#  Two Infinite Families of Solutions for Singular Superlinear Equations on
Exterior Domains

Abdeali Diwan, Joseph Iaia
University of North Texas

###### Abstract

In this paper, we study radial solutions of $\Delta u+K(|x|)f(u)=0$ in the
exterior of the ball of radius $R>0$ in $\mathbb{R}^{N}$ with $N>2$ where $f$
grows superlinearly at infinity and is singular at 0 with
$f\sim\frac{1}{|u|^{q-1}u}$ where $0<q<1.$ We also assume
$K(r)\sim|r|^{-\alpha}$ for large $r$ and establish the existence of two
infinite families of solutions when $N+q(N-2)<\alpha<2(N-1).$

##  1 Introduction

In this paper, we are interested in radial solutions of

$\Delta u+K(|x|)f(u)=0\text{ on }\mathbb{R}^{N}\backslash B_{R},$ (1)
$u|_{\partial{B_{R}}}=0,\lim_{|x|\to\infty}u(x)=0$ (2)

when $N>2$ and where $B_{R}$ is the ball of radius $R>0$ centered at the
origin. Assuming $u(x)=u(|x|)=u(r)$ this becomes

$u^{\prime\prime}+\frac{N-1}{r}u^{\prime}+K(r)f(u)=0\text{ \ for }r>R,$ (3)
$u(R)=0,\lim_{r\to\infty}u(r)=0.$ (4)

We make the following assumptions on $f$. We assume
$f:\mathbb{R}\backslash\\{0\\}\to\mathbb{R}$ is odd, locally Lipschitz, and

$f(u)=|u|^{p-1}u+g(u)\text{ for }p>1\text{ where
}\lim_{|u|\to\infty}\frac{g(u)}{|u|^{p}}=0,$ (H1)
$f(u)=\frac{1}{|u|^{q-1}u}+g_{1}(u)\text{ for }0<q<1\text{ and }g_{1}\text{ is
locally Lipschitz with }g_{1}(0)=0.$ (H2)

We also define $F(u)=\int_{0}^{u}f(t)\,dt$. Since $f$ is integrable at $0$ it
follows that $F$ is continuous and $F(0)=0$. In addition, since $f$ is odd
then $F$ is even and since $f(u)>0$ for $u>0$ it follows that $F(u)>0$ for
$u\neq 0$.

We also assume $K>0$ and $K^{\prime}$ are continuous on $[R,\infty)$ and there
are positive constants $K_{0},K_{1},$ and $\alpha$ such that

$K_{0}r^{-\alpha}\leq K\leq K_{1}r^{-\alpha},\
2(N-1)>-\frac{rK^{\prime}}{K},\text{ and
}\lim_{r\to\infty}\frac{rK^{\prime}}{K}=-\alpha\text{ with
}N+q(N-2)<\alpha<2(N-1).$ (H3)

Main Theorem: Under the assumptions (H1)-(H3) there exist two infinite
families of sign-changing solutions to (3)-(4) on $[R,\infty).$

Numerous papers have proved the existence of positive solutions of these
equations with various nonlinearities $f(u)$ and for various functions
$K(|x|)\sim|x|^{-\alpha}$ with $\alpha>0$. See for example [4]-[5]. We will
prove the existence of two infinite families of sign-changing solutions.

##  2 Preliminaries

Let $R>0$ and define $R_{1}=R^{2-N}$. We begin our analysis by first making
the change of variables $u(r)=v(r^{2-N})=v(t)$ to (3)-(4) and obtaining

$v^{\prime\prime}(t)+h(t)f(v(t))=0\text{ for }0<t<R_{1},$ (5)
$v(0)=0,v(R_{1})=0$ (6)

where

$h(t)=\frac{t^{\frac{2(N-1)}{2-N}}K(t^{\frac{1}{2-N}})}{(N-2)^{2}}.$ (7)

It follows from (H3) that $h^{\prime}<0$ and $\lim\limits_{t\to
0^{+}}\frac{th^{\prime}}{h}=-\tilde{\alpha}$ where
$\tilde{\alpha}=\frac{2(N-1)-\alpha}{N-2}$. In addition, (H3) implies there
exist positive constants $h_{0}$ and $h_{1}$ such that

$h_{0}t^{-\tilde{\alpha}}\leq h\leq h_{1}t^{-\tilde{\alpha}}\text{ on
}(0,R_{1}].$ (8)

Also from (H3) it follows that

$0<\tilde{\alpha}<1-q<1.$ (9)

We now attempt to solve the initial value problem

$v^{\prime\prime}+h(t)f(v)=0\text{ for }0<t<R_{1},$ (10) $v(0)=0,\
v^{\prime}(0)=a>0$ (11)

and then try to find values of $a$ so that $v(R_{1})=0$.

##  3  Existence of Solutions

3.1. Integral Equation Formulation

First, we prove the existence of a solution of equations (10)-(11) on
$[0,\epsilon]$ for some $\epsilon>0$. In order to do this, we reformulate
(10)-(11) as an appropriate integral equation. Assuming $v$ is a solution of
(10)-(11), then integrating (10) on $(0,t)$ and using (11) gives

$v^{\prime}+\int_{0}^{t}h(s)f(v)\,ds=a.$ (12)

Integrating again on $(0,t)$ gives

$v+\int_{0}^{t}\int_{0}^{s}h(x)f(v)\,dx\,ds=at.$ (13)

To ensure the integral in (12)-(13) is defined, we observe it follows from
(11) that for sufficiently small $t>0$

$\frac{at}{2}<v<at.$

Furthermore, from conditions (H1)-(H2), it follows that there exists a
constant $f_{1}>0$ such that $0<f(v)\leq f_{1}(v^{-q}+v^{p})$ for $v>0$. Also,
it follows from this and (8) that for $v>0$

$0<h(t)f(v(t))\leq
f_{1}h_{1}(t^{-\tilde{\alpha}}v^{-q}+t^{-\tilde{\alpha}}v^{p})\leq
f_{1}h_{1}\left(\left(\frac{a}{2}\right)^{-q}t^{-\tilde{\alpha}-q}+a^{p}t^{-\tilde{\alpha}+p}\right).$
(14)

It follows from (9) and (H1) that $1-\tilde{\alpha}-q>0$ and
$1-\tilde{\alpha}+p>0$. Thus it follows from (14) that $h(t)f(v)$ is
integrable near $t=0$, and thus the integrals in (12) and (13) are defined and
are continuous functions.

3.2. Contraction Mapping Principle

Using (H2), we see that (13) is equivalent to

$v+\int_{0}^{t}\int_{0}^{s}h(x)\left(\frac{1}{v^{q}}+g_{1}(v)\right)\,dx\,ds=at.$
(15)

Substituting $v=tw$ into (15) gives

$w=a-\frac{1}{t}\int_{0}^{t}\int_{0}^{s}h(x)\left(\frac{1}{x^{q}w^{q}}+g_{1}(xw)\right)\,dx\,ds.$
(16)

We also define $w(0)=a$.

Next let $\epsilon>0$, $a>0$, and

$X=\\{w\in C[0,\epsilon]\,|\,w(0)=a\text{ and }\|w-a\|\leq\frac{a}{2}\\}$

where $C[0,\epsilon]$ is the set of continuous functions on $[0,\epsilon]$ and
$\|\cdot\|$ is the supremum norm, i.e.

$\|w\|=\sup_{[0,\epsilon]}|w(x)|.$

It is well-known that $(X,\|\cdot\|)$ is a Banach space [1]. We now define
$T:X\to C[0,\epsilon]$

$Tw=a-\frac{1}{t}\int_{0}^{t}\int_{0}^{s}h(x)\left(\frac{1}{x^{q}w^{q}}+g_{1}(xw)\right)\,dx\,ds.$

It can be shown that $T:X\to X$ and $T$ is a contraction for $\epsilon$
sufficiently small, say for $\epsilon=\epsilon_{0}$. These details are carried
out in [2].

The contraction mapping principle then ensures the existence of a unique $w$
in $X$ such that $Tw=w$, by choosing any function $w_{1}$ in $X$ and then
defining

$w_{n+1}=T(w_{n}).$

Using the fact that $T$ is a contraction it follows that $\\{w_{n}\\}$ forms a
Cauchy sequence in $X$ and hence converges to some $w\in X$ with

$w=\lim\limits_{n\to\infty}w_{n+1}=\lim\limits_{n\to\infty}T(w_{n})=Tw.$

Thus $w$ is a fixed point of $T$ on $[0,\epsilon_{0}]$.

Finally letting $v=tw$ then we see that $v$ is a solution of (10)-(11) on
$[0,\epsilon_{0}].$

We henceforth denote this solution as $v_{a}$.

3.3. Existence and Uniqueness of Solutions on $[0,R_{1}]$

We now show that the solution $v_{a}$ is defined on all of $[0,R_{1}].$ Let

$E=\frac{1}{2}v_{a}^{\prime 2}+hF(v_{a})$ (17)

and notice it follows from (10) that

$E^{\prime}=h^{\prime}F(v_{a})\leq 0.$ (18)

Hence it follows that $E$ is nonincreasing and so integrating this on $[0,t]$
gives

$\frac{1}{2}v_{a}^{\prime 2}+hF(v_{a})\leq\frac{1}{2}a^{2}.$ (19)

Since $F(v_{a})\geq 0$ it follows from (19) that $v_{a}$ and $v_{a}^{\prime}$
are defined and bounded on the largest interval of existence
$[0,T]\subset[0,R_{1}]$ where they are defined. Since this upper bound is
independent of $T$ it follows then that $v_{a}$ and $v_{a}^{\prime}$ are
defined and continuous on all of $[0,R_{1}].$ In addition, it follows that
$v_{a}$ and $v_{a}^{\prime}$ vary continuously with $a>0$.

##  4  Behavior of Solutions for Small Values of a

Lemma 4.1: Assume (H1)-(H3) and suppose $v_{a}$ satisfies (10)-(11). If $a>0$
is sufficiently small, then $v_{a}$ has a first zero, $z_{a,1}$, with
$0<z_{a,1}<R_{1}$ and $\lim\limits_{a\to 0^{+}}z_{a,1}=0$. In addition,
$\lim\limits_{a\to 0^{+}}|v_{a}^{\prime}(z_{a,1})|=0$.

Proof: Let us suppose by way of contradiction that $|v_{a}|>0$ on $(0,R_{1}]$.
Since $v_{a}(0)=0$ and $v_{a}^{\prime}(0)=a>0$ it follows that $v_{a}$ is
positive and since $f(v_{a})>0$ when $v_{a}>0$ it follows from (10) that
$v_{a}^{\prime\prime}<0$. Inegrating this twice on $(0,t)$ gives

$0<v_{a}<at$ (20)

and therefore

$\frac{1}{v_{a}^{q}}>\frac{1}{a^{q}t^{q}}.$ (21)

It follows from (H1) and (H2) that $f\to\infty$ as $v\to\infty$ and since
$g_{1}$ is continuous it follows that $g_{1}(v_{a})\geq-C$ for some constant
$C>0$. Using this and (21) in (12) gives:

$v_{a}^{\prime}+\int_{0}^{t}\left(\frac{h_{0}x^{-\tilde{\alpha}-q}}{{a^{q}}}-Cx^{-\tilde{\alpha}}\right)\,dx\leq
v_{a}^{\prime}+\int_{0}^{t}h(s)\left(\frac{1}{v_{a}^{q}}+g_{1}(v_{a})\right)\,ds=a.$

Carrying out the integration on the left-hand side and using (9) it follows
then that

$v_{a}^{\prime}+\frac{h_{0}t^{1-\tilde{\alpha}-q}}{a^{q}(1-\tilde{\alpha}-q)}-\frac{Ct^{1-\tilde{\alpha}}}{1-\tilde{\alpha}}\leq
a.$

Integrating again on $(0,t)$ and rewriting gives

$v_{a}\leq
at-\frac{h_{0}t^{2-\tilde{\alpha}-q}}{a^{q}(1-\tilde{\alpha}-q)(2-\tilde{\alpha}-q)}+\frac{Ct^{2-\tilde{\alpha}}}{(1-\tilde{\alpha})(2-\tilde{\alpha})}.$
(22)

Notice for fixed positive $t$ that if we choose $a$ sufficiently small then
the right hand-side of (22) becomes negative contradicting the assumption that
$v_{a}>0$. Thus $v_{a}$ has a first zero, $z_{a,1}$, for $a>0$ sufficiently
small. Then evaluating (22) at $t=z_{a,1}$ and rewriting gives:

$z_{a,1}^{1-\tilde{\alpha}-q}\leq\frac{a^{q+1}+Ca^{q}z_{a,1}^{1-\tilde{\alpha}}}{h_{0}}\leq\frac{a^{q+1}+Ca^{q}R_{1}^{1-\tilde{\alpha}}}{h_{0}}\to
0\text{ as }a\to 0^{+}.$ (23)

In addition, it follows from (19) that $v_{a}^{\prime 2}(z_{a,1})\leq a^{2}$
and hence $\lim\limits_{a\to 0^{+}}|v_{a}^{\prime}(z_{a,1})|=0.$

This completes the proof of the lemma. ∎

In a similar way we can show $v_{a}$ has any finite number of zeros if $a>0$
is sufficiently small and $\lim\limits_{a\to 0^{+}}|v_{a}^{\prime}(z_{a})|=0$
at each zero, $z_{a}$, of $v_{a}$.

##  5  Behavior of Solutions for Large Values of a

Lemma 5.1: Assume (H1)-(H3) and suppose $v_{a}$ satisfies (10)-(11). Then
$\lim\limits_{a\to\infty}\max\limits_{[0,R_{1}]}v_{a}=\infty.$

Proof: Recall $F(u)=\int_{0}^{u}f(t)\,dt$ and suppose now by way of
contradiction that $v_{a}\leq C_{0}$ where $C_{0}$ is independent of $a$.
Using (20) and (H1)-(H2) then it follows that there are constants
$C_{1},C_{2}$ and $C_{3}$ so that

$F(v_{a})\leq
C_{1}v_{a}^{1-q}+C_{2}v_{a}^{p+1}=v_{a}^{1-q}\left(C_{1}+C_{2}v_{a}^{p-q}\right)\leq
a^{1-q}t^{1-q}\left(C_{1}+C_{2}C_{1}^{p-q}\right)\leq C_{3}a^{1-q}t^{1-q}.$
(24)

Next we see that it follows from (17)-(18) that

$\frac{1}{2}v_{a}^{\prime
2}+hF(v_{a})+\int_{0}^{t}(-h^{\prime}(s))F(v_{a})\,ds=\frac{1}{2}a^{2}.$ (25)

Using (8) and (24) we see

$hF(v_{a})+\int_{0}^{t}(-h^{\prime}(s))F(v_{a})\,ds\leq
C_{3}h_{1}a^{1-q}t^{-\tilde{\alpha}}t^{1-q}+\int_{0}^{t}2\tilde{\alpha}C_{3}h_{1}a^{1-q}s^{-\tilde{\alpha}-1}s^{1-q}\,ds=C_{4}a^{1-q}t^{1-\tilde{\alpha}-q}$

where $C_{4}=h_{1}(C_{3}+2\tilde{\alpha})$. Using this in (25) we obtain

$|v_{a}^{\prime}|\geq
a^{\frac{1-q}{2}}\sqrt{a^{1+q}-C_{4}R_{1}^{1-\tilde{\alpha}-q}}.$

Thus we see for sufficiently large $a$ that the right-hand side of the above
is positive and so $v_{a}$ is increasing for $a$ sufficiently large. Then
integrating on $(0,t)$ gives

$C_{0}\geq v_{a}\geq
a^{\frac{1-q}{2}}\left(\sqrt{a^{1+q}-C_{4}R_{1}^{1-\tilde{\alpha}-q}}\right)t.$

Thus for fixed $t>0$ and $a$ sufficiently large we see that the right-hand
side of the above gets larger than $C_{0}$ and so we obtain a contradiction.
Thus the lemma holds. ∎

Lemma 5.2: Assume (H1)-(H3) and suppose $v_{a}$ satisfies (10)-(11). Then
$v_{a}$ has a local maximum, $M_{a}$, if $a>0$ is sufficiently large. In
addition, $M_{a}\to 0$ and $|v_{a}(M_{a})|\to\infty$ as $a\to\infty$.

Proof: Assume by way of contradiction that $v_{a}$ is increasing on
$[0,R_{1}]$ for all $a>0$. Then it follows from Lemma 5.1 and for any
$t_{0}>0$ that $\min\limits_{[t_{0},R_{1}]}v_{a}(t)=v_{a}(t_{0})\to\infty$ as
$a\to\infty$. Then we see

$\min\limits_{[t_{0},R_{1}]}\frac{h(t)f(v_{a}(t))}{v_{a}(t)}\geq
h(R_{1})\min\limits_{[t_{0},R_{1}]}\frac{f(v_{a}(t))}{v_{a}(t)}=C_{a}\to\infty\text{
as }a\to\infty.$

Rewriting (10) we obtain

$v_{a}^{\prime\prime}+\left(\frac{h(t)f(v_{a})}{v_{a}}\right)v_{a}=0.$ (26)

Next, let $t_{0}>0$ and consider

$y^{\prime\prime}+C_{a}y=0$ (27)

with $y(t_{0})=v_{a}(t_{0})>0$, $y^{\prime}(t_{0})=v_{a}^{\prime}(t_{0})>0$.
Now multiply (26) by $y$, (27) by $v_{a}$, and subtract to obtain

$(yv_{a}^{\prime}-y^{\prime}v_{a})^{\prime}+\left(\frac{h(t)f(v_{a})}{v_{a}}-C_{a}\right)yv_{a}=0.$
(28)

We know that $y$ is a linear combination of $\sin(\sqrt{C_{a}}t)$ and
$\cos(\sqrt{C_{a}}t)$ and as is well-known, any interval of length
$\frac{\pi}{\sqrt{C_{a}}}$ has a zero of $y$. Next we choose $a$ sufficiently
large so that $\left(\frac{h(t)f(v_{a})}{v_{a}}-C_{a}\right)>0$ on
$[0,R_{1}]$. Then by assumption we have $v_{a}^{\prime}>0$ and $v_{a}>0$ on
$(t_{0},t_{0}+\frac{\pi}{\sqrt{C_{a}}})$ and so integrating (28) on
$(t_{0},t)$ gives $yv_{a}^{\prime}-y^{\prime}v_{a}<0$ on $(t_{0},t).$ It
follows then that $(\frac{y}{v_{a}})^{\prime}<0$ and so we integrate on
$(t_{0},t)$ to get $v_{a}<y$ on $(t_{0},t_{0}+\frac{\pi}{\sqrt{C_{a}}})$.
Also, as mentioned above $y$ has a smallest zero, $Z$, on
$(t_{0},t_{0}+\frac{\pi}{\sqrt{C_{a}}})$ and so we see $v_{a}(Z)<y(Z)=0$. Now
since $v_{a}(t_{0})>0$, $v_{a}^{\prime}(t_{0})>0$, and $v_{a}(Z)<0$ it follows
that $v_{a}$ must have a local maximum on $(t_{0},Z)$ which contradicts our
assumption. Thus there is an $M_{a}\in(0,R_{1})$ at which $v_{a}$ has a local
maximum if $a$ is sufficiently large.

It follows from Lemma 5.1 that $v_{a}(M_{a})\to\infty$ as $a\to\infty$.

Next integrating (10) on $(t,M_{a})$

$v_{a}^{\prime}=\int_{t}^{M_{a}}hf(v_{a})\,ds\geq
c_{0}\int_{t}^{M_{a}}hv_{a}^{p}\,ds\geq c_{0}v_{a}^{p}\int_{t}^{M_{a}}h\,ds.$

Dividing by $v_{a}^{p}$ and integrating again gives

$\frac{v_{a}^{1-p}(M_{a})}{p-1}\geq\frac{v_{a}^{1-p}(M_{a})-v^{1-p}(t)}{p-1}\geq
c_{0}\int_{t}^{M_{a}}\int_{s}^{M_{a}}h\,dx\,ds.$

Now as mentioned above, the left-hand side goes to zero as $a\to\infty$ and so
this is only possible in the above inequality if $M_{a}\to 0$ as $a\to\infty$.
∎

Lemma 5.3: Assume (H1)-(H3) and suppose $v_{a}$ satisfies (10)-(11). Then
$v_{a}$ has a zero $z_{a}$ if $a>0$ is sufficiently large. Also $z_{a}\to 0$
and $|v_{a}^{\prime}(z_{a})|\to\infty$ as $a\to\infty$.

Proof: Assume by way of contradiction that $v_{a}(t)>0$ on $(M_{a},R_{1})$.
Now let $E_{1}=\frac{1}{2}\frac{v_{a}^{\prime 2}}{h}+F(v_{a}).$ It then
follows that $E_{1}^{\prime}=-\frac{v_{a}^{\prime 2}h^{\prime}}{h^{2}}\geq 0$
and so

$\frac{1}{2}\frac{v_{a}^{\prime 2}}{h}+F(v_{a})\geq F(v_{a}(M_{a}))\text{ on
}(M_{a},R_{1}).$

Rewriting, integrating on $(M_{a},t)$, and changing variables gives

$\int_{v_{a}(t)}^{M_{a}}\frac{1}{\sqrt{2}\sqrt{F(v_{a}(M_{a}))-F(s)}}\,ds=\int_{M_{a}}^{t}\frac{-v_{a}^{\prime}(s)}{\sqrt{2}\sqrt{F(v_{a}(M_{a}))-F(v_{a})}}\,ds\geq\int_{M_{a}}^{t}\sqrt{h}\,ds.$

Assuming $v_{a}(t)>0$ on $(0,R_{1})$ it follows then that

$\int_{0}^{M_{a}}\frac{1}{\sqrt{2}\sqrt{F(v_{a}(M_{a}))-F(s)}}\,ds\geq\int_{M_{a}}^{R_{1}}\sqrt{h}\,ds.$

After another change of variables we obtain

$\frac{v_{a}(M_{a})}{\sqrt{2}\sqrt{F(v_{a}(M_{a}))}}\int_{0}^{1}\frac{1}{\sqrt{1-\frac{F(v_{a}(M_{a})s)}{F(v_{a}(M_{a}))}}}\,ds\geq\int_{M_{a}}^{R_{1}}\sqrt{h}\,ds.$
(29)

However, it follows that since $f$ is superlinear then
$\frac{v_{a}(M_{a})}{\sqrt{F(v_{a}(M_{a}))}}\to 0$ as $a\to\infty$ and by the
dominated convergence theorem

$\int_{0}^{1}\frac{1}{\sqrt{1-\frac{F(v_{a}(M_{a})s)}{F(v_{a}(M_{a}))}}}\,ds\to\int_{0}^{1}\frac{1}{\sqrt{1-s^{p+1}}}\,ds$

hence the left-hand side of (29) goes to 0 as $a\to\infty$ and thus this
forces $\int_{0}^{R_{1}}\sqrt{h}\,ds=0$ which is false. Thus it must be the
case that $v_{a}$ has a zero, $z_{a}$, on $(M_{a},R_{1})$. Further, evaluating
the above integral at $t=z_{a}$ gives

$\int_{0}^{M_{a}}\frac{1}{\sqrt{2}\sqrt{F(v_{a}(M_{a}))-F(s)}}\,ds\geq\int_{M_{a}}^{z_{a}}\sqrt{h}\,ds.$

Since the left-hand side goes to 0 as $a\to\infty$ it follows then that
$z_{a}-M_{a}\to 0$ as $a\to\infty$ and since $M_{a}\to 0$ by Lemma 5.2 it
follows that $z_{a}\to 0$ as $a\to\infty$.

It follows from this then that if $M_{a}<z_{a}$ then
$F(v_{a}(M_{a})=E_{1}(M_{a})\leq E_{1}(z_{a})=\frac{1}{2}\frac{v_{a}^{\prime
2}(z_{a})}{h(z_{a})}\leq\frac{1}{2}\frac{v_{a}^{\prime 2}(z_{a})}{h(R_{1})}$
and since the left-hand side goes to infinity by Lemma 5.2 it follows then
that $|v_{a}^{\prime}(z_{a})|\to\infty$ as $a\to\infty$. ∎

In a similar way we can show $v_{a}$ has any finite number of zeros if $a>0$
is sufficiently large and
$\lim\limits_{a\to\infty}|v_{a}^{\prime}(z_{a})|=\infty$ at each zero $z_{a}$
and $z_{a}\to 0$ as $a\to\infty$.

##  6 Proof of the Main Theorem

Lemma 6.1: Assume (H1)-(H3) and suppose $v_{a}$ satisfies (10)-(11). If
$z_{a}>0$ and $v_{a}(z_{a})=0$ then $v_{a}^{\prime}(z_{a})\neq 0$. In
addition, $v_{a}$ has a finite number of zeros on $[0,R_{1}]$ for each $a>0$.

Proof: Recall from above that $E_{1}=\frac{1}{2}\frac{v_{a}^{\prime
2}}{h}+F(v_{a})$ and that $E_{1}^{\prime}=-\frac{v_{a}^{\prime
2}h^{\prime}}{h^{2}}\geq 0$. Hence integrating on $(0,t)$ we see that
$\frac{1}{2}\frac{v_{a}^{\prime 2}}{h}+F(v_{a})>0$ for $t>0$. It follows from
this then that if $v_{a}(z_{a})=0$ and $z_{a}>0$ then
$E_{1}(z_{a})=\frac{1}{2}\frac{v_{a}^{\prime 2}(z_{a})}{h(z_{a})}>0$ and thus
$v_{a}^{\prime}(z_{a})\neq 0$.

It follows from this that $v_{a}$ cannot have an infinite number of zeros for
if so then there would exist $z_{k}$ such that $0<z_{k}<z_{k+1}\leq R_{1}$
such that $v_{a}(z_{k})=0$ and there would also exist local extrema, $M_{k}$,
with $z_{k}<M_{k}<z_{k+1}\leq R_{1}$ and $v_{a}^{\prime}(M_{k})=0$. But then
the $z_{k}$ are increasing and bounded above so there is a $z^{*}\leq R_{1}$
such that $z_{k}\to z^{*}\leq R_{1}$. Also $M_{k}\to z^{*}$ and it follows
from continuity that $v_{a}(z^{*})=v_{a}^{\prime}(z^{*})=0$ contradicting the
first part of the lemma. ∎Proof of the Main Theorem Let $n$ be a nonnegative
integer. Let

$S_{n}=\\{a|v_{a}>0\text{ has exactly }n\text{ zeros on }(0,R_{1})\\}.$

It follows from Lemma 6.1 that there is smallest $n_{0}\geq 0$ such that
$S_{n_{0}}$ is nonempty.

Then let

$a_{n_{0}}=\sup S_{n_{0}}.$

It can then be show that $v_{a_{n_{0}}}$ has exactly $n_{0}$ zeros on
$(0,R_{1})$ and $v_{a_{n_{0}}}(R_{1})=0.$ See [2] for details.

Then by continuity with respect to $a$ it follows that if $a>a_{n_{0}}$ and
$a$ is close to $a_{n_{0}}$ that $v_{a}$ will have exactly $n_{0}+1$ zeros on
$(0,R_{1})$ and thus $S_{n_{0}+1}$ is nonempty. Then letting
$a_{{n_{0}}+1}=\sup S_{n_{0}+1}$ we can similarly show $v_{a_{n_{0}}}$ has
exactly $n_{0}+1$ zeros on $(0,R_{1})$ and $v_{a_{n_{0}}+1}(R_{1})=0.$
Continuing in this way we can show that there are an infinite number of
solutions.

In a similar way we can let

$a_{n_{0}^{*}}=\inf S_{n_{0}}$

and proceed to get a second infinite family of solutions.

Finally we set $u_{a_{n}}(r)=v_{a_{n}}(r^{2-N})$ and
$u_{a_{n^{*}}}(r)=v_{a_{n^{*}}}(r^{2-N})$ thus we obtain two infinite families
of solutions $u_{a_{n}}$ and $u_{a_{n}^{*}}$. This completes the proof of the
Main Theorem.. ∎

## References

  * [1] L. Evans, Partial Differential Equations, 2nd ed, Graduate Studies in Mathematics, Vol 19, AMS, 2010.
  * [2] J. Iaia, Existence of Two Infinite Families of Solutions for Singular Superlinear Equations on Exterior Domains, Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 06, pp. 1-14.
  * [3] J. Iaia, Existence and Nonexistence of Sign-changing Solutions for Singular Superlinear Equations on Exterior Domains, Communications on Pure and Applied Analysis, 22(11) 3218-3232, 2023.
  * [4] E.K. Lee, R. Shivaji, B. Son, “Positive radial solutions to classes of singular problems on the exterior of a ball,” Journal of Mathematical Analysis and Applications, 434, No. 2, 1597-1611, 2016.
  * [5] E. Lee, L. Sankar, R. Shivaji, “Positive solutions for infinite semipositone problems on exterior domains,” Differential and Integral Equations, Volume 24, Number 9/10, 861-875, 2011.

University of North Texas
P.O. Box 311430

Denton, Texas 76203

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