# On multiple positive ground state solutions for a mean curvature equation in
Minkowski space

Ruyun Maa, Yanqiong Lub Tianlan Chenc,
a,b,cDepartment of Mathematics, Northwest Normal University, Lanzhou 730070, P
R China

###### Abstract

In this paper, we show how changes in the sign of nonlinearity leads to
multiple radial ground state solutions of the mean curvature equation
$\nabla\cdot\Big{[}\frac{\nabla u}{\sqrt{1-|\nabla u|^{2}}}\Big{]}+\lambda
f(u)=0\ \ \text{in}\ \mathbb{R}^{N}$ for sufficiently large $\lambda$ with
$N\geq 2$.

000E-mail<EMAIL_ADDRESS>(R. Ma<EMAIL_ADDRESS>(Y. Lu),
chentianlan511@126.com(T. Chen). 000∗Supported by the NSFC (No.11361054),
SRFDP(No.20126203110004) and Gansu provincial National Science Foundation of
China (No.1208RJZA258).

Keywords. Ground state solution, multiplicity, radial solution, multiplicity

MR(2010) 39A12, 34B18

##  1 Introduction

Hypersurfaces of prescribed mean curvature in Minkowski space are of interest
in differential geometry and in general relativity. In this paper, we are
concerned with the existence and multiplicity of such a kind of hypersurfaces
which are graphs of the solution of the following problem

$\displaystyle\nabla\cdot\Big{[}\frac{\nabla u}{\sqrt{1-|\nabla
u|^{2}}}\Big{]}+\lambda f(u)=0,\ \ \ \ \text{in}\ \mathbb{R}^{N},$ $None$
$\displaystyle\ u(x)>0,\ \qquad\qquad\qquad\qquad\qquad\ \text{in}\
\mathbb{R}^{N},$ $\displaystyle\ u(x)\to 0,\ \qquad\qquad\qquad\qquad\qquad\
\text{as}\ |x|\to\infty,$

where $f:\mathbb{R}\to\mathbb{R}$ is a local Lipschitz function with $f(0)=0$,
$\lambda>0$ is a parameter and $N\geq 2$.

The differential operator we are considering has been deeply studied in the
recent years, in nonlinear equations on bounded domains with various type of
boundary conditions (see [1-5] and the references within) and in the whole
$\mathbb{R}^{N}$ (see [6,7]).

The radial solutions which only depend on $r=|x|$ of (1) satisfy the following
ODE

$\displaystyle\Big{(}\frac{u^{\prime}}{\sqrt{1-(u^{\prime})^{2}}}\Big{)}^{\prime}+\frac{N-1}{r}\frac{u^{\prime}}{\sqrt{1-u^{\prime
2}}}+\lambda f(u)=0,$ $None$ $\displaystyle u(0)=\zeta,\ \ u^{\prime}(0)=0,$

where $u\in C^{2}([0,+\infty])$ is now a function of $r=|x|$ alone, and
$\zeta$ has to be determined in order to have

$\lim_{r\to\infty}u(r)=0.$ $None$

The existence of the positive solution of (1) can be interpreted in this
context as the existence of a ground state solution.

Recently, Azzollini [7] proves the existence of a ground state solutions of
(1) with $\lambda=1$ by the shooting method under the assumptions:

(f1) $f(0)=0$,

(f2) $f:[0,+\infty)\to\mathbb{R}$ is locally Lipschitz,

(f3) $\exists\ \alpha:=\inf\\{\xi>0|\,f(\xi)\geq 0\\}>0$,

(f4) (if $N\geq 3$), $\lim\limits_{s\to\alpha^{+}}\frac{f(s)}{s-\alpha}>0$,

(f5) $\exists\ \gamma>0$ such that $F(\gamma):=\int^{\gamma}_{0}f(s)ds>0$,

(f6) $f(\xi)>0$ in $(\alpha,\xi_{0}]$, where
$\xi_{0}:=\inf\\{\xi\in(0,\infty)\,|\,F(\xi)>0\\}.$

He proved the following

Theorem A. ([7, Theorem 0.1]) If

$\bullet$ $N\geq 3$ and $f$ satisfies (f1)-(f6),

$\bullet$ $N=2$ and $f$ satisfies (f1)-(f3), (f5) and (f6),

then (1) has a radially decreasing solution with $\lambda=1$.

The shooting argument has been used in the past to find ground state solutions
to various types of equations. For examples, Berestycki, Lions and Peletier
[8] study the existence of a ground state solution of the Laplace equation

$\Delta u+f(u)=0\ \ \ \text{in}\ \mathbb{R}^{N}$ $None$

with $N\geq 2$. And the case $N=1$, Berestycki and Lions [9] find the
sufficient and necessary condition for the existence of the unique solution of
the problem (4). Peletier and Serrin [10] are concerned with the existence of
a ground state solution of the following prescribed mean curvature equation

$\nabla\cdot\Big{[}\frac{\nabla u}{\sqrt{1+|\nabla u|^{2}}}\Big{]}-\lambda
u+u^{q}=0.$

The shooting method consists in studying the profile of the solution of (2) as
the initial value $\zeta$ varies into an interval. The main ideas is to
exclude the cases in which for a finite $R>0$ either $u$ or $u^{\prime}$
vanishes.

On the other hand, Dávila del Pino and Guerra [11] find the problem

$\Delta u-u+u^{p}+\lambda u^{q}=0\ \ \ \text{in}\ \mathbb{R}^{N}$

has at least three positive decaying radial solutions if $N=3,1<q<3$, $q<p<5$
is taken sufficiently close to $5$ and $\lambda$ is fixed sufficiently large.

Naturally, what is really interesting is to find the conditions which permit
to multiple ground state solutions of (1). Motivated above papers [6-11], this
paper devotes to studying how changes in the sign of $f(s)$ leads to multiple
positive radial solutions of (1).

We make the following assumptions:

(A1) $f:[0,+\infty)\to\mathbb{R}$ is locally Lipschitz with $f(0)=0$;

(A2) there exists $2n$ real numbers
$0=:\beta_{0}<\alpha_{1}<\beta_{1}<\cdots<\alpha_{n}<\beta_{n}<\infty$ such
that for $i\in\\{1,\cdots,n\\}$,

$f(s)<0,\ \ \ s\in(\beta_{i-1},\alpha_{i});\ \ \ \ \ \ f(s)>0,\ \ \
s\in(\alpha_{i},\beta_{i});$

$(A2)^{\prime}$ there exists $2n-1$ real numbers
$0<\alpha_{1}<\beta_{1}<\cdots<\alpha_{n}<\infty$ such that for
$i\in\\{1,\cdots,n-1\\}$,

$f(s)<0,\ \ \ s\in(\beta_{i-1},\alpha_{i});\ \ \ \ \ \ f(s)>0,\ \ \
s\in(\alpha_{i},\beta_{i});\ \ \ \ \ \ f(s)>0,\ \ \ s\in(\alpha_{n},\infty);$

(A3) let $F(u):=\int^{u}_{0}f(s)ds$. Then for each $i\in\\{1,2,\cdots n\\}$,
there exists $\xi_{i}\in(\alpha_{i},\beta_{i})$ such that $F(\xi_{i})=0$;

(A4) for each $i\in\\{1,2,\cdots n\\}$, $F(\beta_{i-1})<F(\beta_{i})$;

(A5) (if $N\geq 3$)
$\lim\limits_{s\to\alpha_{i}^{+}}\frac{f(s)}{s-\alpha_{i}}>0,\ i=1,2,\cdots,n$
and $\lim\limits_{s\to\beta_{j}^{+}}\frac{f(s)}{\beta_{j}-s}>0,\
j=1,2,\cdots,n-1$.

In the sequel, we will suppose that $f$ is extended in $\mathbb{R}$ by setting
$f(s)=0$ if $s\leq 0$. Clearly, $f$ is locally Lipschitz continuous on
$\mathbb{R}$. The main result of the paper is the following:

Theorem 1. If

$\bullet$ $N\geq 3$ and $f$ satisfies (A1)-(A5),

$\bullet$ $N=2$ and $f$ satisfies (A1)-(A4),

then (1) has $n$ distinct radially decreasing solutions for $\lambda>0$ is
sufficiently large .

Remark 1. Note that Theorem 1 is Theorem A in the case $n=1$ and $\lambda=1$.

Remark 2. If we replace (A2) with (A2)′, then the result of Theorem 1 is also
true by a similar argument with obvious changes.

Remark 3. We exhibit some examples of functions $f$ satisfying our
assumptions: consider the function

$f(s)=(s^{2}-19s+18)(12s^{2}-s^{3}-27s).$

By a simple computation, we can get that $f$ satisfies (A1)-(A5) with
$\alpha_{1}=1,\alpha_{2}=9,\beta_{1}=3,\beta_{2}=18$. From Theorem 1, there
exist numbers $\zeta_{i}\in(\xi_{i},\beta_{i})$, $i=1,2$ such that for
sufficiently large $\lambda$, the problem (1) has two distinct positive,
decaying radial solutions.

##  2 Proof of the main result

Since we are interested in the multiplicity of ground state solutions of (1),
we aim to find $n$ distinct numbers $\zeta_{i}\in(\xi_{i},\beta_{i})$,
$i=1,2,\cdots,n$ such that for $\lambda>0$ is sufficiently large , the
solution $u_{i}\in C^{2}(\mathbb{R}_{+})$ of the IVP:

$\displaystyle\Big{(}\frac{u^{\prime}}{\sqrt{1-(u^{\prime})^{2}}}\Big{)}^{\prime}+\frac{N-1}{r}\frac{u^{\prime}}{\sqrt{1-u^{\prime
2}}}+\lambda f(u)=0,$ $None$ $\displaystyle u(0)=\zeta_{i},\ \
u^{\prime}(0)=0$

has the properties: $u_{i}(r)>0$ for $r\in[0,\infty)$, $u_{i}^{\prime}(r)<0$
for $r\in(0,\infty)$ and

$\underset{r\to+\infty}{\lim}u_{i}(r)=0.$ $None$

Observe that the solution of $(5)_{i}$ satisfies the equation

$(r^{N-1}\phi^{\prime}(u^{\prime}))^{\prime}=-r^{N-1}\lambda f(u),$ $None$

where $\phi(s):=1-\sqrt{1-s^{2}}$ for $s\in[-1,1]$. It is easy to verify that
$\phi^{\prime}:(-1,1)\to\mathbb{R}$ is an increasing diffeomorphism. Set
$\delta>0$ and denote by $C:=C([0,\infty),\mathbb{R})$ and by
$C_{\delta}:=C([0,\delta],\mathbb{R})$. Define the following operators

$S:C\to C,\ \ \ \
Su(r):=\left\\{\begin{array}[]{ll}-\frac{1}{r^{N-1}}\int_{0}^{r}t^{N-1}u(t)dt,&\text{if}\
r>0,\\\ 0,&\text{if}\ r=0,\\\ \end{array}\right.$

and $K:C\to C,\ K(u)(r)=\int_{0}^{r}u(t)dt.$

For every $\zeta_{i}\in\mathbb{R}$, define the translation operator
$T_{\zeta_{i}}:C\to C$ such that $T_{\zeta_{i}}(u)=\zeta_{i}+u$. Moreover,
consider the Nemytskii operators associated to $f$ and $(\phi^{\prime})^{-1}$,

$N_{f}:C\to C,\ \ \ \ N_{f}(u)(r)=f(u(r)),$ $N_{(\phi^{\prime})^{-1}}:C\to C,\
\ \ \ N_{(\phi^{\prime})^{-1}}(u)(r)=(\phi^{\prime})^{-1}(u(r)).$

Set $\rho_{i}>0$ and denote with $B_{\rho_{i}}:=\\{u\in
C_{\delta}\,|\,\|u\|_{\infty}\leq\rho_{i}\\}$. We set the following fixed
point problem: for any $\zeta_{i}\in\mathbb{R}$ we want to find
$u\in\zeta_{i}+B_{\rho_{i}}$ such that

$u=T_{\zeta_{i}}\circ K\circ N_{(\phi^{\prime})^{-1}}\circ S\circ(\lambda
N_{f}(u)).$ $None$

Since $(\phi^{\prime})^{-1}$ and $f$ are respectively Lipschitz and locally
Lipschitz, Banach-Caccioppoli fixed point theorem guarantees the existence of
a sufficiently small $\delta>0$ such that the function
$u_{i}(\lambda):=u(\zeta_{i},r)\in\zeta_{i}+B_{\rho_{i}}$ is a solution of
(8). It is easy to see that $u_{i}$ is a local solution of the Cauchy problem
$(5)_{i}$.

Let $R_{\zeta_{k}}>0$ be such that $[0,R_{\zeta_{k}})$ is the maximal interval
where the function $u_{k}$ is defined, here $k=1,2,\cdots,n$. Multiplying
$(5)_{k}$ by $u^{\prime}_{k}$ and integrating over $(0,r)$ we obtain the
following equality for any $r\in(0,R_{k})$:

$H(u^{\prime}_{k}(r))+(N-1)\int_{0}^{r}\frac{[u^{\prime}_{k}(s)]^{2}}{s\sqrt{1-[u^{\prime}_{k}(s)]^{2}}}=\lambda[F(\zeta_{k})-F(u_{k}(r))],$
$None$

where $H(t):=\frac{1-\sqrt{1-t^{2}}}{\sqrt{1-t^{2}}}$.

For each $k\in\\{1,2,\cdots,n\\}$, let $I_{k}=(\alpha_{k},\beta_{k})$, and
take $\zeta_{k}\in I_{k}$. By (A2) and (A4), for every
$s\in(\beta_{k-1},\beta_{k}]$, we have $F(s)\geq F(\alpha_{k})$. Thus from
(9), we deduce that $H(u^{\prime}(r))$ is bounded as far as $\beta_{k-1}\leq
u\leq\beta_{k}$. Obviously, since $f(u_{k}(0))=f(\zeta_{k})>0$, from
Eq.$(5)_{k}$ we deduce that $u^{\prime\prime}_{k}(0)<0$ and this implies that
there exists $\sigma>0$ such that

$u^{\prime}_{k}(r)<0\ \ \ \text{and}\ \ 0<u_{k}(r)<\zeta_{k}\ \ \ \text{for}\
0<r<\sigma.$

Set

$\bar{R}_{\zeta_{k}}:=\left\\{\begin{array}[]{ll}\inf\\{r\in(0,R_{\zeta_{k}})\,|\,u^{\prime}_{k}(r)\geq
0\\},&\text{if}\ u^{\prime}_{k}(r)=0\ \ \text{for some}\
r\in(0,{R}_{\zeta_{k}}),\\\ +\infty,&\text{otherwise}.\\\ \end{array}\right.$
$None$

From [7, Remark 1.1], it follows that
$0<\sigma\leq\bar{R}_{\zeta_{k}}\leq+\infty$ and for every
$r\in(0,\bar{R}_{\zeta_{k}})$,

$\exists\ \varepsilon>0\ \text{such that for any}\
r\in(0,\bar{R}_{\zeta_{k}}),\ |u_{k}^{\prime}(r)|\leq 1-\varepsilon.$ $None$

In particular, $\bar{R}_{\zeta_{k}}=+\infty$ implies
${R}_{\zeta_{k}}=+\infty$.

Define the following two classes of intervals

$I_{k}^{+}:=\big{\\{}\zeta_{k}\in I_{k}\,\big{|}\,\exists\ R^{\prime}_{k}\leq
R_{\zeta_{k}}\ \text{such that}\;u_{k}(r)>0,\;u^{\prime}_{k}(r)<0\ \text{for}\
r\in(0,R^{\prime}_{k}),u^{\prime}_{k}(R^{\prime}_{k})=0\big{\\}},$

and

$I_{k}^{-}:=\big{\\{}\zeta_{k}\in I_{k}\,\big{|}\,\exists\ R^{\prime}_{k}\leq
R_{\zeta_{k}}\ \text{such that}\;u_{k}(r)>0,\;u^{\prime}_{k}(r)<0\ \text{for}\
r\in(0,R^{\prime}_{k}),u_{k}(R^{\prime}_{k})=0\big{\\}}.$

We will prove that the sets $I_{k}^{+}$ and $I_{k}^{-}$ are non-empty,
disjoint and open, $k=1,2,\cdots n.$ Moreover, $I_{k}^{+}$ and $I_{k}^{-}$ do
not cover $I_{k}$.

Lemma 2.1. Assume that $R_{\zeta_{k}}=+\infty$. For any fixed $\lambda>0$ and
$k\in\\{1,2,\cdots,n\\}$, $\zeta_{k}\in(0,\infty)$ be such that $u_{k}(r)>0$
for all $r\geq 0$ and $u^{\prime}_{k}(r)<0$ for all $r>0$. Then the number
$l=\underset{r\to\infty}{\lim}u_{k}(r)$ satisfies

$f(l)=0.$

Furthermore, if $f$ satisfies (A2) and (A5), then $l=0$.

Proof. Clearly, there exists $l=\lim\limits_{r\to+\infty}u(r)\geq 0$. By
$(5)_{k}$ and (11), we imply that

$\lim\limits_{r\to+\infty}\Big{(}\frac{u^{\prime}_{k}(r)}{\sqrt{1-[u^{\prime}_{k}(r)]^{2}}}\Big{)}=-\lambda
f(l).$ $None$

Suppose that $f(l)\neq 0$, say $f(l)>0$. By simple computations, together with
(11) and (12), we deduce that, definitively,
$u^{\prime\prime}_{k}(r)<-\delta<0$ for some $\delta>0$. Of course this is not
possible because of (11). Therefore, $f(l)=0$.

Now, we claim that $l=0$.

To this end, we only need to prove that for $k=1$, $l\neq\alpha_{1}$ and for
each $k\in\\{2,\cdots,n\\}$, $l\neq\alpha_{i},\ i=1,2,\cdots,k,\
\l\neq\beta_{j}$, $j=1,2,\cdots,k-1$. We divide into three steps.

Step 1. We show that for $k=1$, $l\neq\alpha_{1}$.

If $N=2$ and, by contraction, $l=\alpha_{1}$. Since for any $r>0$,
$\alpha_{1}<u_{1}(r)<\beta_{1}$, from (7) we deduce that
$r\phi^{\prime}(u^{\prime}_{1}(r))$ is decreasing in $[0,+\infty)$ and then,
in particular, there exist $R_{1}>0$ and $\delta>0$ such that for any
$r>R_{1}$, we have $\phi^{\prime}(u^{\prime}_{1}(r))<-\frac{\delta}{r}$. By
(11) we infer that, for some $M_{1}>0$, we have
$M_{1}u^{\prime}_{1}(r)\leq\phi^{\prime}(u^{\prime}_{1}(r))$ and then

$u^{\prime}_{1}(r)\leq-\frac{\delta}{M_{1}r}\ \ \ \text{for any}\ \ r>R_{1}.$

Integrating in $(R_{1},r)$ we obtain

$u_{1}(r)\leq
u_{1}(R_{1})-\frac{\delta}{M_{1}}\ln\Big{(}\frac{r}{R_{1}}\Big{)}\to-\infty\ \
\text{as}\ r\to+\infty,$

which contradicts $l=\alpha_{1}$.

If $N\geq 3$, and suppose on the contrary that $l=\alpha_{1}$, then computing
in $(5)_{1}$, we have that the following equality holds in $(0,+\infty)$:

$\frac{u^{\prime\prime}_{1}}{[1-(u^{\prime}_{1})^{2}]^{\frac{3}{2}}}=-\frac{N-1}{r}\frac{u^{\prime}_{1}}{\sqrt{1-(u^{\prime}_{1})^{2}}}-\lambda
f(u_{1}).$

Taking into account (11), there exists $\delta>0$ such that
$\delta\leq\sqrt{1-(u^{\prime}_{1})^{2}}\leq 1$. We deduce that

$u^{\prime\prime}_{1}=-\frac{N-1}{r}u^{\prime}_{1}[1-(u^{\prime}_{1})^{2}]-\lambda
f(u_{1})[1-(u^{\prime}_{1})^{2}]^{\frac{3}{2}}\leq-\frac{N-1}{r}u^{\prime}_{1}-\delta^{3}\lambda
f(u_{1}),$ $None$

where we have used the fact that $u^{\prime}_{1}<0$ and $f(u_{1})>0$. Now we
proceed as in [7,8], repeating the arguments for completeness. If we set
$v=r^{\frac{N-1}{2}}(u_{1}-\alpha_{1})$, by (13) we get the following estimate

$v^{\prime\prime}\leq\Big{\\{}\frac{(N-1)(N-3)}{4r^{2}}-\delta^{3}\lambda\frac{f(u_{1})}{u_{1}-\alpha_{1}}\Big{\\}}v\\\
$ $None$

from which, in view of (A5), we deduce that $v^{\prime\prime}$ is definitively
negative. Now, since $v^{\prime}$ is definitively decreasing, certainly there
exists $L=\lim\limits_{r\to+\infty}v^{\prime}(r)<+\infty.$

However, $L$ cannot be negative, since otherwise
$\lim\limits_{r\to+\infty}v(r)=-\infty$, this is a contradiction. On the other
hand, if $L\geq 0$, then we deduce that $v$ is definitively increasing and
then there exists $R_{1}>0$ such that for any $r>R_{1}$, we have
$v(r)>v(R_{1})$. From (14) we infer that, for some positive constant $C$,
$v^{\prime\prime}(r)\leq-C<0$ definitively and this implies
$L=\lim\limits_{r\to+\infty}v^{\prime}(r)=-\infty$, again a contradiction.

Step 2. we show that for $k=2$, $l\neq\alpha_{1},\ \alpha_{2}$ and
$l\neq\beta_{1}$.

By a similar argument as step 1 with $u_{2}(r)$ instead of $u_{1}(r)$, and
$v_{2,i}=r^{\frac{N-1}{2}}[u_{2}(r)-\alpha_{i}]$, $i=1,\ 2$ instead of $v$, we
can deduce that $l\neq\alpha_{i},\ i=1,2$. Notice that when $N=2$, we prove
$l\neq\alpha_{1}$, by contradiction, suppose
$\lim\limits_{r\to+\infty}u_{2}(r)=\alpha_{1}$, this implies that there exists
$R_{2}>0$ large enough such that $\alpha_{1}\leq u_{2}(r)\leq\beta_{1}$ for
$r>R_{2}$, by a same argument as step 1, which deduce a contradiction. So, we
only need to show $l\neq\beta_{1}$.

Suppose on the contrary that $l=\beta_{1}$. If $N=2$, then it follows from
$\lim\limits_{r\to+\infty}u_{2}(r)=\beta_{1}$ that there exists
$\tilde{R}_{2}>0$ large enough such that for any $r>\tilde{R}_{2}$,
$\beta_{1}<u_{2}(r)<\alpha_{2}$. If $N=2$, and $l=\beta_{1}$, from (7) we
deduce that $r\phi^{\prime}(u^{\prime}(r))$ is increasing in
$[\tilde{R}_{2},+\infty)$ and then, in particular, there exist
$R_{2}>\tilde{R}_{2}$ and $\delta_{1}>0$ such that for any $r>R_{2}$, we have
$\phi^{\prime}(u^{\prime}(r))>\frac{\delta_{1}}{r}$. By (11) we infer that,
for some $M_{2}>0$, we have
$M_{2}u^{\prime}(r)\geq\phi^{\prime}(u^{\prime}(r))$ and then

$u^{\prime}(r)\geq\frac{\delta_{1}}{M_{2}r}\ \ \ \text{for any}\ \ r>R_{2}.$

Integrating in $(R_{2},r)$ we obtain

$u(r)\geq u(R_{2})+\frac{\delta_{1}}{M_{2}}\ln(\frac{r}{R_{2}})\to+\infty\ \
\text{as}\ r\to+\infty,$

which contradicts $l=\beta_{1}$.

If $N\geq 3$, then computing in $(5)_{2}$, we have that the following equality
holds in $(0,+\infty)$:

$\frac{u^{\prime\prime}_{2}}{[1-(u^{\prime}_{2})^{2}]^{\frac{3}{2}}}=-\frac{N-1}{r}\frac{u^{\prime}_{2}}{\sqrt{1-(u^{\prime}_{2})^{2}}}-\lambda
f(u_{2}).$

Taking into account (11), there exists $\delta_{2}>0$ such that
$\delta_{2}\leq\sqrt{1-(u^{\prime}_{2})^{2}}\leq 1$. We deduce that

$u^{\prime\prime}_{2}=-\frac{N-1}{r}u^{\prime}_{2}[1-(u^{\prime}_{2})^{2}]-\lambda
f(u_{2})[1-(u^{\prime}_{2})^{2}]^{\frac{3}{2}}\geq-\delta_{2}^{2}\frac{N-1}{r}u^{\prime}_{2}-\delta_{2}^{3}\lambda
f(u_{2}),$ $None$

where we have used the fact that $u^{\prime}_{2}<0$ and $f(u_{2})<0$ on
$[\tilde{R}_{2},\infty)$.

Let $w(r)=r^{\frac{\delta^{2}(N-1)}{2}}(\beta_{1}-u(r))$. It follows that

$w^{\prime}(r)=\frac{\delta^{2}(N-1)}{2}r^{\frac{\delta^{2}(N-1)-2}{2}}(\beta_{1}-u(r))-r^{\frac{\delta^{2}(N-1)}{2}}u^{\prime}(r),$
$w^{\prime\prime}(r)=-r^{\frac{\delta^{2}(N-1)}{2}}u^{\prime\prime}(r)-\delta^{2}(N-1)r^{\frac{\delta^{2}(N-1)-2}{2}}u^{\prime}(r)+\frac{\delta^{2}(N-1)[\delta^{2}(N-1)-2]}{4r^{2}}r^{\frac{\delta^{2}(N-1)}{2}}(\beta_{1}-u(r)).$

This together with inequality (15) implies that

$\displaystyle w^{\prime\prime}(r)\leq$ $\displaystyle
r^{\frac{\delta^{2}(N-1)}{2}}[\delta^{2}\frac{N-1}{r}u^{\prime}(r)+\delta^{3}\lambda
f(u)]-\delta^{2}(N-1)r^{\frac{\delta^{2}(N-1)-2}{2}}u^{\prime}(r)$ $None$
$\displaystyle+\frac{\delta^{2}(N-1)[\delta^{2}(N-1)-2]}{4r^{2}}r^{\frac{\delta^{2}(N-1)}{2}}(\beta_{1}-u(r))$
$\displaystyle=$
$\displaystyle\delta^{2}\frac{N-1}{r}r^{\frac{\delta^{2}(N-1)}{2}}u^{\prime}(r)-\delta^{2}\frac{N-1}{r}r^{\frac{\delta^{2}(N-1)}{2}}u^{\prime}(r)$
$\displaystyle+[\delta^{3}\lambda\frac{f(u)}{\beta_{1}-u}+\frac{\delta^{2}(N-1)[\delta^{2}(N-1)-2]}{4r^{2}}]w(r)$
$\displaystyle=$
$\displaystyle[\delta^{3}\lambda\frac{f(u)}{\beta_{1}-u}+\frac{\delta^{2}(N-1)[\delta^{2}(N-1)-2]}{4r^{2}}]w(r),$

from which, in view of (A5), we deduce that $w^{\prime\prime}$ is definitively
negative. Now, since $w^{\prime}$ is definitively decreasing, certainly there
exists $L=\lim\limits_{r\to+\infty}w^{\prime}(r)<+\infty.$ However, $L$ cannot
be positive, since otherwise $\lim\limits_{r\to+\infty}w(r)=+\infty$, this is
a contradiction. On the other hand, if $L\leq 0$, then we deduce that there
exists $R_{2}>0$ such that

$w^{\prime}(r)\leq 0,\ \ \ \ \ r>R_{2},$

and $w$ is definitively decreasing. Hence, there exist two constants $R_{*}$
and $\sigma$ with $R_{\ast}>R_{2}$ and $\sigma>0$, such that

$w(r)<0,\ w^{\prime\prime}(r)\leq\sigma^{2}w(r),\ \ \ \ \ \
r\in[R_{\ast},+\infty).$

Set $b_{1}:=w(R_{\ast}),b_{2}:=w^{\prime}(R_{\ast})$. Then $b_{1}<0,b_{2}\leq
0$. Let us consider the initial value problem

$x^{\prime\prime}(r)=\sigma^{2}x(r),\ r\in(R_{\ast},\infty),\ \ \
x(R_{\ast})=b_{1},\ \ x^{\prime}(R_{\ast})=b_{2}.$

Its unique solution can be explicitly given by

$x(r)=\frac{b_{1}-b_{2}}{2}e^{-\sigma(r-R_{\ast})}+\frac{b_{1}+b_{2}}{2}e^{\sigma(r-R_{\ast})},\
\ r\in(R_{\ast},\infty).$

Let $z(r)=x(r)-w(r)$. Then

$z^{\prime\prime}(r)\geq\sigma^{2}z(r),\ \ r\in(R_{\ast},\infty),\ \ \
z(R_{\ast})=0,\ \ z^{\prime}(R_{\ast})=0.$

Let

$M(r):=z^{\prime\prime}(r)-\sigma^{2}z(r),\ \ \ \ r\in(R_{\ast},\infty).$

Then

$z^{\prime\prime}(r)-\sigma^{2}z(r)=M(r),\ \ r\in(R_{\ast},\infty),\ \ \
z(R_{\ast})=0,\ \ z^{\prime}(R_{\ast})=0,$

which has a unique solution

$z(r)=\frac{1}{2\sigma}\int^{r}_{R_{\ast}}[e^{\sigma(r-s)}-e^{-\sigma(r-s)}]M(s)ds,\
\ \ \ \ \ r\in(R_{\ast},\infty).$

Obviously, $z(r)\geq 0$ for $r\in(R_{\ast},\infty)$, which implies

$x(r)\geq w(r),\ \ \ \ \ r\in(R_{\ast},\infty),$

i. e.

$\frac{b_{1}-b_{2}}{2}e^{-\sigma(r-R_{\ast})}+\frac{b_{1}+b_{2}}{2}e^{\sigma(r-R_{\ast})}\geq
r^{\frac{\delta^{2}(N-1)}{2}}(\beta_{1}-u(r)),\ \ \ \ r\in(R_{\ast},\infty).$

However, this is impossible since

$\frac{b_{1}+b_{2}}{2}<0,\ \ \beta_{1}-\zeta_{2}<\beta_{1}-u(r)\leq 0,\ \ \
\lim_{r\to\infty}\frac{e^{\sigma(r-R_{\ast})}}{r^{\frac{\delta^{2}(N-1)}{2}}}=+\infty.$

Therefore, $l\neq\beta_{1}$.

Step 3. We claim that for each $k\in\\{3,4,\cdots,n\\}$, $l\neq\alpha_{i},\
i=1,2,\cdots,k$ and $l\neq\beta_{j}$, $j=1,2,\cdots,k-1$.

Toward this end, we only need to repeat the arguments of step 1 and step 2
with $u_{k}(r)$ instead of $u_{1}(r)$, and
$v_{k,i}=r^{\frac{N-1}{2}}[u_{k}(r)-\alpha_{i}]$, $i=1,\ 2,\cdots,k$ instead
of $v$, and the proof of $l\neq\beta_{1}$ with $u_{k}$ instead of $u_{2}$ and
$w_{k,j}(r)=r^{\frac{N-1}{2}}[\beta_{j}-u_{k}(r)],\ j=1,2,\cdots,k-1$ instead
of $w$. $\Box$

Lemma 2.2. For any fixed $\lambda>0$ and let $k\in\\{1,2,\cdots,n\\}$,
$I_{k}^{+}\neq\emptyset$.

Proof. Let $\zeta_{k}\in(\alpha_{k},\xi_{k}]$. By (A2) and (A3),
$F(\zeta_{k})<0$. Because of (9) and the definition of $\xi_{k}$, it is clear
to see that $F(u(r))<F(\zeta_{k})<0$ for any $r\in(0,R_{\zeta_{k}})$. As a
consequence, by the fact $f(\zeta_{k})>0$ in $(\alpha_{k},\xi_{k}]$ we have
that there exists $m_{k}>0$ such that

$0<m_{k}<u_{k}(r)<\zeta_{k}.$ $None$

Suppose on the contrary that $\zeta_{k}\not\in I_{k}^{+}$, then
$\bar{R}_{\zeta_{k}}=+\infty$ implies $R_{\zeta_{k}}=+\infty$. So
$u^{\prime}(r)<0$ for any $r>0$, by Lemma 2.1 we get a contradiction with
(17). $\Box$

Next, we will prove that $I_{k}^{-}$ is not empty, we need some preliminary
results.

For each $i\in\\{1,2,\cdots,n\\}$, consider the problem

$\displaystyle\nabla\cdot\Big{[}\frac{\nabla u}{\sqrt{1-|\nabla
u|^{2}}}\Big{]}+\lambda f(u)=0,\ \ \ \ \text{in}\ B_{\rho},$ $None$
$\displaystyle\ u=0,\ \qquad\qquad\qquad\qquad\qquad\quad\ \ \text{on}\
\partial B_{\rho}.$

Recall the definition of $\beta_{i}$, we replace $f$ in (18) by

$f_{i}(s)=\left\\{\begin{array}[]{ll}f(s),&\text{if}\ s\leq\beta_{i},\\\
f(\beta_{i}),&\text{if}\ s>\beta_{i}.\\\ \end{array}\right.$ $None$

As in [3, 7], we use a variational approach to (18).

Set $W_{\rho}:=W^{1,\infty}((0,\rho),\mathbb{R})$. It is well known that
$W_{\rho}\hookrightarrow C_{\rho}$. Define

$K:=\\{u\in W_{\rho}\,|\,\|u^{\prime}\|_{\infty}\leq 1,\ \ u(\rho)=0\\}$

and

$\Psi(u):=\left\\{\begin{array}[]{ll}\int_{0}^{\rho}r^{N-1}\Big{(}1-\sqrt{1-(u^{\prime})^{2}}\Big{)}dr,&\text{if}\
u\in K,\\\ +\infty,&\text{if}\ u\in W_{\rho}\backslash K.\\\
\end{array}\right.$

For any $u\in W_{\rho}$, we set

$J_{i}(\lambda,u):=\Psi(u)-\lambda\int_{0}^{\rho}r^{N-1}F_{i}(u)dr.$

It is easy to verify that the functional $J_{i}(\lambda,\cdot)$ is a Szulkin’s
functional (see [12]) so that, by [12, Proposition 1.1], we have that if $u\in
W_{\rho}$ is a local minimum of $J_{i}(\lambda,\cdot)$, then it is a Szulkin
critical point and for any $v\in K$ it solves the inequality

$\int_{0}^{\rho}r^{N-1}(\phi(v^{\prime})-\phi(u^{\prime}))dr-\lambda\int_{0}^{\rho}r^{N-1}f_{i}(u)(v-u)dr\geq
0,$ $None$

where we recall that $\phi$ is defined in (7). By a similar argument from [7,
13], we obtain the following lemma.

Lemma 2.3. For all $\lambda>0$, if $v_{i}(\lambda,\cdot)\in K$ is a local
minimum for $J_{i}(\lambda,\cdot)$, then $v_{i}(\lambda,|x|)$ is a classical
solution of (18) for each $i\in\\{1,2,\cdots,n\\}$.

Lemma 2.4. For all $\lambda>0$, $\forall\ \rho>0$ and for each
$i\in\\{1,2,\cdots,n\\}$, there exists $v_{i}(\lambda,\cdot)\in K$ such that
$J_{i}(\lambda,\cdot)$ attains its local minimum at $v_{i}(\lambda,\cdot)$.

Moreover, $v_{i}(\lambda,\cdot)$ is a classical nontrivial solution of (18)
and satisfies $0\leq v_{i}(\lambda,\cdot)\leq\beta_{i}$.

Proof. As a first step, we show that $J_{i}(\lambda,\cdot)$ is bounded below
and achieves its infimum.

Observe that $\forall\ v\in K,\|v\|_{\infty}\leq\rho$. As a consequence, it is
easy to see that $J_{i}(\lambda,\cdot)$ is bounded below. Consider
$\\{v_{i,k}\\}_{k=1}^{\infty}\in W_{\rho}$ a minimizing sequence. Of course we
can assume $v_{i,k}\in K$ for any $k\geq 1$. By the Ascoli Arzelà theorem,
there exists a subsequence, relabeled $\\{v_{i,k}\\}_{k=1}^{\infty}$, and a
continuous function $v_{i}^{\ast}$ such that

$v_{i,k}\to v_{i}^{\ast}\ \ \ \text{uniformly in}\ \ [0,\rho].$ $None$

To prove that $v_{i}^{\ast}$ is in $K$, we just observe that, for any
$x,y\in[0,\rho]$ with $x\neq y$, we have

$\lim\limits_{k}\frac{v_{i,k}(x)-v_{i,k}(y)}{x-y}=\frac{v_{i}^{\ast}(x)-v_{i}^{\ast}(y)}{x-y},$

and then also $v_{i}^{\ast}$ has Lipschitz constant $1$. By (21) and [13,
Lemma 1], it deduce that
$\Psi(v_{i}^{\ast})\leq\liminf\limits_{k}\Psi(v_{i,k}).$ Then, again by (21),
we have

$J_{i}(\lambda,v_{i}^{\ast})\leq c_{i,0},$

where $c_{i,0}=\inf\limits_{v\in W_{\rho}}J_{i}(\lambda,v)$.

Now we claim that if $\rho>0$ is sufficiently large, then $c_{i,0}<0.$
Consider the following function defined for $\rho>2\gamma_{i}$,

$\omega_{\rho}(r)=\left\\{\begin{array}[]{cc}\gamma_{i},&\text{in}\ \
[0,\rho-2\gamma_{i}],\\\ \frac{\rho-r}{2},&\text{in}\ \
[\rho-2\gamma_{i},\rho].\\\ \end{array}\right.$

Of course $\omega_{\rho}\in K$. Moreover

$\displaystyle J_{i}(\lambda,\omega_{\rho})\leq$
$\displaystyle\frac{1}{2}\int_{\rho-2\gamma_{i}}^{\rho}(2-\sqrt{3})s^{N-1}ds-F(\gamma_{i})\frac{(\rho-2\gamma_{i})^{N}}{N}+\frac{1}{N}\max\limits_{0\leq
s\leq\gamma_{i}}|F(s)|[(\rho)^{N}-(\rho-2\gamma_{i})^{N}]$ $\displaystyle\leq$
$\displaystyle
C_{1}[\rho^{N}-(\rho-2\gamma_{i})^{N}]-\frac{F(\gamma_{i})(\rho-2\gamma_{i})^{N}}{N}$
$\displaystyle\leq$ $\displaystyle C_{2}\rho^{N-1}-C_{3}\rho^{N}<0\ \ \
\text{as}\ \rho>2\gamma_{i}\ \ \text{sufficiently large},$

where $C_{1},\ C_{2}$ and $C_{3}$ are suitable positive constants. The claim
is an obvious consequence of the previous chain of inequalities. This together
with Lemma 2.3 yields the conclusion. $\Box$

We will use a similar method in [14, Lemma 2.5] to obtain an important lemma.

Lemma 2.5. If $\lambda>0$ is sufficiently large, then
$\sup\\{v_{i+1}(\lambda,r)\,|\,r\in B_{\rho}\\}>\beta_{i}$ and consequently,
$v_{i+1}(\lambda,\cdot)\neq v_{i}(\lambda,\cdot)$, $i\in\\{1,2,\cdots,n-1\\}$.

Proof. To this end, we only need to show that there exists $w\in K$ such that
$J_{i+1}(\lambda,w)<J_{i+1}(\lambda,v)$ for all $v\in K$ satisfying $0\leq
v\leq\beta_{i}$.

First of all, we show that for $\lambda>0$ is sufficiently large, then
$\sup\\{v_{2}(\lambda,r)\,|\,r\in B_{\rho}\\}>\beta_{1}$ and subsequently
$v_{2}(\lambda,\cdot)\neq v_{1}(\lambda,\cdot)$.

Let $\varrho_{0}=\inf\\{F(\beta_{2})-F(v(r)):r\in\bar{B}_{\rho}\ \text{and}\
0\leq v\leq\beta_{1}\\}$. Then $\varrho_{0}>0$ as $f$ satisfies the condition
(A4). If $v\in K$ satisfies $0\leq v\leq\beta_{1}$, then

$\displaystyle\int_{0}^{\rho}r^{N-1}F_{2}(v(r))dr=\int_{0}^{\rho}r^{N-1}F(u(r))dr$
$None$ $\displaystyle\leq$
$\displaystyle\int_{0}^{\rho}r^{N-1}F(\beta_{2})dr-\frac{\rho^{N}}{N}\varrho_{0}=F(\beta_{2})\frac{\rho^{N}}{N}-\varrho_{0}\frac{\rho^{N}}{N}.$

On the other hand, let $\rho>2\beta_{2}$, consider the following function

$w_{\rho}(r)=\left\\{\begin{array}[]{cc}\beta_{2},&\text{in}\ \
[0,\rho-2\beta_{2}],\\\ \frac{\rho-r}{2},&\text{in}\ \
[\rho-2\beta_{2},\rho].\\\ \end{array}\right.$

Obviously, $w_{\rho}\in K$ and

$\displaystyle\int_{0}^{\rho}r^{N-1}F(w_{\rho}(r))dr=$
$\displaystyle\int_{0}^{\rho-2\beta_{2}}r^{N-1}F(\beta_{2})dr+\int_{\rho-2\beta_{2}}^{\rho}r^{N-1}F\big{(}\frac{\rho-r}{2}\big{)}dr$
$None$ $\displaystyle=$
$\displaystyle\int_{0}^{\rho}r^{N-1}F(\beta_{2})dr-\int_{\rho-2\beta_{2}}^{\rho}r^{N-1}F(\beta_{2})dr+\int_{\rho-2\beta_{2}}^{\rho}r^{N-1}F\big{(}\frac{\rho-r}{2}\big{)}dr$
$\displaystyle\geq$ $\displaystyle
F(\beta_{2})\frac{\rho^{N}}{N}-2\sup\limits_{u\in[0,\beta_{2}]}|F(u)|\frac{\rho^{N}-(\rho-2\beta_{2})^{N}}{N}.$

By (22) and (23) we can choose and fix $\rho>2\beta_{2}$ sufficiently large so
that

$\displaystyle\int_{0}^{\rho}r^{N-1}F(w_{\rho}(r))dr-\int_{0}^{\rho}r^{N-1}F(v(r))dr$
$\displaystyle\geq$
$\displaystyle\int_{0}^{\rho}r^{N-1}F(w_{\rho}(r))dr-\int_{0}^{\rho}r^{N-1}F(v(r))dr$
$\displaystyle\geq$
$\displaystyle\varrho_{0}\frac{\rho^{N}}{N}-2\sup\limits_{v\in[0,\beta_{2}]}|F(v)|\frac{\rho^{N}-(\rho-2\beta_{2})^{N}}{N}$
$\displaystyle\geq$ $\displaystyle
C_{4}\rho^{N}-C_{5}\rho^{N-1}>0,\qquad\forall\ 0\leq v\leq\beta_{1},$

here $C_{4},\,C_{5}$ are suitable positive constants. Thus, there exists
$\sigma_{1}>0$ such that

$\int_{0}^{\rho}r^{N-1}F(w_{\rho}(r))dr-\int_{0}^{\rho}r^{N-1}F(v(r))dr>\sigma_{1}$

for all $0\leq v\leq\beta_{1}$. Moreover, for such $\rho>2\beta_{2}$, it
follows that

$\displaystyle J_{2}(\lambda,w_{\rho})-J_{2}(\lambda,v)$ $\displaystyle=$
$\displaystyle\int_{0}^{\rho}r^{N-1}[1-\sqrt{1-(w^{\prime}_{\rho}(r))^{2}}]dr-\int_{0}^{\rho}r^{N-1}[1-\sqrt{1-(v^{\prime}(r))^{2}}]dr$
$\displaystyle-\lambda\int_{0}^{\rho}r^{N-1}[F(w_{\rho}(r))-F(v(r))]dr$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{\rho-2\beta_{2}}^{\rho}(2-\sqrt{3})r^{N-1}dr-\lambda\int_{0}^{\rho}r^{N-1}[F(w_{\rho}(r))-F(v(r))]dr$
$\displaystyle\leq$
$\displaystyle\frac{2-\sqrt{3}}{2N}[\rho^{N}-(\rho-2\beta_{2})^{N}]-\lambda\sigma_{1}$
$\displaystyle\leq$ $\displaystyle 0\ \ \ \ \text{for\ $\lambda$\ sufficiently
large}.$

Hence, for such $\lambda$, the local minimum of $J_{2}(\lambda,\cdot)$ cannot
be attained at any $v\in W_{\rho}$ such that $0\leq v\leq\beta_{1}$.
Therefore, $\sup\\{v_{2}(\lambda,r)\,|\,r\in B_{\rho}\\}>\beta_{1}$ and so
$v_{2}(\lambda,\cdot)\neq v_{1}(\lambda,\cdot)$.

By the same argument with obvious changes, we can obtain that
$\sup\\{v_{i+1}(\lambda,r)\,|\,r\in B_{\rho}\\}>\beta_{i}$ and
$v_{i+1}(\lambda,\cdot)\neq v_{i}(\lambda,\cdot)$, $i\in\\{1,2,\cdots,n-1\\}$
for $\lambda$ sufficiently large. $\Box$

From Lemma 2.3 to Lemma 2.5, it deduce that for any fixed $\rho>0$ large
enough and $k\in\\{2,3,\cdots,n\\}$, the problem (18) with $f_{k}$ instead of
$f$ has $k$ distinct nontrivial solutions and the $k$-th solution $v_{k}$
satisfying $\sup\\{v_{k}(\lambda,r)\,|\,r\in B_{\rho}\\}>\beta_{k-1}$ with
$\lambda>0$ sufficiently large.

Lemma 2.6 Let $\lambda>0$ be sufficiently large and $i\in\\{1,2,\cdots,n\\}$.
Then $I_{i}^{-}\neq\emptyset$.

Proof From Lemma 2.4, it follows that $J_{i}(\lambda,\cdot)$ is bounded below
and achieves its infimum. Moreover, if $\rho>0$ is sufficiently large, then
$c_{i,0}<0.$

Now choose $\rho_{i}>0$ large enough such that there exists
$u_{i}:=v_{i}(\lambda,\cdot)\in K_{i}$ satisfying $J_{i}(u_{i})=c_{i,0}<0$ and
$\sup\\{u_{i}(r)\,|\,r\in B_{\rho_{i}}\\}>\beta_{i-1}$. Set
$\tilde{\zeta}_{i}=u_{i}(0)$. Then the value
$\tilde{\zeta}_{i}\in(\alpha_{i},\beta_{i})$. Indeed, by Lemma 2.3 and Lemma
2.5, $u_{i}(|\cdot|)$ is a classical solution of (18) with $\rho_{i}$ instead
of $\rho$, and then $u_{i}$ is a local solution of $(5)_{i}$ with
$\zeta_{i}=\tilde{\zeta}_{i}$ and $f_{i}$ instead of $f$. If
$\tilde{\zeta}_{i}\leq\alpha_{i}$, then
$\tilde{\zeta}_{i}\in(\beta_{i-1},\alpha_{i}]$ such that
$F(\tilde{\zeta}_{i})\leq 0$ leads to an obvious contradiction to (9) computed
in $r=\rho_{i}$. On the other hand, $\tilde{\zeta}_{i}$ can not be greater
than $\beta_{i}$, since in this case, by (19), the unique solution of the
Cauchy problem $(5)_{i}$ would be the constant function
$u_{i}(r)=\tilde{\zeta}_{i}$.

By contradiction, suppose that $\tilde{\zeta}_{i}\not\in I_{i}^{-}$. Since we
can assume $u_{i}(r)>0$ in $[0,\rho_{i})$, otherwise we consider the function
$u_{i}$ restricted to the interval $[0,R_{i}^{\prime})$, where
$R_{i}^{\prime}:=\inf\\{r>0\,|\,u_{i}(r)=0\\}$, our contradiction assumption
implies that $\bar{R}_{\tilde{\zeta}_{i}}\in(0,\rho_{i})$ (the definition of
$\bar{R}_{\tilde{\zeta}_{i}}$ is given in (10)).

Computing (9) for $r=\bar{R}_{\tilde{\zeta}_{i}}$ and for $r=\rho_{i}$, we
respectively have

$(N-1)\int_{0}^{\bar{R}_{\tilde{\zeta}_{i}}}\frac{(u^{\prime}_{i}(s))^{2}}{s\sqrt{1-(u^{\prime}_{i}(s))^{2}}}ds=\lambda[F(\tilde{\zeta}_{i})-F(u_{i}(\bar{R}_{\tilde{\zeta}_{i}}))],$
$None$
$H(u^{\prime}_{i}(\rho_{i}))+(N-1)\int_{0}^{\rho_{i}}\frac{(u^{\prime}_{i}(s))^{2}}{s\sqrt{1-(u^{\prime}_{i}(s))^{2}}}ds=\lambda
F(\tilde{\zeta}_{i}).$ $None$

Subtracting (24) from (25), we obtain

$H(u^{\prime}_{i}(\rho_{i}))+(N-1)\int_{\bar{R}_{\tilde{\zeta}_{i}}}^{\rho_{i}}\frac{(u^{\prime}_{i}(s))^{2}}{s\sqrt{1-(u^{\prime}_{i}(s))^{2}}}ds=\lambda
F(u_{i}(\bar{R}_{\tilde{\zeta}_{i}})),$

which implies that $F(u_{i}(\bar{R}_{\tilde{\zeta}_{i}}))>0$.

Since $u_{i}^{\prime}(r)<0$ for any $r\in(0,\bar{R}_{\tilde{\zeta}_{i}})$, we
have that $u_{i}^{\prime\prime}(\bar{R}_{\tilde{\zeta}_{i}})\geq 0$ and then
from the equation of $(5)_{i}$, it follows that
$f(u_{i}(\bar{R}_{\tilde{\zeta}_{i}}))\leq 0$. Since $f$ is positive in
$I_{i}$ and
$0<u_{i}(\bar{R}_{\tilde{\zeta}_{i}})<\tilde{\zeta}_{i}<\beta_{i}$, certainly
$u_{i}(\bar{R}_{\tilde{\zeta}_{i}})\in(\beta_{i-1},\alpha_{i}]$. From this we
deduce that $F(u_{i}(\bar{R}_{\tilde{\zeta}_{i}}))<0$ and then the
contradiction is obtained. $\Box$

Lemma 2.7 For any fixed $\lambda>0$ sufficiently large and let
$k\in\\{1,2,\cdots,n\\}$, $I_{k}^{-}$ and $I_{k}^{+}$ are open and disjoint.

Proof By contradiction, suppose $\bar{\zeta}_{k}\in I_{k}^{+}\cap I_{k}^{-}$.
Then, since the solution of $(5)_{k}$ with ${\zeta}_{k}=\bar{\zeta}_{k}$ is
such that
$u_{k}(R^{\prime}_{{\zeta}_{k}})=u^{\prime}_{k}(R^{\prime}_{{\zeta}_{k}})=0$,
by uniqueness theorem, $u=0$ is the unique solution of the Cauchy problem

$\displaystyle\Big{(}\frac{u^{\prime}_{k}}{\sqrt{1-(u^{\prime}_{k})^{2}}}\Big{)}^{\prime}+\frac{N-1}{r}\frac{u^{\prime}_{k}}{\sqrt{1-(u^{\prime}_{k})^{2}}}+\lambda
f(u)=0,$ $\displaystyle u(R^{\prime}_{{\zeta}_{k}})=0,\ \ \ \
u^{\prime}(R^{\prime}_{{\zeta}_{k}})=0.$

Finally, by continuous dependence on the initial datum, it is easy to see that
$I_{k}^{+}$ and $I_{k}^{-}$ are open sets. $\Box$

By Lemma 2.2, Lemma 2.6 and Lemma 2.7, for $\lambda>0$ is sufficiently large,
we can take $\zeta_{k}\in I_{k}\backslash(I_{k}^{+}\cup I_{k}^{-})$ such that
$u_{k}(r)$ is defined on $[0,\infty)$ and, $u_{i}\neq u_{j},\ \ i\neq j$. By
Lemma 2.1, $\lim\limits_{r\to+\infty}u_{k}(r)=0$. As a consequence, the
problem (1) has $n$ distinct positive, decaying radial solution $u_{k}$,
$k=1,2,\cdots,n$.

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