\prime })}\frac{1}{\bar{\varphi }(s)}ds}\right|=\left|{\displaystyle {\int }_{t^{\prime }}^{t_0}\frac{1}{\bar{\varphi }(x^{\prime }(t))}d{x}^{\prime }(t)}\right|\\ =\left|{\displaystyle {\int }_{t^{\prime }}^{t_0}\frac{x^{″}(t)}{\bar{\varphi }(x^{\prime }(t))}dt}\right|=\left|{\displaystyle {\int }_{t^{\prime }}^{t_0}\frac{g(t,x(t),x^{\prime }(t))}{\bar{\varphi }(x^{\prime }(t))}dt}\right|\\ ={\displaystyle {\int }_{t^{\prime }}^{t_0}\frac{\bar{\psi }(t)\bar{\varphi }(x^{\prime }(t))}{\bar{\varphi }(x^{\prime }(t))}dt}={\displaystyle {\int }_{t^{\prime }}^{t_0}\bar{\psi }}(t)dt=\left|\right|\bar{\psi }|{|}_1.\end{array}$

This contradicts to (2.7).

Hence $|f(t,x(t),x^{\prime }(t))|\le \bar{N}(t)$ , which together with $u\in [\alpha ,\beta ]$ guarantees that

$g(t,x(t),x^{\prime }(t))=f(t,x(t),x^{\prime }(t)),\forall t\in (0,1).$

Step 5. We claim that $x(t)$ satisfies (2.1).

Since $|x^{\prime }{|}_{\infty }\le R$ and $\alpha (t)\le x(t)\le \beta (t)$ , by (2.8), (2.10) and (2.12), we have

$\{\begin{array}{l}-x^{″}(t)=\max \left\{\min \right\{f(t,x(t),x^{\prime }(t)),\bar{N}(t)\}=f(t,x(t),x^{\prime }(t)),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ),\hfill \end{array}$

that is, $x(t)$ is a $C^1[0,1]$ solution of (2.1). The proof is complete. $\square$

Now we consider the following problem

$\{\begin{array}{l}-x^{″}(t)=f(t,x(t)),t\in (0,1),\hfill \\ x(0)=0,x(\eta )=ax(1),\hfill \end{array}$ (2.13)

where $\eta \in (0,1)$ , $0<a<1$ and $f\in [0,1]\times ℝ\times ℝ$.

Now we give the definitions of lower and upper solutions for problem (2.13).

Definition 2.2. (see [16]) A function $\alpha (t)$ is called a lower solution to the problem (2.13), if $\alpha (t)\in C[0,1]\cap C^2(0,1)$ and satisfies

$\{\begin{array}{l}-{\alpha }^{″}(t)\le f(t,\alpha (t)),t\in (0,1),\hfill \\ \alpha (0)\le 0,\alpha (1)\le a\alpha (\eta ).\hfill \end{array}$ (2.14)

Upper solution is defined by reversing the above inequality signs in problem (2.14).

By Theorem 2.1, we have following result.

Corollary 2.1. Suppose that there exists a lower solution $\alpha (t)$ and an upper solution $\beta (t)$ of problem (2.1) such that $\alpha (t)\le \beta (t)$ , $t\in [0,1]$ and there exists $F\in L^1[0,1]$ such that $|f(t,x)|\le F(t)$ for all $(t,x)\in D_{\alpha }^{\beta }$. Then the problem (2.13) has at least one C[0,1] solution $x(t)$ satisfies $\alpha (t)\le x(t)\le \beta (t)$ , $t\in [0,1]$.

Remark 2.1: This result can be found in [15]. So our theorem improves the works in the previous literature.

Lemma 2.3. Suppose that $f:(0,1)\times [0,+\infty )\to ℝ$ is a continuous functions such that $s^{-1}f(t,s)$ is strictly decreasing for $s>0$ at each $t\in (0,1)$. Let $w,v\in C[0,1]\cap C^2(0,1)$ satisfies:

1) $w^{″}+f(t,w)\le 0\le v^{″}+f(t,v)$ , $t\in (0,1)$ ;

2) $w,v>0$ , $t\in (0,1)$ and $w(0)\ge v(0)$ , $w(1)\ge aw(\eta )$ , $v(1)\le av(\eta )$ ;

3) $v^{″}\in L^1[0,1]$.

Then $w(t)\ge v(t)$ , $t\in [0,1]$.

Proof. By $v^{″}\in L^1(0,1)$ , we know that $v^{\prime }(0+)$ and $v^{\prime }(1-)$ exist and then $v\in C^1[0,1]$.

Suppose conversely $v(t)\cancel{\le }w(t)$ on [0,1]. We may assume without loss of generality that there exists $t_0\in (0,1)$ such that $v(t_0)-w(t_0)=\underset{0\le t\le 1}{\max }(v(t)-w(t))>0$. Let

$t_{*}=\inf \left\{t_1\right|0\le t_1<t_0,v(t)>w(t),t\in (t_1,t_0)\},$

$t^{*}=\sup \left\{t_2\right|t_0\le t_2<1,v(t)>w(t),t\in (t_0,t_2)\}.$

It’s obvious that $0\le t_{*}<t^{*}\le 1$ and $v(t_{*})=w(t_{*})$ , $v^{\prime }(t_{*}+)\ge D^{+}w(t_{*}+)$ , where $D^{+}$ denote Dini derivatives.

For $t^{*}\le 1$ , there are three cases.

1) $t^{*}<1$. Then $v(t^{*})=w(t^{*})$ , $v^{\prime }(t^{*})\le w^{\prime }(t^{*})$ , $v(t)>w(t)$ for all $t\in (t_{*},t^{*})$.

2) $t^{*}=1$ and $v(t^{*})=w(t^{*})$ , $v^{\prime }(t^{*}-)\le D_{-}w(t^{*}-)$ , $v(t)>w(t)$ for all $t\in (t_{*},t^{*})$ , where $D_{-}$ denotes Dini derivatives.

3) $t^{*}=1$ and $v(t^{*})>w(t^{*})$ , $v(t)>w(t)$ for all $t\in (t_{*},t^{*}]$. Since $v(1)-w(1)\le a(v(\eta )-w(\eta ))<v(\eta )-w(\eta )$ , then there is $t^{\prime }\in [\eta ,1]$ such that

$v(t^{\prime })-w(t^{\prime })>0,{(v(t^{\prime })-w(t^{\prime }))}^{\prime }<0.$

Combining above (1), (2) and (3), there is a $t^{\prime }>t_{*}$ such that

$v(t_{*})=w(t_{*}),v^{\prime }(t_{*}+)\ge D^{+}w(t_{*}+),v(t^{\prime })\ge w(t^{\prime }),v^{\prime }(t^{\prime }-)\le D_{-}w(t^{\prime }-),$

and

$v(t)>w(t),\forall t\in (t_{*},t^{\prime }).$

Let $y(t)=v^{\prime }(t)w(t)-w^{\prime }(t)v(t)$ , $t\in (t_{*},t^{\prime })$. Then we have

$\underset{t\to t_{*}+}{\lim }\inf y(t)\ge 0\ge \underset{t\to t^{\prime }-}{\lim }\sup y(t).$ (2.15)

On the other hand,

$\begin{array}{c}{y}^{\prime }(t)=w(t)v^{″}(t)-w^{″}(t)v(t)\\ =-w(t)f(t,v(t)+v(t)f(t,w(t))\\ =w(t)v(t)(\frac{f(t,w(t))}{w(t)}-\frac{f(t,v(t))}{v(t)})\\ \ge 0\end{array}$

for $t\in (t_{*},t^{\prime })$ and $y^{\prime }(t)\cancel{\equiv }0$ on $(\alpha ,\beta )$. This implies $y(t^{\prime })>y(t_{*})$. This contradicts (2.15), so $v(t)\le w(t)$. The proof is complete. $\square$

By analogous methods in [19], we establish the following maximal theorem, which can be used in the proof of the uniqueness of positive solutions.

Lemma 2.4. (maximal theorem) Suppose that $0<\eta <1$ , and $F=\{x\in C[0,1]\cap C^2(0,1),x(1)-ax(\eta )\ge 0,x(0)\ge 0\}$ , if $x(t)\in F$ such that $-x\text{'}\text{'}(t)\ge 0$ for $t\in (0,1)$ , then $x(t)\ge 0$ for $t\in [0,1]$.

3. Proofs of Main Theorems

In this section, we’ll always assume that $f(t,x)=\lambda x^p-K(t)x^{-q}$.

(A) The proof of Theorem 1.1.

Proof.

1) We consider the problem

$\{\begin{array}{l}-x^{″}(t)+K(t)x^{-q}(t)=\lambda x^p(t),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ),\hfill \end{array}$ (3.1.1)

where $0<q,p<1$ , $K\in C[0,1]$ , $K_{*}>0$ , $0<a<1$ , $0<\eta <1$ and $\lambda$ is a positive parameter.

In [19], when $f(t,x)$ is increasing in x, the problem

$\{\begin{array}{l}-x^{″}(t)=f(t,x),t\in (0,1),\hfill \\ x(0)=ax(\eta ),x(1)=0\hfill \end{array}$

has an unique $C^1[0,1]$ positive solution. From that, suppose that $x_{*}(t)$ is an unique $C^1[0,1]$ positive solution of the problem

$\{\begin{array}{l}-x^{″}(t)=x^p(t),t\in (0,1),\hfill \\ x(0)=0,x(1)=ax(\eta ),\hfill \end{array}$ (3.1.2)

where $0<a<1$ , $0<\eta <1$.

Set $\beta (t)={\lambda }^{\frac{1}{1-p}}{x}_{*}(t)$. Then

$\begin{array}{c}-{\beta }^{″}(t)+K(t){\beta }^{-q}(t)={\lambda }^{\frac{1}{1-p}}{x}_{*}(t)+K(t){\lambda }^{\frac{-q}{1-p}}{x}_{*}^{-q}(t)\\ >{\lambda }^{\frac{1}{1-p}}{x}_{*}(t)+K_{*}{\lambda }^{\frac{-q}{1-p}}{x}_{*}^{-q}(t)\\ >{\lambda }^{\frac{1}{1-p}}{x}_{*}^p(t),\end{array}$

$\lambda {\beta }^p(t)={\lambda }^{\frac{1}{1-p}}{x}_{*}^p(t).$

Thus $-{\beta }^{″}(t)+K(t){\beta }^{-q}(t)>\lambda {\beta }^p(t).$ Combining it with (3.1.2) we obtain

$\{\begin{array}{l}-{\beta }^{″}(t)+K(t){\beta }^{-q}(t)>\lambda {\beta }^p(t),t\in (0,1),\hfill \\ \beta (0)=0,\beta (1)=a\beta (\eta ).\hfill \end{array}$

Consequently, $\beta (t)$ is a upper solution of (3.1.1).

Set $\alpha (t)=M{\phi }_1^{\frac{2}{1+q}}$ , where M is a positive constant and ${\phi }_1$ is the first eigenfunction. Then

$\begin{array}{c}-\alpha \text{'}\text{'}(t)+K(t){\alpha }^{-q}(t)=-\frac{2M}{1+q}{\phi }_1{}^{{}_{{}^{\frac{1-q}{1+q}}}}(t){\phi }_1\text{'}\text{'}(t)+\frac{K(t)}{M^q{\phi }_1{}^{{}_{{}^{\frac{2q}{1+q}}}}}-\frac{2(1-q)M|{\phi }_1\text{'}{|}^2}{{(1+q)}^2{\phi }_1{}^{\frac{2q}{1+q}}}\\ =\frac{2\lambda M}{1+q}{\phi }_1{}^{{}_{{}^{\frac{2}{1+q}}}}+\frac{K(t)}{M^q}{\phi }_1{}^{{}_{{}^{\frac{2q}{1+q}}}}-\frac{2(1-q)M|{\phi }_1\text{'}{|}^2}{{(1+q)}^2{\phi }_1{}^{\frac{2q}{1+q}}}\\ <\frac{2\lambda M}{{\phi }_1{}^{{}_{{}^{\frac{2}{1+q}}}}}+\frac{K^{*}}{M^q}{\phi }_1{}^{{}_{{}^{\frac{2q}{1+q}}}}-\frac{2(1-q)M|{\phi }_1\text{'}{|}^2}{{(1+q)}^2{\phi }_1{}^{\frac{2q}{1+q}}}.\end{array}$

By Lemma 2.1 we have ${\phi }_1(t)=\sin (\sqrt{{\lambda }_1}t)$ , ${\phi }_1(t)=\sqrt{{\lambda }_1}\cos (\sqrt{{\lambda }_1}t)$. Thus there exists ${\delta }_0>0$ and $b\in (0,1)$ such that

$|{\phi }_1{}^{\text{'}}(t)|=|\sqrt{{\lambda }_1}\cos (\sqrt{{\lambda }_1}t)|>{\delta }_0,t\in [0,b),$

$|{\phi }_1(t)|=|\sin (\sqrt{{\lambda }_1}t)|>{\delta }_0,t\in [b,1].$

a) On $[0,b)$ , choosing $M\ge M_1={[\frac{{(1+q)}^2{K}^{*}}{2(1-q){\delta }_0^2}]}^{\frac{1}{1+q}}$ , then we have

$\frac{K^{*}}{M^q{\phi }_1{}^{\frac{2q}{1+q}}}\le \frac{{\lambda }_{1}M}{1+q}{\phi }_1{}^{\frac{2}{1+q}}$.

b) On $[b,1]$ , choosing $M\ge M_2={[\frac{{(1+q)}^2{K}^{*}}{2(1-q){\delta }_0^2}]}^{\frac{1}{1+q}}$ , then we have

$\frac{K^{*}}{M^q{\phi }_1^{\frac{2q}{1+q}}}\le \frac{{\lambda }_{1}M}{1+q}{\phi }_1^{\frac{2}{1+q}}.$

Fixing $M=\max \{M_1,M_2\}$ , then

$-{\alpha }^{″}(t)+K(t){\alpha }^{-q}(t)\le \frac{3{\lambda }_{1}M}{1+q}{\phi }_1^{\frac{2}{1+q}}$

and

$\lambda {\alpha }^p(t)=\lambda M^p{\phi }_1^{\frac{2q}{1+q}}.$