# Symmetry and monotonicity results for solutions of semilinear PDEs in
sector-like domains

Francesca Gladiali and Antonio Greco

###### Abstract

In this paper we consider semilinear PDEs, with a convex nonlinearity, in a
sector-like domain. Using cylindrical coordinates $(r,\theta,z)$, we
investigate the shape of solutions whose derivative in $\theta$ vanishes at
the boundary. We prove that any solution with Morse index less than two must
be either independent of $\theta$ or strictly monotone with respect to
$\theta$. In the special case of a planar domain, the result holds in a
circular sector as well as in an annular, and it can also be extended to a
rectangular domain. The corresponding problem in higher dimensions is also
considered, as well as an extension to unbounded domains. The proof is based
on a rotating-plane argument: a convenient manifold is introduced in order to
avoid overlapping the domain with its reflected image in the case when its
opening is larger than $\pi$.

##  1 Introduction

In this paper we consider solutions of semilinear elliptic PDEs in a sector-
like domain $\Omega_{0\beta}\subset\mathbb{R}^{N}$, $N\geq 2$. To describe the
results, for every $x=(x_{1},\ldots,x_{N})\in\mathbb{R}^{N}$ we define
$d(x)=\sqrt{x_{1}^{2}+x_{2}^{2}\,}$ and
$z=(x_{3},\ldots,x_{N})\in\mathbb{R}^{N-2}$, and we denote by $\Upsilon$ the
$(N-2)$-dimensional subspace $\Upsilon:=\\{x\in\mathbb{R}^{N}:d(x)=0\\}$.
Observe that each point $x\in\mathbb{R}^{N}\setminus\Upsilon$ can be
represented in cylindrical coordinates as $i(x)=(r,\theta,z)$, where $r=d(x)$,
and $\theta=\theta(x)\in[0,2\pi)$ is uniquely determined by

$\begin{cases}x_{1}=r\,\cos\theta,\\\ x_{2}=r\,\sin\theta.\end{cases}$

If $N=2$, then $(r,\theta)$ are the polar coordinates of $x$, the set
$\Upsilon$ contains just the origin, and the symbol $z$ should be ignored. In
general, a rotation of $\mathbb{R}^{N}$ about the origin is a linear mapping
$x\mapsto Ax$ associated to an orthogonal matrix $A$. In the present paper we
deal with cylindrical rotations, i.e., the particular rotations whose matrix
$A=A^{N}_{\theta}$ is given by

$A^{2}_{\theta}=\begin{pmatrix}\cos\theta&\sin\theta\\\
-\sin\theta&\cos\theta\end{pmatrix}\quad\mbox{and}\quad
A^{N}_{\theta}=\begin{pmatrix}A^{2}_{\theta}&0\\\ 0&I^{N-2}\\\
\end{pmatrix}\\!,\ N\geq 3,$

where $I^{N-2}$ is the $(N-2)$-dimensional identity matrix. In the first part
of the paper we deal with bounded domains. More precisely, let $\Omega$ be a
bounded, Lipschitz domain which is invariant under cylindrical rotations in
the sense that $x\in\Omega$ if and only if $A_{\theta}\,x\in\Omega$ for every
$\theta\in[0,2\pi)$. For $\theta_{1},\theta_{2}\in[0,2\pi)$ satisfying
$\theta_{1}<\theta_{2}$ we introduce the bounded open sector

$\Omega_{\theta_{1}\theta_{2}}:=\\{\,x\in\Omega\setminus\Upsilon:\theta(x)\in(\theta_{1},\theta_{2})\,\\}$

whose boundary is made of the open, flat surfaces
$\Gamma_{\\!\theta_{i}}:=\\{\,x\in\Omega\setminus\Upsilon:\theta(x)=\theta_{i}\,\\}$,
$i=1,2$, the torical surface
$\gamma_{\theta_{1}\theta_{2}}:=\\{\,x\in\partial\Omega\setminus\Upsilon:\theta(x)\in[\theta_{1},\theta_{2}]\,\\}$
and the (possibly empty) set $\gamma:=\overline{\Omega}\cap\Upsilon$. The
typical examples of the domain $\Omega_{\theta_{1}\theta_{2}}$ are: a sector
of a sphere, a sector of a cylinder, a sector of an annulus, a sector of a
torus, a sector of a cone. But also more complicated domains as cylinders (or
spheres) with cavities can be considered for $\Omega$, and even tori with a
torical, coaxial tunnel inside. Observe that $\gamma$ may be empty (which is
the case, for instance, when $\Omega$ is a torus), and can be disconnected as
in the case when $\Omega$ is a cylinder in $\mathbb{R}^{3}$ with a cavity. Now
fix $\beta\in(0,2\pi)$ and consider the boundary-value problem

$\left\\{\begin{array}[]{ll}-\Delta u=f(d(x),z,u)&\text{ in
}\Omega_{0\beta},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr\frac{\partial
u}{\partial\theta}=0&\text{ on
$\gamma_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$,}\end{array}\right.$
(1.1)

where $\Delta$ is the Laplace operator, and $f(r,z,u)$ is locally Hölder
continuous on $[0,+\infty)\times\mathbb{R}^{N-1}$ together with
$f^{\prime}=\partial f/\partial u$. Observe that $u_{\theta}=\frac{\partial
u}{\partial\theta}$ is the normal derivative of $u$ on
$\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$ while it is the tangential derivative on
$\gamma_{0\beta}$. Let $\cal X$ be the set of all functions $u\in
C^{2}(\Omega_{0\beta})\cap C^{0}(\overline{\Omega}_{0\beta})\cap
H^{1}(\Omega_{0\beta})$ such that $u_{\theta}\in H^{1}(\Omega_{0\beta})$. We
investigate functions $u\in\cal X$ satisfying (1.1) pointwise. Using [16,
Prop. 6.1], it can be proved that the solution $u$ is smooth, namely $C^{2}$,
in $\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$. The assumption
that $u_{\theta}\in H^{1}(\Omega_{0\beta})$ is introduced to avoid
technicalities and it is satisfied, for instance, if $u\in
W^{2,2}(\Omega_{0\beta})$. Such an assumption is dropped in the last part of
the paper, where we use a truncation technique to deal with unbounded domains
(see, in particular, (8.1)-(8.2)). The boundary condition $u_{\theta}=0$ on
$\gamma_{0\beta}$ can be interpreted in the sense of traces. However, since
$u$ is continuous up to there, the condition implies that the values of $u$ on
$\gamma_{0\beta}$ are constant in $\theta$. Thus, (1.1) is a shorthand for the
following mixed problem of Dirichlet-Neumann type:

$\left\\{\begin{array}[]{ll}-\Delta u=f(d(x),z,u)&\text{ in
}\Omega_{0\beta},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr u=g(d(x),z)&\text{
on }\gamma_{0\beta},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr\frac{\partial
u}{\partial\nu}=0&\text{ on
}\Gamma_{\\!0}\cup\Gamma_{\\!\beta}\end{array}\right.$ (1.2)

where $g(r,z)$ is a continuous function, and $\nu$ denotes the outward unit
normal to $\partial\Omega_{0\beta}$. If $u$ is a solution to (1.1), we denote
by $I$ the identity operator and by

$L_{u}:=-\Delta-f^{\prime}(d(x),z,u(x))\,I$

the linearized operator at $u$. Furthermore,

$Q_{D}(v):=\int_{D}\Big{(}|\nabla
v(x)|^{2}-f^{\prime}(d(x),z,u(x))\,v^{2}(x)\Big{)}\,dx,\quad v\in H^{1}(D)$
(1.3)

is the quadratic form associated to $L_{u}$ in an open subset
$D\subset\Omega_{0\beta}$. In fact, the operator $L_{u}$ and the quadratic
form $Q_{D}$ are well defined for every $u\in L^{\infty}(\Omega_{0\beta})$. If
$\Gamma$ is a sufficiently smooth (possibly disconnected) part of $\partial
D$, we denote by

$H^{1}_{\Gamma}(D):=\\{v\in H^{1}(D):v=0\text{ on }\Gamma\\}$

the Sobolev space of square-summable functions having square-summable weak
derivatives and vanishing trace along $\Gamma$, and, for $i\geq 1$, we denote
by $\lambda_{i}(H^{1}_{\Gamma}(D))$ the eigenvalues of the operator $L_{u}$ in
$H^{1}_{\Gamma}(D)$. We will assume that the solution $u$ to (1.1) satisfies
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$. Our main result
in bounded domains is the following:

######  Theorem 1.1.

Let $u\in\cal X$ be a solution to (1.1), where $f(r,z,u)$ is convex with
respect to $u$.

  1. 1.

If $\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ then $u$ is
constant with respect to the angular variable $\theta$.

  2. 2.

If
$\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))<0\leq\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))$
then $u$ is either independent of $\theta$ or strictly monotone with respect
to $\theta$ in $\Omega_{0\beta}$. In the last case, $u$ has a nonvanishing
derivative $u_{\theta}$, the first Dirichlet eigenvalue
$\lambda_{1}(H^{1}_{0}(\Omega_{0\beta}))$ equals zero and $u_{\theta}$ is a
corresponding eigenfunction.

In Section 4 we state a corresponding result for the case when
$\Omega_{0\beta}$ is a cylindrical domain, which is in some sense the
asymptotic shape of a sector-like domain when the opening $\beta$ is small.
All the occurrences in Theorem 1.1 can happen and we will provide some
examples in Section 5. In particular we show that if the first Dirichlet
eigenvalue $\lambda_{1}(H^{1}_{0}(\Omega_{0\beta}))=0$ then $u_{\theta}$ may
well vanish identically. We also show that if
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))<0$ it can happen that
the solution $u$ is neither independent of $\theta$ nor monotone with respect
to $\theta$. Starting from Section 6 we extend the result to unbounded
domains: a precise statement is given in Theorem 6.4.

######  Remark 1.2.

In the statement of Theorem 1.1 it is enough to require that $f(r,z,s)$ is
convex with respect to $s$ as long as $s\in(m,M)$, where
$m=\inf\limits_{\Omega_{0\beta}}u(x)$ and
$M=\sup\limits_{\Omega_{0\beta}}u(x)$. This assumption has been introduced in
the paper [15] in order to compare the quadratic form $Q_{D}(v)$ with the
quadratic form associated to the equation satisfied by the difference of $u$
and its reflection with respect to some $\Gamma_{\\!\theta}$ (as we also do in
Section 2).
In the case of solutions to (1.2) the assumption
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ is equivalent to
consider solutions of Morse index $1$ and it satisfied by any ground state
solution and any solution of mountain pass type. It is needed in the proof to
start the “rotating plane” method in place of the positiveness of the solution
$u$ which is usually required in the moving plane procedure. Moreover a bound
on the Morse index usually implies an $L^{\infty}$-bound of the solution as
observed in [2] in the case of a Dirichlet problem under some growth condition
on the nonlinear term $f$.
Some comments on the assumption $u_{\theta}\in H^{1}(\Omega_{0\beta})$, which
is needed in the proof of Lemma 3.2 and Lemma 3.3, and consequently enters in
Theorem 1.1. By standard regularity theory, any weak solution $u\in
H^{1}(\Omega_{0\beta})$ of (1.2) belongs to the smoothness class
$C^{2}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})\cap
C^{0}(\overline{\Omega}_{0\beta})$. Furthermore, at least in the case when
$\gamma=\emptyset$, if $\Omega\in C^{2,\alpha}$ and $g(d(x),z)\in
C^{2,\alpha}$ then $u\in C^{2,\alpha}(\overline{\Omega}_{0\beta})$ showing
that $u_{\theta}\in H^{1}(\Omega_{0\beta})$.
In the particular case of dimension $N=2$, the domain $\Omega_{0\beta}$ is a
sector of the annulus or a sector of the disc so that $\gamma_{0\beta}$ is
always smooth. In the case of the annulus then the assumption $u_{\theta}\in
H^{1}(\Omega_{0\beta})$ is satisfied if $g(r)$ belongs to $C^{2,\alpha}$. In
the case of the disc, instead, we can use [14] and [1] to gain the $W^{2,2}$
regularity of the solution $u$ when the domain is convex.
Some previous results on cylindrical symmetry of low Morse index solutions for
a mixed problem can be found in [7] where, differently from our case, the
equation holds in the whole of $\Omega$ and a nonlinear Neumann condition is
imposed on a subset of $\partial\Omega$. Finally we recall that in the case
when $\Omega_{0\beta}$ is a sector of the circle and $\beta<\pi$, the radial
symmetry of the positive solution to (1.2) with $g(d(x),z)=0$ was proved in
[4] using an involved version of the moving plane method.

##  2 Preliminary results

The proof of Theorem 1.1 is based on the rotating plane method (rotating line,
in two dimensions). The method was previously used in [15, 17] to prove
Schwarz symmetry for solutions of low Morse index to the Dirichlet problem in
radially symmetric bounded domains. To be more precise, we use the reflection
$\sigma_{\alpha}$ with respect to $\Gamma_{\\!\alpha}$, where $\alpha$ is an
angular parameter in the given interval $(0,\beta)$. In the special case when
$\alpha\in(0,\pi)$ we may easily define
$\sigma_{\alpha}\colon\Omega_{0,2\alpha}\to\Omega_{0,2\alpha}$ by prescribing
that $i(\sigma_{\alpha}(x))=(r,\,2\alpha-\theta,\,z)$ when
$i(x)=(r,\theta,z)$. However, in order to simplify and generalize the
subsequent arguments, it is convenient to extend the definition of
$\sigma_{\alpha}$ to the general case when $\alpha\in\mathbb{R}$. To this aim
we consider the Riemannian manifold ${\cal
M}:=\\{(r,t,z):r\in(0,+\infty),t\in\mathbb{R},z\in\mathbb{R}^{N-2}\\}$ endowed
with the flat metric whose first fundamental form is
$ds^{2}=dr^{2}+r^{2}\,dt^{2}+dz^{2}$. Every sector-like domain
$\Omega_{\theta_{1}\theta_{2}}\subset\mathbb{R}^{N}$ is isometrically embedded
into $\cal M$ by the mapping $i\colon\Omega_{\theta_{1}\theta_{2}}\to\cal M$
$i(x)=(r,\theta,z)$. We will identify $\Omega_{\theta_{1}\theta_{2}}$ with its
image $i(\Omega_{\theta_{1}\theta_{2}})$ and, more generally, we will use the
notation $\Omega_{\theta_{1}\theta_{2}}$ to denote the submanifold

$\Omega_{\theta_{1}\theta_{2}}=\\{\,(r,t,z)\in{\cal
M}:t\in(\theta_{1},\theta_{2}),(r,0,z)\in\Omega\,\\}$

for every $\theta_{1},\theta_{2}\in\mathbb{R}$ with $\theta_{1}<\theta_{2}$.
Now, for any given $\alpha\in\mathbb{R}$, we define the function
$\sigma_{\alpha}\colon{\cal M}\to{\cal M}$ by letting
$\sigma_{\alpha}(r,t,z)=(r,\,2\alpha-t,\,z)$. Thus, for instance, if we take
$\alpha\in(\pi,2\pi)$ and apply $\sigma_{\alpha}$ to some point $(r,\theta,z)$
with $\theta\in(0,\,2\alpha-2\pi)$, we obtain $(r,\,2\alpha-\theta,\,z)$ where
$2\alpha-\theta>2\pi$. The last point, as an element of $\cal M$, is distinct
from the point $(r,\,2\alpha-\theta-2\pi,\,z)$. We denote still by $x$ an
arbitrary point $(r,t,z)\in\cal M$, for shortness, and we define the function
$d\colon{\cal M}\to(0,+\infty)$ by letting $d(x)=r$. Thus, any function $u$
defined in $\Omega_{0\beta}\cup\,\Gamma_{\\!0}\cup\,\Gamma_{\\!\beta}$ can be
extended to the submanifold $\Omega_{-\beta,2\beta}$ by letting
$u(x):=u(\sigma_{0}(x))$ for $x\in\Omega_{-\beta,0}$ and
$u(x):=u(\sigma_{\beta}(x))$ for $x\in\Omega_{\beta,2\beta}$. If $u$ is a
solution to (1.1) and we extend it as above, the extended function (still
denoted by $u$) satisfies $-\Delta u=f(d(x),z,u)$ not only in
$\Omega_{0\beta}$, but also in $\Omega_{-\beta,0}$ and in
$\Omega_{\beta,2\beta}$. To be precise, here $\Delta$ denotes the Beltrami-
Laplace operator on $\cal M$, which reduces to the standard Laplacian because
$\cal M$ is flat. Moreover, since $f$ is Hölder continuous, standard
regularity results imply that $u$ is regular through $\Gamma_{\\!0}$ and
$\Gamma_{\\!\beta}$, and satisfies the equation in all of
$\Omega_{-\beta,2\beta}$. Hence for every $\alpha\in(0,\beta)$, the function
$w_{\alpha}\colon\Omega_{0\beta}\to\mathbb{R}$ given by
$w_{\alpha}(x):=u(\sigma_{\alpha}(x))-u(x)$ is well defined and satisfies

$-\Delta w_{\alpha}-c_{\alpha}(x)\,w_{\alpha}=0\ \text{ in }\Omega_{0\beta},$
(2.1)

where

$c_{\alpha}(x):=\int_{0}^{1}f^{\prime}\big{(}d(x),\,z,\,t\,u(\sigma_{\alpha}(x))+(1-t)\,u(x)\big{)}\,dt,$

together with

$w_{\alpha}=0\text{ on }\gamma_{0\beta}\cup\,\Gamma_{\\!\alpha}.$ (2.2)

In the sequel, the monotonicity of a solution $u$ of (1.1) will be detected
through the sign of $w_{\alpha}$: this motivates our interest in the function
spaces $H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha})$
and
$H^{1}_{\gamma_{\alpha\beta}\cup\,\Gamma_{\\!\alpha}}(\Omega_{\alpha\beta})$.
We have:

######  Lemma 2.1 (Splitting lemma).

Let $u\in L^{\infty}(\Omega_{0\beta})$. Then for every $\alpha\in(0,\beta)$ we
have

$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\leq\max\Big{\\{}\,\lambda_{1}(H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha})),\
\lambda_{1}(H^{1}_{\gamma_{\alpha\beta}\cup\,\Gamma_{\\!\alpha}}(\Omega_{\alpha\beta}))\,\Big{\\}}.$

###### Proof.

Let $\varphi_{0\alpha},\varphi_{\alpha\beta}$ be the eigenfunctions associated
to the eigenvalues
$\lambda_{1}(H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha}))$
and
$\lambda_{1}(H^{1}_{\gamma_{\alpha\beta}\cup\,\Gamma_{\\!\alpha}}(\Omega_{\alpha\beta}))$,
respectively. Without loss of generality we assume
$\|\varphi_{0\alpha}\|_{\Omega_{0\alpha}}=\|\varphi_{\alpha\beta}\|_{\Omega_{\alpha\beta}}=1$,
where $\|\cdot\|_{D}$ denotes the norm in $L^{2}(D)$. Since
$\varphi_{0\alpha},\varphi_{\alpha\beta}$ satisfy a Dirichlet boundary
condition on $\Gamma_{\\!\alpha}$, we may extend them to the whole
$\Omega_{0\beta}$ by letting $\varphi_{0\alpha}:=0$ in $\Omega_{\alpha\beta}$
and $\varphi_{\alpha\beta}:=0$ in $\Omega_{0\alpha}$. The extended functions
are orthogonal in $H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})$ since their
supports are disjoint, and we may define the two-dimensional linear subspace
$W_{0}=\\{\,\phi\in
H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}):\phi=a\,\varphi_{0\alpha}+b\,\varphi_{\alpha\beta}\mbox{
for }a,b\in\mathbb{R}\,\\}$. Moreover, for every
$\phi=a\,\varphi_{0\alpha}+b\,\varphi_{\alpha\beta}\not\equiv 0$ we have

$\displaystyle\frac{\,Q_{\Omega_{0\beta}}(\phi)\,}{\|\phi\|^{2}_{\Omega_{0\beta}}}$
$\displaystyle=\frac{\,a^{2}\,Q_{\Omega_{0\alpha}}(\varphi_{0\alpha})+b^{2}\,Q_{\Omega_{\alpha\beta}}(\varphi_{\alpha\beta})\,}{a^{2}+b^{2}}$
$\displaystyle\leq\max\Big{\\{}\,Q_{\Omega_{0\alpha}}(\varphi_{0\alpha}),\
Q_{\Omega_{\alpha\beta}}(\varphi_{\alpha\beta})\,\Big{\\}}$
$\displaystyle=\max\Big{\\{}\,\lambda_{1}(H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha})),\
\lambda_{1}(H^{1}_{\gamma_{\alpha\beta}\cup\,\Gamma_{\\!\alpha}}(\Omega_{\alpha\beta}))\,\Big{\\}}.$

The variational formulation of the eigenvalues (see [6, Theorem 1.42 (iii)]),
sometimes called Poincaré’s minimax characterization, ensures that

$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})))=\inf_{\stackrel{{\scriptstyle
W\subset H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})}}{{{\rm dim}(W)=2}}}\
\max_{\phi\in
W\setminus\\{0\\}}\frac{\,Q_{\Omega_{0\beta}}(\phi)\,}{\|\phi\|^{2}_{\Omega_{0\beta}}},$

and the conclusion follows immediately. ∎

######  Corollary 2.2.

Assume $\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ for some
$u\in L^{\infty}(\Omega_{0\beta})$, and let $\alpha\in(0,\beta)$. Then either
$\lambda_{1}(H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha}))\geq
0$ or
$\lambda_{1}(H^{1}_{\gamma_{\alpha\beta}\cup\,\Gamma_{\\!\alpha}}(\Omega_{\alpha\beta}))\geq
0$.

The positivity of the first eigenvalue is a necessary and sufficient condition
in order that the weak maximum principle holds: see, for instance, [6, Theorem
1.50]. When the first eigenvalue is allowed to vanish, we have a sign-
preservation property for weak supersolutions of the Dirichlet problem
associated to the operator $L_{u}$, as well as to the corresponding mixed
problem. A weak supersolution of the Dirichlet problem for $L_{u}$ in
$\Omega_{0\alpha}$ is a function $v\in H^{1}(\Omega_{0\alpha})$ such that
$v\geq 0$ almost everywhere on $\partial\Omega_{0\alpha}$ and the inequality

$\int_{\Omega_{0\alpha}}\Big{(}\nabla
v\,\nabla\varphi-f^{\prime}(d(x),\,z,\,u)\,v\,\varphi\Big{)}dx\geq 0$ (2.3)

holds for every nonnegative $\varphi\in H^{1}_{0}(\Omega_{0\alpha})$. Recall,
further, that a weak supersolution of the mixed problem for $L_{u}$ in
$\Omega_{0\alpha}$, with Neumann condition on $\Gamma_{\\!0}$, is a function
$v\in H^{1}(\Omega_{0\alpha})$ such that $v\geq 0$ a.e. on
$\gamma_{0\alpha}\cup\Gamma_{\\!\alpha}$ and the inequality (2.3) holds for
every nonnegative $\varphi\in
H^{1}_{\gamma_{0\alpha}\cup\Gamma_{\\!\alpha}}(\Omega_{0\alpha})$ [6, p. 15].
A smooth function $v$ satisfying $L_{u}\,v\geq 0$ pointwise in
$\Omega_{0\alpha}$ together with $v\geq 0$ on
$\gamma_{0\alpha}\cup\Gamma_{\\!\alpha}$ and $\partial v/\partial\nu\geq 0$ on
$\Gamma_{\\!0}$ is a also a weak supersolution of the mixed problem.

######  Lemma 2.3 (Sign preservation).

Assume
$\lambda_{1}(H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha}))\allowbreak\geq
0$ for some $u\in L^{\infty}(\Omega_{0\beta})$ and $\alpha\in(0,\beta)$. Let
$v\in C^{1}(\Omega_{0\alpha}\cup\Gamma_{\\!0})\cap H^{1}(\Omega_{0\alpha})$ be
a weak supersolution of the Dirichlet problem associated to the operator
$L_{u}$ in $\Omega_{0\alpha}$, or a weak supersolution of the corresponding
mixed problem with Neumann condition on $\Gamma_{0}$.

  1. 1.

If $v$ satisfies $v=0$ a.e. on $\Gamma_{\\!\alpha}$ without being identically
zero in $\Omega_{0\alpha}$, then either $v>0$ in $\Omega_{0\alpha}$ or $v<0$
in $\Omega_{0\alpha}$.

  2. 2.

If $v>0$ a.e. on $\Gamma_{\\!\alpha}$, then $v>0$ in $\Omega_{0\alpha}$.

###### Proof.

The negative part $v^{-}(x)=\min\\{\,v(x),0\,\\}\leq 0$ satisfies $v^{-}=0$
a.e. on $\gamma_{0\alpha}\cup\Gamma_{\\!\alpha}$. Furthermore, if $v$ is a
weak supersolution of the Dirichlet problem, then $v^{-}=0$ on $\Gamma_{\\!0}$
as well. In both cases, we may take $\varphi=-v^{-}$ in (2.3) and get
$Q_{\Omega_{0\alpha}}(v^{-})\leq 0$. Since

$0\leq\lambda_{1}(H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha}))=\inf\left\\{\left.\frac{Q_{\Omega_{0\alpha}}(\varphi)}{\|\varphi\|^{2}_{\Omega_{0\alpha}}}\
\right|\ \varphi\in
H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha})\setminus\\{0\\}\right\\}$
(2.4)

we deduce that either $v^{-}$ vanishes identically in $\Omega_{0\alpha}$, or
we may let $\varphi=v^{-}$ to minimize the quotient in (2.4). In the last case
$v^{-}$ is a first eigenfunction of $L_{u}$ in
$H^{1}_{\gamma_{0\alpha}\cup\,\Gamma_{\\!\alpha}}(\Omega_{0\alpha})$ and
therefore we must have $v^{-}<0$ in all of $\Omega_{0\alpha}$ (see [6, Theorem
1.42 (vi)]). If, instead, $v^{-}$ vanishes identically, then $v\geq 0$ in
$\Omega_{0\alpha}$ and by the strong maximum principle (see [6, Theorem 1.28])
we have either $v\equiv 0$ or $v>0$ in $\Omega_{0\alpha}$. The two claims
follow at once. ∎

The starting point of our rotating plane argument, developed in Section 3, is
the case when $\alpha=\frac{\beta}{2}$. The function $w_{\frac{\beta}{2}}$,
which is odd with respect to $\Gamma_{\\!\frac{\beta}{2}}$, satisfies
$w_{\frac{\beta}{2}}=0$ on $\gamma_{0\beta}\cup\,\Gamma_{\\!\frac{\beta}{2}}$.
Therefore the following corollary applies:

######  Corollary 2.4.

Assume $u\in\cal X$ is a solution to (1.1) such that
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ and assume
$f(r,z,u)$ is convex with respect to $u$. If $w_{\frac{\beta}{2}}$ does not
vanish identically in $\Omega_{0\beta}$, then either $w_{\frac{\beta}{2}}>0$
in $\Omega_{0\frac{\beta}{2}}$ or $w_{\frac{\beta}{2}}>0$ in
$\Omega_{\frac{\beta}{2}\beta}$.

###### Proof.

By Corollary 2.2 we have that one among
$\lambda_{1}(H^{1}_{\gamma_{0\frac{\beta}{2}}\cup\,\Gamma_{\\!\frac{\beta}{2}}}(\Omega_{0\frac{\beta}{2}}))$
and
$\lambda_{1}(H^{1}_{\gamma_{\frac{\beta}{2}\beta}\cup\,\Gamma_{\\!\frac{\beta}{2}}}(\Omega_{\frac{\beta}{2}\beta}))$
is nonnegative. Suppose that
$\lambda_{1}(H^{1}_{\gamma_{0\frac{\beta}{2}}\cup\,\Gamma_{\\!\frac{\beta}{2}}}(\Omega_{0\frac{\beta}{2}}))\geq
0$, the other case being analogous. By the convexity of $f$, the function
$v=w_{\frac{\beta}{2}}\in
C^{2}(\Omega_{0\frac{\beta}{2}}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\frac{\beta}{2}})\cap
C^{0}(\overline{\Omega}_{0\frac{\beta}{2}})\cap
H^{1}(\Omega_{0\frac{\beta}{2}})$ satisfies $L_{u}\,v\geq 0$ in
$\Omega_{0\frac{\beta}{2}}$. Moreover $v$ vanishes on
$\gamma_{0\frac{\beta}{2}}\cup\Gamma_{\\!\frac{\beta}{2}}$ and satisfies
$\frac{\partial v}{\partial\nu}=-\frac{\partial v}{\partial\theta}=0$ on
$\Gamma_{\\!0}$. Taking into account that $w_{\frac{\beta}{2}}$ is odd with
respect to $\Gamma_{\\!\frac{\beta}{2}}$, the conclusion follows by letting
$\alpha=\frac{\beta}{2}$ in Claim 1 of Lemma 2.3.∎

##  3 Proof of Theorem 1.1

The proof of Theorem 1.1 is based on a fine interplay of the derivative
$u_{\theta}$ and the function $w_{\alpha}$ introduced in Section 2. In fact,
the strict monotonicity of $u$ with respect to $\theta$ in $\Omega_{0\beta}$
holds if and only if $w_{\alpha}>0$ in $\Omega_{0\alpha}$ for every
$\alpha\in(0,\beta)$. By contrast, symmetry of $u$ with respect to
$\Gamma_{\\!\frac{\beta}{2}}$, and constancy with respect to $\theta$ as a
special case, occur when $w_{\frac{\beta}{2}}$ vanishes identically. First of
all we observe

######  Proposition 3.1.

If $u\in C^{2}(\Omega_{0\beta})$ satisfies the equation in (1.1) pointwise,
then the derivative $u_{\theta}$ still belongs to $C^{2}(\Omega_{0\beta})$ and
is a classical solution to

$L_{u}\,u_{\theta}=0\ \ \text{ in }\Omega_{0\beta}.$ (3.1)

###### Proof.

As usual, we call test function any smooth function $\varphi\in
C^{\infty}(\Omega_{0\beta})$ compactly supported in $\Omega_{0\beta}$. The set
of all test functions is denoted by $C^{\infty}_{0}(\Omega_{0\beta})$. To
prove the claim, we differentiate the equality

$\int_{\Omega_{0\beta}}\nabla
u\,\nabla\varphi\,dx=\int_{\Omega_{0\beta}}\textstyle
f(d(x),z,u)\,\varphi\,dx$

with respect to $\theta$. Since $\varphi_{\theta}$ is still a test function,
the terms containing $\varphi_{\theta}$ cancel each other, and we obtain

$\int_{\Omega_{0\beta}}\nabla
u_{\theta}\,\nabla\varphi\,dx=\int_{\Omega_{0\beta}}\textstyle
f^{\prime}(d(x),z,u)\,u_{\theta}\,\varphi\,dx.$ (3.2)

Hence $u_{\theta}$ is a weak solution of $L_{u}\,u_{\theta}=0$. Since
$f^{\prime}$ is Hölder continuous by assumption, $u_{\theta}$ is a classical
solution, as claimed. ∎

To proceed further, observe that by Corollary 2.4 the function
$w_{\frac{\beta}{2}}(x)$ does not change sign in $\Omega_{0\frac{\beta}{2}}$
nor in $\Omega_{\frac{\beta}{2}\beta}$. We consider the case when
$w_{\frac{\beta}{2}}(x)\equiv 0$ in $\Omega_{0\beta}$ first:

######  Lemma 3.2.

Assume $u\in\cal X$ solves (1.1) and satisfies
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$. If
$w_{\frac{\beta}{2}}(x)\equiv 0$ in $\Omega_{0\beta}$ then $u$ is constant
with respect to $\theta$.

###### Proof.

If $w_{\frac{\beta}{2}}(x)\equiv 0$ then $u$ is symmetric with respect to
$\Gamma_{\\!\frac{\beta}{2}}$ and the linearized operator $L_{u}$ is invariant
under reflections about $\Gamma_{\\!\frac{\beta}{2}}$. In particular, the
first eigenfunction $\varphi_{1}$ in
$H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})$ is symmetric with respect to
$\Gamma_{\\!\frac{\beta}{2}}$. Furthermore, since $u_{\theta}\in
H^{1}_{0}(\Omega_{0\beta})$, we may let $\varphi$ in (3.2) approach
$u_{\theta}$ in $H^{1}(\Omega_{0\beta})$ and deduce
$Q_{\Omega_{0\beta}}(u_{\theta})=0$. In order to prove the lemma it is enough
to show that $u_{\theta}$ vanishes identically in $\Omega_{0\beta}$. The
argument is by contradiction. Observe that $\varphi_{1}\perp u_{\theta}$,
i.e., $\varphi_{1}$ is orthogonal to $u_{\theta}$ in $L^{2}(\Omega_{0\beta})$,
because $u_{\theta}$ is odd with respect to $\Gamma_{\\!\frac{\beta}{2}}$.
Furthermore, by assumption we have

$0\leq\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))=\inf\left\\{\left.\frac{\,Q_{\Omega_{0\beta}}(\varphi)\,}{\|\varphi\|^{2}_{\Omega_{0\beta}}}\
\right|\ \varphi_{1}\perp\varphi\in
H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})\setminus\\{0\\}\right\\}.$

If $u_{\theta}\not\equiv 0$, we see that the infimum is achieved at
$\varphi=u_{\theta}$ because $Q_{\Omega_{0\beta}}(u_{\theta})=0$. But then
$u_{\theta}$, being a minimizer of the Rayleigh quotient in
$H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})$ under the constraint
$\varphi_{1}\perp\varphi$, should satisfy the Neumann conditions $\partial
u_{\theta}/\partial\nu=0$ on $\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$. If this
were the case, we could extend $u_{\theta}$ to $\Omega_{-\beta,\beta}$ by
letting

$\tilde{u}_{\theta}(x)=\begin{cases}u_{\theta}(x),&x\in\Omega_{0\beta}\cup\Gamma_{\\!0};\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0,&x\in\Omega_{-\beta,0},\end{cases}$
(3.3)

thus obtaining a function $\tilde{u}_{\theta}\in C^{1}(\Omega_{-\beta,\beta})$
that satisfies $L_{u}\,\tilde{u}_{\theta}=0$ in the weak sense in
$\Omega_{-\beta,\beta}$ and vanishes identically in $\Omega_{-\beta,0}$. Since
we are assuming that $u_{\theta}$ does not vanish identically in
$\Omega_{0\beta}$, this is in contrast with the unique continuation property
[18, p. 519], and the lemma follows. ∎

Next we consider the case when $w_{\frac{\beta}{2}}>0$ in
$\Omega_{0\frac{\beta}{2}}$. The case when $w_{\frac{\beta}{2}}>0$ in
$\Omega_{\frac{\beta}{2}\beta}$ is similar. Taking Corollary 2.2 into account,
we may assume
$\lambda_{1}(H^{1}_{\gamma_{0\frac{\beta}{2}}\cup\,\Gamma_{\\!\frac{\beta}{2}}}(\Omega_{0\frac{\beta}{2}}))\geq
0$ without loss of generality. Let us show that the inequality $u_{\theta}>0$
holds:

######  Lemma 3.3.

Assume $u\in\cal X$ is a solution to (1.1) such that
$\lambda_{1}(H^{1}_{\gamma_{0\frac{\beta}{2}}\cup\,\Gamma_{\\!\frac{\beta}{2}}}(\Omega_{0\frac{\beta}{2}}))\geq
0$. If $w_{\frac{\beta}{2}}(x)>0$ in $\Omega_{0\frac{\beta}{2}}$ then

$u_{\theta}>0\ \ \text{in\/
}\Omega_{0\frac{\beta}{2}}\cup\Gamma_{\\!\frac{\beta}{2}}.$ (3.4)

###### Proof.

Arguing as in the proof of Lemma 3.2 we see that the angular derivative
$u_{\theta}$ (which cannot be odd, now, because $u$ is not symmetric about
$\Gamma_{\\!\frac{\beta}{2}}$) satisfies (3.1) together with the boundary
condition $u_{\theta}=0$ on $\gamma_{0\beta}\cup\Gamma_{0}\cup\Gamma_{\beta}$.
Hence $u_{\theta}\in H^{1}_{0}(\Omega_{0\beta})\cap
C^{1}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$. Let us prove
that $u_{\theta}>0$ on $\Gamma_{\\!\frac{\beta}{2}}$. By (2.1),
$w_{\frac{\beta}{2}}$ satisfies $-\Delta
w_{\frac{\beta}{2}}-c_{\frac{\beta}{2}}(x)\,w_{\frac{\beta}{2}}=0$ in
$\Omega_{0\frac{\beta}{2}}$. Since $w_{\frac{\beta}{2}}(x)>0$ in
$\Omega_{0\frac{\beta}{2}}$ by assumption, and $w=0$ on
$\Gamma_{\\!\frac{\beta}{2}}$ by definition, the Hopf boundary lemma [6,
Theorem 1.28] implies that $\frac{\partial w}{\partial\nu}<0$ on
$\Gamma_{\\!\frac{\beta}{2}}$. Here $\nu$ is the outward unit normal to
$\Omega_{0\frac{\beta}{2}}$ at $\Gamma_{\\!\frac{\beta}{2}}$. This means

$\left.\frac{\partial
w}{\partial\nu}\right|_{\Gamma_{\\!\frac{\beta}{2}}}=-2\left.\frac{\partial
u}{\partial\nu}\right|_{\Gamma_{\\!\frac{\beta}{2}}}=-2\left.\frac{\partial
u}{\partial\theta}\right|_{\Gamma_{\\!\frac{\beta}{2}}}<0.$ (3.5)

Hence $u_{\theta}>0$ on $\Gamma_{\\!\frac{\beta}{2}}$, as claimed, and the
conclusion follows from Claim 2 of Lemma 2.3. ∎

We are now ready to apply the rotating plane method.

###### Proof of Theorem 1.1.

We assume that $\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$
and prove that either $u_{\theta}\equiv 0$ or $u_{\theta}$ keeps its sign in
$\Omega_{0\beta}$. In the last case, from (3.1) and the boundary condition it
follows immediately that $u_{\theta}$ is a first Dirichlet eigenfunction of
$L_{u}$ in $\Omega_{0\beta}$ and therefore
$\lambda_{1}(H^{1}_{0}(\Omega_{0\beta}))=0$. At the end of the proof, a final
observation shows that if
$\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ then $u$ is
constant in $\theta$, which completes the proof of the theorem.

If $w_{\frac{\beta}{2}}\equiv 0$ in $\Omega_{0\beta}$ then $u$ is independent
of $\theta$ by Lemma 3.2. Otherwise, by Corollary 2.4, either
$w_{\frac{\beta}{2}}>0$ in $\Omega_{0\frac{\beta}{2}}$ or
$w_{\frac{\beta}{2}}>0$ in $\Omega_{\frac{\beta}{2}\beta}$. By Corollary 2.2
we may assume
$\lambda_{1}(H^{1}_{\gamma_{0\frac{\beta}{2}}\cup\,\Gamma_{\\!\frac{\beta}{2}}}(\Omega_{0\frac{\beta}{2}}))\geq
0$ and $w_{\frac{\beta}{2}}>0$ in $\Omega_{0\frac{\beta}{2}}$ without loss of
generality (indeed, if $w_{\frac{\beta}{2}}<0$ in $\Omega_{0\frac{\beta}{2}}$
the argument is similar). Let us prove that $u_{\theta}>0$ in
$\Omega_{0\beta}$. Denote by $A^{\prime}$ the set of all
$\alpha^{\prime}\in[\frac{\beta}{2},\beta]$ having the property that for every
$\alpha\in(0,\alpha^{\prime})$ the function $w_{\alpha}$ is positive in
$\Omega_{0\alpha}$.

Part I. The set $A^{\prime}$ contains $\frac{\beta}{2}$ and therefore it is
not empty. To see this, we have to prove that $u(x)<u(\sigma_{\alpha}(x))$ for
every $\alpha\in(0,\frac{\beta}{2})$ and $x\in\Omega_{0\alpha}$, which is
equivalent to say that $u(r,\theta_{1},z)<u(r,\theta_{2},z)$ for
$0<\theta_{1}<\theta_{2}$ such that $\theta_{1}+\theta_{2}<\beta$. If
$\theta_{2}\leq\frac{\beta}{2}$ the conclusion follows from (3.4). Otherwise
we must have $\theta_{1}<\beta-\theta_{2}<\frac{\beta}{2}<\theta_{2}$ and we
argue as follows. Let $x_{i}$, for $i=1,2$, be the point whose cylindrical
coordinates are $(r,\theta_{i},z)$, and observe that the coordinates of
$\sigma_{\frac{\beta}{2}}(x_{2})$ are $(r,\,\beta-\theta_{2},\,z)$. We may
write $u(\sigma_{\frac{\beta}{2}}(x_{2}))<u(x_{2})$ because we are considering
$w_{\frac{\beta}{2}}>0$ in $\Omega_{0\frac{\beta}{2}}$ (hence
$w_{\frac{\beta}{2}}<0$ in $\Omega_{\frac{\beta}{2}\beta}$). Furthermore we
have $u(x_{1})<u(\sigma_{\frac{\beta}{2}}(x_{2}))$ because of (3.4), and
therefore $\frac{\beta}{2}\in A^{\prime}$ as claimed. We note for later usage
that the preceding argument still holds when $\theta_{1}=0$, hence

$w_{\alpha}>0$ on $\Gamma_{\\!0}$ for every $\alpha\in(0,\frac{\beta}{2})$.
(3.6)

Part II. The set $A^{\prime}$ is a closed subinterval of
$[\frac{\beta}{2},\beta]$. Indeed, by the definition of $A^{\prime}$ it
follows that if $\alpha^{\prime},\alpha^{\prime\prime}$ satisfy
$\frac{\beta}{2}\leq\alpha^{\prime}\leq\alpha^{\prime\prime}$ and
$\alpha^{\prime\prime}\in A^{\prime}$ then $\alpha^{\prime}\in A^{\prime}$,
hence $A^{\prime}$ is an interval. It also follows immediately that
$A^{\prime}$ is closed, and we may write
$A^{\prime}=[\frac{\beta}{2},\tilde{\alpha}]$ for some
$\tilde{\alpha}\in[\frac{\beta}{2},\beta]$.

Part III. The derivative $u_{\theta}$ is positive in the sector
$\Omega_{0\tilde{\alpha}}$. To check this, we verify that $u_{\theta}>0$ along
$\Gamma_{\\!\alpha}$ for each $\alpha\in(0,\tilde{\alpha})$. The argument is
the same as in Lemma 3.3: since $w_{\alpha}$ is positive in $\Omega_{0\alpha}$
by the definition of $A^{\prime}$, and satisfies (2.1), by the Hopf boundary-
point lemma it must also satisfy $\partial w_{\alpha}/\partial\nu<0$ along
$\Gamma_{\\!\alpha}$, where $w_{\alpha}=0$. Hence $u_{\theta}>0$ along
$\Gamma_{\\!\alpha}$, as claimed.

Part IV. If $\tilde{\alpha}<\beta$ then $w_{\tilde{\alpha}}>0$ in
$\Omega_{0\tilde{\alpha}}$. Indeed, by continuity we may write
$w_{\tilde{\alpha}}\geq 0$ in $\Omega_{0\tilde{\alpha}}$, and by the convexity
of $f$ it follows that $w_{\tilde{\alpha}}$ satisfies the inequality
$L_{u}\,w_{\tilde{\alpha}}\geq 0$. Hence either $w_{\tilde{\alpha}}>0$ or
$w_{\tilde{\alpha}}\equiv 0$ in $\Omega_{0\tilde{\alpha}}$ by the strong
maximum principle. Let us exclude the second case. If
$w_{\tilde{\alpha}}\equiv 0$ in $\Omega_{0\tilde{\alpha}}$ then
$\tilde{\alpha}>\frac{\beta}{2}$ because we are assuming
$w_{\frac{\beta}{2}}>0$ in $\Omega_{0\frac{\beta}{2}}$. Furthermore, by
differentiating the equality $u(r,\,2\tilde{\alpha}-\theta,\,z)=u(r,\theta,z)$
we obtain $u_{\theta}(r,\,2\tilde{\alpha}-\theta,\,z)=-u_{\theta}(r,\theta,z)$
for $(r,\theta,z)\in\Omega_{0\tilde{\alpha}}$, which yields at
$\tilde{\theta}=2\tilde{\alpha}-\beta\in(0,\tilde{\alpha})$

$0=\left.\frac{\partial
u}{\partial\nu}\right|_{\Gamma_{\\!\beta}}=u_{\theta}(r,\beta,z)=-u_{\theta}(r,\tilde{\theta},z),$

contradicting Part III. Hence we can write the implication
$\tilde{\alpha}<\beta\implies w_{\tilde{\alpha}}>0$ in
$\Omega_{0\tilde{\alpha}}$.

Part V. Recall that for every $\alpha\in(0,\beta)$ the function $w_{\alpha}$
is well defined in $\Omega_{0\alpha}$ and satisfies (2.1)-(2.2) together with
(3.6). To complete the boundary conditions, let us check that

$w_{\alpha}>0$ on $\Gamma_{\\!0}$ for every
$\alpha\in[\frac{\beta}{2},\beta)$, (3.7)

which is equivalent to $u(r,0,z)<u(r,2\alpha,z)$. By the definition of the
extended function $u$, we have $u(r,2\alpha,z)=u(r,\,2\beta-2\alpha,\,z)$. The
point $(r,\,2\beta-2\alpha,\,z)$ can be rewritten as
$\sigma_{\alpha^{\prime}}(r,0,z)$ by choosing
$\alpha^{\prime}=\beta-\alpha>0$. If $\alpha\in(\frac{\beta}{2},\beta)$ then
$\alpha^{\prime}<\frac{\beta}{2}$ and the inequality
$u(r,0,z)<u(\sigma_{\alpha^{\prime}}(r,0,z))$ follows from (3.6). Finally,
since $w_{\frac{\beta}{2}}$ is positive in $\Omega_{0\frac{\beta}{2}}$ by
assumption and satisfies $\partial w_{\frac{\beta}{2}}/\partial\nu=0$ on
$\Gamma_{\\!0}$, the Hopf boundary point lemma prevents it from vanishing
there.

Part VI. The second endpoint $\tilde{\alpha}\geq\frac{\beta}{2}$ of the
interval $A^{\prime}$ is in fact $\beta$. To prove this, we show that if
$\tilde{\alpha}<\beta$ then for some $\varepsilon_{0}>0$ and for every
$\alpha\in[\tilde{\alpha},\,\tilde{\alpha}+\varepsilon_{0})$ we have
$w_{\alpha}>0$ in $\Omega_{0\alpha}$, which contradicts the definition of
$\tilde{\alpha}$. In order to reach our goal we need a suitable maximum
principle. Observe, to begin with, that the sup-norm $\|c_{\alpha}\|_{\infty}$
is bounded uniformly with respect to $\alpha$. Therefore, by the weak maximum
principle in small domains [6, Theorem 1.20] there exists $\delta>0$
(independent of $\alpha$) such that if an open subset
$D\subset\Omega_{0\alpha}$ satisfies $|D|<\delta$, and if $w_{\alpha}\geq 0$
on $\partial D$, then $w_{\alpha}\geq 0$ in all of $D$. Accordingly, let us
fix a (nonempty) compact subset $K\subset\Omega_{0\tilde{\alpha}}$ such that
$|\Omega_{0\tilde{\alpha}}\setminus K|<\frac{\delta}{2}$ and define

$\eta=\min_{K}w_{\tilde{\alpha}}.$

Recall that $\eta>0$ by Part IV. Then, for a convenient $\varepsilon_{0}>0$ we
achieve that for all
$\alpha\in[\tilde{\alpha},\allowbreak\,\tilde{\alpha}+\varepsilon_{0})$ the
open set $D_{\alpha}=\Omega_{0\alpha}\setminus K$ satisfies
$|D_{\alpha}|<\delta$ and

$w_{\alpha}>\frac{\eta}{2}$ in $K$.

Taking into account the boundary conditions established in Part V, we may
write $w_{\alpha}\geq 0$ on $\partial D_{\alpha}$ and therefore, by the weak
maximum principle in small domains, the inequality $w_{\alpha}\geq 0$ holds in
$D_{\alpha}$, and consequently in all of $\Omega_{0\alpha}$. More precisely,
since $w_{\alpha}>0$ in $K$, by the strong maximum principle we obtain
$w_{\alpha}>0$ in $\Omega_{0\alpha}$ for every
$\alpha\in[\tilde{\alpha},\,\tilde{\alpha}+\varepsilon_{0})$. This
contradiction proves that $\tilde{\alpha}=\beta$, as claimed.

Conclusion. By Part VI, the solution $u$ is strictly increasing with respect
to $\theta$ in $\Omega_{0\beta}$, and by Part III it satisfies $u_{\theta}>0$.
We have thus proved that if
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ then either
$u_{\theta}$ vanishes identically in $\Omega_{0\beta}$ or it keeps its sign
there. In order to complete the proof, it suffices to check that if
$\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ the last case
cannot occur, i.e., $u_{\theta}$ does not keep its sign in $\Omega_{0\beta}$.
In fact, if this were the case, we would have
$0\leq\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\leq\lambda_{1}(H^{1}_{0}(\Omega_{0\beta}))=0$
by the comparison of Dirichlet and Neumann eigenvalues (see, for instance, [5,
Theorem 1, p. 408] or [12, Example 11.13]). This shows that $u_{\theta}$ would
also minimize the Rayleigh quotient in
$H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})$ and therefore it should satisfy the
Neumann boundary condition $\partial u_{\theta}/\partial\nu=0$ on
$\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$. But then $u_{\theta}$ would extend as in
(3.3) to a $C^{1}$-eigenfunction in $\Omega_{-\beta,\beta}$ vanishing
identically in $\Omega_{-\beta,0}$, and this is impossible by the unique
continuation property. ∎

##  4 Cylindrical domains

When the opening of a sector-like domain tends to zero, the domain is
asymptotic to a cylinder. To manage with this case we have to arrange the
notation. Changes are specific to the present section. Here we denote by
$\omega$ a Lipschitz, bounded, $(N-1)$-dimensional domain lying in the
hyperplane $x_{1}=0$ and by $\Omega_{\theta_{1}\theta_{2}}$ the cylinder
$(\theta_{1},\theta_{2})\times\omega$, where
$\theta_{1},\theta_{2}\in\mathbb{R}$ satisfy $\theta_{1}<\theta_{2}$. Given
$\beta\in(0,+\infty)$, we let $\mathcal{X}$ be the set of functions $u\in
C^{2}(\Omega_{0\beta})\cap C^{0}(\overline{\Omega}_{0\beta})\cap
H^{1}(\Omega_{0\beta})$ such that $\frac{\partial u}{\partial x_{1}}\in
H^{1}(\Omega_{0\beta})$ and we consider functions $u\in\mathcal{X}$ that
satisfy, pointwise, the boundary-value problem:

$\left\\{\begin{array}[]{ll}-\Delta u=f(z,u)&\text{ in }\Omega_{0\beta},\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr\frac{\partial u}{\partial
x_{1}}=0&\text{ on $\partial\Omega_{0\beta}$, }\end{array}\right.$ (4.1)

where $z=(x_{2},\ldots,x_{N})\in\mathbb{R}^{N-1}$ and $f(z,u)$ is locally
Hölder continuous in $\mathbb{R}^{N}$ together with its derivative
$f^{\prime}=\partial f/\partial u$. Letting
$\Gamma_{\\!\alpha}=\\{\alpha\\}\times\omega$ and
$\gamma_{\theta_{1}\theta_{2}}=[\theta_{1},\theta_{2}]\times\partial\omega$,
problem (4.1) is equivalent to the following mixed boundary-value problem:

$\left\\{\begin{array}[]{ll}-\Delta u=f(z,u)&\text{ in }\Omega_{0\beta},\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr u=g(z)&\text{ on }\gamma_{0\beta},\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr\frac{\partial u}{\partial\nu}=0&\text{
on }\Gamma_{\\!0}\cup\Gamma_{\\!\beta}\end{array}\right.$

where $g$ is continuous on $\gamma_{0\beta}$. The preceding results can be
extended to such a case. We have:

######  Theorem 4.1.

Let $u\in\mathcal{X}$ be a solution to (4.1), where $f(z,u)$ is convex with
respect to $u$.

  1. 1.

If $\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ then $u$ is
constant with respect to $x_{1}$.

  2. 2.

If
$\lambda_{1}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))<0\leq\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))$
then $u$ is either constant with respect to $x_{1}$ or strictly monotone with
respect to $x_{1}$ in $\Omega_{0\beta}$. In the last case, the derivative
$u_{1}=\partial u/\partial x_{1}$ does not vanish in $\Omega_{0\beta}$, the
first Dirichlet eigenvalue $\lambda_{1}(H^{1}_{0}(\Omega_{0\beta}))$ equals
zero and $u_{1}$ is a corresponding eigenfunction.

The argument is identical to the proof of Theorem 1.1 found in Section 3,
keeping in mind that in the present case the reflection $\sigma_{\alpha}$ is
obviously defined as $\sigma_{\alpha}(x_{1},z)=(2\alpha-x_{1},\,z)$, and every
occurrence of $\theta$ has to be replaced with $x_{1}$. Since
$\gamma_{0\beta}$ and $\Gamma_{\alpha}$ intersects orthogonally we can obtain
that $\partial u/\partial x_{1}\in H^{1}(\Omega_{0\beta})$ using [16,
Proposition 6.1] when $\omega$ is smooth.

##  5 Examples

In this section we produce some examples of solutions to (1.1) that are not
constant in $\theta$ and satisfy Claim 2 of Theorem 1.1. In the first examples
$\Omega_{0\beta}$ is a sector of the unit disc $B_{1}\subset\mathbb{R}^{2}$.

######  Example 5.1.

Let $N=2$ and $\Omega=B_{1}$. We consider the eigenvalue problem

$\left\\{\begin{array}[]{ll}-\Delta\psi=\lambda\psi&\text{ in
}\Omega_{0\beta},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr\psi=0&\text{ on
}\gamma_{0\beta},\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr\frac{\partial\psi}{\partial\nu}=0&\text{ on
}\Gamma_{\\!0}\cup\Gamma_{\\!\beta}\end{array}\right.$ (5.1)

for $0<\beta<2\pi$. If we neglect to give the eigenvalues an increasing order,
the solutions $(\lambda,\psi)$ of (5.1) are given by

$\begin{array}[]{ll}\psi_{nk}(r,\theta)=J_{s_{n}}\big{(}j_{s_{n},k}\,r\big{)}\,\cos(s_{n}\,\theta)\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr\lambda_{nk}=j^{2}_{s_{n},k}\end{array}$

for $n=0,1,2,\dots$ and $k=1,2,\dots$ where $J_{s_{n}}$ is the Bessel function
of order $s_{n}=\frac{n\pi}{\beta}$, and $j_{s_{n},k}$ is the $k$-th positive
zero of $J_{s_{n}}$. Each $\psi_{nk}$ can be considered as a solution of (1.1)
or (1.2) where $f(r,z,u)=\lambda_{nk}\,u$, and the corresponding linearized
operator is $L_{\psi_{nk}}=-\Delta-\lambda_{nk}\,I$. Of course, if $\psi_{nk}$
is the $i$-th eigenfunction of problem (5.1) for some positive integer $i$,
then the $i$-th eigenvalue of $L_{\psi_{nk}}$ in
$H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})$ vanishes. The first eigenvalue of
problem (5.1) is $\lambda_{01}=j^{2}_{0,1}$ and the first eigenfunction is
radial and given by

$\psi_{01}(r,\theta)=J_{0}\big{(}j_{0,1}\,r\big{)}.$

Thus, we have an instance of problem (1.1) or (1.2) where
$\lambda_{1}(H_{\gamma_{0\beta}}(\Omega_{0\beta}))=0$ and a corresponding
solution $u=\psi_{01}(r,\theta)$ which is radial, as stated in Theorem 1.1,
Claim 1. In order to find the second eigenvalue and eigenfunction of (5.1) we
have to compare $j_{0,2}\approx 5.5201$ and $j_{\frac{\pi}{\beta},1}$ whose
value depends on the angle $\beta$. It can be seen that if
$\beta=\frac{\pi}{2}$ then $J_{\frac{\pi}{\beta}}=J_{2}$ and
$j_{\frac{\pi}{\beta},1}=j_{2,1}\approx 5.1356<j_{0,2}$, while if
$\beta=\frac{\pi}{3}$ then $J_{\frac{\pi}{\beta}}=J_{3}$ and
$j_{\frac{\pi}{\beta},1}=j_{3,1}\approx 6.3802>j_{0,2}$. Next, using that
$j_{s,1}$ is a continuous and strictly increasing function of the variable $s$
(see for instance [8, pp. 67-68]), it follows that there exists a unique angle
$\widehat{\beta}\in(\frac{\pi}{3},\frac{\pi}{2})$ such that
$j_{\frac{\pi}{\widehat{\beta}},1}=j_{0,2}$. Moreover one has
$j_{\frac{\pi}{\beta},1}<j_{0,2}$ for $\beta>\widehat{\beta}$ while
$j_{\frac{\pi}{\beta},1}>j_{0,2}$ for $0<\beta<\widehat{\beta}$. The value of
$\widehat{\beta}$ can be estimated numerically obtaining
$\widehat{\beta}\approx 1.3629$. Reassuming, for $\beta>\widehat{\beta}$ the
second eigenvalue of (5.1) is $\lambda_{11}=j^{2}_{\frac{\pi}{\beta},1}$, and
the second eigenfunction

$\psi_{11}(r,\theta)=J_{\frac{\pi}{\beta}}\big{(}j_{\frac{\pi}{\beta},1}\,r\big{)}\,\cos\big{(}\textstyle\frac{\pi}{\beta}\,\theta\big{)}$

is nonradial and strictly monotone in the angular variable in
$\Omega_{0\beta}$. Let us observe that the second eigenfunction $\psi_{11}$
satisfies the assumptions of Theorem 1.1, Claim 2 since the second eigenvalue
of the linearized operator
$L_{\psi_{11}}=-\Delta-j^{2}_{\frac{\pi}{\beta},1}\,I$ in
$H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta})$ vanishes. For
$\beta<\widehat{\beta}$ the second eigenvalue of problem (5.1) is
$\lambda_{02}=j^{2}_{0,2}$ and the second eigenfunction is radial and given by

$\psi_{02}(r,\theta)=J_{0}(j_{0,2}\,r),$

while for $\beta=\widehat{\beta}$ the second eigenvalue has multiplicity two
and the eigenspace is spanned by $\psi_{02}=J_{0}(j_{0,2}\,r)$ and
$\psi_{\widehat{s}1}=J_{\widehat{s}}\big{(}j_{\widehat{s},1}\,r\big{)}\,\cos\big{(}\widehat{s}\,\theta\big{)}$,
where $\widehat{s}=\pi/\widehat{\beta}$. Thus, letting
$m=j^{2}_{0,2}=j^{2}_{\widehat{s},1}$ and $f(r,z,u)=mu$ we obtain an instance
of problem (1.1) or (1.2) where
$\lambda_{2}(H^{1}_{\gamma_{0\widehat{\beta}}}(\Omega_{0\widehat{\beta}}))=0$
and there are both a radial solution and a solution strictly monotone in
$\theta$ thus implying that
$\lambda_{1}(H^{1}_{0}(\Omega_{0\widehat{\beta}})=0$. Since we are in the
linear case, then $L_{\psi_{02}}=L_{\psi_{\widehat{s}1}}$ showing that even if
$\frac{\partial\psi_{02}}{\partial\theta}\equiv 0$ the corresponding Dirichlet
eigenvalue $\lambda_{1}(H^{1}_{0}(\Omega_{0\widehat{\beta}}))$ vanishes.

Figure 1: A solution monotone in $\theta$ (left) and a radial solution (right)
in $\Omega_{0\widehat{\beta}}$ from different viewpoints

Next we want to show that, if we allow the solution $u$ in (1.1) or in (1.2)
to have a higher Morse index, namely
$\lambda_{2}(H^{1}_{\gamma_{0}\beta}(\Omega_{0\beta}))<0$ then $u$ can be
nonradial without being strictly monotone in $\theta$ in $\Omega_{0\beta}$.
Indeed if we consider the third eigenvalue and eigenfunction of (5.1) in the
case when $\beta>\widehat{\beta}$ we have to compare $j_{0,2}\approx 5.5201$
and $j_{\frac{2\pi}{\beta},1}$ (that corresponds to $n=2$) and it is easy to
see that when $\beta=\pi$ then $j_{\frac{2\pi}{\beta},1}=j_{2,1}\approx
5.1356<j_{0,2}$. As before we can say that there exists
$\widetilde{\beta}>\widehat{\beta}$ such that when $\beta>\widetilde{\beta}$
the third eigenvalue of (5.1) is $\lambda_{2,1}=j^{2}_{\frac{2\pi}{\beta},1}$
and the third eigenfunction is

$\psi_{21}(r,\theta)=J_{\frac{2\pi}{\beta}}\big{(}j_{\frac{2\pi}{\beta},1}\,r\big{)}\,\cos\big{(}\textstyle\frac{2\pi}{\beta}\,\theta\big{)}$

which is symmetric with respect to the bisector of $\Omega_{0\beta}$, strictly
monotone with respect to $\theta$ in $\Omega_{0\frac{\beta}{2}}$ and in
$\Omega_{\frac{\beta}{2}\beta}$ and satisfies
$\lambda_{3}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))=0$.
We conjecture that this behavior should be the only new one when one passes
from solutions to (1.1) with
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$ to solutions
with
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))<0\leq\lambda_{3}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))$.

######  Example 5.2.

Next we consider a semilinear equation. Let us recall the Hénon problem

$\left\\{\begin{array}[]{ll}-\Delta u=|x|^{\alpha}\,u^{p},\ u>0&\text{ in
}B_{1},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr u=0&\text{ on }\partial
B_{1},\\\ \end{array}\right.$ (5.2)

where $B_{1}$ is the unit ball in $\mathbb{R}^{N}$, $\alpha>0$ and $p>1$. It
has been proved in [19] that for any $p>1$ there exists $\alpha^{*}(p)$ such
that for any $\alpha>\alpha^{*}(p)$ the ground state of (5.2) is nonradial.
When $p>2$, ground states have a symmetry axis (that we can assume, up to a
rotation, is the $x_{1}$-axis) and they are increasing (or decreasing) in the
polar angle from this axis, namely in $\theta$ (see [3] and [17]). If we call
$\bar{u}$ the restriction to the half ball
$\Omega_{0\pi}=B_{1}\cap\\{\,x_{2}>0\,\\}$ of a ground state $u$ to (5.2), we
have that $\bar{u}$ satisfies

$\left\\{\begin{array}[]{ll}-\Delta\bar{u}=|x|^{\alpha}\,\bar{u}^{p}&\text{ in
}\Omega_{0\pi},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr\bar{u}=0&\text{ on
}\gamma_{0\pi},\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr\frac{\partial\bar{u}}{\partial\nu}=0&\text{ on
}\Gamma_{\\!0}\cup\Gamma_{\\!\pi}\end{array}\right.$ (5.3)

and, when $\alpha$ is large enough, it is strictly monotone in $\theta$.
Moreover $\bar{u}$ satisfies the assumption of Theorem 1.1, namely
$\lambda_{2}(H^{1}_{\gamma_{0\pi}}(\Omega_{0\pi}))\geq 0$. Indeed if
$\lambda_{2}(H^{1}_{\gamma_{0\pi}}(\Omega_{0\pi}))<0$ then the linearized
operator has two negative eigenvalues with corresponding eigenfunctions
orthogonal in $L^{2}(\Omega_{0\pi})$. We can extend the eigenfunctions by
symmetry about the hyperplane $\\{\,x_{2}=0\,\\}$ thus obtaining two
$L^{2}$-orthogonal eigenfunctions of the linearization to (5.2) associated
with negative eigenvalues. This is not possibile since the linearized operator
associated to a ground state $u$ can have only one negative eigenvalue, hence
we must have $\lambda_{2}(H^{1}_{\gamma_{0\pi}}(\Omega_{0\pi}))\geq 0$. In the
special case when $N=2$, letting $v(\rho,\phi)=c\,\bar{u}(r,\theta)$ with
$c=\big{(}\frac{\pi}{\beta}\big{)}^{\frac{2}{p-1}}$ for
$r=\rho^{\frac{\pi}{\beta}}$ and $\theta=\frac{\pi}{\beta}\,\phi$,
$\beta\in(0,2\pi)$, a straightforward computation shows that $v$ solves

$\left\\{\begin{array}[]{ll}-\Delta
v=|x|^{(\alpha+2)\frac{\pi}{\beta}-2}\,v^{p}&\text{ in }\Omega_{0\beta},\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr v=0&\text{ on }\gamma_{0\beta},\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr\frac{\partial v}{\partial\nu}=0&\text{
on }\Gamma_{\\!0}\cup\Gamma_{\\!\beta}\end{array}\right.$

giving an example of monotone solution in a sector of any amplitude $\beta$.
This solution satisfies
$\lambda_{2}(H^{1}_{\gamma_{0\beta}}(\Omega_{0\beta}))\geq 0$. Indeed if
$\varphi_{1}(\rho,\phi)$ and $\varphi_{2}(\rho,\phi)$ are two
$L^{2}(\Omega_{0\beta})$-orthogonal functions that make negative the quadratic
form associated with the linearization to this last equation at the solution
$v$, then the functions $\bar{\varphi}_{1}(r,\theta)=\varphi_{1}(\rho,\phi)$
and $\bar{\varphi}_{2}(r,\theta)=\varphi_{2}(\rho,\phi)$ are two
$L^{2}(\Omega_{0\pi})$-orthogonal functions that makes negative the quadratic
form associated with the linearization to (5.3) at the solution $\bar{u}$ and
this is not possible.

######  Example 5.3.

Here we consider the case of an annulus $A$ of $\mathbb{R}^{2}$. In [9,
Proposition 1.4] it has been proved that for any $k=1,2,\dots$, there exists
an exponent $p_{k}>1$ such that the least energy solution of the Lane-Emden
problem

$\left\\{\begin{array}[]{ll}-\Delta u=u^{p}&\text{ in }A,\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr u=0&\text{ on }\partial A\end{array}\right.$

in the space of $H^{1}$ functions that are $\frac{2\pi}{k}$ periodic in
$\theta$ is nonradial when $p>p_{k}$. Moreover, by [10], up to a rotation,
these solutions are strictly monotone in $\theta$ in the sector
$\Omega_{0\frac{\pi}{k}}$ and, by symmetry reasons, they satisfy a Neumann
boundary condition on $\Gamma_{\\!0}\cup\Gamma_{\\!\frac{\pi}{k}}$. Finally,
since they are least energy solutions, they satisfy
$\lambda_{2}(H^{1}_{\gamma_{0\frac{\pi}{k}}}(\Omega_{0\frac{\pi}{k}}))\geq 0$.

##  6 Symmetry and monotonicity in unbounded domains

Hereafter we consider the case when $\Omega_{0\beta}$ is not bounded. To keep
the presentation simple and to deal with some difficulties coming from the
unboudeness of the domain, we focus on the special case when the domain of the
problem is a sector of the $N$-dimensional Euclidean space with an angular
amplitude $\beta\in(0,2\pi)$. More generally, we also consider the set
difference of such a sector and the infinite cylinder $d(x)\leq r_{1}$, where
$d(x)=\sqrt{x_{1}^{2}+x_{2}^{2}}$. To be precise, choosing
$r_{1}\in[0,+\infty)$ and $\theta_{1},\theta_{2}\in[0,2\pi)$, with
$\theta_{1}<\theta_{2}$, we denote by $\Omega_{\theta_{1}\theta_{2}}$ the
domain

$\Omega_{\theta_{1}\theta_{2}}=\\{\,x\in\mathbb{R}^{N}:d(x)>r_{1},\
\theta(x)\in(\theta_{1},\theta_{2})\,\\}$

whose boundary contains the open, flat subsets
$\Gamma_{\\!\theta_{i}}=\\{\,x\in\mathbb{R}^{N}:d(x)>r_{1},\
\theta(x)=\theta_{i}\,\\}$, $i=1,2$. Moreover when $r_{1}>0$ we may write
$\partial\Omega_{\theta_{1}\theta_{2}}=\Gamma_{\\!\theta_{1}}\cup\Gamma_{\\!\theta_{2}}\cup\gamma_{\theta_{1}\theta_{2}}$
where $\gamma_{\theta_{1}\theta_{2}}$ is the closed subset

$\gamma_{\theta_{1}\theta_{2}}:=\\{\,x\in\mathbb{R}^{N}:d(x)=r_{1},\
\theta(x)\in[\theta_{1},\theta_{2}]\,\\}.$

When $r_{1}=0$, instead, we have
$\partial\Omega_{\theta_{1}\theta_{2}}=\Gamma_{\\!\theta_{1}}\cup\Gamma_{\\!\theta_{2}}\cup\Upsilon$
and $\gamma_{\theta_{1}\theta_{2}}=\emptyset$. In this part of the paper we
deal with problem (1.1) where the domain $\Omega_{0\beta}$ is unbounded and
defined as above. The function $f(r,z,u)$ is subject to the same conditions
stated in the Introduction. Let $\cal Y$ be the set of all functions $u\in
C^{2}(\Omega_{0\beta})\cap C^{0}(\overline{\Omega}_{0\beta})$ such that
$u_{\theta}\in C^{0}(\overline{\Omega}_{0\beta}\setminus\Upsilon)$ and having
a square-summable gradient $\nabla u\in
L^{2}(\Omega_{0\beta},\mathbb{R}^{N})$. Here we investigate functions
$u\in\cal Y$ satisfying (1.1) pointwise and having a small Morse index, whose
definition is recalled below. By standard regularity theory, any solution
$u\in\cal Y$ belongs to
$C^{2}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$. As mentioned
in the Introduction, problem (1.1) can be rewritten as (1.2). Of course, if
$u\in\cal Y$ then for every $R>r_{1}$ we have $u\in H^{1}(\Omega_{0\beta}\cap
B_{R})$, where $B_{R}\subset\mathbb{R}^{N}$ denotes the ball with radius $R$
centered at the origin. Now we do not assume that $u_{\theta}$ belongs to
$H^{1}(\Omega_{0\beta}\cap B_{R})$, but we require $u_{\theta}$ to extend
continuously up to the set $d(x)=r_{1}$ when $r_{1}>0$. Roughly speaking, the
reason is that in order to derive a conclusion on $u_{\theta}$ from a bound on
the Morse index (see below) we have to approximate $u_{\theta}$ with some
compactly supported function: hence $u_{\theta}$ must vanish on
$\gamma_{0\beta}$ either in the sense of traces or as a continuous function.
In this part of the paper we assume we are in the second case: the assumption
is used, for instance, in the proof of (8.9). To proceed further, for any
subset $D\subset\overline{\Omega}_{0\beta}$ we denote by $C^{1}_{0}(D)$ the
set of functions $\varphi\in C^{1}(D)$ having a compact support contained in
$D$.

######  Definition 6.1.

We say that a solution $u$ to (1.1)

  * -

is stable, and has Morse index $m(u)=0$, if $Q_{\Omega_{0\beta}}(\varphi)\geq
0$ for every $\varphi\in
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$.

  * -

has Morse index $m(u)=M\geq 1$ if $M$ is the maximal dimension of a vector
subspace $X\subset
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ such that
$Q_{\Omega_{0\beta}}(\varphi)<0$ for every $\varphi\in X\setminus\\{0\\}$.

We will investigate solutions with $m(u)\leq 1$. By definition, if $m(u)=0$
then the bilinear form

${\cal
B}(\varphi,\psi):=\int_{\Omega_{0\beta}}\Big{(}\nabla\varphi\,\nabla\psi-f^{\prime}(d(x),z,u)\,\varphi\,\psi\Big{)}\,dx$
(6.1)

is positive semidefinite in
$C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})\times
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$, and
therefore the Cauchy-Schwarz inequality holds:

$0\leq\big{(}{\cal B}(\varphi,\psi)\big{)}^{2}\leq
Q_{\Omega_{0\beta}}(\varphi)\,Q_{\Omega_{0\beta}}(\psi).$ (6.2)

Since $u$ is locally bounded, so is $f^{\prime}(d(x),z,u)$ and the definition
of ${\cal B}(\varphi,\psi)$ extends to $H^{1}_{\gamma_{0\beta}\cup\partial
B_{R}}(D)\times H^{1}_{\gamma_{0\beta}\cup\partial B_{R}}(D)$, where
$D=\Omega_{0\beta}\cap B_{R}$, $R>r_{1}$, and
$H^{1}_{\gamma_{0\beta}\cup\partial B_{R}}(D)$ denotes the completion of
$C^{1}_{0}(\left(\Omega_{0\beta}\cup\Gamma_{0}\cup\Gamma_{\beta}\right)\cap
B_{R})$ in the $H^{1}$-norm. Of course, positive semidefiniteness is preserved
under such an extension.

######  Remark 6.2.

The integral in (6.1) makes sense even if we replace $\varphi$ with some
function $v$ having an unbounded support, because the only contribution comes
from the bounded set $\mathop{\rm supp}\psi$: this will often be done in the
sequel in order to prove that ${\cal B}(v,\psi)=0$ for all $\psi\in
C^{\infty}_{0}(D)$, i.e., $v$ is locally a weak solution of $L_{u}\,v=0$. In
such a case, however, $Q_{\Omega_{0\beta}}(v)$ is undefined, in general, and
(6.2) makes no sense: this difficulty is overcome by means of a truncation
technique inspired by [11] (see Section 9 for details).

######  Remark 6.3.

If $m(u)=1$ then there exists a one-dimensional subspace
$X=\\{\,\lambda\varphi_{1}:\lambda\in\mathbb{R}\,\\}\subset
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ such that
$Q_{\Omega_{0\beta}}(\lambda\varphi_{1})<0$ for every $\lambda\neq 0$. In such
a case, we have

$Q_{\Omega_{0\beta}}(\varphi)\geq 0$ for every $\varphi\in
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ such that
$\mathop{\rm supp}\varphi\cap\mathop{\rm supp}\varphi_{1}=\emptyset$.

Indeed, if there were $\varphi_{2}\in
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ supported in
$\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta}\setminus\mathop{\rm
supp}\varphi_{1}$ and such that $Q_{\Omega_{0\beta}}(\varphi_{2})<0$, we would
immediately construct the two dimensional subspace
$W=\\{\,a\varphi_{1}+b\varphi_{2}:a,b\in\mathbb{R}\,\\}\subset
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ such that
$Q_{\Omega_{0\beta}}(\varphi)<0$ for every $\varphi\in W\setminus\\{0\\}$,
thus contradicting $m(u)=1$. The argument extends to the more general case
when $m(u)<+\infty$: see, for instance, [11, Remark 2.7].

The main result in this part of the paper is the extension of Theorem 1.1 to
the case of the unbounded domain $\Omega_{0\beta}$ defined just before. In
particular we will prove the following:

######  Theorem 6.4.

Assume $u\in\cal Y$ is a solution to (1.1) with Morse index $m(u)\leq 1$.
Assume further $f(r,z,u)$ is convex with respect to $u$.

  1. 1.

If $m(u)=0$ then $u$ is constant with respect to the angular variable
$\theta$.

  2. 2.

If $m(u)=1$ then $u$ is either independent of $\theta$ or strictly monotone
with respect to $\theta$ in $\Omega_{0\beta}$, in which case $u$ has a
nonvanishing derivative $u_{\theta}$ and $\inf_{\psi\in
C^{1}_{0}(\Omega_{0\beta})}Q_{\Omega_{0\beta}}(\psi)=0$.

The rest of the paper is devoted to prove Theorem 6.4. First, in Section 7 we
introduce a suitable function space to which the solutions belong and we prove
that some sequences of boundary integrals should converge to zero, which is
useful to manage the boundary term coming from an integration by parts of the
function $u_{\theta}$ or the function $w_{\alpha}$ over a bounded subset of
$\Omega_{0\beta}$. In Section 8 we prove a sign preservation property, in
Section 9 we extend the splitting lemma and in Section 10 we conclude the
proof.
We remark that the most difficult point to treat is the extension of Part VI
of the proof of Theorem 1.1. Indeed we cannot use the maximum principle in
small domains and we need to introduce a suitable torus $T$, invariant under
cylindrical rotations, to overcome this difficulty.

##  7 A suitable function space

As mentioned before, we consider (possibly unbounded) solutions to (1.1)
belonging (in particular) to the set $E=\\{\,v\in C^{2}(\Omega_{0\beta})\cap
C^{1}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})\cap
C^{0}(\overline{\Omega}_{0\beta}):\nabla v\in
L^{2}(\Omega_{0\beta},\mathbb{R}^{N})\,\\}$. We collect here some properties
of the space $E$ and its subspaces $E_{\alpha}$, where for
$\alpha\in(0,\beta]$, we denote by $E_{\alpha}$ the subset of all $v\in E$
such that $v=0$ on $\Gamma_{\\!\alpha}$. Note that the function
$w_{\frac{\beta}{2}}$ associated to a solution $u\in E$ obviously belongs to
$E_{\frac{\beta}{2}}$. For $\alpha\in(0,\beta]$ we let $\Sigma_{\alpha R}$ be
the surface $\Omega_{0\alpha}\cap\partial B_{R}$, where $R>r_{1}$. By
Tonelli’s theorem, for every $v\in E$ and almost every $R>r_{1}$ we have

$\int_{\Sigma_{\beta R}}|\nabla v|^{2}\,d\Sigma<+\infty.$

Of course, if $N=2$ then $\Sigma_{\beta R}$ reduces to an arc, $v\in
C^{1}(\overline{\Sigma}_{\beta R})$ and the integral above is finite for every
$R>r_{1}$.

######  Lemma 7.1.

Let $\alpha\in(0,\beta]$, $\Sigma_{\alpha R}=\Omega_{0\alpha}\cap\partial
B_{R}$ and $\nu$ be the outward normal to the sphere $\partial B_{R}$. For
every $v\in E_{\alpha}$ and almost every $R>r_{1}$ the following inequalities
hold:

$\displaystyle\sqrt{\lambda_{1}(r_{1},R)\,}\,\left|\int_{\Sigma_{\alpha
R}}v\,\frac{\,\partial v\,}{\partial\nu}\,d\Sigma\right|$
$\displaystyle\leq\int_{\Sigma_{\alpha R}}|\nabla v|^{2}\,d\Sigma,$ (7.1)
$\displaystyle\lambda_{1}(r_{1},R)\,\int_{\Sigma_{\alpha R}}v^{2}\,d\Sigma$
$\displaystyle\leq\int_{\Sigma_{\alpha R}}|\nabla v|^{2}\,d\Sigma,$ (7.2)

where $\lambda_{1}(r_{1},R)>0$ is given by

$\lambda_{1}(r_{1},R)=\inf_{\genfrac{}{}{0.0pt}{1}{w\in H^{1}(\Sigma_{\alpha
R})\setminus\\{0\\}}{w=0\mbox{ \scriptsize on }\Gamma_{\\!\alpha}\cap\partial
B_{R}}}\frac{\,\int_{\Sigma_{\alpha
R}}|\nabla_{\\!\tau\,}w|^{2}\,d\Sigma\,}{\int_{\Sigma_{\alpha
R}}w^{2}\,d\Sigma}$

and $\nabla_{\\!\tau\,}w$ denotes the intrinsic gradient of $w$ over the
surface $\Sigma_{\alpha R}$.

###### Proof.

By the Cauchy-Schwarz inequality, for almost every $R>r_{1}$ we may write

$\left|\int_{\Sigma_{\alpha R}}v\,\frac{\,\partial
v\,}{\partial\nu}\,d\Sigma\right|\leq\left(\int_{\Sigma_{\alpha
R}}v^{2}\,d\Sigma\right)^{\\!\frac{1}{2}}\left(\int_{\Sigma_{\alpha
R}}\Big{(}\frac{\,\partial
v\,}{\,\partial\nu\,}\Big{)}^{\\!2}\,d\Sigma\right)^{\\!\frac{1}{2}}\\!\\!.$
(7.3)

By Poincaré’s inequality, and recalling that $\nabla
v=\nabla_{\\!\tau\,}v+\frac{\,\partial v\,}{\,\partial\nu\,}\,\nu$, the first
integral in the right-hand side is estimated as follows:

$\displaystyle\lambda_{1}(r_{1},R)\,\int_{\Sigma_{\alpha R}}v^{2}\,d\Sigma$
$\displaystyle\leq\int_{\Sigma_{\alpha R}}|\nabla_{\\!\tau\,}v|^{2}\,d\Sigma$
$\displaystyle\leq\int_{\Sigma_{\alpha R}}|\nabla v|^{2}\,d\Sigma,$

which yields (7.2). Furthermore, we obviously have

$\int_{\Sigma_{\alpha R}}\Big{(}\frac{\,\partial
v\,}{\,\partial\nu\,}\Big{)}^{\\!2}\,d\Sigma\leq\int_{\Sigma_{\alpha
R}}|\nabla v|^{2}\,d\Sigma$

and (7.1) follows by multiplying the last two estimates. ∎

Inequalities (7.1)-(7.2) enter in the proof of the following result:

######  Lemma 7.2.

Let the domain $\Omega_{0\beta}$ and the function spaces $E$ and $E_{\alpha}$
be defined as above.

  1. 1.

For every $v\in E_{\alpha}$ and $R_{0}>r_{1}$, the iterated integral

$\int_{R_{0}}^{+\infty}\bigg{(}\frac{1}{\,R^{2}\,}\int_{\Sigma_{\alpha
R}}v^{2}\,d\Sigma\bigg{)}dR$

converges to a finite value, hence it is infinitesimal as $R_{0}\to+\infty$.

  2. 2.

For every $v\in E_{\alpha}$ there exists a sequence $R_{k}\to+\infty$ such
that

$\lim_{k\to+\infty}\int_{\Sigma_{\alpha R_{k}}}v\,\frac{\,\partial
v\,}{\partial\nu}\,d\Sigma=0.$ (7.4)

  3. 3.

For every $v\in E$ such that $v_{\theta}\in
C^{1}(\overline{\Omega}_{0\alpha})$ there exists a sequence $R_{k}\to+\infty$
such that the negative part $v_{\theta}^{-}=\min\\{\,0,v_{\theta}\,\\}\leq 0$
satisfies

$\lim_{k\to+\infty}\int_{\Sigma_{\alpha
R_{k}}}v_{\theta}^{-}\,\frac{\,\partial
v_{\theta}^{-}\,}{\partial\nu}\,d\Sigma=0.$ (7.5)

###### Proof.

1\. By the scaling property of the eigenvalues we may write

$\Big{(}\frac{R}{\,R_{0}\,}\Big{)}^{\\!2}\,\lambda_{1}(r_{1},R)=\lambda_{1}\Big{(}\frac{\,R_{0}\,}{R}\,r_{1},\,R_{0}\Big{)}\stackrel{{\scriptstyle
R\to+\infty}}{{\longrightarrow}}\lambda_{1}(0,R_{0})>0.$ (7.6)

We note in passing that recent advances on spectral asymptotics are found, for
instance, in [13] and in the references cited there. From (7.6) it follows
that there exists $\varepsilon_{0}>0$ such that
$\lambda_{1}(r_{1},R)\geq\varepsilon_{0}/R^{2}$ for $R$ large. This and (7.2)
imply

$\displaystyle\frac{\,\varepsilon_{0}\,}{\,R^{2}\,}\int_{\Sigma_{\alpha
R}}v^{2}\,d\Sigma\leq\int_{\Sigma_{\alpha R}}|\nabla v|^{2}\,d\Sigma.$

Taking into account that the right-hand side is summable by assumption over
the interval $(r_{1},+\infty)$, Claim 1 follows.

2\. If the claim were false, then by (7.1) there would be $R_{0}>r_{1}$ and
$\varepsilon_{0}>0$ such that for almost every $R\geq R_{0}$

$\int_{\Sigma_{\alpha R}}|\nabla
v|^{2}\,d\Sigma\geq\varepsilon_{0}\sqrt{\lambda_{1}(r_{1},R)\,}.$

Taking the asymptotics (7.6) into account, for almost every $R$ large enough
we may write

$\int_{\Sigma_{\alpha R}}|\nabla
v|^{2}\,d\Sigma\geq\frac{\,\varepsilon_{0}\,}{2}\,\frac{\,R_{0}\,}{R}\,\sqrt{\lambda_{1}(0,R_{0})\,}.$

Integrating both sides in $dR$ we would get $\|\nabla
v\|_{L^{2}(\Omega_{0\alpha})}=+\infty$, which is in contrast with the
definition of the function space $E_{\alpha}$.

3\. Since $\nabla v\in L^{2}(\Omega_{0\beta})$, the function

$f(R)=\int_{\Omega_{0\alpha}\cap B_{R}}(v_{\theta}^{-})^{2}\,dx$

has a finite limit for $R\to+\infty$. By differentiation we find

$f^{\prime}(R)=\int_{\Sigma_{\alpha R}}(v_{\theta}^{-})^{2}\,d\Sigma.$

Differentiating once more yields

$f^{\prime\prime}(R)=\int_{\Sigma_{\alpha R}}2v_{\theta}^{-}\,\frac{\,\partial
v_{\theta}^{-}\,}{\partial\nu}\,d\Sigma+\frac{\,N-1\,}{R}\,f^{\prime}(R),$

which may be rewritten as

$\Big{(}\frac{\,f^{\prime}(R)\,}{\,R^{N-1}\,}\Big{)}^{\prime}=\frac{1}{\,R^{N-1}\,}\int_{\Sigma_{\alpha
R}}2v_{\theta}^{-}\,\frac{\,\partial v_{\theta}^{-}\,}{\partial\nu}\,d\Sigma.$

If the last integral were larger than $\varepsilon_{0}>0$ for $R\geq R_{0}$,
then, integrating both sides over the interval $(R_{0},R)$, we would get

$\frac{\,f^{\prime}(R)\,}{\,R^{N-1}\,}=C+\varepsilon_{0}\int_{R_{0}}^{R}\frac{dr}{\,r^{N-1}\,}$

which implies $f^{\prime}(R)\to+\infty$ as $R\to+\infty$. But this is
impossible because $f(R)$ converges to a finite limit. A similar contradiction
arises if we assume that the integral in (7.5) keeps smaller than
$-\varepsilon_{0}<0$. ∎

We note in passing that (7.4) continues to hold in the special case when $N=2$
even though $v$ does not vanish along $\Gamma_{\\!\alpha}$. However, the proof
is different:

######  Proposition 7.3.

In the special case when $N=2$, for every $v\in E$ and every
$\alpha\in(0,\beta]$ there exists a sequence $R_{k}\to+\infty$ such that (7.4)
holds.

###### Proof.

By the Cauchy-Schwarz inequality we may still write (7.3), which holds now for
every $R>r_{1}$ because $v\in C^{1}(\overline{\Omega}_{0\beta}\setminus
B_{R})$. However, the Poincaré inequality fails because of the lack of
boundary conditions. Nevertheless, the first integral in the right-hand side
is estimated as follows. Fix $r_{0}>r_{1}$, so that the function
$v(r_{0},\theta)$ of the variable $\theta$ belongs to $C^{1}([0,\alpha])$. By
the fundamental theorem of calculus we may write

$\displaystyle|v(R,\theta)-v(r_{0},\theta)|$
$\displaystyle\leq\int_{r_{0}}^{R}\Big{|}\frac{\,\partial v\,}{\,\partial
r\,}\Big{|}\,dr$
$\displaystyle\leq\left(\int_{r_{0}}^{R}\frac{1}{\,r\,}\,dr\right)^{\\!\frac{1}{2}}\left(\int_{r_{0}}^{R}r\,\Big{(}\frac{\,\partial
v\,}{\,\partial r\,}\Big{)}^{\\!2}\,dr\right)^{\\!\frac{1}{2}}\\!\\!.$

Recall that the area element in $\mathbb{R}^{2}$ is $r\,dr\,d\theta$.
Therefore, since $\nabla v\in L^{2}(\Omega_{0\alpha})$, the last integral
converges almost everywhere as $R\to+\infty$ to some function
$\psi(\theta)<+\infty$ which is summable over the interval $(0,\alpha)$. Hence
$v^{2}(R,\theta)\leq C\,\psi(\theta)\log R$ for almost every $\theta$ and for
large $R$, with a convenient constant $C$. By plugging this estimate into
(7.3) we obtain

$\left|\int_{\Sigma_{\alpha R}}v\,\frac{\,\partial
v\,}{\partial\nu}\,d\Sigma\right|\leq C\,(R\,\log
R)^{\frac{1}{2}}\left(\int_{\Sigma_{\alpha R}}|\nabla
v|^{2}\,d\Sigma\right)^{\\!\frac{1}{2}}$

where we have redefined the constant $C$. If the claim were false, there would
be $R_{0}>r_{0}$ and $\varepsilon_{0}>0$ such that for every $R\geq R_{0}$

$\int_{\Sigma_{\alpha R}}|\nabla
v|^{2}\,d\Sigma\geq\frac{\varepsilon_{0}}{\,R\,\log R\,}.$

Integrating both sides over the interval $[R_{0},+\infty)$ we would get
$\|\nabla v\|_{L^{2}(\Omega_{0\alpha})}=+\infty$, which is in contrast with
the definition of the function space $E$. ∎

######  Remark 7.4.

In general, (7.4) fails for $v\in E$ and $N\geq 3$. For instance, we may take
$r_{1}>0$, $\varepsilon\in(0,\,\frac{N}{2}-1)$ and

$v(x)=1-\frac{1}{\,|x|^{\frac{N}{2}-1+\varepsilon}\,},\ \mbox{hence}\ \nabla
v(x)=\frac{\,\frac{N}{2}-1+\varepsilon\,}{R^{\frac{N}{2}+\varepsilon}}\,\nu\
\mbox{when $x\in\Sigma_{\beta R}$.}$

In this example we have $v\in E$ because $\varepsilon>0$. However,

$\int_{\Sigma_{\beta R}}v\,\frac{\,\partial
v\,}{\partial\nu}\,d\Sigma=\Big{(}1-\frac{1}{\,R^{\frac{N}{2}-1+\varepsilon}\,}\Big{)}\,\Big{(}\frac{N}{2}-1+\varepsilon\Big{)}\,R^{\frac{N}{2}-1-\varepsilon}\,|\Omega_{0\beta}\cap\partial
B_{1}|\to+\infty.$

######  Remark 7.5.

A function $v\in E_{\alpha}\subset E$ may well be unbounded. To see this,
recall that for every $N\geq 2$ there exists a function $\varphi\in
C^{\infty}_{0}(B_{1})$ such that $\|\nabla\varphi\|_{L^{2}(B_{1})}$ is
arbitrarily small and $\sup|\varphi|$ is arbitrarily large. Then we may take a
sequence of centers $x_{k}\in\Omega_{0\alpha}$ and a sequence of radii
$r_{k}>0$ such that $|x_{k}|\to+\infty$ and the balls
$B_{k}=B(x_{k},r_{k})\subset\Omega_{0\alpha}$ are pairwise disjoint. For each
$k$ we fix $\varphi_{k}\in C^{\infty}_{0}(B_{k})$ such that
$\|\nabla\varphi_{k}\|^{2}_{L^{2}(B_{k})}\leq 2^{-k}$ and
$\sup|\varphi_{k}|\geq k$. The series

$v(x)=\sum_{k=1}^{+\infty}\varphi_{k}(x)$

trivially converges in $E_{\alpha}$ because for every $x$ it has at most one
nonzero term, and it is apparent that $v$ is unbounded and satisfies

$\|\nabla v\|^{2}_{L^{2}(\Omega_{0\alpha})}\leq\sum_{k=1}^{+\infty}2^{-k}=1.$

######  Remark 7.6.

If $N=2$ there exist radial, unbounded functions in the space $E$. For
instance, we may take $v(x)=\log\log|x|$ for $|x|>1$: in this case $\nabla
v=(|x|\,\log|x|)^{-1}\,\frac{x}{|x|}$ and therefore $\nabla v\in
L^{2}(\Omega_{0\beta})$ provided that $r_{1}>1$. If, instead, $N\geq 3$, every
radial function $v\in E$ is bounded. To see this, let us write $v(x)=f(|x|)$.
In the special case when $r_{1}=0$, the surface integral

$I(R)=\int_{\Sigma_{\beta R}}|\nabla v|^{2}\,d\Sigma$

equals $\frac{\beta}{2\pi}\,N\omega_{N}\,\psi(R)$, where
$\psi(R)=R^{N-1}\,(f^{\prime}(R))^{2}$ and $N\omega_{N}=|\partial B_{1}|$ is
the Hausdorff measure of the unit spherical surface in $\mathbb{R}^{N}$.
Hence, $\psi(R)$ must be summable over the interval $(0,+\infty)$ because
$\nabla v\in L^{2}(\Omega_{0\beta})$. By the Cauchy-Schwarz inequality, for
every $R>1$ we may write

$\displaystyle|f(R)-f(1)|=\left|\int_{1}^{R}f^{\prime}(R)\,dR\right|\leq\int_{1}^{R}|f^{\prime}(R)|\,dR$
$\displaystyle=\int_{1}^{R}\Big{(}\frac{\,\psi(R)\,}{\,R^{N-1}\,}\Big{)}^{\\!\frac{1}{2}}\,dR$
$\displaystyle\leq\left(\int_{1}^{R}\psi(R)\,dR\right)^{\\!\\!\frac{1}{2}}\left(\int_{1}^{R}\frac{1}{\,R^{N-1}\,}\right)^{\\!\\!\frac{1}{2}},$

which shows that the right-hand side keeps bounded as $R\to+\infty$. In the
general case when $r_{1}\geq 0$ the integral $I(R)$ is asymptotic to
$\psi(R)$, and the conclusion follows similarly.

##  8 Sign preservation in unbounded domains

In this section we prepare two fundamental lemmas which replace Lemma 2.3 in
the case when the domain is unbounded. In order to prove the first result, we
need to turn the negative part $v^{-}$ of a supersolution of the linear
equation $L_{u}\,v=0$ into a function with bounded support, thus allowing to
use the Cauchy-Schwarz inequality (6.2). Following [11], this is accomplished
by a family of cut-off functions $\xi_{R}\in C^{\infty}_{0}(\mathbb{R}^{N})$,
$R>0$, such that

$\mbox{$0\leq\xi_{R}(x)\leq 1$ and
$|\nabla\xi_{R}(x)|<\displaystyle\frac{2}{\,R\ }$ for every
$x\in\mathbb{R}^{N}$,}\quad\xi_{R}(x)=\begin{cases}1,&\text{ if }|x|<R,\\\
0,&\text{ if }|x|>2R.\end{cases}$ (8.1)

Later on we will need a further family of cut-off functions in order to deal
with possible singularities at $d(x)=r_{1}$, i.e., the case when $u\not\in
W^{2,2}(\Omega_{0\beta}\cap B_{R})$, $R>r_{1}$. Hence we denote by
$\zeta_{\delta}(x)\in C^{\infty}_{0}(\mathbb{R}^{N})$, $\delta>0$, a function
depending only on $d(x)$ and such that

$\mbox{$0\leq\zeta_{\delta}(x)\leq 1$ and
$|\nabla\zeta_{\delta}(x)|<\displaystyle\frac{3}{\,\delta\,}$ for all
$x\in\mathbb{R}^{N}$,}\quad\zeta_{\delta}(x)=\begin{cases}0,&\text{ if
$d(x)<r_{1}+\delta/2$,}\\\ 1,&\text{ if $d(x)>r_{1}+\delta$.}\end{cases}$
(8.2)

######  Lemma 8.1 (Sign preservation in unbounded domains).

Fix $\alpha\in(0,\beta)$ and let $K$ be a (possibly empty) compact subset of
$\mathbb{R}^{N}$. Suppose that the set difference
$D_{\alpha}=\Omega_{0\alpha}\setminus K$ is connected, and let $v\in
E_{\alpha}$ satisfy $L_{u}\,v\geq 0$ in $D_{\alpha}$ for some $u\in
L^{\infty}(\Omega_{0\alpha})$. Define $\tilde{\Gamma}_{\\!0}=\partial
D_{\alpha}\cap\Gamma_{\\!0}$, and assume

$\inf_{\varphi\in
C^{1}_{0}(D_{\alpha}\cup\tilde{\Gamma}_{\\!0})}Q_{\Omega_{0\alpha}}(\varphi)\geq
0.$ (8.3)

Assume, further, that $v\geq 0$ on $\partial D_{\alpha}\setminus\Gamma_{\\!0}$
together with $v\geq 0$ or $\partial v/\partial\nu\geq 0$ on
$\tilde{\Gamma}_{\\!0}$. If $v$ does not vanish identically in $D_{\alpha}$,
then either $v>0$ in $D_{\alpha}$ or $v<0$ there.

###### Proof.

Let $r_{K}=\max_{x\in K}d(x)$ and define $R_{0}=\max\\{\,r_{1},r_{K}\,\\}$. By
(8.3), for every $R>R_{0}$ the bilinear form (6.1) is positive semidefinite in
the Cartesian product $H^{1}_{\partial D\setminus\Gamma_{0}}(D)\times
H^{1}_{\partial D\setminus\Gamma_{0}}(D)$, where $D=D_{\alpha}\cap B_{2R}$,
and the Cauchy-Schwarz inequality (6.2) holds. Let us check that the negative
part $v^{-}(x)=\min\\{\ 0,\,v(x)\,\\}\leq 0$ is a weak solution of

$L_{u}\,v^{-}=0,$ (8.4)

i.e., ${\cal B}(v^{-},\,\psi)=0$ for every $\psi\in
C^{\infty}_{0}(D_{\alpha})$. By making $R$ larger we may achieve that
$\mathop{\rm supp}\psi\subset B_{R}$, hence $\psi\in C^{\infty}_{0}(D)$.
Letting $\varphi(x)=\xi_{R}(x)\,v(x)\in H^{1}_{\partial
D\setminus\Gamma_{0}}(D)$ in (6.2) we obtain

$0\leq\big{(}{\cal B}(v^{-},\psi)\big{)}^{2}=\big{(}{\cal
B}(\xi_{R}\,v^{-},\psi)\big{)}^{2}\leq
Q_{\Omega_{0\alpha}}(\xi_{R}\,v^{-})\,Q_{\Omega_{0\alpha}}(\psi).$ (8.5)

The value of $Q_{\Omega_{0\alpha}}(\xi_{R}\,v^{-})$ may depend on $R$: however
we claim that it is infinitesimal as $R\to+\infty$. Indeed, by differentiation
we find

$\displaystyle|\nabla\xi_{R}|^{2}\,\varphi^{2}+\nabla(\xi_{R}^{2}\,\varphi)\,\nabla\varphi$
$\displaystyle=|\nabla\xi_{R}|^{2}\,\varphi^{2}+2\xi_{R}\,\varphi\,\nabla\xi_{R}\,\nabla\varphi+\xi_{R}^{2}\,|\nabla\varphi|^{2}$
$\displaystyle=|\nabla(\xi_{R}\,\varphi)|^{2},$ (8.6)

an identity that will be repeatedly used in the sequel, with different choices
of $\varphi$. Taking into account the boundary conditions $v^{-}=0$ on
$\partial D\setminus\Gamma_{\\!0}$ and $v^{-}\,\frac{\partial
v^{-}}{\partial\nu}\leq 0$ on $\partial D\cap\Gamma_{\\!0}$, multiplication of
the inequality $L_{u}\,v\geq 0$ by $\xi_{R}^{2}\,v^{-}$ and integration over
$D$ yields

$\int_{D}\Big{\\{}\nabla(\xi_{R}^{2}\,v^{-})\,\nabla
v^{-}-f^{\prime}(d(x),z,u)\,\xi_{R}^{2}\,(v^{-})^{2}\Big{\\}}\,dx\leq 0.$

Letting $\varphi=v^{-}$ in (8.6), and using (8.1), this is turned into

$0\leq
Q_{\Omega_{0\alpha}}(\xi_{R}\,v^{-})\leq\int_{D}|\nabla\xi_{R}|^{2}\,(v^{-})^{2}\,dx\leq\frac{4}{\,R^{2}\,}\int_{R}^{2R}\bigg{(}\int_{\Sigma_{\alpha
r}}(v^{-})^{2}\,d\Sigma\bigg{)}\,dr,$ (8.7)

where $\Sigma_{\alpha r}=\Omega_{0\alpha}\cap\partial B_{r}$. To manage with
the last integral, note that

$\displaystyle\frac{1}{\,(2R)^{2}\,}\int_{R}^{2R}\bigg{(}\int_{\Sigma_{\alpha
r}}(v^{-})^{2}\,d\Sigma\bigg{)}\,dr$
$\displaystyle\leq\int_{R}^{2R}\bigg{(}\frac{1}{\,r^{2}\,}\int_{\Sigma_{\alpha
r}}(v^{-})^{2}\,d\Sigma\bigg{)}\,dr$
$\displaystyle\leq\int_{R}^{+\infty}\bigg{(}\frac{1}{\,r^{2}\,}\int_{\Sigma_{\alpha
r}}(v^{-})^{2}\,d\Sigma\bigg{)}\,dr,$

therefore by Claim 1 of Lemma 7.2 the right-hand side of (8.7) is
infinitesimal. But then (8.5) implies ${\cal B}(v^{-},\psi)=0$. Since $\psi$
is arbitrary, $v^{-}$ is a weak solution of equation (8.4), as claimed. Now
the unique continuation property implies that either $v^{-}\equiv 0$ or
$v^{-}<0$ in $D_{\alpha}$. In the last case we obviously have $v<0$ in
$D_{\alpha}$. If, instead, $v^{-}\equiv 0$ in $D_{\alpha}$, then $v\geq 0$,
and since $L_{u}\,v\geq 0$ in $D_{\alpha}$, the strong maximum principle
implies that either $v\equiv 0$ or $v>0$ in the connected set $D_{\alpha}$. ∎

######  Remark 8.2.

If we restrict to the case when $v\geq 0$ on $\Gamma_{\\!0}$, we may replace
assumption (8.3) with the weaker requirement

$\inf_{\varphi\in C^{1}_{0}(D_{\alpha})}Q_{\Omega_{0\alpha}}(\varphi)\geq 0.$

In the sequel we will need to apply a similar argument to $u_{\theta}$ and
$u^{-}_{\theta}$, using a double truncation. Therefore we prepare:

######  Proposition 8.3.

Assume $u\in\cal Y$ is a solution to (1.1), and define
$u^{-}_{\theta}(x)=\min\\{\,u_{\theta}(x),\,0\,\\}$. Let
$\xi_{R},\zeta_{\delta}$ be the cut-off functions in (8.1)-(8.2). For every
$\varepsilon>0$ there exist $\delta>0$ and $R>r_{1}$ such that
$Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})\in[0,\varepsilon)$.
If, furthermore, $u^{-}_{\theta}=0$ on $\Gamma_{\\!\frac{\beta}{2}}$ then we
may also achieve
$Q_{\Omega_{0\frac{\beta}{2}}}(\zeta_{\delta}\,\xi_{R}\,u^{-}_{\theta})\in[0,\varepsilon)$.

###### Proof.

First of all, we obtain

$\int_{\Omega_{0\beta}}|\nabla\xi_{R}|^{2}\,u_{\theta}^{2}\,dx<\frac{\varepsilon}{\,2\,}$
(8.8)

by taking $R$ large. Indeed, we have

$\int_{\Omega_{0\beta}}|\nabla\xi_{R}|^{2}\,u_{\theta}^{2}\,dx\leq\frac{4}{\,R^{2}\,}\int_{R}^{2R}\bigg{(}\int_{\Sigma_{\beta
r}}u_{\theta}^{2}\,d\Sigma\bigg{)}\,dr,$

where $\Sigma_{\beta r}=\Omega_{0\beta}\cap\partial B_{r}$. Since
$|u_{\theta}(x)|\leq d(x)\,|\nabla u(x)|\leq r\,|\nabla u|$ on $\partial
B_{r}$, we may write

$\displaystyle\frac{1}{\,4R^{2}\,}\int_{R}^{2R}\bigg{(}\int_{\Sigma_{\beta
r}}u_{\theta}^{2}\,d\Sigma\bigg{)}\,dr$
$\displaystyle\leq\frac{1}{\,4R^{2}\,}\int_{R}^{2R}r^{2}\,\bigg{(}\int_{\Sigma_{\beta
r}}|\nabla u|^{2}\,d\Sigma\bigg{)}\,dr$
$\displaystyle\leq\int_{R}^{+\infty}\bigg{(}\int_{\Sigma_{\beta r}}|\nabla
u|^{2}\,d\Sigma\bigg{)}\,dr,$

where the right-hand side is infinitesimal as $R\to+\infty$. Hence we may
choose $R$ so that (8.8) holds. Next we observe that the function
$u_{\theta}$, which is uniformly continuous in the compact set
$\overline{\Omega}_{0\beta}\cap\overline{B}_{2R}$, converges uniformly to zero
when $d(x)\to r_{1}$. Keeping this in mind, and letting $D_{\delta
R}=\\{\,x\in\Omega_{0\beta}:\zeta_{\delta}(x)\,\xi_{R}(x)>0\,\\}$, we may
check that

$\lim_{\delta\to 0}\int_{D_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,u_{\theta}^{2}\,dx=\int_{\Omega_{0\beta}}|\nabla\xi_{R}|^{2}\,u_{\theta}^{2}\,dx.$
(8.9)

To see this, choose $\mu>0$ and let $\delta$ be so small that
$u_{\theta}^{2}<\mu$ in the set $S_{\delta R}=\\{\,x\in\Omega_{0\beta}\cap
B_{2R}:d(x)<r_{1}+\delta\,\\}$. Since
$\nabla(\zeta_{\delta}\,\xi_{R})=\zeta_{\delta}\,\nabla\xi_{R}+\xi_{R}\,\nabla\zeta_{\delta}$,
we have

$\displaystyle\Big{|}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}-|\nabla\xi_{R}|^{2}\Big{|}$
$\displaystyle=\Big{|}(\zeta_{\delta}^{2}-1)\,|\nabla\xi_{R}|^{2}+2\,\zeta_{\delta}\,\xi_{R}\,\nabla\zeta_{\delta}\,\nabla\xi_{R}+\xi_{R}^{2}\,|\nabla\zeta_{\delta}|^{2}\Big{|}$
$\displaystyle\leq
2\,|\nabla\xi_{R}|^{2}+2\,|\nabla\zeta_{\delta}|^{2}\leq\frac{8}{\,R^{2}\,}+\frac{\,18\,}{\,\delta^{2}\,}$

and the difference of the two preceding integrals is estimated as follows:

$\displaystyle\left|\int_{D_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,u_{\theta}^{2}\,dx-\int_{\Omega_{0\beta}}|\nabla\xi_{R}|^{2}\,u_{\theta}^{2}\,dx\right|\leq\mu\,|S_{\delta
R}|\,\Big{(}\frac{8}{\,R^{2}\,}+\frac{\,18\,}{\,\delta^{2}\,}\Big{)}.$

Since $R$ is kept fixed, we have $|S_{\delta R}|\leq C\,\delta^{2}$ for some
constant $C$, hence

$\left|\int_{D_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,u_{\theta}^{2}\,dx-\int_{\Omega_{0\beta}}|\nabla\xi_{R}|^{2}\,u_{\theta}^{2}\,dx\right|\leq\mu\,C\,\Big{(}\frac{\,8\,\delta^{2}\,}{\,R^{2}\,}+18\Big{)}.$

Finally, since $\mu$ is arbitrary, (8.9) follows. Recalling (8.8), we conclude
that

$\int_{D_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,u_{\theta}^{2}\,dx<\varepsilon$
(8.10)

for convenient $R$ and $\delta$. Now, multiplying the equation
$L_{u}\,u_{\theta}=0$ (see Proposition 3.1) by
$(\zeta_{\delta}\,\xi_{R})^{2}\,u_{\theta}$ and integrating by parts over
$\Omega_{0\beta}\cap B_{2R}$ we obtain

$\int_{D_{\delta
R}}\Big{\\{}\nabla\big{(}(\zeta_{\delta}\,\xi_{R})^{2}\,u_{\theta}\big{)}\,\nabla
u_{\theta}-(\zeta_{\delta}\,\xi_{R})^{2}\,u_{\theta}^{2}\,f^{\prime}(d(x),z,u)\Big{\\}}\,dx=0.$

Replacing $\xi_{R}$ with $\zeta_{\delta}\,\xi_{R}$ in (8.6) and letting
$\varphi=u_{\theta}$ (see also (8.7)) this is turned into

$Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})=\int_{D_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,u_{\theta}^{2}\,dx.$ (8.11)

Hence
$Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})\in[0,\varepsilon)$
for a convenient choice of $\delta$. To prove the second claim it suffices to
multiply the equation $L_{u}\,u_{\theta}=0$ by
$(\zeta_{\delta}\,\xi_{R})^{2}\,u_{\theta}^{-}$ and integrate by parts over
$\Omega_{0\frac{\beta}{2}}\cap B_{2R}$. Since $u^{-}_{\theta}=0$ on
$\Gamma_{\\!\frac{\beta}{2}}$ by assumption, we obtain

$\int_{D}\Big{\\{}\nabla\big{(}(\zeta_{\delta}\,\xi_{R})^{2}\,u_{\theta}^{-}\big{)}\,\nabla
u_{\theta}^{-}-(\zeta_{\delta}\,\xi_{R})^{2}\,(u_{\theta}^{-})^{2}\,f^{\prime}(d(x),z,u)\Big{\\}}\,dx=0,$

and therefore (see (8.11))

$Q_{\Omega_{0\frac{\beta}{2}}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta}^{-})=\int_{D^{\prime}_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,(u_{\theta}^{-})^{2}\,dx\leq\int_{D_{\delta
R}}|\nabla(\zeta_{\delta}\,\xi_{R})|^{2}\,u_{\theta}^{2}\,dx,$

where $D^{\prime}_{\delta
R}=\\{\,x\in\Omega_{0\frac{\beta}{2}}:\zeta_{\delta}(x)\,\xi_{R}(x)>0\,\\}$.
The last integral is made arbitrarily small by a convenient choice of $\delta$
and $R$ (see (8.10)), and the proof is complete. ∎

##  9 Preliminaries in the unbounded half-sector

The counterpart of Corollary 2.2 in an unbounded domain is the following

######  Lemma 9.1 (Splitting lemma for unbounded domains).

Assume $u\in\cal Y$ is a solution to (1.1) with Morse index $m(u)\leq 1$, and
let $\alpha\in(0,\beta)$. Then either

$\inf_{\varphi\in
C^{1}_{0}(\Omega_{0\alpha}\cup\Gamma_{\\!0})}Q_{\Omega_{0\alpha}}(\varphi)\geq
0$ (9.1)

or

$\inf_{\varphi\in
C^{1}_{0}(\Omega_{\alpha\beta}\cup\Gamma_{\\!\beta})}Q_{\Omega_{\alpha\beta}}(\varphi)\geq
0.$

###### Proof.

If the thesis is not true there exist two functions
$\varphi_{0\alpha},\varphi_{\alpha\beta}$ supported in
$\Omega_{0\alpha}\cup\Gamma_{\\!0}$ and
$\Omega_{\alpha\beta}\cup\Gamma_{\\!\beta}$, respectively, such that
$Q_{\Omega_{0\alpha}}(\varphi_{0\alpha}),Q_{\Omega_{\alpha\beta}}(\varphi_{\alpha\beta})<0$.
Since they both vanish in a neighborhood of $\Gamma_{\\!\alpha}$ we may extend
them to the whole of $\Omega_{0\beta}$ as in Lemma 2.1. The extended functions
belong to $C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$
and are linearly independent. Then, as in the proof of Lemma 2.1 it is easy to
prove that $Q_{\Omega_{0\beta}}(\phi)<0$ for every
$\phi=a\,\varphi_{0\alpha}+b\,\varphi_{\alpha\beta}$ with $(a,b)\neq(0,0)$
contradicting the fact that the Morse index of $u$ in $\Omega_{0\beta}$ is at
most one. ∎

Next we state a sign preservation property for $w_{\frac{\beta}{2}}$ which
extends Corollary 2.4 to unbounded domains.

######  Lemma 9.2 (Sign preservation in the unbounded half-sector).

Assume $u\in\cal Y$ is a solution to (1.1) with Morse index $m(u)\leq 1$.
Assume further $f(r,z,u)$ is convex with respect to $u$. If
$w_{\frac{\beta}{2}}$ does not vanish identically in
$\Omega_{0\frac{\beta}{2}}$, then either $w_{\frac{\beta}{2}}>0$ in
$\Omega_{0\frac{\beta}{2}}$ or $w_{\frac{\beta}{2}}<0$ there.

###### Proof.

Letting $\alpha=\frac{\beta}{2}$ in Lemma 9.1, we may suppose

$\inf_{\varphi\in
C^{1}_{0}(\Omega_{0\frac{\beta}{2}}\cup\Gamma_{\\!0})}Q_{\Omega_{0\frac{\beta}{2}}}(\varphi)\geq
0.$ (9.2)

Thus, assumption (8.3) is satisfied, and the claim follows from Lemma 8.1 with
$K=\emptyset$, since, by the convexity of $f$, $w_{\frac{\beta}{2}}$ satisfies
$L_{u}w_{\frac{\beta}{2}}\geq 0$. ∎

We can now prove the following counterpart of Lemma 3.2 for unbounded domains:

######  Lemma 9.3.

Assume $u\in\cal Y$ is a solution to (1.1) with Morse index $m(u)\leq 1$. If
$w_{\frac{\beta}{2}}\equiv 0$ in $\Omega_{0\beta}$ then $u$ is constant with
respect to $\theta$.

###### Proof.

Since $w_{\frac{\beta}{2}}\equiv 0$ in $\Omega_{0\beta}$, the solution $u$ is
symmetric about $\Gamma_{\\!\frac{\beta}{2}}$, as well as
$f^{\prime}(d(x),z,u)$, while the derivative $u_{\theta}$ is odd with respect
to $\Gamma_{\\!\frac{\beta}{2}}$ in the sense that
$u_{\theta}(x)=-u_{\theta}(\sigma_{\frac{\beta}{2}}(x))$ in $\Omega_{0\beta}$.
We already know that $u_{\theta}$ satisfies the equation in (3.1) and vanishes
on $\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$. Our goal is to prove that $\partial
u_{\theta}/\partial\nu$ also vanishes there.

Part I. We start by showing that $Q_{\Omega_{0\beta}}(\varphi_{1})\geq 0$ for
every $\varphi_{1}$ in the vector space $C^{1,\rm
odd}_{0}(\Omega_{0\beta}\allowbreak\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ of
the odd elements of
$C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$. In fact,
if $\varphi_{1}$ vanishes identically the claim is obvious. Otherwise we
consider the absolute value $\varphi_{2}(x)=|\varphi_{1}(x)|$, which is
symmetric about $\Gamma_{\\!\frac{\beta}{2}}$ and therefore

$\int_{\Omega_{0\beta}}f^{\prime}(d(x),z,u)\,\varphi_{1}\,\varphi_{2}\,dx=0.$

Moreover, we have $\nabla\varphi_{2}=\nabla\varphi_{1}$ in the set
$\Omega^{+}_{0\beta}=\\{\,x\in\Omega_{0\beta}:\varphi_{1}(x)>0\,\\}$, and
$\nabla\varphi_{2}=-\nabla\varphi_{1}$ in
$\Omega^{-}_{0\beta}=\\{\,x\in\Omega_{0\beta}:\varphi_{1}(x)<0\,\\}$. Since
$\Omega^{-}_{0\beta}=\sigma_{\frac{\beta}{2}}(\Omega^{+}_{0\beta})$ by the
oddity of $\varphi_{1}$, we find

$\int_{\Omega_{0\beta}}\nabla\varphi_{1}\,\nabla\varphi_{2}\,dx=\int_{\Omega^{+}_{0\beta}}|\nabla\varphi_{1}|^{2}\,dx-\int_{\Omega^{-}_{0\beta}}|\nabla\varphi_{1}|^{2}\,dx=0.$

Thus, for every $(a,b)\neq(0,0)$ we may compute

$\displaystyle Q_{\Omega_{0\beta}}(a\,\varphi_{1}+b\,\varphi_{2})$
$\displaystyle=\int_{\Omega_{0\beta}}\Big{(}|a\,\nabla\varphi_{1}+b\,\nabla\varphi_{2}|^{2}-(a\,\varphi_{1}+b\,\varphi_{2})^{2}\,f^{\prime}(d(x),z,u)\Big{)}\,dx$
$\displaystyle=a^{2}\,Q_{\Omega_{0\beta}}(\varphi_{1})+b^{2}\,Q_{\Omega_{0\beta}}(\varphi_{2})=(a^{2}+b^{2})\,Q_{\Omega_{0\beta}}(\varphi_{1}).$

Since $\varphi_{1},\varphi_{2}$ are linearly independent, and since $m(u)\leq
1$ by assumption, we must have $Q_{\Omega_{0\beta}}(\varphi_{1})\geq 0$, as
claimed.

Part II. We claim that ${\cal B}(u_{\theta},\psi)=0$ for every $\psi\in
C^{1,\rm odd}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$,
which is not obvious because $\psi$ is allowed to take nonzero values on
$\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$. However, by Part I the bilinear form
${\cal B}$ is positive semidefinite in $C^{1,\rm
odd}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})\times C^{1,\rm
odd}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ and therefore
the Cauchy-Schwarz inequality (6.2) holds. Without loss of generality we may
assume that the cut-off functions $\xi_{R}(x)$ are symmetric with respect to
$\Gamma_{\\!\frac{\beta}{2}}$: thus, $\xi_{R}(x)\,u_{\theta}(x)$ is odd and
has a compact support. Furthermore, since $f^{\prime}(r,z,u)$ is locally
Hölder continuous and $u\in\cal Y$, $u_{\theta}$ is in
$C^{1}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})\cap
C^{0}(\overline{\Omega}_{0\beta}\setminus\Upsilon)$: however, its gradient may
fail to belong to $L^{2}(\Omega_{0\beta}\cap B_{2R})$, due to the possible
singularity at $d(x)=r_{1}$. To overcome this difficulty, we use the cut-off
functions $\zeta_{\delta}(x)$ introduced in (8.2). The doubly truncated
function $\zeta_{\delta}\,\xi_{R}\,u_{\theta}$ is still odd, has a compact
support, and belongs to $C^{1,\rm
odd}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$. Thus, we may
write

$0\leq\big{(}{\cal
B}(\zeta_{\delta}\,\xi_{R}\,u_{\theta},\,\psi)\big{)}^{2}\leq
Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})\,Q_{\Omega_{0\beta}}(\psi),$

where $\psi\in C^{1,\rm
odd}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$. Keeping
$\psi$ fixed, we take $\delta$ small and $R$ large to achieve
$\zeta_{\delta}\,\xi_{R}=1$ on the set $\mathop{\rm supp}\psi$. Consequently,
the preceding inequalities reduce to

$0\leq\big{(}{\cal B}(u_{\theta},\,\psi)\big{)}^{2}\leq
Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})\,Q_{\Omega_{0\beta}}(\psi).$

By Proposition 8.3, the term
$Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})$ in the right-hand
side becomes arbitrarily small by a convenient choice of $R$ and $\delta$,
which implies ${\cal B}(u_{\theta},\,\psi)=0$.

Part III. Let us verify that $u_{\theta}$ satisfies the Neumann condition on
$\Gamma_{\\!0}$. Recall that ${\cal B}(u_{\theta},\,\psi)=0$ by Part II, and
take $R>r_{1}$ such that $\mathop{\rm supp}\psi\subset B_{R}$. Multiplying the
equation $L_{u}\,u_{\theta}=0$ by $\psi\in C^{1,\rm
odd}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ and
integrating by parts over $\Omega_{0\beta}\cap B_{R}$ we obtain

$0={\cal
B}(u_{\theta},\psi)=\int_{\Gamma_{\\!0}\cup\Gamma_{\\!\beta}}\psi\,\frac{\,\partial
u_{\theta}\,}{\partial\nu}\,d\Sigma=2\int_{\Gamma_{\\!0}}\psi\,\frac{\,\partial
u_{\theta}\,}{\partial\nu}\,d\Sigma,$

where the equality holds because $\psi$ vanishes in a neighborhood of
$\gamma_{0\beta}$ as well as on $\Omega_{0\beta}\cap\partial B_{R}$ and
$\psi\,\frac{\,\partial u_{\theta}\,}{\partial\nu}$ is symmetric with respect
to $\Gamma_{\\!\frac{\beta}{2}}$. Since $\psi$ can take any value on
$\Gamma_{\\!0}$ then $\frac{\,\partial u_{\theta}\,}{\partial\nu}=0$ on
$\Gamma_{0}$ as claimed.

Conclusion. The derivative $u_{\theta}$ can be extended to zero in the domain
$\Omega_{-\frac{\beta}{2},0}$ (cf. (3.3)), thus obtaining a $C^{1}$ weak
solution of $L_{u}\,u_{\theta}=0$ in
$\Omega_{-\frac{\beta}{2},\frac{\beta}{2}}$. By the unique continuation
property, $u_{\theta}$ must vanish identically, and therefore $u$ is constant
with respect to $\theta$. ∎

When the domain $\Omega_{0\beta}$ is unbounded, in place of Lemma 3.3 we have:

######  Lemma 9.4.

Assume $u\in\cal Y$ is a solution to (1.1) with Morse index $m(u)\leq 1$.
Assume further that (9.2) holds. If $w_{\frac{\beta}{2}}(x)>0$ in
$\Omega_{0\frac{\beta}{2}}$ then

$u_{\theta}>0\ \ \text{in\/
}\Omega_{0\frac{\beta}{2}}\cup\Gamma_{\\!\frac{\beta}{2}}.$ (9.3)

###### Proof.

Let us check that the negative part
$u_{\theta}^{-}=\min\\{\,0,u_{\theta}\,\\}\leq 0$ is a weak solution of
$L_{u}\,u_{\theta}^{-}=0$ in $\Omega_{0\frac{\beta}{2}}$, i.e., ${\cal
B}(u_{\theta}^{-},\psi)=0$ for all $\psi\in
C^{\infty}_{0}(\Omega_{0\frac{\beta}{2}})$. Assumption (9.2) implies that the
bilinear form ${\cal B}$ is positive semidefinite on
$C^{1}_{0}(\Omega_{0\frac{\beta}{2}}\cup\Gamma_{\\!0})\times
C^{1}_{0}(\Omega_{0\frac{\beta}{2}}\cup\Gamma_{\\!0})$. Hence we pick $\psi\in
C^{\infty}_{0}(\Omega_{0\frac{\beta}{2}})$ and consequently $R>r_{1}$ such
that $\mathop{\rm supp}\psi\subset B_{R}$, and also $\delta>0$ such that

$\inf_{x\in\mathop{\rm supp}\psi}d(x)>r_{1}+\delta.$ (9.4)

As in the proof of Lemma 3.3, the assumption $w_{\frac{\beta}{2}}>0$ in
$\Omega_{0\frac{\beta}{2}}$ implies $u_{\theta}>0$ on
$\Gamma_{\\!\frac{\beta}{2}}$ (see (3.5)), hence $u_{\theta}^{-}=0$ on
$\gamma_{0\frac{\beta}{2}}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\frac{\beta}{2}}$.
Furthermore, the doubly truncated function
$\zeta_{\delta}\,\xi_{R}\,u_{\theta}^{-}$ has a square-summable gradient and
therefore belongs to $H^{1}_{0}(D)$, where $D=\Omega_{0\frac{\beta}{2}}\cap
B_{2R}$. The Cauchy-Schwarz inequality (6.2) yields

$0\leq\big{(}{\cal B}(u_{\theta}^{-},\psi)\big{)}^{2}=\big{(}{\cal
B}(\zeta_{\delta}\,\xi_{R}\,u_{\theta}^{-},\,\psi)\big{)}^{2}\leq
Q_{\Omega_{0\frac{\beta}{2}}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta}^{-})\,Q_{\Omega_{0\frac{\beta}{2}}}(\psi).$

By Proposition 8.3, the term
$Q_{\Omega_{0\frac{\beta}{2}}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta}^{-})$
becomes arbitrarily small by a convenient choice of $R$ and $\delta$, which
implies ${\cal B}(u_{\theta}^{-},\psi)=0$ and therefore $u_{\theta}^{-}$ is a
weak solution of $L_{u}\,u_{\theta}^{-}=0$ in $\Omega_{0\frac{\beta}{2}}$, as
claimed. By the unique continuation property, this implies that either
$u_{\theta}^{-}\equiv 0$ or $u_{\theta}^{-}<0$ in $\Omega_{0\frac{\beta}{2}}$.
But since $u_{\theta}>0$ on $\Gamma_{\\!\frac{\beta}{2}}$, we must have
$u_{\theta}^{-}\equiv 0$ and therefore $u_{\theta}\geq 0$ in
$\Omega_{0\frac{\beta}{2}}$. The unique continuation property of $u_{\theta}$
proves (9.3). ∎

##  10 Proof of Theorem 6.4

###### Proof of Theorem 6.4.

If $w_{\frac{\beta}{2}}\equiv 0$, then $u$ is independent of $\theta$ by Lemma
9.3. If, instead, $w_{\frac{\beta}{2}}\not\equiv 0$, then Lemma 9.2 implies
that either $w_{\frac{\beta}{2}}>0$ in $\Omega_{0\frac{\beta}{2}}$ or
$w_{\frac{\beta}{2}}>0$ in $\Omega_{\frac{\beta}{2}\beta}$. By Lemma 9.1 we
may assume that (9.1) holds and $w_{\frac{\beta}{2}}>0$ in
$\Omega_{0\frac{\beta}{2}}$. We then define the set $A^{\prime}$ as in the
proof of Theorem 1.1 and, arguing as in Part I with (9.3) instead of (3.4), we
get that $A^{\prime}$ contains $\frac{\beta}{2}$. The fact that $A^{\prime}$
is a closed subinterval of $\left[\frac{\beta}{2},\beta\right]$ follows as in
Part II. Next, exactly as in Part III, the Hopf boundary point lemma implies
that $u_{\theta}>0$ in $\Omega_{0,\tilde{\alpha}}$ if
$A^{\prime}=\left[\frac{\beta}{2},\tilde{\alpha}\right]$,
$\tilde{\alpha}<\beta$. In such a case we can repeat exactly Part IV getting
that

$w_{\tilde{\alpha}}>0$ in $\Omega_{0\tilde{\alpha}}$.

If the Morse index $m(u)$ is $1$, there must be $\varphi_{1}\in
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ such that
$Q_{\Omega_{0\beta}}(\varphi_{1})<0$ and we may take a torus $T$ containing
$\mathop{\rm supp}\varphi_{1}$. By a torus we mean an open subset
$T\subset\mathbb{R}^{N}$, invariant under cylindrical rotations, whose
intersection with the hyperplane $x_{1}=0$ is an $N-1$ dimensional ball lying
in the half-plane $x_{1}=0$, $x_{2}>r_{1}+\varepsilon$ for some
$\varepsilon>0$. The last condition implies that $d(x)\geq
r_{1}+\varepsilon>r_{1}$ for all $x\in\overline{T}$. In the case when $m(u)=0$
we take any torus $T$ as defined above. Note that when $N=2$ the tori are just
annuli. Since $w_{\tilde{\alpha}}$ and $u_{\theta}$ are positive in
$\Omega_{0\tilde{\alpha}}$, we obviously have

$w_{\tilde{\alpha}},u_{\theta}>0$ in
$\Omega_{0\tilde{\alpha}}\cap\overline{T}$. (10.1)

Part V. By the same argument used to derive (3.6) and (3.7) we see that
$w_{\alpha}>0$ on $\Gamma_{\\!0}$ for every $\alpha\in(0,\beta)$.

Part VI. We claim that $w_{\alpha}$ keeps positive in
$G_{\alpha}=(\Omega_{0\alpha}\cup{\Gamma_{\\!0}})\cap\overline{T}$ for every
$\alpha>\tilde{\alpha}$ such that the difference $\alpha-\tilde{\alpha}$ is
sufficiently small. This can be proved by contradiction: suppose that there
exists a sequence $\alpha_{n}\searrow\tilde{\alpha}$ such that
$w_{\alpha_{n}}$ becomes negative somewhere in $G_{\alpha_{n}}$, and let
$x_{n}\in\overline{\Omega}_{0\alpha_{n}}\cap\overline{T}$ be a point where
$w_{\alpha_{n}}$ takes its negative minimum. Of course,
$x_{n}\not\in\Gamma_{\\!\alpha_{n}}$ because $w_{\alpha_{n}}=0$ there, and
$x_{n}\not\in\Gamma_{\\!0}$ by Part V. By compactness there exists a
subsequence, still denoted by $(x_{n})$, converging to some
$x_{0}\in\overline{\Omega}_{0\alpha}\cap\overline{T}$ with
$w_{\tilde{\alpha}}(x_{0})=0$. Since $w_{\tilde{\alpha}}>0$ on $\Gamma_{\\!0}$
by Part V, and by (10.1), we must have

$x_{0}\in\Gamma_{\tilde{\alpha}}\cap\overline{T}$. (10.2)

Furthermore, the gradient $\nabla w_{\alpha_{n}}$ vanishes at $x_{n}$ in the
case when $x_{n}\in\Omega_{0\alpha_{n}}\cap T$, and is directed towards the
interior of $T$ if $x_{n}\in\Omega_{0\alpha_{n}}\cap\partial T$. Hence, we may
write

$\partial w_{\alpha_{n}}/\partial\theta=0$ at $x_{n}$ for every $n$.

Passing to the limit, this property should be inherited by $\partial
w_{\tilde{\alpha}}/\partial\theta$ at $x_{0}$. However, by (10.2) and by the
Hopf boundary point lemma we have $\partial
w_{\tilde{\alpha}}/\partial\theta<0$ at $x_{0}$, a contradiction. Hence,
$w_{\alpha}$ must be positive in $G_{\alpha}$ for every
$\alpha>\tilde{\alpha}$ such that the difference $\alpha-\tilde{\alpha}$ is
sufficiently small, as claimed.

Part VII. For every $\alpha$ as above, $w_{\alpha}$ is in fact positive in all
of $\Omega_{0\alpha}$. To see this, we let $K=\overline{T}$ in Lemma 8.1,
hence $D_{\alpha}=\Omega_{0\alpha}\setminus\overline{T}$. The quadratic form
$Q_{\Omega_{0\alpha}}$ is positive semidefinite in
$C^{1}_{0}(\Omega_{0\alpha}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\alpha}\setminus\overline{T})$
by Remark 6.3, and keeps positive semidefinite in
$H^{1}_{\gamma_{0\alpha}\cup\partial T\cup\partial B_{R}}(D_{\alpha}\cap
B_{R})$ for $R>\max_{x\in\overline{T}}d(x)$. Taking into account the boundary
conditions $w_{\alpha}>0$ on $(\Omega_{0\alpha}\cap\partial
T)\cup\Gamma_{\\!0}$ and $w_{\alpha}=0$ on
$\gamma_{0\alpha}\cup\Gamma_{\\!\alpha}$, by Lemma 8.1 it follows that
$w_{\alpha}$ is positive in any connected component of $D_{\alpha}$ and
consequently in $\Omega_{0\alpha}$.

Conclusion. We have thus seen that if we suppose $\tilde{\alpha}<\beta$ we
obtain the positivity of $w_{\alpha}$ in $\Omega_{0\alpha}$ for every
$\alpha>\tilde{\alpha}$ such that the difference $\alpha-\tilde{\alpha}$ is
sufficiently small, contradicting the definition of $\tilde{\alpha}$. Hence,
under the assumption that $w_{\frac{\beta}{2}}>0$ in
$~{}\Omega_{0\frac{\beta}{2}}$, we must have $\tilde{\alpha}=\beta$ and
therefore the solution $u$ is strictly increasing with respect to $\theta$ in
$\Omega_{0\beta}$. In this case, by Part III we also have $u_{\theta}>0$ in
$\Omega_{0\beta}$ and we can apply Lemma 2.1 in [11] in $\Omega_{0\beta}$
getting that $\inf Q_{\Omega_{0\beta}}(\psi)\geq 0$ when $\psi\in
C^{1}_{0}(\Omega_{0\beta})$. By Proposition 8.3, the term
$Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})$ becomes arbitrarily
small by a convenient choice of $R$ and $\delta$ showing that $\inf_{\psi\in
C^{1}_{0}(\Omega_{0\beta})}Q_{\Omega_{0\beta}}(\psi)=0$. In summary, we have
shown that if $m(u)\leq 1$ then either $u_{\theta}$ vanishes identically in
$\Omega_{0\beta}$ or it keeps its sign there. Now, in order to prove Claim 1
of the statement, it suffices to check that if $m(u)=0$ then $u_{\theta}$
vanishes identically. To this purpose, take $\psi\in
C^{1}_{0}(\Omega_{0\beta}\cup\Gamma_{\\!0}\cup\Gamma_{\\!\beta})$ and consider
the doubly truncated function $\zeta_{\delta}\,\xi_{R}\,u_{\theta}$ with $R$
such that $\mathop{\rm supp}\psi\subset B_{R}$ and $\delta$ satisfying (9.4).
Since $Q_{\Omega_{0\beta}}$ is positive semidefinite by assumption, and since
$\zeta_{\delta}\,\xi_{R}\,u_{\theta}$ belongs to
$H^{1}_{\gamma_{0\beta}\cup\partial B_{2R}}(D)$, where $D=\Omega_{0\beta}\cap
B_{2R}$, the Cauchy-Schwarz inequality yields

$0\leq\big{(}{\cal B}(u_{\theta},\,\psi)\big{)}^{2}=\big{(}{\cal
B}(\zeta_{\delta}\,\xi_{R}\,u_{\theta},\,\psi)\big{)}^{2}\leq
Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})\,Q_{\Omega_{0\beta}}(\psi).$

By Proposition 8.3, the term
$Q_{\Omega_{0\beta}}(\zeta_{\delta}\,\xi_{R}\,u_{\theta})$ becomes arbitrarily
small by a convenient choice of $R$ and $\delta$. On the other side,
multiplying the equation $L_{u}\,u_{\theta}=0$ by $\psi$ and integrating by
parts we obtain

${\cal
B}_{\Omega_{0\beta}}(u_{\theta},\psi)=\int_{\Gamma_{\\!0}\cup\Gamma_{\\!\beta}}\psi\,\frac{\,\partial
u_{\theta}\,}{\,\partial\nu\,}\,d\Sigma,$

hence the integral must vanish. Since $\psi$ can take any value on
$\Gamma_{\\!0}\cup\Gamma_{\\!\beta}$, it follows that $\partial
u_{\theta}/\partial\nu=0$ there, whence $u_{\theta}$ can be extended by zero
to $\Omega_{-\beta,0}$ as in (3.3), thus obtaining a $C^{1}$ weak solution
$\tilde{u}_{\theta}$ of $L_{u}\,\tilde{u}_{\theta}=0$ in
$\Omega_{-\beta,\beta}$. Since $\tilde{u}_{\theta}$ vanishes identically in
$\Omega_{-\beta,0}$, by the unique continuation property we must have
$u_{\theta}=0$ in $\Omega_{0\beta}$, hence $u$ is independent of $\theta$, and
the proof is complete. ∎

## Acknowledgements

The authors are members of the Gruppo Nazionale per l’Analisi Matematica, la
Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta
Matematica (INdAM). F. Gladiali is supported by Uniss, Fondazione di Sardegna,
annualità 2017. A. Greco is partially supported by the research project
Evolutive and stationary Partial Differential Equations with a focus on bio-
mathematics, funded by Fondazione di Sardegna, annualità 2019. This paper is
dedicated to the city of Alghero (Sardinia, Italy) where the authors
repeatedly met to carry their work on.

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Francesca Gladiali | Antonio Greco
---|---
Dipartimento di Chimica e Farmacia | Dipartimento di Matematica e Informatica
Università degli Studi di Sassari | Università degli Studi di Cagliari
Italy | Italy
e-mail<EMAIL_ADDRESS>| e-mail<EMAIL_ADDRESS>