# Some Existence Results for a Singular Elliptic Problem via Bifurcation
Theory

Claudio Saccon 111Dipartimento di Matematica, Largo Pontecorvo 5, I56127 Pisa,
ITALY, e-mail<EMAIL_ADDRESS>
##  1 Introduction

The aim of this paper is to prove existence of solutions pairs $({\lambda},u)$
of the singular problem:

$\begin{cases}-\Delta u={\lambda}u-\dfrac{1}{u}&\mbox{ in }\Omega,\\\
u>0&\mbox{ in }\Omega,\\\ \nabla u\cdot\nu=0&\mbox{ on
}\partial\Omega\end{cases}$ (P)

where $\Omega$ is a bounded smooth open subset of N and $\nu$ denotes the unit
normal defined on $\partial\Omega$ – we are therefore considering a Neumann
boundary condition. This research is motivated by the increasing interest on
singular problem that has been showing up in the literature in recent years
(starting more or less from the year 2000). If we confine ourselves on
elliptic problems, studying Problem P seems a natural continuation of the
paper [5] by Montenegro Silva, where a two solutions result is proved for the
Dirichlet problem with the semilinear term
${\lambda}u^{p}-\dfrac{1}{u^{\alpha}}$, with $0<p<1$ and $0<\alpha<1$. In our
case the singular term has the exponent $\alpha=1$. Notice that the minus sign
in front of the singular term makes the problem quite more challenging than
the plus sign. In the latter problem indeed one can exploit the convexity of
the corresponding term in the Euler-Lagrange functional, which allows to treat
any positive esponent $\alpha$ (using suitable tricks, see e.g. [1]). If one
makes some tests in the radial case is not difficult to realize that the
Dirichlet problem has no reasonable solutions for $\alpha=1$. Actually
solutions starting from zero are forced to stick at zero and not allowed to
emerge (in contrast with the $\alpha<1$ case). On this respect see Remark
(3.9). For this reason we are lead to consider the corresponding Neumann
problem P.

In this alternative setting we need to mention the paper [3] by Del Pino,
Manásevich, and Montero which deals with the ODE case ($N=1$) in the periodic
case with a more general, non autonomous, singular term $f(u,x)$ (singular in
$u$ and $T$-periodic in $x$). Using topological degree arguments the authors
prove for instance that the equation:

$-\ddot{u}={\lambda}u-\frac{1}{u^{\alpha}}\qquad,\qquad u(x)>0\quad,\qquad
u(x+T)=u(x),$

where $\alpha\geq 1$ and $h$ is $T$-periodic, has a solution provided
${\lambda}\neq\dfrac{\mu_{k}}{4}$ for all $k$. Here $\mu_{k}$ denote the
eigenvalues of a suitable linearized problem which arises in a natural way
from the problem. In the variational case the results of [3] can be seen as
deriving from the existence of two global bifurcation branches originating
from the pairs $(\mu_{k}/2,1/\sqrt{\mu_{k}/2})$ (the second element of the
pair is a constant function).

Following this idea we prove a local bifurcation result for P, (see Theorem
(2.1)) showing that there exist two bifurcation branches of solutions,
emanating from the pairs $(\mu_{k}/2,1/\sqrt{\mu_{k}/2})$. In Theorem (3.3) of
section 3 we consider the radial case and, exploiting a continuation argument
for the nodal regions of the solutions, we can prove that one of the two
branches is global and bounded in ${\lambda}$.

##  2 A local bifurcation result

Let $\Omega$ be a bounded open subset of N with smooth boundary.

######  2.1 Theorem.

Let $\hat{\mu}>0$ be an eigenvalue of the following Neumann problem:

$\begin{cases}-\Delta u=\mu u&\mbox{ in }\Omega,\\\ \nabla u\cdot\nu=0&\mbox{
on }\partial\Omega,\end{cases}$ (1)

($\nu$ denotes the normal to $\partial\Omega$).

Then there exists $\rho_{0}>$ such that for all $\rho\in]0,\rho_{0}[$ there
exist two distinct pairs $({\lambda}_{1,\rho},u_{1,\rho})$ and
$({\lambda}_{2,\rho},u_{2,\rho})$ such that, for $i=1,2$:

$({\lambda}_{i,\rho,u_{i,\rho}})\mbox{ are solutions of {\rm P}}\ ,\
\int_{\Omega}\left(u_{i,\rho}-\frac{1}{\sqrt{{\lambda}_{i,\rho}}}\right)^{2}dx=\rho^{2}\
,\ {\lambda}_{i,\rho}\xrightarrow{\rho\to 0}\frac{\mu_{k}}{2}.$

###### Proof.

We start by introducing some changes of variables. First of all notice that,
for all ${\lambda}>0$, Problem P has the constant solution
$u(x)=\dfrac{1}{\sqrt{{\lambda}}}$. If we seek for solutions of the form
$u=\dfrac{1}{\sqrt{{\lambda}}}+z$ we easily end up with the equivalent problem
on $z$:

$\begin{cases}-\Delta z=2{\lambda}z-h_{\lambda}(z)&\mbox{ in }\Omega,\\\
\sqrt{{\lambda}}z>-1&\mbox{ in }\Omega,\\\ \nabla z\cdot\nu=0&\mbox{ on
}\partial\Omega\end{cases}$ (P0)

where
$h_{{\lambda}}:\left]-\dfrac{1}{\sqrt{{\lambda}}},+\infty\right[\to\real$ is
defined by:

$h_{\lambda}(s)=\frac{{\lambda}\sqrt{{\lambda}}s^{2}}{1+\sqrt{{\lambda}}s}.$

Now we consider another simple transformation: $v:=\sqrt{{\lambda}}z$, so that
P0 turns out to be equivalent to:

$\begin{cases}-\Delta v={\lambda}\left(2v-h_{1}(v)\right)&\mbox{ in
}\Omega,\\\ v>-1&\mbox{ in }\Omega,\\\ \nabla v\cdot\nu=0&\mbox{ on
}\partial\Omega\end{cases}$ (P1)

Choose $s_{0}\in]-1,0[$ (for instance $s_{0}:=1/2$) and define
$\tilde{h}_{1}:\real\to\real$ by:

$\tilde{h}_{1}(s)(=\tilde{h}_{1,s_{0}}(s)):=\begin{cases}h_{1}(s)&\mbox{ if
}s\geq s_{0},\\\ h_{1}(s_{0})+h_{1}^{\prime}(s_{0})(s-s_{0})\\\
\qquad+\dfrac{h_{1}^{\prime\prime}(s_{0})}{2}(s-s_{0})^{2}&\mbox{ if }s\leq
s_{0}.\end{cases}$ (2)

Then $\tilde{h}_{1}\in\mathcal{C}^{2}(\real\times\real)$ and
$\tilde{h}_{1}^{\prime}(0)=h_{1}^{\prime\prime}(0)=0$. Let
$\tilde{H}_{1}:\real\to\real$ denote the primitive function for
$\tilde{h}_{1}$ (i.e. $\tilde{H}_{1}^{\prime}=\tilde{h}_{1}$) such that
$\tilde{H}_{1}(0)=0$. Moreover we consider the functional
$\tilde{I}:W^{1,2}(\Omega)\to\real$ defined by

$\tilde{I}(v):=Q(v)+\tilde{\mathcal{H}}_{1}(v)$

where:

$Q(v):=\frac{1}{2}\int_{\Omega}|\nabla
v|^{2}\,dx-\int_{\Omega}v^{2}\,dx\quad,\quad\tilde{\mathcal{H}}_{1}(v):=\int_{\Omega}\tilde{H}_{1}(v)\,dx.$

Using a classical bifurcation risult for potential operators (see e.g. [8]) we
get that there exists $\rho_{0}>0$ such that for all $\rho\in]0,\rho_{0}[$
there are two distinct pairs $({\lambda}_{1,\rho},v_{1,\rho})$ and
$({\lambda}_{2,\rho},v_{2,\rho})$ which solve:

$\begin{cases}-\Delta v={\lambda}\left(2v-\tilde{h}_{1}(v)\right)&\mbox{ in
}\Omega,\\\ \nabla v\cdot\nu=0&\mbox{ on }\partial\Omega\end{cases}$ (3)

and such that

$\int_{\Omega}v_{i,\rho}^{2}\,dx=\rho^{2}\quad,\quad\lim_{\rho\to
0}{\lambda}_{i,\rho}=\frac{\mu_{k}}{2}\qquad i=1,2.$

Now using a standard regularity argument we can find a constant $K$ such that,
for any solution $w$ of (3):

$\|v\|_{\infty}\leq\|2v-\tilde{h}_{1}(v)\|_{2}\leq K\|v\|_{2}.$

Then, for $\rho_{0}$ small, $v_{i,\rho}>s_{0}$, $i=1,2$, so $v_{i,\rho}$
actually solve P1. Going backwards and setting
$u_{i,\rho}:=\dfrac{1}{\sqrt{{\lambda}_{i,\rho}}}+\sqrt{{\lambda}_{i,\rho}}v_{i,\rho}$,
we find the desired solutions of (P).

∎

##  3 A global bifurcation result for radial solutions

We consider the case $N=2$ and
$\Omega=B(0,R)=\left\\{x\in{}^{2}{\,:\,}\|X\|<R\right\\}$. We look for radial
solutions i.e. $z(x,y)=w(\|(x,y)\|)$. Actually with similar arguments we could
have considered the general case $N\geq 2$. Given $R>0$, it is therefore
convenient to introduce the Hilbert space:

$E:=\left\\{w:[0,R]\to\real{\,:\,}\int_{0}^{R}\rho\dot{w}^{2}\,d\rho<+\infty\right\\}$

endowed with
$\displaystyle{(v,w)_{E}:=\int_{0}^{R}\rho\dot{v}\dot{w}\,d\rho+\int_{0}^{R}\rho
vw\,d\rho}$ and for ${\Lambda_{j}}>0$ the set:

$W_{\lambda}:=\left\\{w\in
E{\,:\,}1+\sqrt{{\lambda}}w(\rho)>0\right\\},\qquad\mathcal{W}:=\left\\{({\lambda},w)\in\real\times
E{\,:\,}w\in W_{\lambda}\right\\}.$

It is clear that $\|w\|_{\infty}\leq constant\|w\|_{E}$ so $W$ is open in $E$
and $\mathcal{W}$ is open in $\real\times E$. As well known the search for
radial solutions leads the equation:

$\left\\{\begin{aligned}
&\ddot{w}+\frac{\dot{w}}{\rho}=-{\lambda}w-\frac{{\lambda}w}{1+\sqrt{{\lambda}}w}=:f_{\lambda}(u),\\\
&\dot{w}(0)=\dot{w}(R)=0.\end{aligned}\right.$ (RP)

By the above we mean that:

$({\lambda},w)\in\mathcal{W},\qquad\int_{0}^{R}\rho\dot{w}\dot{\delta}\,d\rho=\int_{0}^{R}\rho
f_{\lambda}(w)\delta\,d\rho\quad\forall v\in E.$ (4)

It is standard to check that “weak solutions”, i.e. solutions to (4) actually
solve (RP) in a classical sense.

It is clear that $({\lambda},0)$ is a solution for (RP) for any
${\lambda}\in\real$. We call “nontrivial ” solution a pair $({\lambda},w)$
with $w\neq 0$ such that (RP) holds.

######  3.1 Remark.

If $({\lambda},w)$ is a nontrivial solution then ${\lambda}>0$. To see this it
suffices to multiply (RP) by $u$ and integrate over $[0,R]$. Actually this
property is true in the general case (not just in the radial problem).

We shall use the following simple inequality.

######  3.2 Remark.

Let $0<a<b<+\infty$. We have:

$\frac{b-a}{b}\leq\ln\left(\frac{b}{a}\right)\leq\frac{b-a}{a}.$ (5)

We have indeed:

$\ln\left(\frac{b}{a}\right)=\ln\left(1+\frac{b-a}{a}\right)\leq\frac{b-a}{a}$

and

$\ln\left(\frac{b}{a}\right)=-\ln\left(\frac{a}{b}\right)=-\ln\left(1+\frac{a-b}{b}\right)\geq-\frac{a-b}{b}=\frac{b-a}{b}.$

Now let us suppose that a solution $({\lambda},w)$ exists so we can find some
properties and estimates on $w$. Arguing as in the proof of Lemma 2.2 in [2]
we have that either $w=0$ or $[0,R]$ can be split as the union of a finite
number of subintervals $[\rho_{1},\rho_{2}]$ where $w$ has one of the
following behaviors:

(A)

$w(r_{1})>0$, $\dot{w}(r_{1})=0$, $\dot{w}<0$ in $]r_{1},r_{2}]$, and
$w(r_{2})=0$;

(B)

$w(r_{1})=0$, $\dot{w}<0$ in $]r_{1},r_{2}]$, $\dot{w}(r_{2})=0$, and
$w(r_{2})<0$;

(C)

$w(r_{1})<0$, $\dot{w}(r_{1})=0$, $\dot{w}>0$ in $]r_{1},r_{2}]$, and
$w(r_{2})=0$;

(D)

$w(r_{1})=0$, $\dot{w}>0$ in $]r_{1},r_{2}]$, $\dot{w}(r_{2})=0$, and
$w(r_{2})>0$.

So let $w:[r_{1},r_{2}]\to\real$ be as in one of the above cases. Multiplying
(RP) by $\dot{w}$ gives

$\frac{1}{2}\ddot{w}+\dot{w}^{\prime}\frac{\dot{w}^{2}}{\rho}=\frac{d}{d\rho}F_{\lambda}(w)$

Let $p:=\dot{w}^{2}$ the previous equation can be written as:

$\frac{1}{2}\dot{p}+\frac{p}{\rho}=\frac{d}{d\rho}F_{\lambda}(w)$

which is equivalent to

$\frac{d}{d\rho}(\rho^{2}p)=2\rho^{2}\frac{d}{d\rho}F_{\lambda}(w)\rho^{2}=2\rho^{2}\frac{d}{d\rho}F_{1}\left(\sqrt{{\lambda}}w\right).$

We integrate between $\rho_{1}$ and $\rho_{2}$, where
$r_{1}\leq\rho_{1}\leq\rho_{2}\leq r_{2}$:

$\rho_{2}^{2}p(\rho_{2})-\rho_{1}^{2}p(\rho_{1})=2\rho_{2}^{2}F(w(\rho_{2}))-2\rho_{1}^{2}F(w(\rho_{1}))-\int_{\rho_{1}}^{\rho_{2}}4{\sigma}F_{\lambda}(w({\sigma}))\,d{\sigma}$

Notice that $F_{\lambda}$ is increasing on
$\left]-\dfrac{1}{\sqrt{{\lambda}}},0\right[$ and decreasing on $]0,+\infty[$,
so:

${\sigma}\mapsto F_{\lambda}(w({\sigma}))\mbox{ is increasing (decreasing) in
cases {\rm A} and {\rm C} (in cases {\rm B} and {\rm D})}$

We hence get, in cases A and C:

$-2(\rho_{2}^{2}-\rho_{1}^{2})F_{\lambda}(w(\rho_{2}))\leq-\int_{\rho_{1}}^{\rho_{2}}4{\sigma}F_{\lambda}(w({\sigma}))\,d{\sigma}\leq-2(\rho_{2}^{2}-\rho_{1}^{2})F_{\lambda}(w(\rho_{1}))$

while in cases B and D:

$-2(\rho_{2}^{2}-\rho_{1}^{2})F_{\lambda}(w(\rho_{1}))\leq-\int_{\rho_{1}}^{\rho_{2}}4{\sigma}F_{\lambda}(w({\sigma}))\,d{\sigma}\leq-2(\rho_{2}^{2}-\rho_{1}^{2})F_{\lambda}(w(\rho_{2}))$

So in cases A and C we have:

$2\rho_{1}^{2}(F_{\lambda}(w(\rho_{2}))-F_{\lambda}(w(\rho_{1}))\leq\rho_{2}^{2}p(\rho_{2})-\rho_{1}^{2}p(\rho_{1})\leq
2\rho_{2}^{2}(F_{\lambda}(w(\rho_{2}))-F_{\lambda}(w(\rho_{1}))$ (6)

and in cases B and D:

$2\rho_{2}^{2}(F_{\lambda}(w(\rho_{2}))-F_{\lambda}(w(\rho_{1}))\leq\rho_{2}^{2}p(\rho_{2})-\rho_{1}^{2}p(\rho_{1})\leq
2\rho_{1}^{2}(F_{\lambda}(w(\rho_{2}))-F_{\lambda}(w(\rho_{1}))$ (7)

Now we estimate $w(\rho)$ \- we need to take into account all the four cases
A,B,C,D.

Case A. We rename $\bar{\rho}:=r_{1}$, $\rho_{0}:=r_{2}$ and let
$h:=w(\bar{\rho})>0$. We use (6) with $\rho_{1}=\bar{\rho}$ and
$\rho_{2}={\sigma}\in[\bar{\rho},\rho_{0}]$:

$2\bar{\rho}^{2}(F_{\lambda}(w({\sigma}))-F_{\lambda}(h))\leq{\sigma}^{2}\dot{w}({\sigma})^{2}\leq
2{\sigma}^{2}(F_{\lambda}(w({\sigma}))-F_{\lambda}(h)).$

Then we take the square root and divide:

$\sqrt{2}\frac{\bar{\rho}}{{\sigma}}\leq\frac{-\dot{w}({\sigma})}{\sqrt{F_{\lambda}(w({\sigma}))-F_{\lambda}(h)}}\leq\sqrt{2}$

and now we integrate between $\bar{\rho}$ and $\rho\in[\bar{\rho},\rho_{0}]$
getting:

$\sqrt{2}\bar{\rho}\ln\left(\frac{\rho}{\bar{\rho}}\right)\leq-\Phi_{{\lambda},h}(w(\rho))+\Phi_{{\lambda},h}(h)\leq\sqrt{2}(\rho-\bar{\rho})$

where $\Phi_{{\lambda},h}:[0,h]\to\real$ is defined by:

$\Phi_{{\lambda},h}(s):=\int_{0}^{s}\frac{d\xi}{\sqrt{F_{\lambda}(\xi)-F_{\lambda}(h)}}$

(it is simple to check the the integral converges at $\xi=h$). So we deduce:

$\Phi_{{\lambda},h}^{-1}\left(\Phi_{{\lambda},h}(h)-\sqrt{2}\left(\rho-\bar{\rho}\right)\right)\leq
w(\rho)\leq\Phi_{{\lambda},h}^{-1}\left(\Phi_{{\lambda},h}(h)-\sqrt{2}\bar{\rho}\ln\left(\frac{\rho}{\bar{\rho}}\right)\right)$

which we prefer to write as

$\Phi_{{\lambda},h}^{-1}\left(\Phi_{{\lambda},h}(h)+\sqrt{2}\left(\bar{\rho}-\rho\right)\right)\leq
w(\rho)\leq\Phi_{{\lambda},h}^{-1}\left(\Phi_{{\lambda},h}(h)+\sqrt{2}\bar{\rho}\ln\left(\frac{\bar{\rho}}{\rho}\right)\right)$
(8)

In particular, taking $\rho=\rho_{0}$, which gives $w(\rho_{0})=0$, (and using
(5)) we have:

$\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}(\rho_{0}-\bar{\rho})\leq\sqrt{2}\bar{\rho}\ln\left(\frac{\rho_{0}}{\bar{\rho}}\right)\leq\Phi_{{\lambda},h}(h)\leq\sqrt{2}\left(\rho_{0}-\bar{\rho}\right).$
(9)

Moreover taking $\rho_{1}=\bar{\rho}$ and $\rho_{2}=\rho_{0}$ in (6) we have:

$\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}\sqrt{-F_{\lambda}(h))}\leq-\dot{w}(\rho_{0})\leq\sqrt{2}\sqrt{-F_{\lambda}(h))}$
(10)

Case B. We rename $\rho_{0}:=r_{1}$, $\bar{\rho}:=r_{2}$ and let
$h:=w(\bar{\rho})<0$. We use (7) with
$\rho_{1}={\sigma}\in[\rho_{0},\bar{\rho}]$ and $\rho_{2}=\bar{\rho}$:

$2\bar{\rho}^{2}(F_{\lambda}(h)-F_{\lambda}(w({\sigma})))\leq-{\sigma}^{2}\dot{w}({\sigma})^{2}\leq
2{\sigma}^{2}(F_{\lambda}(h)-F_{\lambda}(w({\sigma}))).$

We change sign and proceed as in case A:

$2{\sigma}^{2}(F_{\lambda}(w({\sigma}))-F_{\lambda}(h))\leq{\sigma}^{2}\dot{w}({\sigma})^{2}\leq
2\bar{\rho}^{2}(F_{\lambda}(w({\sigma}))-F_{\lambda}(h))).$

Take the square root and divide:

$\sqrt{2}\leq\frac{-\dot{w}({\sigma})}{\sqrt{F_{\lambda}(w({\sigma}))-F_{\lambda}(h)}}\leq\sqrt{2}\frac{\bar{\rho}}{{\sigma}}.$

Integrate on $[\rho,\bar{\rho}_{0}]$:

$\sqrt{2}(\bar{\rho}-\rho)\leq-\Phi_{{\lambda},h}(h)+\Phi_{{\lambda},h}(w(\rho))\leq\sqrt{2}\bar{\rho}\ln\left(\frac{\bar{\rho}}{\rho}\right)$

defining $\Phi_{{\lambda},h}:[h,0]\to\real$ as in case A. Applying
$\Phi_{{\lambda},h}^{-1}$ we get that (8) holds in case B too. In particular,
taking $\rho=\rho_{0}$ (and using (5)):

$\sqrt{2}(\bar{\rho}-\rho_{0})\leq-\Phi_{{\lambda},h}(h)\leq\sqrt{2}\bar{\rho}\ln\left(\frac{\bar{\rho}}{\rho_{0}}\right)\leq\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}(\bar{\rho}-\rho_{0})$
(11)

and taking $\rho_{1}=\rho_{0}$ and $\rho_{2}=\bar{\rho}$ in (7) we have:

$\sqrt{2}\sqrt{-F_{\lambda}(h))}\leq-\dot{w}(\rho_{0})\leq\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}\sqrt{-F_{\lambda}(h))}$
(12)

Case C. We rename $\bar{\rho}:=r_{1}$, $\rho_{0}:=r_{2}$ end let
$h:=w(\bar{\rho})<0$. Using (6) with $\rho_{1}=\bar{\rho}$ and
$\rho_{2}={\sigma}\in[\bar{\rho},\rho_{0}]$ we obtain the same inequality of
case A. After taking the square root and dividing:

$\sqrt{2}\frac{\bar{\rho}}{{\sigma}}\leq\frac{\dot{w}({\sigma})}{\sqrt{F_{\lambda}(w({\sigma}))-F_{\lambda}(h)}}\leq\sqrt{2}.$

We integrate between $\bar{\rho}$ and $\rho\in[\bar{\rho},\rho_{0}]$ getting:

$\sqrt{2}\bar{\rho}\ln\left(\frac{\rho}{\bar{\rho}}\right)\leq\Phi_{{\lambda},h}(w(\rho))-\Phi_{{\lambda},h}(h)\leq\sqrt{2}(\rho-\bar{\rho})$

with $\Phi_{{\lambda},h}:[h,0]\to\real$ defined as above. So we deduce:

$\Phi_{{\lambda},h}^{-1}\left(\Phi_{{\lambda},h}(h)+\sqrt{2}\bar{\rho}\ln\left(\frac{\rho}{\bar{\rho}}\right)\right)\leq
w(\rho)\leq\Phi_{{\lambda},h}^{-1}\left(\Phi_{{\lambda},h}(h)+\sqrt{2}\left(\rho-\bar{\rho}\right)\right)$
(13)

In particular, taking $\rho=\rho_{0}$ (and using (5)):

$\sqrt{2}\frac{\bar{\rho}}{\bar{\rho}_{0}}(\rho_{0}-\bar{\rho})\leq\sqrt{2}\bar{\rho}\ln\left(\frac{\rho_{0}}{\bar{\rho}}\right)\leq-\Phi_{{\lambda},h}(h)\leq\sqrt{2}\left(\rho_{0}-\bar{\rho}\right).$
(14)

Moreover taking $\rho_{1}=\bar{\rho}$ and $\rho_{2}=\rho_{0}$ in (6) we have:

$\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}\sqrt{-F_{\lambda}(h))}\leq\dot{w}(\rho_{0})\leq\sqrt{2}\sqrt{-F_{\lambda}(h))}$
(15)

Case D. We rename $\rho_{0}:=r_{1}$, $\bar{\rho}:=r_{2}$ and let
$h:=w(\bar{\rho})>0$. Using (7) with
$\rho_{1}={\sigma}\in[\rho_{0},\bar{\rho}]$ and $\rho_{2}=\bar{\rho}$ we
obtain the same inequalities of case B. When we take the square root and
divide:

$\sqrt{2}\leq\frac{\dot{w}({\sigma})}{\sqrt{F_{\lambda}(w({\sigma}))-F_{\lambda}(h)}}\leq\sqrt{2}\frac{\bar{\rho}}{{\sigma}}.$

Integrate on $[\rho,\bar{\rho}_{0}]$:

$\sqrt{2}(\bar{\rho}-\rho)\leq\Phi_{{\lambda},h}(h)-\Phi_{{\lambda},h}(w(\rho))\leq\sqrt{2}\bar{\rho}\ln\left(\frac{\bar{\rho}}{\rho}\right)$

with the usual definition of $\Phi_{{\lambda},h}:[h,0]\to\real$. Applying
$\Phi_{{\lambda},h}^{-1}$ we obtain that (13) holds in case D too. In
particular, taking $\rho=\rho_{0}$ (and using (5)):

$\sqrt{2}\left(\bar{\rho}-\rho_{0}\right)\leq\Phi_{{\lambda},h}(h)\leq\sqrt{2}\bar{\rho}\ln\left(\frac{\bar{\rho}}{\bar{\rho}_{0}}\right)\leq\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}(\bar{\rho}-\rho_{0})$
(16)

and taking $\rho_{1}=\rho_{0}$ and $\rho_{2}=\bar{\rho}_{0}$ in (7) we have:

$\sqrt{2}\sqrt{-F_{\lambda}(h))}\leq\dot{w}(\rho_{0})\leq\sqrt{2}\frac{\bar{\rho}}{\rho_{0}}\sqrt{-F_{\lambda}(h))}$
(17)

Now we have:

$\sqrt{2}\Phi_{{\lambda},h}(h)=\\\
\int_{0}^{h}\frac{d\xi}{\sqrt{F(\sqrt{{\lambda}}\xi)-F(\sqrt{{\lambda}}h)}}=\int_{0}^{1}\frac{h\,d{\sigma}}{\sqrt{F({\sigma}\sqrt{{\lambda}}h)-F(\sqrt{{\lambda}}h)}}=\frac{1}{\sqrt{{\lambda}}}\bar{\Phi}(\sqrt{{\lambda}}h)$

where:

$\bar{\Phi}(s):=\int_{0}^{1}\frac{s\,d{\sigma}}{\sqrt{F({\sigma}s)-F(s)}}=\mathrm{sgn}(s)\int_{0}^{1}\sqrt{\frac{s^{2}}{F({\sigma}s)-F(s)}}\,d{\sigma}.$

With simple computations:

$\lim_{s\to
0}\frac{s^{2}}{F({\sigma}s)-F(s)}=\frac{1}{1-{\sigma}^{2}},\qquad\lim_{s\to+\infty}\frac{s^{2}}{F({\sigma}s)-F(s)}=\frac{2}{1-{\sigma}^{2}},$

and

$\lim_{s\to-1^{-}}\frac{s^{2}}{F({\sigma}s)-F(s)}=0$

So we deduce that

$\displaystyle\lim_{h\to
0^{+}}\Phi_{{\lambda},h}(h)=\frac{\pi}{2\sqrt{2{\lambda}}},\quad\lim_{h\to+\infty}\Phi_{{\lambda},h}(h)=\frac{\pi}{2\sqrt{{\lambda}}},$
(18) $\displaystyle\lim_{h\to
0^{-}}\Phi_{{\lambda},h}(h)=-\frac{\pi}{2\sqrt{2{\lambda}}},\quad\lim_{h\to-1^{+}}\Phi_{{\lambda},h}(h)=0.$
(19)

To state the main result we need some notation, which we take from [2, 6]. For
$k\in\mathbb{N}$, $k\geq 1$, we consider

$\displaystyle\mathcal{S}:=\left\\{({\lambda},w)\in\mathcal{W}{\,:\,}{\lambda},w)\mbox{
is a solution to \eqref{eqn:radial-equation}}\right\\}$
$\displaystyle\mathcal{S}^{+}_{k}:=\left\\{({\lambda},w)\in\mathcal{S}{\,:\,}\mbox{
$w$ has $k$ nodes in $]0,R[$},w(0)>0\right\\},$
$\displaystyle\mathcal{S}^{-}_{k}:=\left\\{({\lambda},w)\in\mathcal{S}{\,:\,}\mbox{
$w$ has $k$ nodes in $]0,R[$},w(0)<0\right\\},$

We also consider the two eigenvalue problems:

$\ddot{w}+\frac{\dot{w}}{\rho}=-\mu w,\qquad\dot{w}(0)=\dot{w}(R)=0.$ (RP*)
$\ddot{v}+\frac{\dot{v}}{\rho}=-\nu v,\qquad\dot{v}(0)=0,{v}(R)=0.$ (RP0)

It is clear that $w\neq 0$ and $\mu\neq 0$ solve (RP*) if and only if, for
some integer $k\geq 1$:

$\mu=\mu_{k}:=\left(\frac{y_{k}}{R}\right)^{2}$ (20)

where $y_{k}$ denotes the $k$-th nontrivial zero of $J_{0}^{\prime}$ and
$J_{0}$ is the first Bessel function, and

$w=\alpha w_{k},\quad\alpha\in\real,\qquad
w_{k}(\rho):=J_{0}\left(\frac{y_{k}}{R}\rho\right).$ (21)

For the sake of completeness we can agree that $\mu_{0}=0$ and
$w_{0}(\rho)=J_{0}(0)$. In the same way $v\neq 0$ and $\nu$ solve (RP0) if and
only if, for some integer $k\geq 1$:

$\nu=\nu_{k}:=\left(\frac{z_{k}}{R}\right)^{2}$ (22)

where $z_{k}$ is the $k$-th zero of $J_{0}$ and

$v=\alpha v_{k},\quad\alpha\in\real,\qquad
v_{k}(\rho):=J_{0}\left(\frac{z_{k}}{R}\rho\right).$ (23)

Notice that $\nu_{k}<\mu_{k}<\nu_{k+1}$ for all $k$.

######  3.3 Theorem.

Let $\mu_{k}>0$ be an eigenvalue for (RP*). Then $\mathcal{S}_{k}^{+}$ is a
connected set and

  * •

$(\mu_{k}/2,0)\in\overline{\mathcal{S}_{k}^{+}}$;

  * •

$\displaystyle{0<\inf\left\\{{\lambda}\in\real{\,:\,}\exists w\in E\mbox{ with
}({\lambda},w)\in\mathcal{S}_{k}^{+}\right\\}}$;

  * •

$\displaystyle{\sup\left\\{{\lambda}\in\real{\,:\,}\exists w\in E\mbox{ with
}({\lambda},w)\in\mathcal{S}_{k}^{+}\right\\}<+\infty}$;

  * •

$\mathcal{S}_{k}^{+}$ is unbounded and contains a sequence
$({\lambda}_{n},w_{n})$ such that $\|w_{n}\|_{E}\to\infty$ and

$\lim_{n\to\infty}{\lambda}_{n}=\begin{cases}\mu_{k/2}&\mbox{ if $k$ is
even},\\\ \nu_{(k+1)/2}&\mbox{ if $k$ is odd}.\end{cases}$ (24)

The proof of (3.3) will be obtained from some preliminary statements.

######  3.4 Remark.

If $({\lambda},w)\in\mathcal{S}^{+}$ ( resp.
$({\lambda},w)\in\mathcal{S}^{+}$), and
$0=\rho_{0}<\rho_{1},\dots,\rho_{k}<\rho_{k+1=R}$, $\rho_{1},\dots,\rho_{k}$
being the nodal points of $w$, then:

$\rho_{i+1}-\rho_{i}\geq\frac{\pi}{4\sqrt{{\lambda}}}\ \mbox{for $i$
even}\quad(\mbox{resp. for $i$ odd}).$ (25)

This is easily seen using the right hand sides of the inequalities (9), (14),
and (18).

######  3.5 Lemma.

For any integer $k$ there esist two constants $\underline{{\lambda}}_{k}$ and
$\overline{{\lambda}}_{k}$ such that

$({\lambda},w)\in\mathcal{S}_{k}^{+}\cup\mathcal{S}_{k}^{-}\Rightarrow
0<\underline{{\lambda}}_{k}\leq{\lambda}\leq\overline{{\lambda}}_{k}<+\infty$
(26)

###### Proof.

Take any subinterval $[r_{1},r_{2}]$ as in cases A–D and consider the first
eigenvalue $\bar{\mu}=\bar{\mu}(r_{1},r_{2})$ for the mixed type boundary
condition:

$\left\\{\begin{aligned} -&(\rho\dot{w})^{\prime}=\mu w\quad\mbox{on
}]r_{1},r_{2}[\\\ &\dot{w}(r_{1})=0,w(r_{2})=0\qquad(\mbox{resp.
}w(r_{1})=0,\dot{w}(r_{2})=0)\end{aligned}\right.$

in cases A,C (resp. cases C,D). We can choose an eigenfunction $\bar{e}$
corresponding to $\bar{\mu}$ so that $z\bar{e}>0$ in $]r_{1},r_{2}[$.
Multiplying (RP) by $\bar{e}$ and integrating over $[r_{1},r_{2}]$ yields:

$\bar{\mu}\int_{r_{1}}^{r_{2}}\rho
z\bar{e}\,d\rho={\lambda}\int_{r_{1}}^{r_{2}}\rho
z\bar{e}\left(1+\frac{1}{1+\sqrt{{\lambda}}z}\right)\,d\rho.$

This implies:

${\lambda}\int_{r_{1}}^{r_{2}}\rho
z\bar{e}\,d\rho\leq\bar{\mu}\int_{r_{1}}^{r_{2}}\rho z\bar{e}\,d\rho\leq
2{\lambda}\int_{r_{1}}^{r_{2}}\rho z\bar{e}\,d\rho$

which gives $\dfrac{\bar{\mu}}{2}\leq{\lambda}\leq\bar{\mu}.$ Now since
$]r_{1},r_{2}[\subset]0,R[$ we have $\bar{\mu}\geq\bar{\mu}[0,R]$. On the
other side since $w$ has $k$ nodal points we can choose $r_{1}$, $r_{2}$ such
that $r_{2}-r_{1}\geq R/k$, which implies
$\bar{\mu}\leq\sup\limits_{b-a=R/k}\bar{\mu}(a,b)<+\infty$. This proves (26) ∎

######  3.6 Lemma.

Let $({\lambda}_{n},w_{n})$ be a sequence in $\mathcal{S}_{k}^{+}$. Then we
can consider $0<\rho_{1,n}<\cdots<\rho_{k,n}<R$ to be the nodes of $w_{n}$ and
set $\rho_{0,n}:=0$, $\rho_{k+1,n}:=R$; in ths way $wn(\rho)>0$ on
$]\rho_{1},\rho_{i+1}[$ if $i$ is even and $wn(\rho)<0$ on
$]\rho_{1},\rho_{i+1}[$ if $i$ is odd. The following facts are equivalent:

(a)

$\lim_{n\to\infty}\sup_{\rho\in[0,R]}w_{n}(\rho)=+\infty;$

(b)

$\lim_{n\to\infty}\inf_{\rho\in[0,R]}(1+{\lambda}_{n}w_{n}(\rho))=0;$

(c)

$\lim_{n\to\infty}\sup_{\rho\in[\rho_{i,n},\rho_{i+1,n}]}w_{n}(\rho)=+\infty\quad\mbox{if
$i$ is even};$

(d)

$\lim_{n\to\infty}\inf_{\rho\in[\rho_{i,n},\rho_{i+1,n}]}(1+{\lambda}_{n}w_{n}(\rho))=0\quad\mbox{if
$i$ is odd};$

(e)

$\lim_{n\to\infty}\rho_{1+1,n}-\rho_{i,n}=0\quad\mbox{if $i$ is odd};$

Moreover, if any of the above holds, then (24) holds.

###### Proof.

We can assume, passing to a subsequence that
${\lambda}_{n}\to\hat{\lambda}\in[\underline{{\lambda}}_{k},\overline{\overline{{\lambda}}_{k}}]$.
First notice that for all $i$ even (corresponding to $w>0$) we have:

$\rho_{i+1,n}-\rho_{i,n}\geq\frac{\pi}{4\sqrt{\underline{{\lambda}}_{k}}}$

as we can infer from (9) or (16) and the behaviour of $\Phi_{{\lambda},h}(h)$
in (18).

Let

$h_{i,n}:=\max_{\rho_{i,n}\leq\rho_{i+1,n}}w(\rho)\mbox{ for $i$ even},\qquad
h_{i,n}:=\min_{\rho_{i,n}\leq\rho_{i+1,n}}w(\rho)\mbox{ for $i$ odd}$

Then for any $i$ even:

$h_{i,n}\to+\infty\Leftrightarrow\Phi_{{\lambda}_{n},h_{i,n}}(h_{i,n})\to\frac{\pi}{2\sqrt{\hat{\lambda}}}\Leftrightarrow\dot{w}(\rho_{i,n})\to+\infty\Leftrightarrow\dot{w}(\rho_{i+1,n})\to-\infty.$

This can be deduced from (18), (10), and (17). In the same way, using (19),
(12), and (15) we get that, for $i$ odd:

$1+\sqrt{{\lambda}_{n}}h_{i,n}\to
0\Leftrightarrow\Phi_{{\lambda}_{n},h_{i,n}}(h_{i,n})\to
0\Leftrightarrow\dot{w}(\rho_{i,n})\to-\infty\Leftrightarrow\dot{w}(\rho_{i+1,n})\to+\infty.$

Now we prove our claims. Let $\bar{i}\in\left\\{0,\dots,k\right\\}$ with
$\bar{i}$ even (resp. odd) and suppose that $h_{\bar{i},n}\to+\infty$ (resp.
$1+\sqrt{{\lambda}_{n}}h_{\bar{i},n}\to 0$). Then
$F_{{\lambda}_{n}}(h_{\bar{i},n})\to+\infty$ (resp.
$F_{{\lambda}_{n}}(h_{\bar{i},n})\to-\infty$) and by (10), (17) ( (12), (15) )
we get that:

$\dot{w}_{n}(\rho_{\bar{i},n})\to+\infty,\
\dot{w}_{n}(\rho_{\bar{i}+1,n})\to-\infty\quad(\dot{w}_{n}(\rho_{\bar{i},n})\to-\infty,\
\dot{w}_{n}(\rho_{\bar{i}+1,n})\to+\infty)$

which in turn implies:

$F_{{\lambda}_{n}}(h_{\bar{i}-1,n})\to-\infty\mbox{ ( resp. }+\infty),\
F_{{\lambda}_{n}}(h_{\bar{i}+1,n})\to-\infty\mbox{ ( resp. }+\infty)$

(with the obvious exceptions when $\bar{i}-1<0$ or $\bar{i}+1>k$). So we get:

$1+\sqrt{{\lambda}_{n}}h_{\bar{i}-1,n}\to 0\ (h_{\bar{i}-1,n}\to+\infty),\
1+\sqrt{{\lambda}_{n}}h_{\bar{i}+1,n}\to 0\ (h_{\bar{i}+1,n}\to+\infty).$

This shows that the property $|F_{{\lambda}}(h_{i,n})|\to+\infty$ “propagates”
from the $i$-th interval to the previous and to the next one. From this it is
easy to deduce that a–d are all equivalent. To prove that they are equivalent
to e just use (9),(11),(14),(16), depending on the case, noticing that
$\rho_{1,n}\geq\dfrac{\pi}{4\pi\underline{{\lambda}}_{k}}$, as from (25) (this
would not be possible if we were considering $\mathcal{S}_{k}^{-}$).

Finally suppose that $({\lambda}_{n},w_{n})$ verifies any of a–e. Then
$\|w_{n}\|_{\infty}\to+\infty$. Let
$\hat{w}_{n}:=\dfrac{w_{n}}{\|w_{n}\|_{\infty}}$. We can suppose that
$\hat{w}_{n}\rightharpoonup\hat{w}$ in $E$ and that:

$\displaystyle\rho_{1,n}\to\rho_{1},\ \rho_{2j-1,n}\to\rho_{j},\
\rho_{2j,n}\to\rho_{j}\mbox{ $1\leq j\leq k/2$},\ \rho_{k,n}\to R\mbox{ if $k$
is odd},$

where $0=\rho_{0}<\rho_{1}<\cdots<\rho_{h}<\rho_{h}+1=R$ and $h=\lfloor
k/2\rfloor$ (so $\rho_{1}=R$ when $k=1$). It is not difficult to prove that
$\hat{w}(\rho)>0$ in $]\rho_{i},\rho_{i+1}[$ if $i=0,\dots,h$,
$\hat{w}(\rho_{1})=\cdots=\hat{w}(\rho_{h})=0$, ${\hat{w}}^{\prime}(0)=0$ and
${\hat{w}}^{\prime}(R)=0$ is $k$ is even while $\hat{w}(R)=0$ is $k$ is odd.
Moreover for any $i=0,\dots,h$:

$-(\rho\hat{w}^{\prime})^{\prime}=\hat{\lambda}\hat{w}\quad\mbox{on
}]\rho_{i},\rho_{i+1}[$

Now we can rearrange $\hat{w}$ defining
$\tilde{w}:=\sum_{j=0}^{h}(-1)^{j}\alpha_{j}\hat{w}\mathbb{1}_{[\rho_{j},\rho_{j+1}]}$,
where $\alpha_{1}=1$ and
$\alpha_{j}\hat{w}^{\prime}_{-}(\rho_{j})=\alpha_{j+1}\hat{w}^{\prime}_{+}(\rho_{j})$,
$j=1,\dots,h$. In this way $(\hat{\lambda},\tilde{w})$ is an eigenvalue –
eigenfunction pair relative for problem (21) if $k$ is even and of (23) if $k$
is odd. Since $\tilde{w}$ has $h=k/2$ nodal points for $k$ even and
$h+1=(k+1)/2$ if $k$ is odd, then (24) holds. ∎

###### Proof of (3.3).

If $\varepsilon{}\in]0,1[$ we set:

$\mathcal{O}_{\varepsilon}{}:=\left\\{({\lambda},w)\in
E{\,:\,}\varepsilon{}<{\lambda}<\varepsilon{}^{-1},1+\sqrt{{\lambda}}w(\rho)>\varepsilon{},w(\rho)<\varepsilon{}^{-1}\
\forall\rho\in[0,R]\right\\}$

Clearly $\mathcal{O}_{\varepsilon}{}$ is an open set with
$\mathcal{O}_{\varepsilon}{}\subset\mathcal{W}$. Moreover
$(\mu_{k/2},0)\in\mathcal{O}_{\varepsilon}{}$ if $\varepsilon{}$ is
sufficently small. Define $\tilde{h}_{{\lambda},\varepsilon{}}$ as in (2) with
$s_{0}=\varepsilon{}$ and let
$\tilde{h}_{{\lambda}}(s):=\tilde{h}_{1}(\sqrt{{\lambda}}s)$. Using [6] we get
there that there exists a pair $({\lambda}_{\varepsilon}{},w_{\varepsilon}{})$
in $\partial\mathcal{O}_{\varepsilon}{}$, with $w_{\varepsilon}{}$ having $k$
nodal points, which solves Problem (RP) with
$\tilde{h}_{\varepsilon{},{\lambda}}:=\tilde{h}_{\varepsilon}{}({\lambda},\cdot)$
instead of $h_{\lambda}$. Since
$({\lambda},w)\in\partial\mathcal{O}_{\varepsilon}{}\Rightarrow\tilde{h}_{\varepsilon}{}({\lambda},w)=h_{\lambda}(w)$,
we get that
$({\lambda}_{\varepsilon}{},w_{\varepsilon}{})\in\mathcal{S}_{k}^{+}$. For
$\varepsilon{}$ small we have
$\varepsilon{}<\underline{{\lambda}}_{k}\leq\overline{{\lambda}}_{k}<\varepsilon{}^{-1}$
so we get
$w_{\varepsilon}{}\in\partial\left\\{1+\sqrt{{\lambda}_{\varepsilon}{}}w>\varepsilon{},w<\varepsilon{}^{-1}\right\\}$
i.e. there exists a point $\rho_{\varepsilon}{}\in[0,R]$ such that

$\mbox{either}\quad
1+\sqrt{{\lambda}_{\varepsilon}{}}w_{\varepsilon}{}(\rho_{\varepsilon}{})=\varepsilon{}\qquad\quad\mbox{or}\quad
w_{\varepsilon}{}(\rho_{\varepsilon}{})=\varepsilon{}^{-1}.$

We can find a sequence $\varepsilon{}_{n}\to 0$ such that the corresponding
$({\lambda}_{n},w_{n}):=({\lambda}_{\varepsilon{}_{n}},w_{\varepsilon{}_{n}})$
verify one of the above properties for all $n\in\mathbb{N}$. If the first one
holds for all $n$, then $({\lambda}_{n},w_{n})$ verifies b of Lemma (3.6); in
the second case $({\lambda}_{n},w_{n})$ verifies a of Lemma (3.6). Then by
Lemma (3.6) $\|w_{n}\|_{\infty}\to\infty$ and (24) holds. This proves the
Theorem. ∎

######  3.7 Remark.

As a consequence of Theorem (3.3) we get that for any $h\geq 1$ integer and
any ${\lambda}$ strictly between ${\lambda}_{h}$ and ${\lambda}_{2h}/2$ there
exists $u$ such that $({\lambda},u)$ solves P. The same is true for all
${\lambda}$ strictly between $\nu_{h}$ and ${\lambda}_{2h-1}/2$

######  3.8 Remark.

The proof of P fails if we follow the bifurcation branch
$({\lambda}_{\rho},w_{\rho})$ with $w_{\rho}(0)<0$. In this case it seems
possible that the branch tends to a point $(\tilde{\lambda},\tilde{w})$ where
$\sqrt{\tilde{\lambda}}\tilde{w}(0)=-1$ (but
$\sqrt{\tilde{\lambda}}\tilde{w}(0)>-1$ for $\rho>0$). This phenomenon, if
true, would be worth studying.

######  3.9 Remark.

The computations of this section show that, if $\Omega$ is the ball, then
there are no solutions for the Dirichlet problem. It is indeed impossible to
construct a (nontrivial) solution $({\lambda},w)$ for RP with $w(R)=0$,

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