ABSTRACT

In this paper, we study the positive radial solutions for elliptic systems to the nonlinear BVP: $\{\begin{array}{l}-\Delta u=\lambda k_1\left(\left|x\right|\right)f_1\left(u,v\right)\text{on}\Omega ,\hfill \\ -\Delta v=\lambda k_2\left(\left|x\right|\right)f_2\left(u,v\right)\text{on}\Omega ,\hfill \\ u\left(x\right)=v\left(x\right)=0\text{on}\left|x\right|\to \infty ,\hfill \\ \frac{\partial u}{\partial \eta }+{\tilde{c}}_1\left(u\right)u=0\text{on}\left|x\right|=r_0,\hfill \\ \frac{\partial v}{\partial \eta }+{\tilde{c}}_2\left(v\right)v=0,\text{on}\left|x\right|=r_0.\hfill \end{array}$, where $\Delta u=div\left(\nabla u\right)$ and $\Delta v=div\left(\nabla v\right)$ are the Laplacian of u, $\lambda$ is a positive parameter, $\Omega =\left\{x\in {ℝ}^n:N>2,\left|x\right|>r_0,r_0>0\right\}$, let $i=\left[1,2\right]$ then $K_i\mathrm{<mo>\; :\; </mo>}\left[r_0\mathrm{,}\infty \right)\to \left(\mathrm{0,}\infty \right)$ is a continuous function such that ${\lim }_{r\to \infty }{k}_i\left(r\right)=0$ and $\frac{\partial }{\partial \eta }$ is The external natural derivative, and ${\tilde{c}}_i\mathrm{<mo>\; :\; </mo>}\left[\mathrm{0,}\infty \right)\to \left(\mathrm{0,}\infty \right)$ is a continuous function. We discuss existence and multiplicity results for classes of f with a) $f_i>0$, b) $f_i<0$, and c) $f_i=0$. We base our presence and multiple outcomes via the Sub-solutions method. We also discuss some unique findings.

Keywords:

Elliptic System, Positive Radial Solution, Exterior Domains, Fixed Point Index

1. Introduction

In reaction diffusion processes, steady states define the long term dynamics. Here we consider a steady state reaction diffusion equation on an exterior domain with a nonlinear boundary condition on the interior boundary. Namely, we study positive radial solutions to:

$\{\begin{array}{l}-\Delta u=\lambda k_1\left(\left|x\right|\right)f_1\left(u,v\right)\text{on}\Omega ,\hfill \\ -\Delta v=\lambda k_2\left(\left|x\right|\right)f_2\left(u,v\right)\text{on}\Omega ,\hfill \\ u\left(x\right)=v\left(x\right)=0\text{on}\left|x\right|\to \infty ,\hfill \\ \frac{\partial u}{\partial \eta }+{\tilde{c}}_1\left(u\right)u=0\text{on}\left|x\right|=r_0,\hfill \\ \frac{\partial v}{\partial \eta }+{\tilde{c}}_2\left(v\right)v=0,\text{on}\left|x\right|=r_0.\hfill \end{array}$ (1.1)

where $\Delta u=div\left(\nabla u\right)$ and $\Delta v=div\left(\nabla v\right)$ are the Laplacian of u, $\lambda$ is a positive parameter, $\Omega =\left\{x\in {ℝ}^n:N>2,\left|x\right|>r_0,r_0>0\right\}$, let $i=\left[1,2\right]$ then $K_i\mathrm{<mo>\; :\; </mo>}\left[r_0\mathrm{,}\infty \right)\to \left(\mathrm{0,}\infty \right)$ is a continuous function such that ${\lim }_{r\to \infty }{k}_i\left(r\right)=0$ and

$\frac{\partial }{\partial \eta }$ is the outward normal derivative, and ${\tilde{c}}_i\mathrm{<mo>\; :\; </mo>}\left[\mathrm{0,}\infty \right)\to \left(\mathrm{0,}\infty \right)$ is a is an non

decreasing (increasing) function. Here the reaction term $f_i\mathrm{<mo>\; :\; </mo>}\left[\mathrm{0,}\infty \right)\times \left[\mathrm{0,}\infty \right)\to R$ is a $C^1$ function. The case when $f_i<0$ (see [1] [2], that the study of positive solutions to such problems is considerably more challenging than in the case $f_i>0$ (positone problems). For a rich history on semipositone problems with Dirichlet boundary conditions on bounded domains, (see [3] - [8], and on domains exterior to a ball, see [9] [10] [11]. Such nonlinear boundary conditions occur very naturally in applications see [12] for a detailed description in a model arising in combustion theory. Recently, the existence of a radial positive solution for (1.1) when $\lambda \gg 1$ has been established in [13], via the method of subsuper solutions. Here we discuss the uniqueness of this radial solution when some additional assumptions hold. In [14], the authors study such a uniqueness result for the case of Dirichlet boundary condition on $\left|x\right|=r_0$. Our focus in this paper is to consider the uniqueness result for semipositone problem when a class of nonlinear boundary condition is satisfied at $\left|x\right|=r_0$. The fact that we have no longer a fixed value of u on $\left|x\right|=r_0$ results in quite a challenge in extending the results in [15].

Namely, we need to establish a detailed behavior of u at $\left|x\right|=r_0$ to achieve our goal. Instead of working directly with (1), we note that the change of

variables $r=\left|x\right|$ and $s={\left(\frac{r}{r_0}\right)}^{2-N}$ transforms (1) into the following boundary value problem:

$\{\begin{array}{l}-u^{″}\left(t\right)=\lambda {\tilde{a}}_1\left(t\right)f_1\left(u\left(t\right),v\left(t\right)\right)t\in \left[0,1\right],\hfill \\ -v^{″}\left(t\right)=\lambda {\tilde{a}}_2\left(t\right)f_2\left(u\left(t\right),v\left(t\right)\right)t\in \left[0,1\right],\hfill \\ \frac{N-2}{r_0}{u}^{\prime }+{\tilde{c}}_1\left(u\left(1\right)\right)u\left(1\right)=0,\hfill \\ \frac{N-2}{r_0}{v}^{\prime }+{\tilde{c}}_2\left(v\left(1\right)\right)v\left(1\right)=0,\hfill \\ u\left(0\right)=v\left(0\right)=0.\hfill \end{array}$ (1.2)

where ${\tilde{a}}_i=\frac{r_0^2}{{\left(2-N\right)}^2}{t}^{\frac{-2\left(N-1\right)}{N-2}}{k}_i\left(r_0{t}^{\frac{1}{2-N}}\right)$. We will only assume $k_i\le \frac{1}{r^{N+\mu }}$ for $r\gg 1$ and for some $\mu \in \left(\mathrm{0,}N-2\right)$. then ${\tilde{a}}_i\in \left(\left(\mathrm{0,1}\right]\mathrm{,}\left(\mathrm{0,}\infty \right)\right)$ could be singular at 0.if $\mu \ge N-2$,${\tilde{a}}_i$ will be nonsingular at 0 and it will be an easier case to study. Note that ${\tilde{a}}_i={\inf }_{t\in \left(0,1\right]}{\tilde{a}}_i\left(t\right)>0$ and there exists a constant $\tilde{d}>0$ such

that ${\tilde{a}}_i\le \frac{\tilde{d}}{t^{\alpha }}$ for all $t\in \left(\mathrm{0,1}\right]$ where $\tilde{\alpha }=\frac{\left(N-2\right)-\mu }{N-2}$. Motivated by the above discussion, in this paper, we will study positive solutions in $C^2\left(\mathrm{0,1}\right)\cap C^1\left[\mathrm{0,1}\right]$ to the following boundary value problems:

$\{\begin{array}{l}-u^{″}\left(t\right)=\lambda a_1\left(t\right)f_1\left(u\left(t\right),v\left(t\right)\right)t\in \left[0,1\right],\hfill \\ -v^{″}\left(t\right)=\lambda a_2\left(t\right)f_2\left(u\left(t\right),v\left(t\right)\right)t\in \left[0,1\right],\hfill \\ u^{\prime }\left(1\right)+c_1\left(u\left(1\right)\right)u\left(1\right)=0,\hfill \\ v^{\prime }\left(1\right)+c_2\left(v\left(1\right)\right)v\left(1\right)=0,\hfill \\ u\left(0\right)=v\left(0\right)=0.\hfill \end{array}$ (1.3)

where $c_i\mathrm{<mo>\; :\; </mo>}\left[\mathrm{0,}\infty \right)\to \left(\mathrm{0,}\infty \right)$ is a continuous function and $a_i\in C\left(\left(\mathrm{0,1}\right]\mathrm{,}\left(\mathrm{0,}\infty \right)\right)$ is such that:

(H1) $a_i={\inf }_{t\in \left(0,1\right]}{a}_i\left(t\right)>0$ ;

(H2) there exists a constant $d>0$ such that $a_i\left(t\right)=\frac{d}{t^{\alpha }}$ for all $t\in \left(\mathrm{0,}\epsilon \right]$ where $a\in \left(\mathrm{0,1}\right)$ and $\epsilon \approx 0$

(H3) $a_i$ is decreasing. We consider various $C^1$ classes of the reaction term $f_i\mathrm{<mo>\; :\; </mo>}\left[\mathrm{0,}\infty \right)\times \left[\mathrm{0,}\infty \right)\to R$ satisfying the following:

(F1) $f_i<0$ and ${\lim }_{s\to \infty }\frac{f_i\left(s\right)}{s}=0$ ; $i=1,2$

(F2) $f_i$ is increasing and ${\lim }_{s\to \infty }{f}_i\left(s\right)=\infty$ ; $i=1,2$

(F3) $f_i$ is concave on $\left[\mathrm{0,}\infty \right)$. $i=1,2$

Theorem 1.1. Assume (H1) - (H3) and (F1) - (F3). Then (1.3) has a unique positive solution for all $\lambda$ sufficiently large.

In Section two we will establish important a priori estimates. We will first recall some important results from [8] where the authors studied the case of Dirichlet boundary condition, or equivalently (1.3) with the boundary condition $t=1$ replaced by $u\left(1\right)=v\left(1\right)=0$. These results do not depend on the boundary condition at $t=1$ and hence it is also true for solutions of (1.3). In view of the readers convenience we include the proofs of these results. In Section three, we prove Theorem 1.1.

2. Advance Estimates

Let $F\left(s\right)={\displaystyle {\int }_0^s}{f}_i\left(t\right)\text{d}s$. Note that there exist unique positive numbers $\beta \mathrm{,}\theta$ such that $f_i\left(\beta \right)=0$ and $F\left(\theta \right)=0$ and $\beta <\theta$

Theorem 2.1. (See [8].) Let $u\mathrm{,}v$ are a positive solution of (1.3). Then u and v has only one interior maximum in $\left(\mathrm{0,1}\right)$, say at $t_m$, depending on $\lambda$, and $u\left(t_m\right)>\theta$,$v\left(t_m\right)>\theta$.

Proof. Let

$\{\begin{array}{l}{E}_1\left(t\right)=\lambda F\left(u\left(t\right)\right)a_1\left(t\right)+\frac{{\left|u^{\prime }\left(t\right)\right|}^2}{2},t\in \left(0,1\right),\hfill \\ E_2\left(t\right)=\lambda F\left(v\left(t\right)\right)a_2\left(t\right)+\frac{{\left|v^{\prime }\left(t\right)\right|}^2}{2},t\in \left(0,1\right)\hfill \end{array}$ (1.4)

then

$\{\begin{array}{l}{E^{\prime }}_1\left(t\right)=\lambda F\left(u\left(t\right)\right){a^{\prime }}_1\left(t\right),t\in \left(0,1\right),\hfill \\ {E^{\prime }}_2\left(t\right)=\lambda F\left(v\left(t\right)\right){a^{\prime }}_2\left(t\right),t\in \left(0,1\right)\hfill \end{array}$ (1.5)

Note that by (H3), ${a^{\prime }}_1\left(t\right)<0$ and ${a^{\prime }}_2\left(t\right)<0$ for all $t\in \left(\mathrm{0,1}\right]$. Hence, $E_1\left(t\right)$ and $E_2\left(t\right)$ are increases when $u\left(t\right)<\theta$,$v\left(t\right)<\theta$ and decreases when $u\left(t\right)>\theta$,$v\left(t\right)>\theta$.

Let $t_m\in \left(\mathrm{0,1}\right)$ be the first point at which u has a local maximum and assume that $u\left(t\right)\le \theta$ and $v\left(t\right)\le \theta$ for all $t\in \left[\mathrm{0,}{t}_m\right]$. Then $E_1\left(t\right)$ and $E_2\left(t\right)$ are increases in $\left[\mathrm{0,}{t}_m\right]$. Now integrating (1.3) from t to $t_m$, for $t<t_m$

$\{\begin{array}{l}{u}^{\prime }\left(t\right)={\displaystyle {\int }_t^{t_m}}\lambda a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\le \lambda f_1\left(\theta \right){\displaystyle {\int }_t^{t_m}}\frac{c_1}{s^{\alpha }}\text{d}s\le \lambda \frac{c_1{f}_1\left(\theta \right)}{1-\alpha }\\ v^{\prime }\left(t\right)={\displaystyle {\int }_t^{t_m}}\lambda a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\le \lambda f_2\left(\theta \right){\displaystyle {\int }_t^{t_m}}\frac{c_2}{s^{\alpha }}\text{d}s\le \lambda \frac{c_2{f}_2\left(\theta \right)}{1-\alpha }\end{array}$ (1.6)

where $c_i>d$ are such that $a_i\left(t\right)\le \frac{c_i}{t^{\alpha }}$ for all $t\in \left(\mathrm{0,1}\right)$ using (H2). Integrating again (1.6) from 0 to t, $t\le t_m$

$\{\begin{array}{l}u\left(t\right)\le {\displaystyle {\int }_0^t}\lambda \frac{c_1{f}_1\left(\theta \right)}{1-\alpha }\text{d}s,C_0=\frac{c_1{f}_1\left(\theta \right)}{1-\alpha }\hfill \\ V\left(t\right)\le {\displaystyle {\int }_0^t}\lambda \frac{c_2{f}_2\left(\theta \right)}{1-\alpha }\text{d}s,C_0=\frac{c_2{f}_2\left(\theta \right)}{1-\alpha }\hfill \end{array}$ (1.7)

Since $f_i$ are continuous, there exists $K_0>0$ such that $\left|F\left(u\right)\right|\le K_{0}u$ and $\left|F\left(v\right)\right|\le K_{0}v$ for all $\left(u\mathrm{,}v\right)\in \left[\mathrm{0,}\theta \right]$. Hence

$\underset{t\to 0^{+}}{\lim }\lambda \left|F\left(u\left(t\right)\right)\right|a_1\left(t\right)\le \underset{t\to 0^{+}}{\lim }\lambda K_{0}u\left(t\right)a_1\left(t\right)\le \underset{t\to 0^{+}}{\lim }{\lambda }^2{K}_0{C}_0{c}_{1}d{t}^{1-\alpha }=0$

$\underset{t\to 0^{+}}{\lim }\lambda \left|F\left(v\left(t\right)\right)\right|a_2\left(t\right)\le \underset{t\to 0^{+}}{\lim }\lambda K_{0}v\left(t\right)a_2\left(t\right)\le \underset{t\to 0^{+}}{\lim }{\lambda }^2{K}_0{C}_0{c}_{2}d{t}^{1-\alpha }=0$

Hence ${\lim }_{t\to 0^{+}}{E}_i\left(t\right)\ge 0$. Since $E_i\left(t\right)$ increases on $\left[\mathrm{0,}{t}_m\right]$,$E_1\left(t_m\right)=\lambda F\left(u\left(t_m\right)\right)a_1\left(t_m\right)>0$ and $E_2\left(t_m\right)=\lambda F\left(v\left(t_m\right)\right)a_2\left(t_m\right)>0$ and which is a contra-diction if $u\left(t_m\right)\le \theta$ and $v\left(t_m\right)\le \theta$. Suppose that u and v has two interior maxima. Then there exists $\tilde{t}\in \left(\tilde{t}\mathrm{,1}\right)$ such that $u^{\prime }\left(\tilde{t}\right)=0$,$v^{\prime }\left(\tilde{t}\right)=0$ and $u^{″}\left(\tilde{t}\right)\ge 0$,$v^{″}\left(\tilde{t}\right)\ge 0$. Since $u^{″}\left(\tilde{t}\right)=\lambda a_1\left(\tilde{t}\right)f_1\left(u\left(\tilde{t}\right)\right)\ge 0$, and

$v^{″}\left(\tilde{t}\right)=\lambda a_2\left(\tilde{t}\right)f_2\left(v\left(\tilde{t}\right)\right)\ge 0$ we have $f_1\left(u\left(\tilde{t}\right)\right)\le 0$ and $f_2\left(v\left(\tilde{t}\right)\right)\le 0$ which implies that $u\left(\tilde{t}\right)\le \beta$ and $u\left(\tilde{t}\right)\le \beta$. Thus $E_i\left(\tilde{t}\right)<0$. Let $t_{\theta }\left(t_m\mathrm{,}\tilde{t}\right)$ such that

$u\left(t_{\theta }\right)=\theta$ and $v\left(t_{\theta }\right)=\theta$. Then $E_1\left(t_{\theta }\right)=\frac{{\left|u^{\prime }\left(t_{\theta }\right)\right|}^2}{2}\ge 0$,$E_2\left(t_{\theta }\right)=\frac{{\left|v^{\prime }\left(t_{\theta }\right)\right|}^2}{2}\ge 0$

and $E_i$ increases in $\left(t_{\theta }\mathrm{,}\tilde{t}\right)$ since $u\left(t\right)<\theta$ and since $v\left(t\right)<\theta$ in $\left(t_{\theta }\mathrm{,}\tilde{t}\right)$. Hence $E_i\left(\tilde{t}\right)>0$, which is a contradiction. Therefore, we have only one interior maximum and that maximum value is larger than $\theta$.

Theorem 2.2. (See [8].) Let u and v be a positive solution of (1.3) and let $t_{\beta }\in \left(\mathrm{0,}{t}_m\right)$ such that $u\left(t_{\beta }\right)=\beta$ and $v\left(t_{\beta }\right)=\beta$. Then $t_{\beta }\le ○\left({\lambda }^{-\frac{1}{2}}\right)$ as $\lambda \to \infty$.

Proof. Let $t_{\frac{\beta }{2}}\in \left(\mathrm{0,}{t}_{\beta }\right)$ be the point such that $u\left(t_{\frac{\beta }{2}}\right)=\frac{\beta }{2}$ and $v\left(t_{\frac{\beta }{2}}\right)=\frac{\beta }{2}$ from Integrating (1.3) from 0 to t for some $t<t_{\frac{\beta }{2}}$

$u^{\prime }\left(t\right)=u^{\prime }\left(0\right)-\lambda {\displaystyle {\int }_0^t}{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\ge \lambda \underset{\_}{a_1}\left(-f_1\left(\frac{\beta }{2}\right)\right)t$

$v^{\prime }\left(t\right)=v^{\prime }\left(0\right)-\lambda {\displaystyle {\int }_0^t}{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\ge \lambda \underset{\_}{a_2}\left(-f_2\left(\frac{\beta }{2}\right)\right)t,$

Integrating the above again from 0 to $t_{\beta }$

$t_{\frac{\beta }{2}}\le {\tilde{c}}_1{\lambda }^{-\frac{1}{2}},t_{\frac{\beta }{2}}\le {\tilde{c}}_2{\lambda }^{-\frac{1}{2}}\text{where}\{\begin{array}{l}{\tilde{c}}_1=\left(\frac{-\beta }{\underset{\_}{a_1}{\left(-f_1\left(\frac{\beta }{2}\right)\right)}^{\frac{1}{2}}}\right)>0\hfill \\ {\tilde{c}}_2=\left(\frac{-\beta }{\underset{\_}{a_2}{\left(-f_2\left(\frac{\beta }{2}\right)\right)}^{\frac{1}{2}}}\right)>0\hfill \end{array}$ (1.8)

By the mean value theorem, there exists a $\tilde{t}\in \left[\mathrm{0,}{t}_{\frac{\beta }{2}}\right]$ such that $u\left(t_{\frac{\beta }{2}}\right)-u\left(0\right)=u^{\prime }\left(\tilde{t}\right)t_{\frac{\beta }{2}}$,$v\left(t_{\frac{\beta }{2}}\right)-v\left(0\right)=v^{\prime }\left(\tilde{t}\right)t_{\frac{\beta }{2}}$ and by (2.5) $u^{\prime }\left(\tilde{t}\right)\ge \frac{\beta }{2{\tilde{c}}_1}{\lambda }^{\frac{1}{2}}$ and $v^{\prime }\left(\tilde{t}\right)\ge \frac{\beta }{2{\tilde{c}}_2}{\lambda }^{\frac{1}{2}}$. Since $u^{\prime }$ and $v^{\prime }$ are increases in $\left[\mathrm{0,}{t}_{\beta }\right]$,

$\{\begin{array}{l}{u}^{\prime }\left(t\right)\ge \frac{\beta }{2{\tilde{c}}_1}{\lambda }^{\frac{1}{2}}\mathrm{,}t\in \left[t_{\frac{\beta }{2}}\mathrm{,}{t}_{\beta }\right]\hfill \\ v^{\prime }\left(t\right)\ge \frac{\beta }{2{\tilde{c}}_2}{\lambda }^{\frac{1}{2}}\mathrm{,}t\in \left[t_{\frac{\beta }{2}}\mathrm{,}{t}_{\beta }\right]\hfill \end{array}$ (1.9)

Integrating (2.6) from $t_{\frac{\beta }{2}}$ to $t_{\beta }$, we have that $\left(t_{\beta }-t_{\frac{\beta }{2}}\right)\le {\tilde{c}}_1{\lambda }^{-\frac{1}{2}}$ and $\left(t_{\beta }-t_{\frac{\beta }{2}}\right)\le {\tilde{c}}_1{\lambda }^{-\frac{1}{2}}$. This implies $t_{\beta }\le ○\left({\lambda }^{-\frac{1}{2}}\right)$ by (2.5). $\square$

Lemma 2.3. Let u and v be a positive solution of (1.3). Then $u\left(1\right)\to \infty$ and $v\left(1\right)\to \infty$ as $\lambda \to \infty$

Proof. We first claim that $u\left(1\right)\ge \frac{\beta +\theta }{2}$ and $v\left(1\right)\ge \frac{\beta +\theta }{2}$ for $\lambda \gg 1$. Assume that $u\left(1\right)<\frac{\beta +\theta }{2}$ and $v\left(1\right)<\frac{\beta +\theta }{2}$ Then there exists a ${\tilde{t}}_{\theta }\in \left(t_m\mathrm{,1}\right)$

such that $u\left({\tilde{t}}_{\theta }\right)=\theta$ and $v\left({\tilde{t}}_{\theta }\right)=\theta$ where $t_m$ is the point at which $u\mathrm{,}v$ achieves are maximum, and $u\left(t_m\right)>\theta$,$v\left(t_m\right)>\theta$ by Lemma 2.1.

From (2.1) and (2.2), $E_1\left({\tilde{t}}_{\theta }\right)=\frac{\left|u^{\prime }\left({\tilde{t}}_{\theta }\right)\right|}{2}>0$,$E_2\left({\tilde{t}}_{\theta }\right)=\frac{\left|v^{\prime }\left({\tilde{t}}_{\theta }\right)\right|}{2}>0$ and $E_i\left(t\right)\ge 0$ on $\left[{\tilde{t}}_{\theta }\mathrm{,1}\right]$ since $u\left(t\right)\le \theta$ and $v\left(t\right)\le \theta$ in $\left[{\tilde{t}}_{\theta }\mathrm{,1}\right]$. Hence we obtain that

$E_1\left(1\right)=\lambda F\left(u\left(1\right)\right)a_1\left(1\right)+\frac{\left|u^{\prime }\left(1\right)\right|}{2}>0$,

$E_2\left(1\right)=\lambda F\left(v\left(1\right)\right)a_2\left(1\right)+\frac{\left|v^{\prime }\left(1\right)\right|}{2}>0$

and from (1.3), we have

$\{\begin{array}{l}{c}_1\left(u\left(1\right)\right)u\left(1\right)=-u^{\prime }\left(1\right)\ge \sqrt{-2\lambda F\left(u\left(1\right)\right)a_1\left(1\right)},\\ c_2\left(v\left(1\right)\right)v\left(1\right)=-v^{\prime }\left(1\right)\ge \sqrt{-2\lambda F\left(v\left(1\right)\right)a_2\left(1\right)}.\end{array}$ (1.10)

This cannot hold unless $u\left(1\right)\to 0$ and $v\left(1\right)\to 0$ as $\lambda \to \infty$. However, rewriting (1.10) as

$\{\begin{array}{l}{c}_1\left(u\left(1\right)\right)u{\left(1\right)}^{\frac{1}{2}}\ge \sqrt{-2\lambda \frac{F\left(u\left(1\right)\right)}{u\left(1\right)}{a}_1\left(1\right)},\\ c_2\left(v\left(1\right)\right)v{\left(1\right)}^{\frac{1}{2}}\ge \sqrt{-2\lambda \frac{F\left(v\left(1\right)\right)}{v\left(1\right)}{a}_2\left(1\right)}.\end{array}$ (1.11)

we obtain a contradiction when $\lambda \gg 1$ since $\frac{F\left(u\left(1\right)\right)}{u\left(1\right)}\to f_1\left(0\right)$,$\frac{F\left(v\left(1\right)\right)}{v\left(1\right)}\to f_2\left(0\right)$ if $u\left(1\right)\to 0$ and $v\left(1\right)\to 0$ as $\lambda \to \infty$. Hence,

$u\left(1\right)\ge \frac{\beta +\theta }{2}\text{and}v\left(1\right)\ge \frac{\beta +\theta }{2}\text{for}\lambda \gg 1$ (1.12)

Next, we claim that $u\left(t_m\right)={\Vert u\Vert }_{\infty }\to \infty$ and $v\left(t_m\right)={\Vert v\Vert }_{\infty }\to \infty$ as $\lambda \to \infty$. Let

$h:=u-\beta$ and $w:=v-\beta$ then $h>0$,$w>0$ in $\left(t_{\beta }\mathrm{,1}\right]$ and satisfies

$\{\begin{array}{l}-h^{″}=\lambda a_1\left(t\right)\frac{f_1\left(u\right)}{u-\beta }h\left(t_{\beta },1\right)\hfill \\ -w^{″}=\lambda a_2\left(t\right)\frac{f_2\left(v\right)}{v-\beta }w\left(t_{\beta },1\right)\hfill \\ h\left(t_{\beta }\right)=w\left(t_{\beta }\right)=0,\hfill \\ h\left(1\right)=u\left(1\right)-\beta >0\hfill \\ w\left(1\right)=v\left(1\right)-\beta >0\hfill \end{array}$ (1.13)

Let $\psi =:\sin \left(\frac{\pi \left(t-t_{\beta }\right)}{t-t_{\beta }}\right)$. Then $\psi$ satisfies:

$\{\begin{array}{l}-{\psi }^{″}=\frac{{\pi }^2}{{\left(t-t_{\beta }\right)}^2}\psi ,\left(t_{\beta },1\right)\hfill \\ \psi \left(t_{\beta }\right)=\psi \left(1\right)=0\hfill \end{array}$ (1.14)

Multiplying (1.13) by $\psi$ and (1.14) by h,and w integrating both from $t_{\beta }$ to 1 and subtracting, we have

${\displaystyle {\int }_{t_{\beta }}^1}\left(h^{″}\psi -{\psi }^{″}h\right)\text{d}t={\displaystyle {\int }_{t_{\beta }}^1}\left(\frac{{\pi }^2}{{\left(1-t_{\beta }\right)}^2}-\lambda \frac{f_1\left(u\right)}{u-\beta }{a}_1\left(t\right)\right)h\psi \text{d}t,$

${\displaystyle {\int }_{t_{\beta }}^1}\left(w^{″}\psi -{\psi }^{″}w\right)\text{d}t={\displaystyle {\int }_{t_{\beta }}^1}\left(\frac{{\pi }^2}{{\left(1-t_{\beta }\right)}^2}-\lambda \frac{f_2\left(v\right)}{v-\beta }{a}_2\left(t\right)\right)w\psi \text{d}t$

Since ${\displaystyle {\int }_{t_{\beta }}^1}\left(h^{″}\psi -{\psi }^{″}h\right)\text{d}t=-{\psi }^{\prime }\left(1\right)h\left(1\right)\left(>0\right)$ and ${\displaystyle {\int }_{t_{\beta }}^1}\left(w^{″}\psi -{\psi }^{″}w\right)\text{d}t=-{\psi }^{\prime }\left(1\right)w\left(1\right)\left(>0\right)$ by integration by parts, we can see that

$\frac{{\pi }^2}{{\left(1-t_{\beta }\right)}^2}>\lambda \frac{f_1\left(u\right)}{u-\beta }{a}_1\left(t\right)\text{and}\frac{{\pi }^2}{{\left(1-t_{\beta }\right)}^2}>\lambda \frac{f_2\left(v\right)}{v-\beta }{a}_2\left(t\right)\text{for}\text{some}t\in \left(t_{\beta },1\right)$ (1.15)

Note that ${\inf }_{\left(0,1\right]}{a}_1\left(t\right)>0$,${\inf }_{\left(0,1\right]}{a}_2\left(t\right)>0$ and from Lemma 2.2 we can assume $\left(1-t_{\beta }\right)>\frac{1}{2}$. Thus (2.11) is only true if $\frac{f_1\left(u\right)}{u-\beta }\to 0$ and $\frac{f_2\left(v\right)}{v-\beta }\to 0$

when $\lambda \gg 1$. By (F1) (F2), we conclude that $u\left(t_m\right)={\Vert u\Vert }_{\infty }$ and $v\left(t_m\right)={\Vert v\Vert }_{\infty }$ as $\lambda \to \infty$. Notice that since $u^{″}<0$ and $v^{″}<0$ in $\left(t_{\beta }\mathrm{,1}\right]$, we have

$u\left(t\right)\ge \frac{u\left(t_m\right)-\beta }{t_m-t_{\beta }}\left(t-t_{\beta }\right)+\beta ,t\in \left[t_{\beta },t_m\right]$

$v\left(t\right)\ge \frac{v\left(t_m\right)-\beta }{t_m-t_{\beta }}\left(t-t_{\beta }\right)+\beta ,t\in \left[t_{\beta },t_m\right]$

$u\left(t\right)\ge \frac{u\left(t_m\right)-\frac{\beta +\theta }{2}}{1-t_m}\left(1-t\right)+\frac{\beta +\theta }{2},t\in \left[t_m,1\right]$

$v\left(t\right)\ge \frac{v\left(t_m\right)-\frac{\beta +\theta }{2}}{1-t_m}\left(1-t\right)+\frac{\beta +\theta }{2},t\in \left[t_m,1\right]$

Since $u\left(t_m\right)\to \infty$,$v\left(t_m\right)\to \infty$ and $t_{\beta }\to 0$ as $\lambda \to \infty$, it is all true that

$v\left(t\right)\ge \frac{\beta +\theta }{2}\text{and}u\left(t\right)\ge \frac{\beta +\theta }{2},\text{in}\left[\frac{1}{4},1\right]\text{for}\lambda \gg 1$ (1.16)

Now we show that $u\left(1\right)\to \infty$ and $v\left(1\right)\to \infty$ as $\lambda \to \infty$. Since $u\mathrm{,}v$ is a solution of (1.3), u and v can be written as: (see Appendix 8.2 in [5] for details)

$u\left(t\right)=\lambda {\displaystyle {\int }_0^1}G\left(t,s\right)a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s-c_1\left(u\left(1\right)\right)u\left(1\right)t$ (1.17)

$v\left(t\right)=\lambda {\displaystyle {\int }_0^1}G\left(t,s\right)a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s-c_2\left(v\left(1\right)\right)v\left(1\right)t$ (1.18)

where

$G\left(t,s\right)=\{\begin{array}{l}s,0\le s\le t\le 1\hfill \\ t,0\le t\le s\le 1\hfill \end{array}$

Let $t=1$. Then from (1.17) and (1.18), we have

$\begin{array}{l}\left[1+c_1\left(u\left(1\right)\right)\right]u\left(1\right)\\ =\lambda \left[{\displaystyle {\int }_0^{t_{\beta }}}G\left(1,s\right)a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{t_{\beta }}^1}G\left(1,s\right)a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\right]\end{array}$ (1.19)

$\begin{array}{l}\left[1+c_2\left(v\left(1\right)\right)\right]v\left(1\right)\\ =\lambda \left[{\displaystyle {\int }_0^{t_{\beta }}}G\left(1,s\right)a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{t_{\beta }}^1}G\left(1,s\right)a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\right]\end{array}$ (1.20)

Then using the fact $G\left(1,s\right)=s$ and $t_{\beta }\to \infty$ as $\lambda \to \infty$, for $\lambda$ large we obtain

$\left[1+c_1\left(u\left(1\right)\right)\right]u\left(1\right)=\lambda \left({\displaystyle {\int }_0^{t_{\beta }}}s{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{t_{\beta }}^1}s{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\right)$

$\left[1+c_2\left(v\left(1\right)\right)\right]v\left(1\right)=\lambda \left({\displaystyle {\int }_0^{t_{\beta }}}s{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{t_{\beta }}^1}s{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\right)$

$\ge \lambda \left({\displaystyle {\int }_0^{t_{\beta }}}s{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{\frac{1}{4}}^1}s{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\right)$

$\ge \lambda \left({\displaystyle {\int }_0^{t_{\beta }}}s{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{\frac{1}{4}}^1}s{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\right)$

$\ge \frac{\lambda }{2}{f}_1\left(\frac{\beta +\theta }{2}\right){\displaystyle {\int }_{\frac{1}{4}}^1}s{a}_1\left(s\right)\text{d}s$

$\ge \frac{\lambda }{2}{f}_2\left(\frac{\beta +\theta }{2}\right){\displaystyle {\int }_{\frac{1}{4}}^1}s{a}_2\left(s\right)\text{d}s$

where the last inequality is obtained by (1.16). Hence, we have

$\left[1+c_1\left(u\left(1\right)\right)\right]u\left(1\right)\ge \lambda K\text{and}\left[1+c_2\left(v\left(1\right)\right)\right]v\left(1\right)\ge \lambda K,$ (1.21)

where $K=\frac{1}{2}{f}_1\left(\frac{\beta +\theta }{2}\right){\displaystyle {\int }_{\frac{1}{4}}^1}s{a}_1\left(s\right)\text{d}s>0$ and $K=\frac{1}{2}{f}_2\left(\frac{\beta +\theta }{2}\right){\displaystyle {\int }_{\frac{1}{4}}^1}s{a}_2\left(s\right)\text{d}s>0$.

Now, from (1.21), clearly $u\left(1\right)\to \infty$,$v\left(1\right)\to \infty$ as $\lambda \to \infty$. $\square$

Lemma 2.4. Let u and v be a positive solution of (1.3). Then there exists $\left[\alpha \mathrm{,}\mu \right]\subset \left[\frac{1}{4}\mathrm{,1}\right]$,$\alpha \ne \mu$, both independent of $\lambda$, such that ${\inf }_{\left[\alpha \mathrm{,}\mu \right]}u\left(t\right)\to \infty$ and ${\inf }_{\left[\alpha \mathrm{,}\mu \right]}v\left(t\right)\to \infty$ as $\lambda \to \infty$

Proof. As $\lambda \to \infty$,$t_m$ may converge to 1 or to any other point in $\left(\mathrm{0,1}\right)$. First we consider the case $t_m\nrightarrow 1$ as $\lambda \to \infty$. Since $u\left(1\right)<u\left(t_m\right)$ and $v\left(1\right)<v\left(t_m\right)$ clearly there exists $\alpha <1$ such that ${\inf }_{\left[\alpha \mathrm{,1}\right]}u\left(t\right)\to \infty$ and ${\inf }_{\left[\alpha \mathrm{,1}\right]}v\left(t\right)\to \infty$ as $\lambda \to \infty$ by Lemma 2.3.

Now, let us consider the case when $t_m\to 1$ as $\lambda \to \infty$. By differentiating (1.17) and (1.18) (or integrating (1.3)), we obtain

$u^{\prime }\left(t\right)=\lambda {\displaystyle {\int }_t^1}{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s-c_1\left(u\left(1\right)\right)u\left(1\right),t\in \left[0,1\right],$

$v^{\prime }\left(t\right)=\lambda {\displaystyle {\int }_t^1}{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s-c_2\left(v\left(1\right)\right)v\left(1\right),t\in \left[0,1\right]$

which gives us that

$u^{\prime }\left(t\right)=\lambda {\displaystyle {\int }_t^1}{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s-c_1\left(u\left(1\right)\right)u\left(1\right)=0,$ (1.22)

$v^{\prime }\left(t\right)=\lambda {\displaystyle {\int }_{t_m}^1}{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s-c_2\left(v\left(1\right)\right)v\left(1\right)=0$ (1.23)

Now we rewrite (1.17) and (1.18) by using (1.22), (1.23) as:

$u\left(t\right)=\lambda {\displaystyle {\int }_0^1}G\left(t,s\right)a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s-\lambda \left({\displaystyle {\int }_{t_m}^1}{a}_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\right)t,$

$v\left(t\right)=\lambda {\displaystyle {\int }_0^1}G\left(t,s\right)a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s-\lambda \left({\displaystyle {\int }_{t_m}^1}{a}_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\right)t$

$\begin{array}{l}=\lambda {\displaystyle {\int }_0^{t_{\beta }}}G\left(s,t\right)a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s+\lambda {\displaystyle {\int }_{t_{\beta }}^{t_m}}G\left(t,s\right)a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s\\ +\lambda {\displaystyle {\int }_{t_m}^1}\left[G\left(t,s\right)-t\right]a_1\left(s\right)f_1\left(u\left(s\right)\right)\text{d}s.\end{array}$

$\begin{array}{l}=\lambda {\displaystyle {\int }_0^{t_{\beta }}}G\left(s,t\right)a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s+\lambda {\displaystyle {\int }_{t_{\beta }}^{t_m}}G\left(t,s\right)a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s\\ +\lambda {\displaystyle {\int }_{t_m}^1}\left[G\left(t,s\right)-t\right]a_2\left(s\right)f_2\left(v\left(s\right)\right)\text{d}s.\end{array}$