Symmetry Breaking for the Planar Lane–Emden Equation with Robin Boundary Conditions

1. Introduction

Let $B_R\subset {\mathbb{R}}^2$ be the open disk of radius $R>0$ centered at the origin. We study

$$\left\{\begin{array}{cc}-\Delta u=u^p\hfill & \mathrm{in}{B}_R,\hfill \\ u>0\hfill & \mathrm{in}{B}_R,\hfill \\ {\partial }_{\nu }u+\beta u=0\hfill & \mathrm{on}\partial B_R,\hfill \end{array}\right.p>1,\beta >0.$$

(1)

We use $\nu$ for the outer unit normal. When no confusion is possible, we refer to (1) simply as problem (1).

For the corresponding Dirichlet problem on a disk, the classical moving-plane theorem forces every positive solution to be radial; uniqueness then follows from the radial initial-value problem [9]. The Robin problem is structurally different because its boundary trace is not prescribed to be constant, so the usual moving-plane initiation at the boundary is unavailable. This distinction is genuine: Celentano, Masiello and Paoli constructed positive nonradial superharmonic solutions for suitable nonlinearities under Robin boundary conditions [2]. Symmetry-breaking phenomena for semilinear elliptic equations have also been studied under general boundary conditions and on annular domains; see, among others, [10,12,13,14].

For the pure-power Robin problem (1), uniqueness in the radial class must be distinguished from uniqueness among all positive solutions. Dai and Fu proved uniqueness for sufficiently small Robin parameter, whereas Fu and Dai obtained nonuniqueness on annular domains for sufficiently large parameter [6,7]. Chen, Grossi and Li explicitly note that arbitrary-parameter uniqueness remains open even when the domain is a ball [3]. In dimension two every finite exponent $p>1$ is Sobolev-subcritical, so this question includes the full exponent range considered here. Related nondegeneracy and local-uniqueness problems for the planar Dirichlet Lane–Emden equation require delicate analysis; see [11] and the references therein.

The papers of Smoller and Wasserman provide the closest general background. Their work on positive solutions and their companion analysis under general boundary conditions establish broad mechanisms by which radial states may lose symmetry, while their later equivariant theory clarifies the role of symmetry in bifurcation [12,13,14]. The present problem, however, requires quantitative information specific to the pure-power Robin branch. The argument below supplies four such ingredients: a global parametrization of the complete positive radial branch by the logarithmic slope of one normalized Lane–Emden profile; an explicit scalar function $F_p$ whose zeros are exactly the first-angular zero modes; a phase–plane estimate producing two sign-changing simple zeros for every $p\ge 12$; and a direct verification of the Fredholm and transversality hypotheses that converts those crossings into positive nonradial branches. Thus the new content is the explicit two-crossing mechanism for the planar pure-power Robin equation, rather than the abstract symmetry-breaking principle itself. To the best of the author’s knowledge, this specific result does not appear in the cited literature.

The purpose of the paper is to establish this quantitative symmetry-breaking mechanism on a planar disk. The ODE and sectorial spectral arguments are developed explicitly, and the functional-analytic step records the kernel, range, and crossing calculations needed for the Crandall–Rabinowitz theorem. The restriction $p\ge 12$ comes from a concrete phase–plane estimate and is not expected to be optimal.

Theorem 1.1 

(Main theorem). Let $p\ge 12$. For the unit disk $B_1\subset {\mathbb{R}}^2$ there exist two parameters

$$0<{\beta }_{p,-}<{\displaystyle \frac{1}{2}}<{\beta }_{p,+}<1$$

such that, for each $\sigma \in \{-,+\}$, the point

$$\left({\beta }_{p,\sigma },U_{{\beta }_{p,\sigma }}\right)$$

on the positive radial branch is a local bifurcation point of positive nonradial solutions of (1).

More precisely, for each $\sigma \in \{-,+\}$ there are $\epsilon >0$ and a $C^1$ curve

$$(-\epsilon ,\epsilon )\ni t⟼\left({\beta }_{\sigma }\left(t\right),u_{\sigma }\left(t\right)\right)$$

of solutions of (1) such that

$${\beta }_{\sigma }\left(0\right)={\beta }_{p,\sigma },u_{\sigma }\left(0\right)=U_{{\beta }_{p,\sigma }},$$

and $u_{\sigma }\left(t\right)$ is positive and nonradial for every sufficiently small $t\ne 0$. Here $U_{\beta }$ denotes the unique positive radial solution at the parameter β.

Corollary 1.2 

(Local multiplicity). Fix $\alpha \in (0,1)$. For each $\sigma \in \{-,+\}$ there are solutions $({\beta }_n,u_n)$ of (1) such that

$${\beta }_n⟶{\beta }_{p,\sigma },u_n⟶U_{{\beta }_{p,\sigma }}\mathrm{in}{C}^{2,\alpha }\left(\overline{B_1}\right),$$

where every $u_n$ is positive and nonradial. At the same parameter ${\beta }_n$, the problem also possesses its unique positive radial solution $U_{{\beta }_n}$. Thus, for each sign, uniqueness among all positive solutions fails along a sequence of Robin parameters converging to ${\beta }_{p,\sigma }$. No assertion is made here that a nonradial solution exists at the limiting parameter ${\beta }_{p,\sigma }$ itself.

Scaling gives the following version on an arbitrary disk.

Corollary 1.3. 

Let $p\ge 12$ and $R>0$. There are

$$0<{\beta }_{p,-}\left(R\right)<{\displaystyle \frac{1}{2R}}<{\beta }_{p,+}\left(R\right)<{\displaystyle \frac{1}{R}}$$

at which positive nonradial solutions bifurcate from the positive radial branch in $B_R$. In fact,

$${\beta }_{p,\pm }\left(R\right)={\displaystyle \frac{{\beta }_{p,\pm }\left(1\right)}{R}}.$$

The proof has four steps. First, the complete positive radial branch is obtained by scaling one normalized Lane–Emden profile. Second, the translation Jacobi field identifies first-angular-mode degeneracy through

$$F_p\left(s\right)=q_p{\left(s\right)}^2-q_p\left(s\right)+s^2{w}_p{\left(s\right)}^{p-1},q_p\left(s\right)=-{\displaystyle \frac{s{w}_p^{\prime }\left(s\right)}{w_p\left(s\right)}}.$$

Third, with $z_p=s^2{w}_p^{p-1}$ and logarithmic time, the pair $(q_p,z_p)$ satisfies

$$\dot{q}=q^2+z,\dot{z}=z\left(2-(p-1)q\right).$$

An explicit differential inequality gives $F_p<0$ at the unique point where $q=1/2$ for $p\ge 12$. We then choose the last positive-to-nonpositive crossing before that point and the first nonpositive-to-positive crossing after it; this choice is essential for proving that both zeros are simple. Finally, a sectorial spectral argument and the Crandall–Rabinowitz theorem [4] yield the two nonradial branches.

The paper is organized as follows. Section 2 parametrizes and proves uniqueness of the radial branch. Section 3 derives the degeneracy function and its phase–plane system. Section 4 establishes the sign estimate and the existence of two simple sign-changing zeros. Section 5 proves spectral simplicity and transversality. Section 6 applies local bifurcation theory, and Section 7 records consequences and open questions.

2. The Normalized Radial Profile

Throughout Sections 2–5 we work on the unit disk and fix $p>1$. Let $w=w_p$ be the solution of

$$w^{\prime \prime }+{\displaystyle \frac{1}{s}}{w}^{\prime }+w^p=0,w\left(0\right)=1,w^{\prime }\left(0\right)=0.$$

(2)

Let $(0,S_p)$ be its maximal interval of positivity.

Lemma 2.1. 

The number $S_p$ is finite, and w extends as a $C^1$ function to $[0,S_p]$. Moreover,

$$w^{\prime }\left(s\right)<0(0<s\le S_p),w\left(S_p\right)=0,w^{\prime }\left(S_p\right)<0.$$

Proof. 

Writing (2) in divergence form gives

$$-{\left(s{w}^{\prime }\left(s\right)\right)}^{\prime }=sw{\left(s\right)}^p,-s{w}^{\prime }\left(s\right)={\int }_0^{s}tw{\left(t\right)}^p\mathrm{d}t.$$

(3)

Hence $w^{\prime }<0$ as long as $w>0$.

Suppose for contradiction that w remains positive on $(0,\infty )$. Define

$$q\left(s\right)=-{\displaystyle \frac{s{w}^{\prime }\left(s\right)}{w\left(s\right)}}.$$

(4)

A direct calculation using (2) yields

$$q^{\prime }\left(s\right)=sw{\left(s\right)}^{p-1}+{\displaystyle \frac{q{\left(s\right)}^2}{s}}>0.$$

(5)

Fix $s_0>0$. Since $q\left(s_0\right)>0$, (5) implies

$${\left({\displaystyle \frac{1}{q}}\right)}^{\prime }=-{\displaystyle \frac{q^{\prime }}{q^2}}\le -{\displaystyle \frac{1}{s}}.$$

Thus

$${\displaystyle \frac{1}{q\left(s\right)}}\le {\displaystyle \frac{1}{q\left(s_0\right)}}-log{\displaystyle \frac{s}{s_0}},$$

which is impossible for all sufficiently large s. Therefore $S_p<\infty$. The final assertion follows from (3):

$$-S_p{w}^{\prime }\left(S_p\right)={\int }_0^{S_p}tw{\left(t\right)}^p\mathrm{d}t>0.$$

□

The logarithmic slope q completely parametrizes the radial Robin branch.

Lemma 2.2. 

The function q defined in (4) is a strictly increasing $C^1$ bijection

$$q:(0,S_p)⟶(0,\infty ).$$

Proof. 

Strict monotonicity follows from (5). The Taylor expansion of (2) at the origin is

$$w\left(s\right)=1-{\displaystyle \frac{s^2}{4}}+{\displaystyle \frac{p{s}^4}{64}}+O\left(s^6\right),$$

(6)

so $q\left(s\right)=s^2/2+O\left(s^4\right)$ and $q\left(s\right)\to 0$ as $s\downarrow 0$. By Lemma 2.1, $w\left(s\right)\to 0$, while $w^{\prime }\left(s\right)\to w^{\prime }\left(S_p\right)<0$, as $s\uparrow S_p$. Hence $q\left(s\right)\to \infty$. □

For $\rho \in (0,S_p)$ set

$$U_{\rho }\left(r\right)={\rho }^{{\displaystyle \frac{2}{p-1}}}w\left(\rho r\right),0\le r\le 1,$$

(7)

and

$$\beta \left(\rho \right)=q\left(\rho \right).$$

(8)

Then $U_{\rho }$ solves $-\Delta U_{\rho }=U_{\rho }^p$, and

$$U_{\rho }^{\prime }\left(1\right)+\beta \left(\rho \right)U_{\rho }\left(1\right)={\rho }^{{\displaystyle \frac{2}{p-1}}}\left(\rho w^{\prime }\left(\rho \right)+q\left(\rho \right)w\left(\rho \right)\right)=0.$$

Proposition 2.3 

(Uniqueness of the radial branch). For every $p>1$ and every $\beta >0$, problem (1) on $B_1$ has exactly one positive radial solution. It is $U_{\rho }$, where $\rho =q^{-1}\left(\beta \right)$.

Proof. 

Every positive radial solution is determined by its central value $a=u\left(0\right)>0$. Scaling the initial-value problem gives

$$u\left(r\right)=aw\left(a^{(p-1)/2}r\right).$$

With $\rho =a^{(p-1)/2}$, the Robin condition is exactly $q\left(\rho \right)=\beta$. Lemma 2.2 gives a unique $\rho$. □

In particular, $\rho \mapsto \beta \left(\rho \right)$ is smooth and strictly increasing. Since $q^{\prime }\left(\rho \right)>0$, the inverse map $\rho =q^{-1}\left(\beta \right)$ is smooth on $(0,\infty )$. Consequently,

$$(0,\infty )\ni \beta ⟼U_{\beta }:=U_{q^{-1}\left(\beta \right)}$$

is a smooth map into $C^{2,\alpha }\left(\overline{B_1}\right)$ locally in $\beta$, for every $\alpha \in (0,1)$.

3. The First Angular Mode and the Degeneracy Function

Define

$$z\left(s\right)=s^{2}w{\left(s\right)}^{p-1},F\left(s\right)=q{\left(s\right)}^2-q\left(s\right)+z\left(s\right).$$

(9)

The relevance of F comes from the translation Jacobi field. Differentiating (2) gives

$$-{\zeta }^{\prime \prime }-{\displaystyle \frac{1}{s}}{\zeta }^{\prime }+{\displaystyle \frac{1}{s^2}}\zeta -p{w}^{p-1}\zeta =0,\zeta =-w^{\prime }>0.$$

(10)

At the radial solution $U_{\rho }$, the linearized Robin operator is

$${\mathcal{L}}_{\rho }\varphi =-\Delta \varphi -p{U}_{\rho }^{p-1}\varphi ,{\partial }_{\nu }\varphi +q\left(\rho \right)\varphi =0.$$

(11)

The $k=1$ angular function $\varphi (r,\theta )=\zeta \left(\rho r\right)cos\theta$ satisfies the interior equation. Its boundary residual is

$$\begin{array}{cc}\hfill \rho {\zeta }^{\prime }\left(\rho \right)+q\left(\rho \right)\zeta \left(\rho \right)& =-\rho w^{\prime \prime }\left(\rho \right)-q\left(\rho \right)w^{\prime }\left(\rho \right)\hfill \\ \hfill & ={\displaystyle \frac{w\left(\rho \right)}{\rho }}\left(q{\left(\rho \right)}^2-q\left(\rho \right)+{\rho }^{2}w{\left(\rho \right)}^{p-1}\right)\hfill \\ \hfill & ={\displaystyle \frac{w\left(\rho \right)}{\rho }}F\left(\rho \right).\hfill \end{array}$$

(12)

The regular solution space of the zero-energy $k=1$ equation is one-dimensional. Indeed, after writing a regular amplitude as $\psi \left(r\right)=rv\left(r\right)$, the equation has the Volterra formulation used in Lemma 5.3; prescribing $v\left(0\right)$ determines the solution uniquely, while regularity forces $v^{\prime }\left(0\right)=0$. Since the translation Jacobi field is nonzero, every regular zero-energy $k=1$ solution is proportional to it. Consequently,

$$F\left(\rho \right)=0$$

is equivalent to the existence of a nontrivial zero mode in the first angular sector, not merely a sufficient condition.

We next establish the phase–plane identities used to locate such zeros. Let

$$\tau =logs,\dot{h}=s{h}^{\prime }\left(s\right),P=p-1.$$

Equations (5) and (9) give

$$\dot{q}=q^2+z,\dot{z}=z(2-Pq).$$

(13)

Lemma 3.1. 

For every $s\in (0,S_p)$,

$$0<z\left(s\right)<2q\left(s\right).$$

(14)

Proof. 

Let $H=z/q$. From (13),

$$\dot{H}=H\left(2-pq-H\right).$$

(15)

The expansion (6) implies

$$H\left(s\right)=2-{\displaystyle \frac{p}{4}}{s}^2+O\left(s^4\right)<2$$

for small positive s. If H reached the value 2 for the first time, then (15) would give $\dot{H}=-2pq<0$ at that point, contradicting a first crossing from below. Hence $H<2$ throughout $(0,S_p)$. □

The function F has useful endpoint behavior:

$$F\left(s\right)={\displaystyle \frac{s^2}{2}}+O\left(s^4\right)>0\mathrm{as}s\downarrow 0,F\left(s\right)⟶+\infty \mathrm{as}s\uparrow S_p.$$

(16)

The first relation follows from (6); the second follows from $q\to \infty$ and $z\to 0$.

4. A Large-Exponent Sign Estimate

Proposition 4.1. 

If $p\ge 12$, then F is negative at the unique point $s_{1/2}\in (0,S_p)$ satisfying

$$q\left(s_{1/2}\right)={\displaystyle \frac{1}{2}}.$$

Proof. 

Set $P=p-1\ge 11$. Since q increases from 0 to ∞, we may regard z as a function of q. From (13),

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}q}}logz={\displaystyle \frac{2-Pq}{q^2+z}}.$$

(17)

For $q\in [2/P,1/2]$, the numerator is nonpositive. Lemma 3.1 gives

$$q^2+z\le q^2+2q\le {\displaystyle \frac{5}{2}}q.$$

Because the numerator in (17) is nonpositive,

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}q}}logz\le {\displaystyle \frac{2}{5}}\left({\displaystyle \frac{2}{q}}-P\right).$$

(18)

At $q=2/P$, Lemma 3.1 gives $z\le 4/P$. Integrating (18) from $2/P$ to $1/2$ yields

$$\begin{array}{cc}\hfill z\left(s_{1/2}\right)& \le {\displaystyle \frac{4}{P}}exp\left\{{\displaystyle \frac{4}{5}}log{\displaystyle \frac{P}{4}}-{\displaystyle \frac{P}{5}}+{\displaystyle \frac{4}{5}}\right\}\hfill \\ \hfill & ={\left({\displaystyle \frac{4}{P}}\right)}^{1/5}exp\left({\displaystyle \frac{4-P}{5}}\right).\hfill \end{array}$$

(19)

For $P\ge 11$, the right-hand side is less than $1/4$. Indeed, after taking the fifth power this is equivalent to

$$4096e^{4-P}<P,$$

which holds at $P=11$ and remains true for all larger P. Therefore

$$F\left(s_{1/2}\right)={\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2}}+z\left(s_{1/2}\right)<0.$$

□

We now select two zeros according to their sign change. Define

$${\rho }_{-}:=sup\{s\in (0,s_{1/2}):F\left(s\right)>0\},{\rho }_{+}:=inf\{s\in (s_{1/2},S_p):F\left(s\right)>0\}.$$

(20)

The sets in (20) are nonempty by (16), and Proposition 4.1 together with continuity shows that

$$0<{\rho }_{-}<s_{1/2}<{\rho }_{+}<S_p,F\left({\rho }_{-}\right)=F\left({\rho }_{+}\right)=0.$$

(21)

Moreover,

$$F\le 0\mathrm{on}({\rho }_{-},s_{1/2}],F\le 0\mathrm{on}[s_{1/2},{\rho }_{+}),$$

(22)

and there are sequences $s_n^{-}\uparrow {\rho }_{-}$ and $s_n^{+}\downarrow {\rho }_{+}$ such that

$$F\left(s_n^{-}\right)>0,F\left(s_n^{+}\right)>0.$$

This sign-changing selection rules out isolated tangential zeros at the two distinguished points.

Lemma 4.2. 

Let $p\ge 12$ and let ${\rho }_{\pm }$ be defined by (20). Then

$$F^{\prime }\left({\rho }_{-}\right)<0,F^{\prime }\left({\rho }_{+}\right)>0.$$

In particular, both zeros are simple.

Proof. 

Along the phase flow,

$$\dot{F}=(2q-1)(q^2+z)+z(2-Pq).$$

At a zero of F, one has $z=q(1-q)>0$, and therefore $0<q<1$. Since also $q^2+z=q$, we obtain

$$\dot{F}=q\left(1-Pq(1-q)\right).$$

(23)

If $F=\dot{F}=0$, differentiating once more along (13) and using $Pq(1-q)=1$ gives

$$\ddot{F}={\displaystyle \frac{q(2q-1)}{1-q}}.$$

(24)

At ${\rho }_{-}$, monotonicity of q and (21) give $q\left({\rho }_{-}\right)<1/2$. If $\dot{F}\left({\rho }_{-}\right)=0$, then (24) yields $\ddot{F}\left({\rho }_{-}\right)<0$. Hence F is strictly negative on both sides of ${\rho }_{-}$ sufficiently close to that point, contradicting the sequence $s_n^{-}\uparrow {\rho }_{-}$ with $F\left(s_n^{-}\right)>0$. Thus $\dot{F}\left({\rho }_{-}\right)\ne 0$. If instead $\dot{F}\left({\rho }_{-}\right)>0$, differentiability and $F\left({\rho }_{-}\right)=0$ would imply $F\left(s\right)<0$ for every $s<{\rho }_{-}$ sufficiently close to ${\rho }_{-}$, again contradicting the same sequence. Therefore $\dot{F}\left({\rho }_{-}\right)<0$.

At ${\rho }_{+}$ one has $q\left({\rho }_{+}\right)>1/2$. If $\dot{F}\left({\rho }_{+}\right)=0$, then (24) gives $\ddot{F}\left({\rho }_{+}\right)>0$, so F is strictly positive on both sides of ${\rho }_{+}$ sufficiently close to that point. This contradicts $F\le 0$ immediately to the left. Hence $\dot{F}\left({\rho }_{+}\right)\ne 0$. If $\dot{F}\left({\rho }_{+}\right)<0$, then $F\left(s\right)>0$ for every $s<{\rho }_{+}$ sufficiently close to ${\rho }_{+}$, contradicting (22). Consequently $\dot{F}\left({\rho }_{+}\right)>0$. Since $\dot{F}=s{F}^{\prime }$ and $s>0$, the claimed signs follow. □

Define

$${\beta }_{p,-}=q\left({\rho }_{-}\right),{\beta }_{p,+}=q\left({\rho }_{+}\right).$$

(25)

Since q is strictly increasing and $q\left(s_{1/2}\right)=1/2$,

$$0<{\beta }_{p,-}<{\displaystyle \frac{1}{2}}<{\beta }_{p,+}.$$

At either zero, $z=q(1-q)>0$, so $0<q<1$. In particular, ${\beta }_{p,+}<1$, and therefore

$$0<{\beta }_{p,-}<{\displaystyle \frac{1}{2}}<{\beta }_{p,+}<1.$$

(26)

Remark 4.3. 

The construction proves the existence of at least two simple first-angular degeneracy points. It does not require, and does not assert, that $F_p$ has exactly two zeros on $(0,S_p)$.

5. Spectral Simplicity and Transversality

We record the spectral facts needed for local bifurcation. Decompose an even-in-$x_2$ function in angular modes $cos\left(k\theta \right)$, $k\ge 0$. In the kth sector, the radial amplitude $\psi$ is governed by

$$-{\psi }^{\prime \prime }-{\displaystyle \frac{1}{r}}{\psi }^{\prime }+{\displaystyle \frac{k^2}{r^2}}\psi -p{U}_{\rho }{\left(r\right)}^{p-1}\psi =\lambda \psi ,{\psi }^{\prime }\left(1\right)+q\left(\rho \right)\psi \left(1\right)=0.$$

(27)

For $k\ge 1$ the natural form domain is

$${\mathcal{H}}_k:=\left\{\psi :{\int }_0^1\left(r|{\psi }^{\prime }{|}^2+{\displaystyle \frac{k^2}{r}}{\psi }^2+r{\psi }^2\right)\mathrm{d}r<\infty \right\},$$

with the boundary trace at $r=1$ understood in the usual Sobolev sense. This is the finite-energy condition at the origin; eigenfunctions in this domain have the classical regular behavior $\psi \left(r\right)=O\left(r^k\right)$. In the $k=0$ sector, regular eigenfunctions are classically characterized by ${\psi }^{\prime }\left(0\right)=0$.

Lemma 5.1. 

If $F\left(\rho \right)=0$, then the kernel of ${\mathcal{L}}_{\rho }$ in the subspace of functions even in $x_2$ is one-dimensional and is generated by

$${\mathsf{\Phi }}_{\rho }(r,\theta )=-w^{\prime }\left(\rho r\right)cos\theta .$$

(28)

Proof. 

By (10) and (12), ${\mathsf{\Phi }}_{\rho }$ belongs to the kernel. Set

$$\xi \left(r\right)=-w^{\prime }\left(\rho r\right).$$

Then $\xi >0$ on $(0,1]$, $\xi \left(r\right)=\rho r/2+O\left(r^3\right)$ at the origin, and $\xi$ satisfies both the zero-energy $k=1$ equation and the Robin condition. For a smooth regular amplitude $\psi$ in the $k=1$ sector, integration by parts using the equation and boundary condition for $\xi$ gives the ground-state identity

$$Q_1\left[\psi \right]={\int }_0^{1}r\xi {\left(r\right)}^2{\left|{\left({\displaystyle \frac{\psi }{\xi }}\right)}^{\prime }\right|}^2\mathrm{d}r.$$

(29)

The boundary term at $r=0$ vanishes because both $\psi$ and $\xi$ are $O\left(r\right)$. At $r=1$, the integration-by-parts contribution combines with the boundary term $q\left(\rho \right)\psi {\left(1\right)}^2$ in $Q_1$ and cancels because ${\xi }^{\prime }\left(1\right)+q\left(\rho \right)\xi \left(1\right)=0$. By density, (29) holds throughout ${\mathcal{H}}_1$. Hence $Q_1\ge 0$, and equality occurs only when $\psi =c\xi$. It follows directly that zero is the principal $k=1$ eigenvalue and that it is simple.

There is no radial kernel. Differentiating the radial family (7) with respect to $\rho$ gives a nonzero regular radial solution of the interior linearized equation. Differentiating

$$U_{\rho }^{\prime }\left(1\right)+q\left(\rho \right)U_{\rho }\left(1\right)=0$$

yields

$${\left({\partial }_{\rho }{U}_{\rho }\right)}^{\prime }\left(1\right)+q\left(\rho \right){\partial }_{\rho }{U}_{\rho }\left(1\right)=-q^{\prime }\left(\rho \right)U_{\rho }\left(1\right)\ne 0.$$

(30)

The regular solution space of the zero-eigenvalue radial ODE is one-dimensional, so no nonzero regular radial solution can satisfy the homogeneous Robin boundary condition.

For $k\ge 1$, let

$$Q_k\left[\psi \right]={\int }_0^1\left(r|{\psi }^{\prime }{|}^2+{\displaystyle \frac{k^2}{r}}{\psi }^2-pr{U}_{\rho }^{p-1}{\psi }^2\right)\mathrm{d}r+q\left(\rho \right)\psi {\left(1\right)}^2$$

(31)

be the quadratic form of (27) on the natural regular sectorial domain. For $k\ge 2$ one has the continuous inclusion ${\mathcal{H}}_k\subset {\mathcal{H}}_1$, because finiteness of the $k^2{r}^{-1}{\psi }^2$ term implies finiteness of the corresponding $k=1$ term. Therefore the nonnegativity of $Q_1$ furnished by (29) applies to every admissible higher-mode amplitude. For every nonzero $\psi \in {\mathcal{H}}_k$ and every $k\ge 2$,

$$\begin{array}{cc}\hfill Q_k\left[\psi \right]& =Q_1\left[\psi \right]+(k^2-1){\int }_0^1{\displaystyle \frac{{\psi }^2}{r}}\mathrm{d}r\hfill \\ \hfill & \ge (k^2-1){\int }_0^{1}r{\psi }^2\mathrm{d}r>0,\hfill \end{array}$$

because $r^{-1}\ge r$ on $(0,1]$. Taking the infimum of the Rayleigh quotient gives

$${\lambda }_1^{\left(k\right)}\left(\rho \right)\ge k^2-1>0,k\ge 2,$$

so zero is not an eigenvalue in any higher angular sector.

In the full space the zero eigenspace is generated by the two translation modes

$$-w^{\prime }\left(\rho r\right)cos\theta ,-w^{\prime }\left(\rho r\right)sin\theta .$$

Indeed, the preceding arguments exclude the radial and all higher-angular sectors, while simplicity in the $k=1$ amplitude problem leaves exactly these two angular factors. Restriction to functions even in $x_2$ removes the sine mode and leaves precisely the one-dimensional span of (28). □

Remark 5.2 

($O\left(2\right)$-equivariant interpretation). The orthogonal group acts on functions by

$$(\gamma ·u)\left(x\right)=u\left({\gamma }^{-1}x\right),\gamma \in O\left(2\right),$$

and the equation and Robin boundary operator are equivariant under this action. If $\kappa (x_1,x_2)=(x_1,-x_2)$, then the space used below is the fixed-point space of κ. The restriction converts the two-dimensional full kernel into the simple kernel generated by the cosine mode, so that the Crandall–Rabinowitz theorem applies there. Rotating a bifurcating representative produces its full $O\left(2\right)$-orbit of solutions. The local uniqueness conclusion supplied by the simple-eigenvalue theorem is therefore asserted only inside the chosen reflection-fixed space, not in the unrestricted function space.

The next two lemmas justify the determinant construction and identify the crossing speed. Normalize the regular $k=1$ solution of (27) by

$$y(r;\rho ,\lambda )=r+O\left(r^3\right)\mathrm{as}r\downarrow 0,$$

and define its boundary determinant

$$D(\rho ,\lambda )=y^{\prime }(1;\rho ,\lambda )+q\left(\rho \right)y(1;\rho ,\lambda ).$$

(32)

At $\lambda =0$,

$$y(r;\rho ,0)={\displaystyle \frac{2}{\rho }}\left(-w^{\prime }\left(\rho r\right)\right).$$

Consequently,

$$D(\rho ,0)={\displaystyle \frac{2w\left(\rho \right)}{{\rho }^2}}F\left(\rho \right).$$

(33)

Lemma 5.3 

(Regular solution and smooth determinant). Fix ${\rho }_0\in (0,S_p)$. In a neighborhood of $({\rho }_0,0)$ there is a unique regular solution of (27) in the $k=1$ sector satisfying

$$y(r;\rho ,\lambda )=r+O\left(r^3\right)(r\downarrow 0).$$

The maps $(\rho ,\lambda )\mapsto y(·;\rho ,\lambda )$ and $(\rho ,\lambda )\mapsto D(\rho ,\lambda )$ are $C^2$ locally, with values in $C^2\left([0,1]\right)$ for $y/r$ and in $\mathbb{R}$ for D. Moreover,

$$y_{\lambda }(r;\rho ,\lambda )=O\left(r^3\right),y_{\lambda }^{\prime }(r;\rho ,\lambda )=O\left(r^2\right)(r\downarrow 0),$$

uniformly on compact parameter sets.

Proof. 

Set $y\left(r\right)=rv\left(r\right)$. The $k=1$ equation becomes

$$v^{\prime \prime }+{\displaystyle \frac{3}{r}}{v}^{\prime }+\left(p{U}_{\rho }{\left(r\right)}^{p-1}+\lambda \right)v=0,v\left(0\right)=1,v^{\prime }\left(0\right)=0.$$

(34)

Equivalently, v is a fixed point of the Volterra equation

$$v\left(r\right)=1-{\int }_0^r{t}^{-3}{\int }_0^t{s}^3\left(p{U}_{\rho }{\left(s\right)}^{p-1}+\lambda \right)v\left(s\right)\mathrm{d}s\mathrm{d}t.$$

(35)

Choose a compact interval $K⋐(0,S_p)$ containing ${\rho }_0$. Since $w>0$ on $[0,maxK]$, there is a constant $c_K>0$ such that

$$U_{\rho }\left(r\right)={\rho }^{2/(p-1)}w\left(\rho r\right)\ge c_K(\rho \in K,0\le r\le 1).$$

Thus the map $(\rho ,r)\mapsto U_{\rho }{\left(r\right)}^{p-1}$ has the required continuous $\rho$-derivatives even when p is not an integer; no differentiation across the origin of the power function is involved. Standard successive approximation for (35), applied also after differentiating with respect to $\rho$ and $\lambda$, then gives a unique solution and $C^2$ dependence on the parameters. The integral equation also gives the expansion

$$v\left(r\right)=1-{\displaystyle \frac{p{U}_{\rho }{\left(0\right)}^{p-1}+\lambda }{8}}{r}^2+O\left(r^4\right),$$

locally uniformly in $(\rho ,\lambda )$. Therefore $y=rv=r+O\left(r^3\right)$, while differentiation in $\lambda$ yields

$$y_{\lambda }(r;\rho ,\lambda )=-{\displaystyle \frac{1}{8}}{r}^3+O\left(r^5\right),y_{\lambda }^{\prime }(r;\rho ,\lambda )=-{\displaystyle \frac{3}{8}}{r}^2+O\left(r^4\right).$$

Since evaluation at $r=1$ is continuous in these spaces and q is smooth on $(0,S_p)$, the determinant (32) is $C^2$. □

Lemma 5.4 

(Transversal eigenvalue crossing). If $F\left({\rho }_{*}\right)=0$ and $F^{\prime }\left({\rho }_{*}\right)\ne 0$, then the simple $k=1$ eigenvalue crosses zero transversally as ρ passes through ${\rho }_{*}$. Equivalently, if ${\lambda }_1^{\left(1\right)}\left(\rho \right)$ denotes this eigenvalue, then

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\rho }}{\lambda }_1^{\left(1\right)}\left({\rho }_{*}\right)\ne 0.$$

Proof. 

By Lemma 5.3, D is $C^1$ near $({\rho }_{*},0)$. Equation (33) and $F\left({\rho }_{*}\right)=0$ give

$${\partial }_{\rho }D({\rho }_{*},0)={\displaystyle \frac{2w\left({\rho }_{*}\right)}{{\rho }_{*}^2}}{F}^{\prime }\left({\rho }_{*}\right)\ne 0.$$

It remains to compute the derivative with respect to $\lambda$. Write

$$y=y(·;{\rho }_{*},0),y_{\lambda }={\partial }_{\lambda }y(·;{\rho }_{*},0),$$

and introduce the Sturm–Liouville expression

$$L_{{\rho }_{*}}\eta :=-{\left(r{\eta }^{\prime }\right)}^{\prime }+\left({\displaystyle \frac{1}{r}}-pr{U}_{{\rho }_{*}}^{p-1}\right)\eta .$$

At the degeneracy point,

$$L_{{\rho }_{*}}y=0,L_{{\rho }_{*}}{y}_{\lambda }=ry.$$

Green’s identity on $(\epsilon ,1)$ gives

$${\int }_{\epsilon }^{1}r{y}^2\mathrm{d}r={\left[r\left(y_{\lambda }{y}^{\prime }-y{y}_{\lambda }^{\prime }\right)\right]}_{r=\epsilon }^{r=1}.$$

The expansions in Lemma 5.3 imply that the term at $r=\epsilon$ tends to zero as $\epsilon \downarrow 0$. At $r=1$, the identities

$$y^{\prime }\left(1\right)+q\left({\rho }_{*}\right)y\left(1\right)=0,{\partial }_{\lambda }D({\rho }_{*},0)=y_{\lambda }^{\prime }\left(1\right)+q\left({\rho }_{*}\right)y_{\lambda }\left(1\right)$$

yield

$$-y\left(1\right){\partial }_{\lambda }D({\rho }_{*},0)={\int }_0^{1}ry{\left(r\right)}^2\mathrm{d}r.$$

(36)

Since $y>0$ on $(0,1]$, both the integral and $y\left(1\right)$ are positive; hence

$${\partial }_{\lambda }D({\rho }_{*},0)<0.$$

The implicit-function theorem applied to $D(\rho ,\lambda )=0$ produces the local simple eigenvalue curve and gives

$${\lambda }^{\prime }\left({\rho }_{*}\right)=-{\displaystyle \frac{D_{\rho }({\rho }_{*},0)}{D_{\lambda }({\rho }_{*},0)}}\ne 0.$$

In fact, $sgn{\lambda }^{\prime }\left({\rho }_{*}\right)=sgn{F}^{\prime }\left({\rho }_{*}\right)$. Thus the first angular eigenvalue crosses downward at ${\rho }_{-}$ and upward at ${\rho }_{+}$. □

6. Bifurcation of Positive Nonradial Solutions

Fix $\alpha \in (0,1)$ and let

$$X=C_{\mathrm{even}}^{2,\alpha }\left(\overline{B_1}\right):=\{u\in C^{2,\alpha }\left(\overline{B_1}\right):u(x_1,x_2)=u(x_1,-x_2)\},$$

$$Y=C_{\mathrm{even}}^{0,\alpha }\left(\overline{B_1}\right)\times C_{\mathrm{even}}^{1,\alpha }(\partial B_1).$$

Introduce the $C^2$ extension

$$g\left(s\right)={\left|s\right|}^{p-1}s,s\in \mathbb{R}.$$

Since $p\ge 12>2$, this extension is $C^2$ on $\mathbb{R}$. It is used only to place the equation on an open Banach-space neighborhood. The solutions obtained on the local bifurcating branches are shown below to remain strictly positive; hence they satisfy $g\left(u\right)=u^p$ and solve the original problem. Define

$$\mathcal{F}(\beta ,u)=\left(-\Delta u-g\left(u\right),{\partial }_{\nu }u+\beta u\right),\mathcal{F}:\mathbb{R}\times X\to Y.$$

(37)

The Nemytskii mapping induced by g and the trace mapping are $C^2$ on these Hölder spaces, so $\mathcal{F}$ is $C^2$. By Proposition 2.3 and the inverse-function observation following it, the radial branch $\beta \mapsto U_{\beta }$ is smooth for $\beta >0$. Set

$$\mathcal{G}(\beta ,v)=\mathcal{F}(\beta ,U_{\beta }+v),(\beta ,v)\in (0,\infty )\times X.$$

Then $\mathcal{G}(\beta ,0)=0$ for every $\beta >0$.

Fix ${\rho }_{*}\in \{{\rho }_{-},{\rho }_{+}\}$ and set ${\beta }_{*}=q\left({\rho }_{*}\right)$. The derivative

$$A_{*}:=D_v\mathcal{G}({\beta }_{*},0):X⟶Y$$

is the elliptic Robin operator

$$A_{*}v=\left(-\Delta v-p{U}_{{\beta }_{*}}^{p-1}v,{\partial }_{\nu }v+{\beta }_{*}v\right).$$

Consider first the Robin Laplacian

$$T_{{\beta }_{*}}v:=\left(-\Delta v,{\partial }_{\nu }v+{\beta }_{*}v\right).$$

Let

$$\tilde{X}=C^{2,\alpha }\left(\overline{B_1}\right),\tilde{Y}=C^{0,\alpha }\left(\overline{B_1}\right)\times C^{1,\alpha }(\partial B_1).$$

For ${\beta }_{*}>0$, the homogeneous problem has only the zero solution: multiplying $-\Delta v=0$ by v and using the Robin condition gives

$${\int }_{B_1}{|\nabla v|}^{2}dx+{\beta }_{*}{\int }_{\partial B_1}{v}^{2}dS=0.$$

Schauder solvability therefore shows that $T_{{\beta }_{*}}:\tilde{X}\to \tilde{Y}$ is an isomorphism; see [1,5,8]. The operator commutes with the reflection $\kappa (x_1,x_2)=(x_1,-x_2)$. If the data are fixed by $\kappa$, then both v and $v\circ \kappa$ solve the same full-space problem, and uniqueness gives $v=v\circ \kappa$. Hence the full-space isomorphism restricts to an isomorphism

$$T_{{\beta }_{*}}:X⟶Y.$$

The perturbation

$$A_{*}-T_{{\beta }_{*}}=\left(-p{U}_{{\beta }_{*}}^{p-1}v,0\right)$$

is compact from X to Y: it factors through the compact embedding $C^{2,\alpha }\left(\overline{B_1}\right)\hookrightarrow C^{0,\alpha }\left(\overline{B_1}\right)$, followed by multiplication by the fixed coefficient $p{U}_{{\beta }_{*}}^{p-1}\in C^{0,\alpha }\left(\overline{B_1}\right)$. Thus $A_{*}$ is Fredholm of index zero. By Lemma 5.1,

$$ker{A}_{*}=span\left\{{\mathsf{\Phi }}_{{\rho }_{*}}\right\}.$$

For $(f,h)\in Y$, introduce

$$\ell (f,h):={\int }_{B_1}{\mathsf{\Phi }}_{{\rho }_{*}}fdx+{\int }_{\partial B_1}{\mathsf{\Phi }}_{{\rho }_{*}}hdS.$$

(38)

If $(f,h)=A_{*}v$, then, since $-\Delta {\mathsf{\Phi }}_{{\rho }_{*}}-p{U}_{{\beta }_{*}}^{p-1}{\mathsf{\Phi }}_{{\rho }_{*}}=0$ and ${\partial }_{\nu }{\mathsf{\Phi }}_{{\rho }_{*}}+{\beta }_{*}{\mathsf{\Phi }}_{{\rho }_{*}}=0$, Green’s second identity gives

$$\begin{array}{cc}\hfill {\int }_{B_1}{\mathsf{\Phi }}_{{\rho }_{*}}fdx& ={\int }_{B_1}\left({\mathsf{\Phi }}_{{\rho }_{*}}(-\Delta v)-v(-\Delta {\mathsf{\Phi }}_{{\rho }_{*}})\right)dx\hfill \\ \hfill & ={\int }_{\partial B_1}\left(v{\partial }_{\nu }{\mathsf{\Phi }}_{{\rho }_{*}}-{\mathsf{\Phi }}_{{\rho }_{*}}{\partial }_{\nu }v\right)dS\hfill \\ \hfill & =-{\int }_{\partial B_1}{\mathsf{\Phi }}_{{\rho }_{*}}hdS.\hfill \end{array}$$

Therefore $Range{A}_{*}\subset ker\ell$. Since $A_{*}$ has index zero and a one-dimensional kernel, its range has codimension one. The functional ℓ is nonzero, for instance $\ell ({\mathsf{\Phi }}_{{\rho }_{*}},0)>0$, and hence

$$Range{A}_{*}=ker\ell .$$

(39)

The determinant construction in Lemma 5.4, followed by the smooth change of parameter $\beta =q\left(\rho \right)$, gives a $C^1$ simple eigenvalue $\lambda \left(\beta \right)$ through zero. Smooth ODE dependence also permits a $C^1$, $L^2\left(B_1\right)$-normalized eigenfunction $\mathsf{\Phi }\left(\beta \right)$. Differentiating both the eigenvalue equation and the Robin boundary condition at ${\beta }_{*}$ yields

$${\lambda }^{\prime }\left({\beta }_{*}\right){\int }_{B_1}{\mathsf{\Phi }}_{{\rho }_{*}}^{2}dx={\int }_{\partial B_1}{\mathsf{\Phi }}_{{\rho }_{*}}^{2}dS-p(p-1){\int }_{B_1}{U}_{{\beta }_{*}}^{p-2}\left({\partial }_{\beta }{U}_{\beta }{|}_{{\beta }_{*}}\right){\mathsf{\Phi }}_{{\rho }_{*}}^{2}dx.$$

(40)

On the other hand,

$$D_{\beta v}\mathcal{G}({\beta }_{*},0){\mathsf{\Phi }}_{{\rho }_{*}}=\left(-p(p-1)U_{{\beta }_{*}}^{p-2}\left({\partial }_{\beta }{U}_{\beta }{|}_{{\beta }_{*}}\right){\mathsf{\Phi }}_{{\rho }_{*}},{\mathsf{\Phi }}_{{\rho }_{*}}\right),$$

so the right-hand side of (40) is exactly $\ell \left(D_{\beta v}\mathcal{G}({\beta }_{*},0){\mathsf{\Phi }}_{{\rho }_{*}}\right)$. By Lemma 5.4 and $q^{\prime }\left({\rho }_{*}\right)>0$,

$${\lambda }^{\prime }\left({\beta }_{*}\right)={\displaystyle \frac{\frac{\mathrm{d}}{\mathrm{d}\rho }{\lambda }_1^{\left(1\right)}\left({\rho }_{*}\right)}{q^{\prime }\left({\rho }_{*}\right)}}\ne 0.$$

Therefore

$$D_{\beta v}\mathcal{G}({\beta }_{*},0){\mathsf{\Phi }}_{{\rho }_{*}}\notin Range{A}_{*},$$

which is precisely the Crandall–Rabinowitz transversality condition [4].

The Crandall–Rabinowitz theorem now gives $\epsilon >0$ and a $C^1$ curve

$$(-\epsilon ,\epsilon )\ni t⟼\left(\beta \left(t\right),v\left(t\right)\right),v\left(t\right)=t{\mathsf{\Phi }}_{{\rho }_{*}}+o\left(t\right)\mathrm{in}X,$$

of nontrivial solutions of $\mathcal{G}(\beta ,v)=0$. Moreover, after shrinking the neighborhood, every even solution near $({\beta }_{*},0)$ belongs either to the trivial radial branch or to this bifurcating curve. In terms of the original unknown,

$$u\left(t\right)=U_{\beta \left(t\right)}+t{\mathsf{\Phi }}_{{\rho }_{*}}+o\left(t\right)\mathrm{in}{C}^{2,\alpha }\left(\overline{B_1}\right).$$

(41)

The projection of $u\left(t\right)-U_{\beta \left(t\right)}$ onto the $cos\theta$ sector equals $t{\mathsf{\Phi }}_{{\rho }_{*}}+o\left(t\right)$ and is nonzero for all sufficiently small $t\ne 0$; hence $u\left(t\right)$ is nonradial. Moreover,

$$m_{*}:=\underset{\overline{B_1}}{min}{U}_{{\beta }_{*}}>0.$$

Continuity of $\beta \mapsto U_{\beta }$ and (41) imply $\parallel u\left(t\right)-U_{{\beta }_{*}}{\parallel }_{C^0}<m_{*}/2$ for sufficiently small $\left|t\right|$, and therefore $u\left(t\right)>0$ on $\overline{B_1}$. Applying this argument at ${\rho }_{-}$ and ${\rho }_{+}$ proves Theorem 1.1; Corollary 1.2 follows by choosing any sequence $t_n\to 0$ with $t_n\ne 0$.

Finally, equivariance implies that $\gamma ·u\left(t\right)$ is a solution for every $\gamma \in O\left(2\right)$. Thus the curve obtained in the reflection-fixed space supplies a representative of a full orbit of nonradial solutions. The assertion that nearby solutions lie on the trivial branch or on the bifurcating curve is used only in X; no corresponding one-curve classification is claimed in the full space, where the kernel has dimension two.

For completeness, the scaling

$$\tilde{u}\left(y\right)=R^{{\displaystyle \frac{2}{p-1}}}u\left(Ry\right)$$

transforms the problem in $B_R$ with parameter $\beta$ into the problem in $B_1$ with parameter $R\beta$. Corollary 1.3 follows.

7. Consequences and Further Questions

Theorem 1.1 disproves, for $p\ge 12$, the assertion that the positive solution is unique for every Robin parameter in a disk. The conclusion is local and should not be read as nonuniqueness for all $\beta >0$. More precisely, for each $\sigma \in \{-,+\}$ there are parameters $\beta \left(t\right)\to {\beta }_{p,\sigma }$ for which a positive nonradial solution coexists with the unique positive radial solution at the same parameter. The local bifurcation theorem alone does not assert that a nonradial solution exists exactly at $\beta ={\beta }_{p,\sigma }$.

The exponent 12 enters only through the explicit bound (19); no sharpness is asserted. The phase–plane reduction suggests the threshold

$$p_{\mathrm{sb}}:=inf\left\{p>1:\underset{0<s<S_p}{inf}{F}_p\left(s\right)<0\right\},$$

(42)

for which the present argument proves $p_{\mathrm{sb}}\le 12$. If the transition at $p_{\mathrm{sb}}$ occurs through an interior tangency, the tangency must satisfy simultaneously

$$F=0,\dot{F}=0,(p-1)q(1-q)=1.$$

Since $q(1-q)\le 1/4$, such a tangency can occur only when $p\ge 5$. Determining $p_{\mathrm{sb}}$ rigorously, describing the global continuation and possible secondary bifurcations of the nonradial branches, and locating higher-angular-mode degeneracies remain open problems.