Positive Radial Symmetric Solutions of Nonlinear Biharmonic Equations in an Annulus

Abstract

This paper discusses the existence of positive radial symmetric solutions of the nonlinear biharmonic equation ${▵}^{2}u=f(u,▵u)$ on an annular domain $\Omega$ in ${\mathbb{R}}^N$ with the Navier boundary conditions ${u|}_{\partial \Omega }=0$ and ${▵u|}_{\partial \Omega }=0$, where $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{-}\to {\mathbb{R}}^{+}$ is a continuous function. We present some some inequality conditions of f to obtain the existence results of positive radial symmetric solutions. These inequality conditions allow $f(\xi ,\eta )$ to have superlinear or sublinear growth on $\xi ,\eta$ as $\left|\right(\xi ,\eta \left)\right|\to 0$ and ∞. Our discussion is mainly based on the fixed-point index theory in cones.

1. Introduction and Main Results

In this paper, we discuss the existence of positive radial symmetric solutions of the nonlinear biharmonic elliptic boundary value problem (BVP)

$$\left\{\begin{array}{c}{▵}^{2}u=f(u,▵u),x\in \Omega ,\hfill \\ u=▵u=0,x\in \partial \Omega \hfill \end{array}\right.$$

(1)

in an annular domain $\Omega =\{x\in {\mathbb{R}}^N:r_1<|x|<r_2\}$, where $N\ge 3$, $0<r_1<r_2<\infty$, $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{-}\to {\mathbb{R}}^{+}$ is a continuous nonlinear function.

The boundary value problem of biharmonic elliptic equations such as BVP (1) arise from the study of traveling waves in suspension bridges [1,2] and the study of the static deflection of an elastic plate in a fluid [3,4]. For the simple case of BVP (1) that the nonlinearity f is without $\Delta u$,

$$\left\{\begin{array}{c}{\Delta }^{2}u=f\left(u\right),x\in \Omega ,\hfill \\ u=▵u=0,x\in \partial \Omega \hfill \end{array}\right.$$

(2)

has been discussed by many researchers; see [5,6,7,8,9,10,11,12] and references therein. Dalmasso [5] obtained uniqueness and positivity results of radial symmetric solutions when $f\left(u\right)={\left|u\right|}^p$ and $\Omega$ is a Ball. In [6,7,8,9,10], some existence results are obtained by applying mountain pass lemma and critical point theory. Recently, the authors of [11,12] using the theory of fixed point index in cones obtained the existence results of a positive solution.

There are some researchers that discussed the case where the right side of the equation is only with a linear term of $\Delta u$

$$\left\{\begin{array}{c}{\Delta }^{2}u=c\Delta u+f\left(u\right),x\in \Omega ,\hfill \\ u=▵u=0,x\in \partial \Omega ,\hfill \end{array}\right.$$

(3)

see [13,14,15,16,17]. The authors of [13,14,15,16,17] mainly applied variational methods and critical point theory to discuss the existence of nontrivial solutions of BVP (3).

For the general case that the nonlinearity f contains $\Delta u$, the existence of solutions has also been discussed by several authors; see [18,19,20,21]. In [18], the existence results of solutions for BVP (1) are acquired by using the method of upper and lower solutions, and in [19,20], the monotonic iterative programs for seeking the solution of BVP (1) are erected. Recently, in [21], the existence and uniqueness results of radial symmetric solutions are obtained by the Leray–Schauder fixed-point theorem and technique of prior estimates. In this paper, we study the general biharmonic elliptic boundary value problem BVP (1) and obtain existence results of positive radial symmetric solutions in the annular domain $\Omega$. For the second order elliptic boundary value problem

$$\left\{\begin{array}{c}-▵u=f\left(u\right),x\in \Omega ,\hfill \\ u=0,x\in \partial \Omega \hfill \end{array}\right.$$

(4)

the existence of positive radial symmetric solutions has been discussed by many authors; see [22,23,24,25,26,27,28,29]. However, there is little research on the existence of positive radial symmetric solutions for the general biharmonic elliptic boundary value problem (1). We will apply the theory of fixed point index in cones to study the existence of positive radial symmetric solutions of BVP (1).

Our main results are involved in the principal eigenvalue ${\lambda }_1$ of the Laplace operator $-\Delta$ radially symmetric eigenvalue problem (EVP)

$$\left\{\begin{array}{c}-\Delta u=\lambda u,x\in \Omega ,\hfill \\ u=0,x\in \partial \Omega ,\hfill \\ u=u\left(\right|x\left|\right),x\in \Omega .\hfill \end{array}\right.$$

(5)

the author of [28] has proved that EVP (5) has a minimum positive real eigenvalue ${\lambda }_1$, see ([28], Lemma 2.4). Our main results are as follows:

Theorem 1.

Let $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{-}\to {\mathbb{R}}^{+}$ be continuous and satisfy the following inequality conditions

(H1)

there exist constants $a_0,b_0\ge 0$ with $\frac{a_0}{{{\lambda }_1}^2}+\frac{b_0}{{\lambda }_1}<1$ and $\delta >0$, such that

$$f(\xi ,\eta )\le a_0\xi -b_0\eta ,(\xi ,\eta )\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-},\left|(\xi ,\eta )\right|<\delta ,$$

(H2)

there exist constants $a_1,b_1\ge 0$ with $\frac{a_1}{{{\lambda }_1}^2}+\frac{b_1}{{\lambda }_1}>1$ and $H>0$, such that

$$f(\xi ,\eta )\ge a_1\xi -b_1\eta ,(\xi ,\eta )\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-},\left|(\xi ,\eta )\right|>H.$$

Then, BVP (1) has a positive radial symmetric solution.

Theorem 2.

Let $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{-}\to {\mathbb{R}}^{+}$ be continuous and satisfy the following inequality conditions:

(H3)

There exist constants $a_0,b_0\ge 0$ with $\frac{a_0}{{{\lambda }_1}^2}+\frac{b_0}{{\lambda }_1}>1$ and $\delta >0$, such that

$$f(\xi ,\eta )\ge a_0\xi -b_0\eta ,(\xi ,\eta )\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-},\left|(\xi ,\eta )\right|<\delta ;$$

(H4)

There exist constants $a_1,b_1\ge 0$ with $\frac{a_1}{{{\lambda }_1}^2}+\frac{b_1}{{\lambda }_1}<1$ and $H>0$, such that

$$f(\xi ,\eta )\le a_1\xi -b_1\eta ,(\xi ,\eta )\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-},\left|(\xi ,\eta )\right|>H.$$

Then, BVP (1) has a positive radial symmetric solution.

The conditions (H1) and (H2), respectively, allow $f\left(\xi \eta \right)$ to have superlinear growth on $\xi$ and $\eta$ as $\left|\right(\xi ,\eta \left)\right|\to 0$ and $\left|\right(\xi ,\eta \left)\right|\to +\infty$, and conditions (H3) and (H4), respectively, allow $f\left(\xi \eta \right)$ to have sublinear growth on $\xi$ and $\eta$ as $\left|\right(\xi ,\eta \left)\right|\to 0$ and $\left|\right(\xi ,\eta \left)\right|\to +\infty$. Consider the following example

$$\left\{\begin{array}{c}{\Delta }^{2}u={a\left|u\right|}^{p-1}u-b{|\Delta u|}^{q-1}\Delta u,x\in \Omega ,\hfill \\ u=▵u=0,x\in \partial \Omega ,\hfill \end{array}\right.$$

(6)

where $a,b,p,$ and $q$ are positive constants. The corresponding nonlinearity is

$$f(\xi ,\eta )={a\left|\xi \right|}^{p-1}\xi -b{\left|\eta \right|}^{q-1}\eta .$$

(7)

We easily verify that when $p,q>1$, f satisfies (H1) and (H2), and by Theorem 1, BVP (6) has at least one positive radial symmetric solution; when $0<p,q<1$, f satisfies (H3) and (H4), and by Theorem 2, BVP (6) has at least one positive radial symmetric solution.

Applying Theorems 1 and 2 to BVP (2), we have

Corollary 1.

Let $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be continuous. If f satisfies one of the following conditions (F1) or (F2):

(F1)

$\underset{\xi \to 0^{+}}{lim\; sup}\frac{f\left(\xi \right)}{\xi }<{{\lambda }_1}^2$, $\underset{\xi \to +\infty }{lim\; inf}\frac{f\left(\xi \right)}{\xi }>{{\lambda }_1}^2$;

(F2)

$\underset{\xi \to 0^{+}}{lim\; inf}\frac{f\left(\xi \right)}{\xi }>{{\lambda }_1}^2$, $\underset{\xi \to +\infty }{lim\; sup}\frac{f\left(\xi \right)}{\xi }<{{\lambda }_1}^2$,

then BVP (2) has at least one positive radial symmetric solution.

Proof. 

BVP (2) is a special case of BVP (1), for $f\left(\xi \right)$, we easily show that

$$\left(\mathrm{F}1\right)⟹\left(\mathrm{H}1\right) \mathrm{and} \left(\mathrm{H}2\right);\left(\mathrm{F}2\right)⟹\left(\mathrm{H}3\right) \mathrm{and} \left(\mathrm{H}4\right).$$

Hence, by Theorems 1 and 2, Corollary 1 is true. □

Corollary 1 is a principal eigenvalue criterion on the existence of positive radial symmetric solutions. For the second-order elliptic boundary value problem (4), it is known, see ([27], Theorem 1.1). However, for BVP (2), it is new.

2. Preliminaries

For radially symmetric solutions of BVP (1) $u\left(\right|x\left|\right)$, writing $r=\left|x\right|$ and setting $v=-▵u$, we have

$$\begin{array}{c}\Delta u={\displaystyle \sum _{i=1}^N}\frac{{\partial }^{2}u}{\partial x_i^2}=u^{\prime \prime }\left(r\right)+\frac{N-1}{r}{u}^{\prime }\left(r\right)=\frac{1}{r^{N-1}}{\left(r^{N-1}{u}^{\prime }\left(r\right)\right)}^{\prime },\hfill \\ \Delta v=v^{\prime \prime }\left(u\right)+\frac{N-1}{r}{v}^{\prime }\left(r\right)=\frac{1}{r^{N-1}}{\left(r^{N-1}{v}^{\prime }\left(r\right)\right)}^{\prime }.\hfill \end{array}$$

(8)

Hence, $\left(u\right(r),v(r\left)\right)$ satisfies the ordinary differential equations system

$$\left\{\begin{array}{c}-{\left(r^{N-1}{u}^{\prime }\left(r\right)\right)}^{\prime }=r^{N-1}v\left(r\right),r\in [r_1,r_2],\hfill \\ -{\left(r^{N-1}{v}^{\prime }\left(r\right)\right)}^{\prime }=r^{N-1}f(u\left(r\right),-v\left(r\right)),r\in [r_1,r_2],\hfill \\ u\left(r_1\right)=u\left(r_2\right)=0,v\left(r_1\right)=v\left(r_2\right)=0.\hfill \end{array}\right.$$

(9)

Conversely, if $\left(u\right(r),v(r\left)\right)$ is a solution of BVP (9), by (8), $u\left(\right|x\left|\right)$ is a radially symmetric solution of BVP (1). We discuss BVP (9) to obtain radial symmetric solutions of BVP (1).

Let $I=[r_1,r_2]$, ${\mathbb{R}}^{+}=[0,+\infty )$ and ${\mathbb{R}}^{-}=(-\infty ,0]$. We use $C\left(I\right)$ to denote the Banach space of all continuous functions $u\left(r\right)$ on I with norm ${\parallel u\parallel }_C={max}_{r\in I}\left|u\left(r\right)\right|$. For given $n\in \mathbb{N}$, we use $C^n\left(I\right)$ to denote the Banach space of all nth-order continuous differentiable functions on I with the norm ${\parallel u\parallel }_{C^n}={max\{\parallel u\parallel }_C,\parallel u^{\prime }{\parallel }_C,\cdots ,\parallel u^{\left(n\right)}{\parallel }_C\}$. $C^{+}\left(I\right)$ denotes the cone of all non-negative functions in $C\left(I\right)$. Let $E_1$ and $E_2$ be two Banach spaces; then, the product space $E_1\times E_2=\left\{(x_1,x_2)\right|x_1\in E_1,x_2\in E_2\}$ is a Banach space endowed with the norm $\parallel (x_1,x_2)\parallel =\parallel x_1\parallel +\parallel x_2\parallel$.

To discuss BVP (9), consider the linear boundary value problem (LBVP) of the second-order ordinary differential equation

$$\left\{\begin{array}{c}-{\left(r^{N-1}{u}^{\prime }\left(r\right)\right)}^{\prime }=h\left(r\right),r\in [r_1,r_2],\hfill \\ u\left(r_1\right)=u\left(r_2\right)=0.\hfill \end{array}\right.$$

(10)

LBVP (10) has been discussed in Reference [28], and the following existence and uniqueness result and the strong positivity result of solutions are shown:

Lemma 1

([28], Lemma 2.1).  For any $h\in C\left(I\right)$, LBVP (10) has a unique solution $u:=Sh\in C^2\left(I\right)$. Furthermore, the solution operator $S:C\left(I\right)\to C\left(I\right)$ is a completely continuous linear operator.

Lemma 2

([28], Lemma 2.3).  If $h\in C^{+}\left(I\right)$, the solution $u=Sh$ of LBVP (10) has the strong positivity estimate:

$$u\left(r\right)\ge {\sigma }_0\left(r\right){\parallel u\parallel }_C,r\in I,$$

(11)

where ${\sigma }_0\in C^{+}\left(I\right)$ is a given function independent of h, and ${\sigma }_0\left(r\right)>0$ for $r\in (r_1,r_2)$.

Moreover, by (2.2) and (2.5) of [28], one can obtain that

$${\sigma }_0\left(r\right)=\frac{\frac{1}{{r_1}^{N-2}}-\frac{1}{r^{N-2}}}{\frac{1}{{r_1}^{N-2}}-\frac{1}{{r_2}^{N-2}}}·\frac{\frac{1}{r^{N-2}}-\frac{1}{{r_2}^{N-2}}}{\frac{1}{{r_1}^{N-2}}-\frac{1}{{r_2}^{N-2}}},r\in I.$$

(12)

Consider EVP (5). Setting $r=\left|x\right|$, by (8), EVP (5) becomes the weighted eigenvalue value problem of the ordinary differential equation

$$\left\{\begin{array}{c}-{\left(r^{N-1}{u}^{\prime }\left(r\right)\right)}^{\prime }=\lambda r^{N-1}u\left(r\right),r\in [r_1,r_2],\hfill \\ u\left(r_1\right)=u\left(r_2\right)=0.\hfill \end{array}\right.$$

(13)

EVP (13) has also been discussed in Reference [28], and the following result is obtained:

Lemma 3

([28], Lemma 2.4).  EVP (13) (or EVP (5)) has a minimum positive real eigenvalue ${\lambda }_1$. Furthermore, ${\lambda }_1$ has a positive unit eigenfunction; that is, there exists ${\varphi }_1\in C^2\left(I\right)\cap C^{+}\left(I\right)$ with $\parallel {\varphi }_1{\parallel }_C=1$, which satisfies the equation

$$\left\{\begin{array}{c}-{\left(r^{N-1}{{\varphi }_1}^{\prime }\left(r\right)\right)}^{\prime }={\lambda }_1{r}^{N-1}{\varphi }_1\left(r\right),r\in [r_1,r_2],\hfill \\ {\varphi }_1\left(r_1\right)={\varphi }_1\left(r_2\right)=0.\hfill \end{array}\right.$$

(14)

Lemma 4.

For every $h\in C^{+}\left(I\right)$, the solution $u=Sh$ of LBVP (10) satisfied

$${\parallel u\parallel }_C\le \frac{{\int }_{r_1}^{r_2}{r}^{N-1}u\left(r\right){\varphi }_1\left(r\right)dr}{{\int }_{r_1}^{r_2}{r}^{N-1}{\sigma }_0\left(r\right){\varphi }_1\left(r\right)dr}.$$

(15)

Proof. 

Multiplying (11) by $r^{N-1}{\varphi }_1\left(r\right)$ and then integrating on I, we can obtain (15). □

Consider BVP (9). We are in the space $E=C\left(I\right)\times C\left(I\right)$ to discuss this problem. Choose a closed convex cone in E by

$$K=C^{+}\left(I\right)\times C^{+}\left(I\right).$$

(16)

Let $f:{\mathbb{R}}^{+}\times {\mathbb{R}}^{-}\to {\mathbb{R}}^{+}$ be continuous. Define a mapping $F:K\to C^{+}\left(I\right)$ by

$$F(u,v)\left(r\right)=f\left(u\right(r),-v(r\left)\right),(u,v)\in K.$$

(17)

Clearly, $F:K\to C^{+}\left(I\right)$ is continuous, and it maps every bounded set in K into a bounded set in $C^{+}\left(I\right)$. Define an operator A by

$$A(u,v)=(S\left(r^{N-1}v\right),S\left(r^{N-1}F(u,v)\right)),(u,v)\in K.$$

(18)

By the complete continuity of $S:C^{+}\left(I\right)\to C^{+}\left(I\right)$ and the continuity of $F:K\to C\left(I\right)$, $A:K\to K$ is completely continuous. If $(u,v)$ is a fixed point of A, then by the definition A

$$u=S\left(r^{N-1}v\right),v=S\left(r^{N-1}F(u,v)\right).$$

(19)

Hence, by the definition of S and F, $(u,v)\in C^2\left(I\right)\times C^2\left(I\right)$ is a solution of BVP (9). Conversely, if $(u,v)\in C^2\left(I\right)\times C^2\left(I\right)$ is a solution of BVP (9), then $u,v$ satisfy (19) and $(u,v)$ is a fixed point of A. If $(u,v)$ is a nonzero fixed point of A, by the first equation of (19), $v\ne 0$, and hence $u\ne 0$. By Lemma 4 and (19), $u\left(r\right),v\left(r\right)>0$ for every $r\in (r_1,r_2)$. Hence, by (8) and (9), $u\left(\right|x\left|\right)$ is a positive radial symmetric solution. We will find the nonzero fixed point of A by using the theory of the fixed point index in cones.

Let E be a Banach space and $K\subset E$ be a closed convex cone in E. Assume D is a bounded open subset of E with boundary $\partial D$, and $K\cap D\ne \varnothing$. Let $A:K\cap \overline{D}\to K$ be a completely continuous mapping. If $Au\ne u$ for any $u\in K\cap \partial$, then the fixed point index $i(A,K\cap D,K):={\mathrm{deg}}_K(I-A,K\cap D,0)$ is well defined, where ${\mathrm{deg}}_K(I-A,K\cap D,0)$ is the topological degree on cone K of $I-A$ on $K\cap D$ at 0. For details, see ([30], Chapter 2). The following two lemmas in ([30], Chapter 2) are needed in our discussion.

Lemma 5.

Let E be a Banach space, $K\subset E$ be a closed convex cone in E, $D\subset E$ be a bounded open subset with $0\in D$, and $A:K\cap \overline{D}\to K$ be a completely continuous mapping. If $\mu Aw\ne w$ for any $w\in K\cap \partial D$ and $0<\mu \le 1$, then $i(A,K\cap D,K)=1$.

Lemma 6.

Let E be a Banach space, $K\subset E$ be a closed convex cone in E, D be a bounded open subset of E and $A:K\cap \overline{D}\to K$ be a completely continuous mapping. If there exists $w_0\in K\setminus \left\{0\right\}$ such that $w-Aw\ne \tau w_0$ for any $w\in K\cap \partial D$ and $\tau \ge 0$, then $i(A,K\cap D,K)=0$.

3. Proofs of the Main Results

Proof of Theorem 1.

Let $E=C\left(I\right)\times C\left(I\right)$, $K=C^{+}\left(I\right)\times C^{+}\left(I\right)$, and $A:K\to K$ be the completely continuous operator defined by (18). Let $0<R_1<R_2<+\infty$ and set

$$D_1=\{(u,v)\in E|\parallel (u,v)\parallel <R_1\},D_2=\{(u,v)\in E|\parallel (u,v)\parallel <R_2\}.$$

(20)

We prove that A has a fixed point in $K\cap (D_2\setminus {\overline{D}}_1)$ when $R_1$ is appropriately small and $R_2$ is appropriately large.

Choose $R_1\in (0,\delta /\sqrt{2})$, where $\delta$ is the positive constant in (H1). We prove that A satisfies the condition of Lemma 5 in $K\cap \partial D_1$, namely

$$\mu A(u,v)\ne (u,v)\forall (u,v)\in K\cap \partial D_1,0<\mu \le 1.$$

(21)

If (21) does not hold, there exist $(u_0,v_0)\in K\cap \partial D_1$ and $0<{\mu }_0\le 1$ such that ${\mu }_{0}A(u_0,v_0)=(u_0,v_0)$. By the definition of A,

$$u_0=S\left({\mu }_0{r}^{N-1}{v}_0\right),v_0=S\left({\mu }_0{r}^{N-1}F(u_0,v_0)\right).$$

(22)

By the definition of S, $u_0$ is the unique solution of LBVP (10) for $h={\mu }_0{r}^{N-1}{v}_0\in C^{+}\left(I\right)$. Hence, $u_0\in C^2\left(I\right)$ satisfies the differential equation

$$\left\{\begin{array}{c}-{\left(r^{N-1}{u}_0^{\prime }\left(r\right)\right)}^{\prime }={\mu }_0{r}^{N-1}{v}_0\left(r\right),r\in [r_1,r_2],\hfill \\ u_0\left(r_1\right)=u_0\left(r_2\right)=0.\hfill \end{array}\right.$$

(23)

Multiplying this equation by ${\varphi }_1\left(r\right)$ and integrating on I, then for the left side using integration by parts and (14), we have

$${\lambda }_1{\int }_{r_1}^{r_2}{r}^{N-1}{u}_0\left(r\right){\varphi }_1\left(r\right)dr={\mu }_0{\int }_{r_1}^{r_2}{r}^{N-1}{v}_0\left(r\right){\varphi }_1\left(r\right)dr.$$

(24)

By the second equation of (22), $v_0$ is the unique solution of LBVP (10) for $h={\mu }_0{r}^{N-1}F(u_0,v_0)\in C^{+}\left(I\right)$. Hence, $v_0\in C^2\left(I\right)$ satisfies the equation

$$\left\{\begin{array}{c}-{\left(r^{N-1}{v}_0^{\prime }\left(r\right)\right)}^{\prime }={\mu }_0{r}^{N-1}f(u_0\left(r\right),-v_0\left(r\right)),r\in [r_1,r_2],\hfill \\ v_0\left(r_1\right)=v_0\left(r_2\right)=0.\hfill \end{array}\right.$$

(25)

For every $r\in I$, since $u_0\in K\cap \partial D_1$, by the definitions of K and $D_1$, we have

$$\begin{array}{c}0\le u_0\left(r\right)\le \parallel u_0{\parallel }_C\le \parallel (u_0,v_0)\parallel =R_1<\delta /\sqrt{2},\hfill \\ 0\le v_0\left(r\right)\le \parallel v_0{\parallel }_C\le \parallel (u_0,v_0)\parallel =R_1<\delta /\sqrt{2}.\hfill \end{array}$$

From this, it follows that

$$(u_0\left(r\right),-v_0\left(r\right))\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-},\left|(u_0\left(r\right),-v_0\left(r\right))\right|<\delta .$$

Hence, by (H1), we have

$$f(u_0\left(r\right),-v_0\left(r\right))\le a_0{u}_0\left(r\right)+b_0{v}_0\left(r\right),r\in I.$$

By this inequality and Equation (25), we obtain that

$$-{\left(r^{N-1}{v_0}^{\prime }\left(r\right)\right)}^{\prime }\le r^{N-1}(a_0{u}_0\left(r\right)+b_0{v}_0\left(r\right)),r\in I.$$

Multiplying this inequality by ${\varphi }_1\left(r\right)$ and integrating on I, for the left side using integration by parts and (14), for the right using (24), we have

$$\begin{array}{cc}{\lambda }_1{\int }_{r_1}^{r_2}{r}^{N-1}{v}_0\left(r\right){\varphi }_1\left(r\right)dr\hfill & \le {\int }_{r_1}^{r_2}{r}^{N-1}(a_0{u}_0\left(r\right)+b_0{v}_0\left(r\right)){\varphi }_1\left(r\right)dr\hfill \\ \hfill & =a_0{\int }_{r_1}^{r_2}{r}^{N-1}{u}_0\left(r\right){\varphi }_1\left(r\right)dr+b_0{\int }_{r_1}^{r_2}{r}^{N-1}{v}_0\left(r\right){\varphi }_1\left(r\right)dr\hfill \\ \hfill & \le \left(\frac{a_0}{{\lambda }_1}+b_0\right){\int }_{r_1}^{r_2}{r}^{N-1}{v}_0\left(r\right){\varphi }_1\left(r\right)dr.\hfill \end{array}$$

(26)

If $v_0=0$, by (22), $u_0=S\left({\mu }_0{r}^{N-1}{v}_0\right)=0$, and $(u_0,v_0)=0$. This contradicts $(u_0,v_0)\in \partial D_1$. Hence, $v_0\ne 0$ and $\parallel v_0{\parallel }_C>0$. By Lemma 2,

$${\int }_{r_1}^{r_2}{r}^{N-1}{v}_0\left(r\right){\varphi }_1\left(r\right)dr\ge {r_1}^{N-1}\parallel v_0{\parallel }_C{\parallel {\varphi }_1\parallel }_C{\int }_{r_1}^{r_2}{\sigma }_0^2\left(r\right)dr>0.$$

From this and (26), it follows that $1\le \frac{a_0}{{{\lambda }_1}^2}+\frac{b_0}{{\lambda }_1}$, which contradicts with the assumption in (H1). Hence, (21) holds, namely, A satisfies the condition of Lemma 5 in $K\cap {\partial D}_1$. By Lemma 5,

$$i(A,K\cap D_1,K)=1.$$

(27)

Choose a positive constant $C_0:=max\left\{\right|f(\xi ,\eta )-(a_1\xi -b_1\eta )|:(\xi ,\eta )\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-},|(\xi ,\eta )|\le H\}+1$. By Condition (H2), we have

$$f(\xi ,\eta )\ge a_1\xi -b_1\eta -C_0,(u,\eta )\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{-}.$$

(28)

Next, we show that when $R_2$ is large enough,

$$i(A,K\cap D_2,K)=0.$$

(29)

For this, choosing $w_0=(0,{\varphi }_1)$, clearly $w_0\in K\setminus \left\{0\right\}$, we prove that A satisfies the condition of Lemma 6 for $w_0$ in $K\cap \partial D_2$, namely

$$(u,v)-\mu A(u,v)\ne \tau (0,{\varphi }_1)\forall (u,v)\in K\cap \partial D_2,\tau \ge 0.$$

(30)

If (30) does not hold, there exist $(u_1,v_1)\in K\cap \partial D_2$ and ${\tau }_1\ge 0$ such that $(u_1,v_1)-A(u_1,v_1)={\tau }_1(0,{\varphi }_1)$. By this and the definition of A, we have

$$u_1=S\left(r^{N-1}{v}_1\right),v_1=S\left(r^{N-1}F(u_1,v_1)\right)+{\tau }_1{\varphi }_1.$$

(31)

By the first equation above, using an argument similar to (24), it can be proven that

$${\lambda }_1{\int }_{r_1}^{r_2}{r}^{N-1}{u}_1\left(r\right){\varphi }_1\left(r\right)dr={\int }_{r_1}^{r_2}{r}^{N-1}{v}_1\left(r\right){\varphi }_1\left(r\right)dr.$$

(32)

By (14), ${\varphi }_1=S\left({\lambda }_1{r}^{N-1}{\varphi }_1\right)$. From this and the second equation of (31), $v_1=S(r^{N-1}$$F(u_1,v_1))+{\tau }_{1}S\left({\lambda }_1{r}^{N-1}{\varphi }_1\right)=S\left(r^{N-1}F(u_1,v_1)+{\tau }_1{\lambda }_1{r}^{N-1}{\varphi }_1\right)$. By the definition of S, $v_1$ is the unique solution of LBVP (10) for $h=r^{N-1}F(u_1,v_1)+{\tau }_1{\lambda }_1{r}^{N-1}{\varphi }_1\in C^{+}\left(I\right)$. This means that $u_1\in C^2\left(I\right)$ satisfies the equation

$$\left\{\begin{array}{c}-{\left(r^{N-1}{v_1}^{\prime }\left(r\right)\right)}^{\prime }=r^{N-1}f(u_1\left(r\right),-v_1\left(r\right))+{\tau }_1{\lambda }_1{r}^{N-1}{\varphi }_1\left(r\right),r\in I,\hfill \\ u_1\left(r_1\right)=u_1\left(r_2\right)=0.\hfill \end{array}\right.$$

(33)

Hence, by (28), we have

$$\begin{array}{cc}\hfill -{\left(r^{N-1}{v_1}^{\prime }\left(r\right)\right)}^{\prime }& =r^{N-1}f(u_1\left(r\right),-v_1\left(r\right))+{\tau }_1{\lambda }_1{r}^{N-1}{\varphi }_1\left(r\right)\hfill \\ \hfill & \ge r^{N-1}f(u_1\left(r\right),-v_1\left(r\right))\hfill \\ \hfill & \ge r^{N-1}(a_1{u}_1\left(r\right)+b_1{v}_1\left(r\right))-C_0{r}^{N-1},r\in I.\hfill \end{array}$$

Multiplying this inequality by ${\varphi }_1\left(r\right)$ and integrating on I, for the left side using integration by parts and (14), for the right side using (32), we have

$$\begin{array}{cc}\hfill {\lambda }_1{\int }_{r_1}^{r_2}{r}^{N-1}{v}_1\left(r\right){\varphi }_1\left(r\right)dr& \ge {\int }_{r_1}^{r_2}{r}^{N-1}(a_1{u}_1\left(r\right)+b_1{v}_1\left(r\right){\varphi }_1\left(r\right)dr-C_0{r}_2^{N-1}(r_2-r_1)\hfill \\ \hfill & =\left(\frac{a_1}{{\lambda }_1}+b_1\right){\int }_{r_1}^{r_2}{r}^{N-1}{v}_1\left(r\right){\varphi }_1\left(r\right)dr-C_0{r}_2^{N-1}(r_2-r_1).\hfill \end{array}$$

From this, it follows that

$${\int }_{r_1}^{r_2}{r}^{N-1}{v}_1\left(r\right){\varphi }_1\left(r\right)dr\le \frac{C_0{r}_2^{N-1}(r_2-r_1)}{{\lambda }_1\left(\frac{a_1}{{{\lambda }_1}^2}+\frac{b_1}{{\lambda }_1}-1\right)}:=C_1.$$

(34)

Hence, by (15) of Lemma 4,

$$\parallel v_1{\parallel }_C\le \frac{{\int }_{r_1}^{r_2}{r}^{N-1}{v}_1\left(r\right){\varphi }_1\left(r\right)dr}{{\int }_{r_1}^{r_2}{r}^{N-1}{\sigma }_0\left(r\right){\varphi }_1\left(r\right)dr}\le \frac{C_1}{{\int }_{r_1}^{r_2}{r}^{N-1}{\sigma }_0\left(r\right){\varphi }_1\left(r\right)dr}:=C_2.$$

(35)

By (31) and (35), we have

$$\parallel u_1{\parallel }_C\le \parallel S\left(r^{N-1}{v}_1\right){\parallel }_C\le \parallel S\parallel ·\parallel r^{N-1}{v}_1{\parallel }_C\le r_2^{N-1}{C}_2\parallel S\parallel :=C_3.$$

(36)

By (35) and (36), we obtain that

$$\parallel (u_1,v_1)\parallel =\parallel u_1{\parallel }_C+{\parallel v_1\parallel }_C\le C_2+C_3:=C_4.$$

(37)

Now, choose $R_2>max\{C_4,\delta \}$. Since $(u_1,v_1)\in K\cap \partial D_2$, by the definition of $D_2$, $\parallel (u_1,v_1)\parallel =R_2>C_4$. This contradicts with (37). Hence, (30) holds: namely, A satisfies the condition of Lemma 6 in $K\cap \partial D_2$. By Lemma 6, (29) holds.

So far, by (27), (29) and the additivity of fixed point index, we have

$$i(A,K\cap (D_2\setminus {\overline{D}}_1),K)=i(A,K\cap D_2,K)-i(A,K\cap D_1,K)=-1.$$

Hence, A has a fixed point $(u^{*},v^{*})\in K\cap (D_2\setminus {\overline{D}}_1)$. Since $(u^{*},v^{*})$ is a nonzero solution of BVP (9), by Lemma 2, $u^{*}\left(r\right)$ and $v^{*}\left(r\right)$ are positive in $(r_1,r_2)$. Hence, $u^{*}\left(\right|x\left|\right)$ is a positive radial symmetric solution of BVP (1). □

Proof of Theorem 2.

Let $K=C^{+}\left(I\right)\times C^{+}\left(I\right)$ and $A:K\to K$ be the completely continuous operator defined by (18). Let $D_1,D_2\subset C\left(I\right)\times C\left(I\right)$ be defined by (20). When $R_1\in (0,\delta /\sqrt{2})$, similar to the proof of (29), we can use Lemma 6 and (H3) to prove that

$$i(A,K\cap D_1,K)=0.$$

(38)

Similar to the proof of (27), we can use Lemma 5 and (H4) to prove that when $R_2$ large enough,

$$i(A,K\cap D_2,K)=1.$$

(39)

By (38), (39) and the additivity of fixed point index, we have

$$i(A,K\cap (D_2\setminus {\overline{D}}_1),K)=i(A,K\cap D_2,K)-i(A,K\cap D_1,K)=1.$$

Hence, A has a fixed point $(\tilde{u},\tilde{v})\in K\cap (D_2\setminus {\overline{D}}_1)$. $(\tilde{u},\tilde{v})$ is a positive solution of BVP (9), and $\tilde{u}\left(\right|x\left|\right)$ is a positive radial symmetric solution of BVP (1). □