Introduction and main results
In this paper, we study partially overdetermined elliptic problems on annular domains with a fixed inner circle and a free outer boundary. We are interested in the parameter values at which nontrivial domains bifurcate from concentric annuli. The classical starting point is Serrin’s overdetermined problem, in which a positive solution has zero Dirichlet data and constant normal derivative on the boundary of a bounded domain. Serrin’s moving-plane argument [18 ] and Weinberger’s proof [20 ] show that the domain must be a ball. For an annular domain whose boundary components carry distinct Dirichlet values, additional boundary data may instead permit nonradial solutions. In [9 ] , the authors investigated this possibility through the spectrum of the linearized boundary operator for a constant vorticity function. Related annular bifurcation results were obtained by Kamburov and Sciaraffia [11 ] , who also considered constant vorticity and imposed overdetermined data on both boundary components. The present paper extends this line of research to a nonconstant vorticity function.
We first mention some related work in this area. For unbounded domains, Schlenk and Sicbaldi [17 ] constructed a smooth family of noncompact domains bifurcating from the straight cylinder B n × R B^{n}\times\mathbb{R} B n × R Δ v + σ v = 0 \Delta v+\sigma v=0 Δ v + σ v = 0 [3 ] found sign-changing solutions. These perturbed domains constructed above are rotationally symmetric and periodic in the axial direction. Ruiz, Sicbaldi and Wu [16 ] extended this cylindrical bifurcation approach to semilinear equations Δ v + g ( v ) = 0 \Delta v + g(v) = 0 Δ v + g ( v ) = 0 [6 ] studied the Schiffer problem (Δ v + σ v = 0 \Delta v + \sigma v = 0 Δ v + σ v = 0 R N × ( R / 2 π Z ) \mathbb{R}^{N} \times(\mathbb{R}/2\pi\mathbb{Z}) R N × ( R /2 π Z ) S 2 \mathbb{S}^{2} S 2
For bounded domains, Ruiz [15 ] showed that the positivity assumption in Serrin’s symmetry theorem cannot simply be omitted. For a suitable C 2 C^{2} C 2 − Δ v = v − ( v + ) 3 -\Delta v = v - (v^{+})^{3} − Δ v = v − ( v + ) 3 [5 ] constructed noncircular doubly connected planar domains bifurcating from annuli that support a nonradial solution of Δ v + μ v = 0 \Delta v + \mu v = 0 Δ v + μv = 0 [21 ] constructed real-analytic branches of noncircular bounded planar domains near a disk, with v = b v = b v = b ∂ ν v = c \partial_{\nu}v = c ∂ ν v = c b b b c c c m ≥ 4 m \ge4 m ≥ 4
These constructions concern, respectively, vanishing Neumann data on annular or curved domains and nonzero constant Cauchy data on a simply connected planar domain. They lead naturally to the question considered here: whether bifurcation can occur for a Bernoulli-type overdetermined elliptic problem, that is to say, whether a domain with a noncircular outer boundary can bifurcate from a concentric annulus when the inner circle is fixed, the two Dirichlet values are prescribed separately, and nonzero Neumann data is imposed only on the outer boundary. Here we will answer this equation combing shape derivatives, spectral estimates, and the Crandall–Rabinowitz bifurcation theorem. More precisely, let λ > 0 \lambda> 0 λ > 0 Ω \Omega Ω B ‾ λ \overline{B}_{\lambda} B λ B λ B_{\lambda} B λ λ \lambda λ A Ω = Ω ∖ B ‾ λ A_{\Omega} = \Omega\setminus\overline{B}_{\lambda} A Ω = Ω ∖ B λ
We consider the system
{ Δ u = f ( u ) in Ω , u = 1 on ∂ B λ , u = 0 on ∂ Ω , ∣ ∇ u ∣ 2 = Q > 0 on ∂ Ω . (1) \begin{cases}
\Delta u=f(u) & \text{in }\Omega,\\
u=1 & \text{on }\partial B_\lambda,\\
u=0 & \text{on }\partial\Omega,\\
\lvert\nabla u\rvert^2=Q>0 & \text{on }\partial\Omega. \tag*{(1)}
\end{cases} ⎩ ⎨ ⎧ Δ u = f ( u ) u = 1 u = 0 ∣ ∇ u ∣ 2 = Q > 0 in Ω , on ∂ B λ , on ∂ Ω , on ∂ Ω. (1) Throughout the analysis, the inner circle remains fixed; that is, λ \lambda λ
∂ Ω R , η = { ( R + η ( θ ) ) ( cos θ , sin θ ) : θ ∈ S 1 } , η ˉ : = 1 2 π ∫ 0 2 π η ( θ ) d θ = 0. (2) \partial\Omega_{R,\eta}=\{(R+\eta(\theta))(\cos\theta,\sin\theta):\theta\in S^1\},\qquad\bar{\eta}:=\frac{1}{2\pi}\int_0^{2\pi}\eta(\theta)\,\mathrm{d}\theta=0. \tag*{(2)} ∂ Ω R , η = {( R + η ( θ )) ( cos θ , sin θ ) : θ ∈ S 1 } , η ˉ := 2 π 1 ∫ 0 2 π η ( θ ) d θ = 0. (2) Here R R R
{ Δ u = − μ u in Ω ∖ B ‾ λ , u = 0 on ∂ Ω , u = 1 on ∂ B λ , ∣ ∇ u ∣ 2 = Q > 0 on ∂ Ω . (3) \begin{cases}
\Delta u=-\mu u & \text{in }\Omega\setminus\overline{B}_\lambda,\\
u=0 & \text{on }\partial\Omega,\\
u=1 & \text{on }\partial B_\lambda,\\
\lvert\nabla u\rvert^2=Q>0 & \text{on }\partial\Omega. \tag*{(3)}
\end{cases} ⎩ ⎨ ⎧ Δ u = − μu u = 0 u = 1 ∣ ∇ u ∣ 2 = Q > 0 in Ω ∖ B λ , on ∂ Ω , on ∂ B λ , on ∂ Ω. (3) For this problem, the reference outer radius is 1 1 1 μ \mu μ [5 ] . They construct noncircular doubly connected domains carrying a nonradial solution of Δ v + μ v = 0 \Delta v+\mu v=0 Δ v + μv = 0 ∂ ν v = 0 \partial_\nu v=0 ∂ ν v = 0 1 1 1 0 0 0 ∣ ∂ ν u ∣ = Q > 0 \lvert\partial_\nu u\rvert=\sqrt{Q}>0 ∣ ∂ ν u ∣ = Q > 0
We then consider the power-law problem for
{ Δ u = u − u p in Ω ∖ B ‾ λ , u > 0 in Ω ∖ B ‾ λ , u = 0 on ∂ Ω , u = 1 on ∂ B λ , ∣ ∇ u ∣ 2 = Q > 0 on ∂ Ω , (4) \begin{cases}
\Delta u=u-u^p & \text{in }\Omega\setminus\overline{B}_\lambda,\\
u>0 & \text{in }\Omega\setminus\overline{B}_\lambda,\\
u=0 & \text{on }\partial\Omega,\\
u=1 & \text{on }\partial B_\lambda,\\
\lvert\nabla u\rvert^2=Q>0 & \text{on }\partial\Omega, \tag*{(4)}
\end{cases} ⎩ ⎨ ⎧ Δ u = u − u p u > 0 u = 0 u = 1 ∣ ∇ u ∣ 2 = Q > 0 in Ω ∖ B λ , in Ω ∖ B λ , on ∂ Ω , on ∂ B λ , on ∂ Ω , (4) where p > 1 p>1 p > 1 R R R Δ u = u − u p \Delta u=u-u^p Δ u = u − u p − Δ u + u − u p = 0 -\Delta u+u-u^p=0 − Δ u + u − u p = 0 [13 ] . Ros, Ruiz and Sicbaldi construct exterior domains admitting positive solutions with zero Dirichlet data and constant normal derivative. Their bifurcation parameter is a radius, which becomes a diffusion coefficient after rescaling, and their spectral analysis is based on the positive radial exterior solution. In (1.4), the domain is bounded and the inner boundary has the nonzero Dirichlet value 1 1 1
For the local analysis, we use the standard Hölder spaces C m , α ( S 1 ) C^{m,\alpha}(\mathbb{S}^{1}) C m , α ( S 1 ) m ≥ 0 m \ge0 m ≥ 0 0 < α < 1 0 < \alpha< 1 0 < α < 1 K ≥ 1 K \ge1 K ≥ 1
X K = { η ∈ C 2 , α ( S 1 ) : η ( − θ ) = η ( θ ) , η ( θ + 2 π / K ) = η ( θ ) , η ‾ = 0 } , X_K=\{\eta\in C^{2,\alpha}(\mathbb{S}^{1}):\eta(-\theta)=\eta(\theta),\ \eta(\theta+2\pi/K)=\eta(\theta),\ \overline{\eta}=0\}, X K = { η ∈ C 2 , α ( S 1 ) : η ( − θ ) = η ( θ ) , η ( θ + 2 π / K ) = η ( θ ) , η = 0 } , and let
Y K = { η ∈ C 1 , α ( S 1 ) : η ( − θ ) = η ( θ ) , η ( θ + 2 π / K ) = η ( θ ) , η ‾ = 0 } . Y_K=\{\eta\in C^{1,\alpha}(\mathbb{S}^{1}):\eta(-\theta)=\eta(\theta),\ \eta(\theta+2\pi/K)=\eta(\theta),\ \overline{\eta}=0\}. Y K = { η ∈ C 1 , α ( S 1 ) : η ( − θ ) = η ( θ ) , η ( θ + 2 π / K ) = η ( θ ) , η = 0 } . The nonzero Fourier modes in these spaces are cos ( n K θ ) \cos(nK\theta) cos ( n K θ ) n ≥ 1 n\ge1 n ≥ 1 λ ∈ ( 0 , 1 ) \lambda\in(0,1) λ ∈ ( 0 , 1 ) K ≥ 1 K\ge1 K ≥ 1 μ K , λ ∗ \mu^*_{K,\lambda} μ K , λ ∗
− h ′ ′ − 1 r h ′ + K 2 r 2 h = μ h , h ( λ ) = 0 , h ′ ( 1 ) + h ( 1 ) = 0 (5) -h^{\prime\prime}-\frac{1}{r}h^\prime+\frac{K^2}{r^2}h=\mu h,\qquad h(\lambda)=0,\qquad h^\prime(1)+h(1)=0 \tag*{(5)} − h ′′ − r 1 h ′ + r 2 K 2 h = μ h , h ( λ ) = 0 , h ′ ( 1 ) + h ( 1 ) = 0 (5) and denote by Σ r a d ( λ ) \Sigma_{\mathrm{rad}}(\lambda) Σ rad ( λ ) − Δ -\Delta − Δ B 1 ∖ B λ ‾ B_1\setminus\overline{B_\lambda} B 1 ∖ B λ
Theorem 1.1. Fix λ ∈ ( 0 , 1 ) \lambda\in(0,1) λ ∈ ( 0 , 1 ) K ≥ 1 K\ge1 K ≥ 1 μ K , λ ∗ ∉ Σ r a d ( λ ) \mu^*_{K,\lambda}\notin\Sigma_{\mathrm{rad}}(\lambda) μ K , λ ∗ ∈ / Σ rad ( λ ) s 0 > 0 s_0>0 s 0 > 0 C ∞ C^\infty C ∞
s ⟼ ( μ ( s ) , η ( s ) , Q ( s ) ) , ∣ s ∣ < s 0 , s\longmapsto(\mu(s),\eta(s),Q(s)),\qquad|s|<s_0, s ⟼ ( μ ( s ) , η ( s ) , Q ( s )) , ∣ s ∣ < s 0 , of solutions to (3) on Ω 1 , η ( s ) ∖ B λ ‾ \Omega_{1,\eta(s)}\setminus\overline{B_\lambda} Ω 1 , η ( s ) ∖ B λ η ( s ) ∈ X K \eta(s)\in X_K η ( s ) ∈ X K
μ ( s ) = μ K , λ ∗ + O ( s 2 ) , η ( s ) = s cos ( K θ ) + O ( s 2 ) , Q ( s ) > 0. \mu(s)=\mu^*_{K,\lambda}+O(s^2),\qquad\eta(s)=s\cos(K\theta)+O(s^2),\qquad Q(s)>0. μ ( s ) = μ K , λ ∗ + O ( s 2 ) , η ( s ) = s cos ( K θ ) + O ( s 2 ) , Q ( s ) > 0. The inner boundary remains ∂ B λ \partial B_\lambda ∂ B λ
For the power-law problem (1.4), we assume
0 < α < min { 1 , p − 1 } . (6) 0 < \alpha< \min\{1,p-1\}. \tag*{(6)} 0 < α < min { 1 , p − 1 } . (6) This restriction gives the Hölder regularity needed for nonintegral powers at the outer boundary. To state the main result on the problem (1.4), we first introduce the constant governing the critical radii. Let V = V p V=V_p V = V p
− V ′ ′ = V p in ( 0 , 1 ) , V ( 0 ) = V ( 1 ) = 0. -V''=V^p \quad\text{in }(0,1), \qquad V(0)=V(1)=0. − V ′′ = V p in ( 0 , 1 ) , V ( 0 ) = V ( 1 ) = 0. We prove its existence and uniqueness in § 3.3 . Denote by ν D N ( p ) \nu_{\mathrm{DN}}(p) ν DN ( p )
− h ′ ′ − p V p − 1 h = ν h , h ( 0 ) = 0 , h ′ ( 1 ) = 0 , (7) -h''-pV^{p-1}h=\nu h,\qquad h(0)=0,\qquad h'(1)=0, \tag*{(7)} − h ′′ − p V p − 1 h = ν h , h ( 0 ) = 0 , h ′ ( 1 ) = 0 , (7) and set
κ p = − ν D N ( p ) . (8) \kappa_p=\sqrt{-\nu_{\mathrm{DN}}(p)}. \tag*{(8)} κ p = − ν DN ( p ) . (8) We shall prove in § 3.3 that ν D N ( p ) < 0 \nu_{\mathrm{DN}}(p)<0 ν DN ( p ) < 0 κ p \kappa_p κ p
Theorem 1.2. Fix λ > 0 \lambda>0 λ > 0 p > 1 p>1 p > 1 α \alpha α K 0 = K 0 ( λ , p ) ≥ 2 K_0=K_0(\lambda,p)\ge2 K 0 = K 0 ( λ , p ) ≥ 2 K ≥ K 0 K\ge K_0 K ≥ K 0 R K > λ R_K>\lambda R K > λ s K > 0 s_K>0 s K > 0 C ∞ C^\infty C ∞
s ⟼ ( R ( s ) , η ( s ) , Q ( s ) ) , ∣ s ∣ < s K , η ( s ) ∈ X K , s\longmapsto(R(s),\eta(s),Q(s)),\qquad|s|<s_K,\qquad\eta(s)\in X_K, s ⟼ ( R ( s ) , η ( s ) , Q ( s )) , ∣ s ∣ < s K , η ( s ) ∈ X K , of solutions to (1.4) on Ω R ( s ) , η ( s ) ∖ B λ ‾ \Omega_{R(s),\eta(s)}\setminus\overline{B_\lambda} Ω R ( s ) , η ( s ) ∖ B λ
R ( s ) = R K + O ( s 2 ) , η ( s ) = s cos ( K θ ) + O ( s 2 ) , Q ( s ) > 0. R(s)=R_K+O(s^2),\qquad\eta(s)=s\cos(K\theta)+O(s^2),\qquad Q(s)>0. R ( s ) = R K + O ( s 2 ) , η ( s ) = s cos ( K θ ) + O ( s 2 ) , Q ( s ) > 0. Moreover, the critical radii satisfy
R K = λ + λ κ p K + o ( K − 1 ) . (9) R_K=\lambda+\frac{\lambda\kappa_p}{K}+o(K^{-1}). \tag*{(9)} R K = λ + K λ κ p + o ( K − 1 ) . (9) Let U K U_K U K s = 0 s=0 s = 0 ( R K , 0 , U K ) (R_K,0,U_K) ( R K , 0 , U K ) X K X_K X K C 2 , α C^{2,\alpha} C 2 , α B R K ∖ B λ ‾ B_{R_K}\setminus\overline{B_\lambda} B R K ∖ B λ R ∗ = R K R_*=R_K R ∗ = R K X K X_K X K
In Theorem 1.2, the inner radius is fixed along each branch, while the mean outer radius varies. The proof constructs the critical radii for this fixed nonlinear equation and verifies the boundary eigenvalue crossing. The nonlinear analysis requires two additional observations. First, the potential in the linearized equation is 1 − p U p − 1 1-pU^{p-1} 1 − p U p − 1 U U U U U U 1 1 1 u − u p u-u^p u − u p
Proposition 1.3. Let U U U
U ′ ′ + r − 1 U ′ = U − U p , U ( λ ) = 1 , U ( R ) = 0 , R > λ . U'' + r^{-1}U' = U-U^p,\qquad U(\lambda)=1,\qquad U(R)=0,\qquad R>\lambda. U ′′ + r − 1 U ′ = U − U p , U ( λ ) = 1 , U ( R ) = 0 , R > λ . Then the boundary-value problem
− h ′ ′ − r − 1 h ′ + ( r − 2 + 1 − p U p − 1 ) h = 0 , h ( λ ) = 0 , h ′ ( R ) + R − 1 h ( R ) = 0 -h''-r^{-1}h' + (r^{-2}+1-pU^{p-1})h=0,\qquad h(\lambda)=0,\qquad h'(R)+R^{-1}h(R)=0 − h ′′ − r − 1 h ′ + ( r − 2 + 1 − p U p − 1 ) h = 0 , h ( λ ) = 0 , h ′ ( R ) + R − 1 h ( R ) = 0 admits only the zero solution. When the corresponding Dirichlet state map is defined, the first harmonic therefore does not belong to the kernel of the linearized Bernoulli operator at this radial solution.
We prove this proposition at the end of Section 3 by using U ′ U' U ′ K ≥ 1 K\geq1 K ≥ 1
Figure 1. Schematic illustration of the domains in the nonlinear bifurcation theorem. The inner circle is fixed, and the outer boundary develops an angular mode.
The remainder of the paper is organized as follows. In Section 2, we prove Theorem 1.1 for the linear-vorticity problem. We introduce the working spaces and radial branch, define the state and Bernoulli equations, and compute the boundary linearization. We then determine its kernel and range and verify the Crandall–Rabinowitz conditions. In Section 3, we follow the same approach for the power-law problem. We construct the large radial solutions from the one-dimensional limiting profile, define the corresponding operators, and use the thin-annulus spectral estimates to verify the bifurcation conditions and prove Theorem 1.2. We conclude this section with the proof of Proposition 1.3.
The linear-vorticity overdetermined problem
In this section, we fix the inner radius λ ∈ ( 0 , 1 ) \lambda\in(0,1) λ ∈ ( 0 , 1 ) 1 1 1
Working spaces and the fixed annulus
Fix 0 < α < 1 0<\alpha<1 0 < α < 1 K ≥ 1 K\geq1 K ≥ 1 A λ = B 1 ∖ B λ ‾ A_\lambda=B_1\setminus\overline{B_\lambda} A λ = B 1 ∖ B λ D K D_K D K θ ↦ − θ \theta\mapsto-\theta θ ↦ − θ θ ↦ θ + 2 π / K \theta\mapsto\theta+2\pi/K θ ↦ θ + 2 π / K C D K m , α ( A λ ‾ ) C^{m,\alpha}_{D_K}(\overline{A_\lambda}) C D K m , α ( A λ )
X K = { η ∈ C D 2 , α ( S 1 ) : η ‾ = 0 } , Y K = { g ∈ C D 1 , α ( S 1 ) : g ‾ = 0 } . (10) X_K=\{\eta\in C_D^{2,\alpha}(\mathbb{S}^1):\overline{\eta}=0\},\qquad
Y_K=\{g\in C_D^{1,\alpha}(\mathbb{S}^1):\overline{g}=0\}. \tag*{(10)} X K = { η ∈ C D 2 , α ( S 1 ) : η = 0 } , Y K = { g ∈ C D 1 , α ( S 1 ) : g = 0 } . (10) For the interior equation, we use
Z K l i n = { w ∈ C D 2 , α ( A λ ‾ ) : w ∣ r = λ = w ∣ r = 1 = 0 } , E K l i n = C D 0 , α ( A λ ‾ ) . Z_K^{\mathrm{lin}}=\{w\in C_D^{2,\alpha}(\overline{A_\lambda}):w|_{r=\lambda}=w|_{r=1}=0\},\qquad
E_K^{\mathrm{lin}}=C_D^{0,\alpha}(\overline{A_\lambda}). Z K lin = { w ∈ C D 2 , α ( A λ ) : w ∣ r = λ = w ∣ r = 1 = 0 } , E K lin = C D 0 , α ( A λ ) . In particular, Z K l i n Z_K^{\mathrm{lin}} Z K lin
Choose χ ∈ C ∞ ( [ λ , 1 ] ) \chi\in C^\infty([\lambda,1]) χ ∈ C ∞ ([ λ , 1 ]) λ \lambda λ η \eta η O K ⊂ X K O_K\subset X_K O K ⊂ X K
T η ( r , θ ) = ( s η ( r , θ ) , θ ) , s η = r + χ ( r ) η ( θ ) . (11) T_\eta(r,\theta)=(s_\eta(r,\theta),\theta),\qquad s_\eta=r+\chi(r)\eta(\theta). \tag*{(11)} T η ( r , θ ) = ( s η ( r , θ ) , θ ) , s η = r + χ ( r ) η ( θ ) . (11) The conditions ∂ r s η > 0 \partial_rs_\eta>0 ∂ r s η > 0 s η > 0 s_\eta>0 s η > 0 ∥ η ∥ C 2 , α \|\eta\|_{C^{2,\alpha}} ∥ η ∥ C 2 , α T η T_\eta T η A λ A_\lambda A λ Ω 1 , η ∖ B λ ‾ \Omega_{1,\eta}\setminus\overline{B_\lambda} Ω 1 , η ∖ B λ
a η = 1 + χ ′ η , b η = χ η θ , D η = a η − 1 ∂ r , E η = ∂ θ − b η a η − 1 ∂ r . a_\eta=1+\chi'\eta,\qquad b_\eta=\chi\eta_\theta,\qquad
D_\eta=a_\eta^{-1}\partial_r,\qquad
E_\eta=\partial_\theta-b_\eta a_\eta^{-1}\partial_r. a η = 1 + χ ′ η , b η = χ η θ , D η = a η − 1 ∂ r , E η = ∂ θ − b η a η − 1 ∂ r . The pullback of the Euclidean Laplacian is
Δ η v = D η 2 v + s η − 1 D η v + s η − 2 E η 2 v . (12) \Delta_\eta v=D_\eta^2v+s_\eta^{-1}D_\eta v+s_\eta^{-2}E_\eta^2v. \tag*{(12)} Δ η v = D η 2 v + s η − 1 D η v + s η − 2 E η 2 v . (12) Here D η 2 D_\eta^2 D η 2 E η 2 E_\eta^2 E η 2 (12) shows that η ↦ Δ η \eta\mapsto\Delta_\eta η ↦ Δ η C ∞ C^\infty C ∞ L ( C 2 , α , C 0 , α ) \mathcal{L}(C^{2,\alpha},C^{0,\alpha}) L ( C 2 , α , C 0 , α ) η \eta η D K D_K D K
The radial branch and its Dirichlet nonresonance
If μ ∉ Σ r a d ( λ ) \mu\notin\Sigma_{\mathrm{rad}}(\lambda) μ ∈ / Σ rad ( λ )
U μ ′ ′ + r − 1 U μ ′ + μ U μ = 0 , U μ ( λ ) = 1 , U μ ( 1 ) = 0 U_\mu''+r^{-1}U_\mu'+\mu U_\mu=0,\qquad U_\mu(\lambda)=1,\qquad U_\mu(1)=0 U μ ′′ + r − 1 U μ ′ + μ U μ = 0 , U μ ( λ ) = 1 , U μ ( 1 ) = 0 has a unique solution. For μ > 0 \mu>0 μ > 0 t = μ t=\sqrt{\mu} t = μ
D 0 ( t ) = J 0 ( t λ ) Y 0 ( t ) − Y 0 ( t λ ) J 0 ( t ) . D_0(t)=J_0(t\lambda)Y_0(t)-Y_0(t\lambda)J_0(t). D 0 ( t ) = J 0 ( t λ ) Y 0 ( t ) − Y 0 ( t λ ) J 0 ( t ) . Then
U μ ( r ) = J 0 ( t r ) Y 0 ( t ) − Y 0 ( t r ) J 0 ( t ) D 0 ( t ) . (13) U_\mu(r)=\frac{J_0(tr)Y_0(t)-Y_0(tr)J_0(t)}{D_0(t)}. \tag*{(13)} U μ ( r ) = D 0 ( t ) J 0 ( t r ) Y 0 ( t ) − Y 0 ( t r ) J 0 ( t ) . (13) The denominator vanishes exactly at a radial Dirichlet eigenvalue. The Bessel Wronskian identity gives
d μ : = U μ ′ ( 1 ) = − 2 π D 0 ( μ ) ≠ 0. (14) d_\mu:=U_\mu'(1)=-\frac{2}{\pi D_0(\sqrt{\mu})}\ne0. \tag*{(14)} d μ := U μ ′ ( 1 ) = − π D 0 ( μ ) 2 = 0. (14) The nonvanishing also follows directly from uniqueness of the radial Cauchy problem: U μ ( 1 ) = U μ ′ ( 1 ) = 0 U_\mu(1)=U_\mu'(1)=0 U μ ( 1 ) = U μ ′ ( 1 ) = 0 U μ ≡ 0 U_\mu\equiv0 U μ ≡ 0 ( μ , 0 , Q μ ) (\mu,0,Q_\mu) ( μ , 0 , Q μ ) Q μ = d μ 2 > 0 Q_\mu=d_\mu^2>0 Q μ = d μ 2 > 0 U μ U_\mu U μ
We next verify that the hypothesis in Theorem 1.1 indeed makes the full Dirichlet state equation invertible in the chosen symmetry class. For j ≥ 1 j \ge1 j ≥ 1 μ j , λ ∗ \mu^{*}_{j,\lambda} μ j , λ ∗ (5) with K K K j j j
μ j , λ ∗ = min h ∈ H 1 ( λ , 1 ) ∖ { 0 } h ( λ ) = 0 ∫ λ 1 ( r ∣ h ′ ∣ 2 + j 2 r ∣ h ∣ 2 ) d r + ∣ h ( 1 ) ∣ 2 ∫ λ 1 r ∣ h ∣ 2 d r . (15) \mu^{*}_{j,\lambda}=\min_{\substack{h\in H^{1}(\lambda,1)\setminus\{0\}\\ h(\lambda)=0}}\frac{\int_{\lambda}^{1}\left(r|h'|^{2}+\frac{j^{2}}{r}|h|^{2}\right)\,\mathrm{d}r+|h(1)|^{2}}{\int_{\lambda}^{1}r|h|^{2}\,\mathrm{d}r}. \tag*{(15)} μ j , λ ∗ = h ∈ H 1 ( λ , 1 ) ∖ { 0 } h ( λ ) = 0 min ∫ λ 1 r ∣ h ∣ 2 d r ∫ λ 1 ( r ∣ h ′ ∣ 2 + r j 2 ∣ h ∣ 2 ) d r + ∣ h ( 1 ) ∣ 2 . (15) In particular, μ j , λ ∗ > 0 \mu^{*}_{j,\lambda}>0 μ j , λ ∗ > 0 ( λ , 1 ] (\lambda,1] ( λ , 1 ]
μ j , λ ∗ ≥ μ ℓ , λ ∗ + j 2 − ℓ 2 ( j > ℓ ) . (16) \mu^{*}_{j,\lambda}\ge\mu^{*}_{\ell,\lambda}+j^{2}-\ell^{2}\qquad(j>\ell). \tag*{(16)} μ j , λ ∗ ≥ μ ℓ , λ ∗ + j 2 − ℓ 2 ( j > ℓ ) . (16) Now we aim to establish radial nonresonance as follows.
Lemma 2.1. At μ ∗ : = μ K , λ ∗ \mu_{*}:=\mu^{*}_{K,\lambda} μ ∗ := μ K , λ ∗
− Δ − μ ∗ : Z K l i n ⟶ E K l i n -\Delta-\mu_{*}:Z^{\mathrm{lin}}_{K}\longrightarrow E^{\mathrm{lin}}_{K} − Δ − μ ∗ : Z K lin ⟶ E K lin is an isomorphism if and only if μ ∗ ∉ Σ r a d ( λ ) \mu_{*}\notin\Sigma_{\mathrm{rad}}(\lambda) μ ∗ ∈ / Σ rad ( λ ) K = 1 K=1 K = 1 λ ∈ ( 0 , 1 ) \lambda\in(0,1) λ ∈ ( 0 , 1 ) K K K
E K = { λ ∈ ( 0 , 1 ) : μ K , λ ∗ ∈ Σ r a d ( λ ) } E_{K}=\{\lambda\in(0,1):\mu^{*}_{K,\lambda}\in\Sigma_{\mathrm{rad}}(\lambda)\} E K = { λ ∈ ( 0 , 1 ) : μ K , λ ∗ ∈ Σ rad ( λ )} is discrete in ( 0 , 1 ) (0,1) ( 0 , 1 ) 1 1 1
Proof. We first compare the mixed and Dirichlet eigenvalues in each nonzero angular mode. Let Λ j D \Lambda^{D}_{j} Λ j D j ≥ 1 j\ge1 j ≥ 1
Λ j D = min h ∈ H 0 1 ( λ , 1 ) ∖ { 0 } ∫ λ 1 ( r ∣ h ′ ∣ 2 + j 2 r ∣ h ∣ 2 ) d r ∫ λ 1 r ∣ h ∣ 2 d r . \Lambda^{D}_{j}=\min_{h\in H^{1}_{0}(\lambda,1)\setminus\{0\}}\frac{\int_{\lambda}^{1}\left(r|h'|^{2}+\frac{j^{2}}{r}|h|^{2}\right)\,\mathrm{d}r}{\int_{\lambda}^{1}r|h|^{2}\,\mathrm{d}r}. Λ j D = h ∈ H 0 1 ( λ , 1 ) ∖ { 0 } min ∫ λ 1 r ∣ h ∣ 2 d r ∫ λ 1 ( r ∣ h ′ ∣ 2 + r j 2 ∣ h ∣ 2 ) d r . Every function in this form domain is admissible in (15) , and the boundary term ∣ h ( 1 ) ∣ 2 |h(1)|^{2} ∣ h ( 1 ) ∣ 2 μ j , λ ∗ ≤ Λ j D \mu^{*}_{j,\lambda}\le\Lambda^{D}_{j} μ j , λ ∗ ≤ Λ j D h h h h ′ ( 1 ) + h ( 1 ) = 0 h'(1)+h(1)=0 h ′ ( 1 ) + h ( 1 ) = 0 h ( 1 ) = 0 h(1)=0 h ( 1 ) = 0 h ′ ( 1 ) = 0 h'(1)=0 h ′ ( 1 ) = 0 r = 1 r=1 r = 1 h ≡ 0 h\equiv0 h ≡ 0 Λ j D > μ j , λ ∗ \Lambda^{D}_{j}>\mu^{*}_{j,\lambda} Λ j D > μ j , λ ∗ (16) gives, for every permitted mode j = n K j=nK j = n K
Λ j D > μ j , λ ∗ ≥ μ K , λ ∗ = μ ∗ . \Lambda^{D}_{j}>\mu^{*}_{j,\lambda}\ge\mu^{*}_{K,\lambda}=\mu_{*}. Λ j D > μ j , λ ∗ ≥ μ K , λ ∗ = μ ∗ . Let w ∈ Z K l i n w\in Z^{\mathrm{lin}}_{K} w ∈ Z K lin ( − Δ − μ ∗ ) w = 0 (-\Delta-\mu_{*})w=0 ( − Δ − μ ∗ ) w = 0 w 0 ( r ) w_{0}(r) w 0 ( r ) cos ( j θ ) \cos(j\theta) cos ( j θ ) w j ( r ) w_{j}(r) w j ( r ) j = n K j=nK j = n K D K D_{K} D K
− w j ′ ′ − r − 1 w j ′ + j 2 r − 2 w j = μ ∗ w j , w j ( λ ) = w j ( 1 ) = 0. -w_{j}''-r^{-1}w_{j}'+j^{2}r^{-2}w_{j}=\mu_{*}w_{j},\qquad w_{j}(\lambda)=w_{j}(1)=0. − w j ′′ − r − 1 w j ′ + j 2 r − 2 w j = μ ∗ w j , w j ( λ ) = w j ( 1 ) = 0. Multiplying by r w j rw_{j} r w j
μ ∗ ∫ λ 1 r ∣ w j ∣ 2 d r = ∫ λ 1 ( r ∣ w j ′ ∣ 2 + j 2 r ∣ w j ∣ 2 ) d r ≥ Λ j D ∫ λ 1 r ∣ w j ∣ 2 d r . \mu_{*}\int_{\lambda}^{1}r|w_{j}|^{2}\,\mathrm{d}r=\int_{\lambda}^{1}\left(r|w_{j}'|^{2}+\frac{j^{2}}{r}|w_{j}|^{2}\right)\,\mathrm{d}r\ge\Lambda^{D}_{j}\int_{\lambda}^{1}r|w_{j}|^{2}\,\mathrm{d}r. μ ∗ ∫ λ 1 r ∣ w j ∣ 2 d r = ∫ λ 1 ( r ∣ w j ′ ∣ 2 + r j 2 ∣ w j ∣ 2 ) d r ≥ Λ j D ∫ λ 1 r ∣ w j ∣ 2 d r . The strict inequality Λ j D > μ ∗ \Lambda_{j}^{D}>\mu_{*} Λ j D > μ ∗ w j = 0 w_{j}=0 w j = 0
− w 0 ′ ′ − r − 1 w 0 ′ = μ ∗ w 0 , w 0 ( λ ) = w 0 ( 1 ) = 0. -w_{0}^{\prime\prime}-r^{-1}w_{0}^{\prime}=\mu_{*}w_{0},\qquad w_{0}(\lambda)=w_{0}(1)=0. − w 0 ′′ − r − 1 w 0 ′ = μ ∗ w 0 , w 0 ( λ ) = w 0 ( 1 ) = 0. This radial problem has only the zero solution precisely when μ ∗ \mu_{*} μ ∗ μ ∗ \mu_{*} μ ∗ Z K l i n Z_{K}^{\mathrm{lin}} Z K lin μ ∗ ∉ Σ r a d ( λ ) \mu_{*}\notin\Sigma_{\mathrm{rad}}(\lambda) μ ∗ ∈ / Σ rad ( λ )
We next pass from this kernel statement to invertibility. The Dirichlet Laplacian is an isomorphism Z K l i n → E K l i n Z_{K}^{\mathrm{lin}}\to E_{K}^{\mathrm{lin}} Z K lin → E K lin [7 ] . Indeed, the usual Dirichlet solution for a D K D_{K} D K D K D_{K} D K J D : Z K l i n → E K l i n J_{D}:Z_{K}^{\mathrm{lin}}\to E_{K}^{\mathrm{lin}} J D : Z K lin → E K lin C 2 , α ( A λ ‾ ) ↪ C 0 , α ( A λ ‾ ) C^{2,\alpha}(\overline{A_{\lambda}})\hookrightarrow C^{0,\alpha}(\overline{A_{\lambda}}) C 2 , α ( A λ ) ↪ C 0 , α ( A λ ) J D J_{D} J D
( − Δ ) − 1 ( − Δ − μ ∗ ) = Id Z K l i n − μ ∗ ( − Δ ) − 1 J D . (-\Delta)^{-1}(-\Delta-\mu_{*})=\operatorname{Id}_{Z_{K}^{\mathrm{lin}}}-\mu_{*}(-\Delta)^{-1}J_{D}. ( − Δ ) − 1 ( − Δ − μ ∗ ) = Id Z K lin − μ ∗ ( − Δ ) − 1 J D . The right-hand side is the identity plus a compact operator, so it is Fredholm of index zero. Thus − Δ − μ ∗ -\Delta-\mu_{*} − Δ − μ ∗
For K = 1 K=1 K = 1 h h h I = ∫ λ 1 h ( t ) d t > 0 I=\int_{\lambda}^{1}h(t)\,\mathrm{d}t>0 I = ∫ λ 1 h ( t ) d t > 0
W ( r ) = I − 1 ∫ r 1 h ( t ) d t . W(r)=I^{-1}\int_{r}^{1}h(t)\,\mathrm{d}t. W ( r ) = I − 1 ∫ r 1 h ( t ) d t . The equation for h h h
( h ′ + h r ) ′ = − μ 1 , λ ∗ h . \left(h^{\prime}+\frac{h}{r}\right)^{\prime}=-\mu_{1,\lambda}^{*}h. ( h ′ + r h ) ′ = − μ 1 , λ ∗ h . Integrating from r r r 1 1 1 h ′ ( 1 ) + h ( 1 ) = 0 h^{\prime}(1)+h(1)=0 h ′ ( 1 ) + h ( 1 ) = 0
h ′ ( r ) + h ( r ) r = μ 1 , λ ∗ ∫ r 1 h ( t ) d t . h^{\prime}(r)+\frac{h(r)}{r}=\mu_{1,\lambda}^{*}\int_{r}^{1}h(t)\,\mathrm{d}t. h ′ ( r ) + r h ( r ) = μ 1 , λ ∗ ∫ r 1 h ( t ) d t . Since W ′ = − h / I W^{\prime}=-h/I W ′ = − h / I W ′ ′ = − h ′ / I W^{\prime\prime}=-h^{\prime}/I W ′′ = − h ′ / I
− W ′ ′ − r − 1 W ′ = h ′ + h / r I = μ 1 , λ ∗ W , W ′ ( λ ) = 0 , W ( 1 ) = 0. -W^{\prime\prime}-r^{-1}W^{\prime}=\frac{h^{\prime}+h/r}{I}=\mu_{1,\lambda}^{*}W,\qquad W^{\prime}(\lambda)=0,\quad W(1)=0. − W ′′ − r − 1 W ′ = I h ′ + h / r = μ 1 , λ ∗ W , W ′ ( λ ) = 0 , W ( 1 ) = 0. Moreover, W ( λ ) = 1 W(\lambda)=1 W ( λ ) = 1 W > 0 W>0 W > 0 [ λ , 1 ) [\lambda,1) [ λ , 1 ) W W W μ 1 , λ ∗ \mu_{1,\lambda}^{*} μ 1 , λ ∗ v v v ν N D \nu_{ND} ν N D
( μ 1 , λ ∗ − ν N D ) ∫ λ 1 r W v d r = 0. (\mu_{1,\lambda}^{*}-\nu_{ND})\int_{\lambda}^{1}rWv\,\mathrm{d}r=0. ( μ 1 , λ ∗ − ν N D ) ∫ λ 1 r W v d r = 0. The integral is positive, so μ 1 , λ ∗ = ν N D \mu_{1,\lambda}^{*}=\nu_{ND} μ 1 , λ ∗ = ν N D
ν N D = min v ∈ H 1 ( λ , 1 ) ∖ { 0 } v ( 1 ) = 0 ∫ λ 1 r ∣ v ′ ∣ 2 d r ∫ λ 1 r ∣ v ∣ 2 d r . \nu_{ND}=\min_{\substack{v\in H^{1}(\lambda,1)\setminus\{0\}\\v(1)=0}}
\frac{\int_{\lambda}^{1}r|v^{\prime}|^{2}\,\mathrm{d}r}
{\int_{\lambda}^{1}r|v|^{2}\,\mathrm{d}r}. ν N D = v ∈ H 1 ( λ , 1 ) ∖ { 0 } v ( 1 ) = 0 min ∫ λ 1 r ∣ v ∣ 2 d r ∫ λ 1 r ∣ v ′ ∣ 2 d r . Adding the condition v ( λ ) = 0 v(\lambda)=0 v ( λ ) = 0 ν D \nu_{D} ν D ν N D ≤ ν D \nu_{ND}\leq\nu_{D} ν N D ≤ ν D v ′ ( λ ) = 0 v'(\lambda)=0 v ′ ( λ ) = 0 v ( λ ) = 0 v(\lambda)=0 v ( λ ) = 0 v = 0 v=0 v = 0 μ 1 , λ ∗ = ν N D < ν D \mu^*_{1,\lambda}=\nu_{ND}<\nu_D μ 1 , λ ∗ = ν N D < ν D ν D \nu_D ν D E 1 = ∅ E_1=\varnothing E 1 = ∅
We now verify the absence of exceptional radii near 1 for every fixed K K K δ = 1 − λ \delta=1-\lambda δ = 1 − λ h ( r ) = sin ( π ( r − λ ) / ( 2 δ ) ) h(r)=\sin(\pi(r-\lambda)/(2\delta)) h ( r ) = sin ( π ( r − λ ) / ( 2 δ ))
∫ λ 1 h 2 d r = δ 2 , ∫ λ 1 ∣ h ′ ∣ 2 d r = π 2 8 δ , h ( 1 ) = 1. \int_\lambda^1 h^2\,dr=\frac{\delta}{2},\qquad\int_\lambda^1 |h'|^2\,dr=\frac{\pi^2}{8\delta},\qquad h(1)=1. ∫ λ 1 h 2 d r = 2 δ , ∫ λ 1 ∣ h ′ ∣ 2 d r = 8 δ π 2 , h ( 1 ) = 1. Since λ ≤ r ≤ 1 \lambda\le r\le1 λ ≤ r ≤ 1
∫ λ 1 r h 2 d r ≥ λ δ 2 , ∫ λ 1 r ∣ h ′ ∣ 2 d r ≤ π 2 8 δ , ∫ λ 1 K 2 r h 2 d r ≤ K 2 δ 2 λ . \int_\lambda^1 rh^2\,dr\ge\frac{\lambda\delta}{2},\qquad\int_\lambda^1 r|h'|^2\,dr\le\frac{\pi^2}{8\delta},\qquad\int_\lambda^1 \frac{K^2}{r}h^2\,dr\le\frac{K^2\delta}{2\lambda}. ∫ λ 1 r h 2 d r ≥ 2 λ δ , ∫ λ 1 r ∣ h ′ ∣ 2 d r ≤ 8 δ π 2 , ∫ λ 1 r K 2 h 2 d r ≤ 2 λ K 2 δ . Substituting these bounds into the quotient gives
μ K , λ ∗ ≤ π 2 4 λ δ 2 + K 2 λ 2 + 2 δ . (17) \mu^*_{K,\lambda}\le\frac{\pi^2}{4\lambda\delta^2}+\frac{K^2}{\lambda^2}+\frac{2}{\delta}. \tag*{(17)} μ K , λ ∗ ≤ 4 λ δ 2 π 2 + λ 2 K 2 + δ 2 . (17) For a function v v v
∫ λ 1 r ∣ ∂ r v ∣ 2 d r ≥ λ ∫ λ 1 ∣ ∂ r v ∣ 2 d r ≥ λ π 2 δ 2 ∫ λ 1 ∣ v ∣ 2 d r ≥ λ π 2 δ 2 ∫ λ 1 r ∣ v ∣ 2 d r . \int_\lambda^1 r|\partial_rv|^2\,dr\ge\lambda\int_\lambda^1|\partial_rv|^2\,dr\ge\frac{\lambda\pi^2}{\delta^2}\int_\lambda^1|v|^2\,dr\ge\frac{\lambda\pi^2}{\delta^2}\int_\lambda^1r|v|^2\,dr. ∫ λ 1 r ∣ ∂ r v ∣ 2 d r ≥ λ ∫ λ 1 ∣ ∂ r v ∣ 2 d r ≥ δ 2 λ π 2 ∫ λ 1 ∣ v ∣ 2 d r ≥ δ 2 λ π 2 ∫ λ 1 r ∣ v ∣ 2 d r . Integrating in θ \theta θ
Λ D ( λ ) ≥ λ π 2 δ 2 . (18) \Lambda_D(\lambda)\ge\frac{\lambda\pi^2}{\delta^2}. \tag*{(18)} Λ D ( λ ) ≥ δ 2 λ π 2 . (18) For fixed K K K δ 2 μ K , λ ∗ \delta^2\mu^*_{K,\lambda} δ 2 μ K , λ ∗ π 2 / 4 \pi^2/4 π 2 /4 λ → 1 \lambda\to1 λ → 1 δ 2 Λ D ( λ ) \delta^2\Lambda_D(\lambda) δ 2 Λ D ( λ ) π 2 \pi^2 π 2 μ K , λ ∗ < Λ D ( λ ) \mu^*_{K,\lambda}<\Lambda_D(\lambda) μ K , λ ∗ < Λ D ( λ ) λ \lambda λ Λ D ( λ ) \Lambda_D(\lambda) Λ D ( λ ) E K E_K E K
Finally, we prove that E K E_K E K ( 0 , 1 ) (0,1) ( 0 , 1 ) h = h ( r ; λ , μ ) h=h(r;\lambda,\mu) h = h ( r ; λ , μ )
− h ′ ′ − r − 1 h ′ + K 2 r − 2 h = μ h , h ( λ ) = 0 , h ′ ( λ ) = 1 , -h''-r^{-1}h'+K^2r^{-2}h=\mu h,\qquad h(\lambda)=0,\qquad h'(\lambda)=1, − h ′′ − r − 1 h ′ + K 2 r − 2 h = μ h , h ( λ ) = 0 , h ′ ( λ ) = 1 , and set B ( λ , μ ) = h ′ ( 1 ) + h ( 1 ) B(\lambda,\mu)=h'(1)+h(1) B ( λ , μ ) = h ′ ( 1 ) + h ( 1 ) [ λ , 1 ] [\lambda,1] [ λ , 1 ] ( λ , μ ) (\lambda,\mu) ( λ , μ ) B B B μ K , λ ∗ \mu^*_{K,\lambda} μ K , λ ∗
With λ \lambda λ h ˙ = ∂ μ h \dot h=\partial_\mu h h ˙ = ∂ μ h
− ( r h ˙ ′ ) ′ + K 2 r h ˙ − μ r h ˙ = r h , h ˙ ( λ ) = h ˙ ′ ( λ ) = 0. -(r\dot h')'+\frac{K^2}{r}\dot h-\mu r\dot h=rh,\qquad\dot h(\lambda)=\dot h'(\lambda)=0. − ( r h ˙ ′ ) ′ + r K 2 h ˙ − μ r h ˙ = r h , h ˙ ( λ ) = h ˙ ′ ( λ ) = 0. Multiplying this equation by h h h h ˙ \dot h h ˙ h h h
∫ λ 1 r h 2 d r = [ r ( h h ˙ ′ − h ˙ h ′ ) ] λ 1 = − h ( 1 ) ( h ′ ( 1 ) + h ˙ ( 1 ) ) . \int_\lambda^1rh^2\,dr=[r(h\dot h'-\dot hh')]_\lambda^1=-h(1)(h'(1)+\dot h(1)). ∫ λ 1 r h 2 d r = [ r ( h h ˙ ′ − h ˙ h ′ ) ] λ 1 = − h ( 1 ) ( h ′ ( 1 ) + h ˙ ( 1 )) . Here the inner term vanishes by the initial data, and the last identity uses h ′ ( 1 ) = − h ( 1 ) h'(1)=-h(1) h ′ ( 1 ) = − h ( 1 )
h ( 1 ) ∂ μ B ( λ , μ ) = − ∫ λ 1 r h ( r ) 2 d r ≠ 0. h(1)\partial_\mu B(\lambda,\mu)=-\int_\lambda^1rh(r)^2\,dr\ne0. h ( 1 ) ∂ μ B ( λ , μ ) = − ∫ λ 1 r h ( r ) 2 d r = 0. In particular, ∂ μ B ≠ 0 \partial_\mu B \ne0 ∂ μ B = 0 μ K , λ ∗ \mu^*_{K,\lambda} μ K , λ ∗ λ ↦ μ K , λ ∗ \lambda\mapsto\mu^*_{K,\lambda} λ ↦ μ K , λ ∗ ( 0 , 1 ) (0,1) ( 0 , 1 )
Since μ K , λ ∗ > 0 \mu^*_{K,\lambda} > 0 μ K , λ ∗ > 0
F K ( λ ) = D 0 ( μ K , λ ∗ ) F_K(\lambda) = D_0\left(\sqrt{\mu^*_{K,\lambda}}\right) F K ( λ ) = D 0 ( μ K , λ ∗ ) is real analytic as well. Here
D 0 ( t ) = J 0 ( t λ ) Y 0 ( t ) − Y 0 ( t λ ) J 0 ( t ) , D_0(t) = J_0(t\lambda)Y_0(t) - Y_0(t\lambda)J_0(t), D 0 ( t ) = J 0 ( t λ ) Y 0 ( t ) − Y 0 ( t λ ) J 0 ( t ) , so its dependence on both λ \lambda λ t > 0 t > 0 t > 0 F K F_K F K E K E_K E K F K F_K F K 1 1 1
The state equation and the Bernoulli operator
Assume henceforth the hypothesis of Theorem 1.1, and write μ ∗ = μ K , λ ∗ \mu_* = \mu^*_{K,\lambda} μ ∗ = μ K , λ ∗ b ( r ) = log r / log λ b(r) = \log r/\log\lambda b ( r ) = log r / log λ b ( λ ) = 1 b(\lambda) = 1 b ( λ ) = 1 b ( 1 ) = 0 b(1) = 0 b ( 1 ) = 0 w ∈ Z K lin w \in Z_K^{\operatorname{lin}} w ∈ Z K lin
F lin ( μ , η , w ) = − Δ η ( b + w ) − μ ( b + w ) ∈ E K lin . (19) \mathcal{F}_{\operatorname{lin}}(\mu,\eta,w) = -\Delta_\eta(b+w)-\mu(b+w) \in E_K^{\operatorname{lin}}. \tag*{(19)} F lin ( μ , η , w ) = − Δ η ( b + w ) − μ ( b + w ) ∈ E K lin . (19) The parameters range over an open neighborhood of ( μ ∗ , 0 , U μ ∗ − b ) (\mu_*,0,U_{\mu_*}-b) ( μ ∗ , 0 , U μ ∗ − b ) R × X K × Z K lin \mathbb{R} \times X_K \times Z_K^{\operatorname{lin}} R × X K × Z K lin (12) , this map is C ∞ C^\infty C ∞
D w F lin ( μ ∗ , 0 , U μ ∗ − b ) = − Δ − μ ∗ . D_w\mathcal{F}_{\operatorname{lin}}(\mu_*,0,U_{\mu_*}-b)=-\Delta-\mu_*. D w F lin ( μ ∗ , 0 , U μ ∗ − b ) = − Δ − μ ∗ . Proposition 2.2. There are an interval I ⊂ ( 0 , ∞ ) I \subset(0,\infty) I ⊂ ( 0 , ∞ ) μ ∗ \mu_* μ ∗ O K ⊂ X K \mathcal{O}_K \subset X_K O K ⊂ X K
{ Δ u + μ u = 0 in Ω 1 , η ∖ B λ ‾ , u = 1 on ∂ B λ , u = 0 on ∂ Ω 1 , η (20) \begin{cases}
\Delta u+\mu u=0 & \text{in } \Omega_{1,\eta}\setminus\overline{B_\lambda},\\
u=1 & \text{on } \partial B_\lambda,\\
u=0 & \text{on } \partial\Omega_{1,\eta} \tag*{(20)}
\end{cases} ⎩ ⎨ ⎧ Δ u + μu = 0 u = 1 u = 0 in Ω 1 , η ∖ B λ , on ∂ B λ , on ∂ Ω 1 , η (20) has a unique D K D_K D K u μ , η u_{\mu,\eta} u μ , η ( μ , η ) ∈ I × O K (\mu,\eta) \in I \times\mathcal{O}_K ( μ , η ) ∈ I × O K v μ , η = u μ , η ∘ T η v_{\mu,\eta}=u_{\mu,\eta}\circ T_\eta v μ , η = u μ , η ∘ T η b + Z K lin b+Z_K^{\operatorname{lin}} b + Z K lin C ∞ C^\infty C ∞ ( μ , η ) (\mu,\eta) ( μ , η ) η = 0 \eta=0 η = 0 U μ U_\mu U μ
Proof. Lemma 2.1 makes D w F lin D_w\mathcal{F}_{\operatorname{lin}} D w F lin w = w ( μ , η ) w=w(\mu,\eta) w = w ( μ , η ) v μ , η = b + w ( μ , η ) v_{\mu,\eta}=b+w(\mu,\eta) v μ , η = b + w ( μ , η ) − Δ η − μ -\Delta_\eta-\mu − Δ η − μ η = 0 \eta=0 η = 0 (13) provides that solution.
Let Π g = g − g ‾ \Pi g=g-\overline{g} Π g = g − g r = 1 + η ( θ ) r=1+\eta(\theta) r = 1 + η ( θ )
ν η = ( 1 + η ) e r − η θ e θ ( 1 + η ) 2 + η θ 2 . \nu_\eta=\frac{(1+\eta)e_r-\eta_\theta e_\theta}{\sqrt{(1+\eta)^2+\eta_\theta^2}}. ν η = ( 1 + η ) 2 + η θ 2 ( 1 + η ) e r − η θ e θ . We define
N lin ( μ , η ) ( θ ) = ∂ ν η u μ , η ( ( 1 + η ( θ ) ) e r ) , N_{\operatorname{lin}}(\mu,\eta)(\theta)=\partial_{\nu_\eta}u_{\mu,\eta}\bigl((1+\eta(\theta))e_r\bigr), N lin ( μ , η ) ( θ ) = ∂ ν η u μ , η ( ( 1 + η ( θ )) e r ) , G lin ( μ , η ) = Π ( N lin ( μ , η ) 2 ) . (21) G_{\operatorname{lin}}(\mu,\eta)=\Pi\left(N_{\operatorname{lin}}(\mu,\eta)^2\right). \tag*{(21)} G lin ( μ , η ) = Π ( N lin ( μ , η ) 2 ) . (21) Since χ = 1 \chi=1 χ = 1 1 1 1
N l i n ( μ , η ) = ( 1 + η θ 2 ( 1 + η ) 2 ) 1 / 2 ∂ r v μ , η ( 1 , ⋅ ) . (22) N_{\mathrm{lin}}(\mu,\eta)=\left(1+\frac{\eta_\theta^2}{(1+\eta)^2}\right)^{1/2}\partial_r v_{\mu,\eta}(1,\cdot). \tag*{(22)} N lin ( μ , η ) = ( 1 + ( 1 + η ) 2 η θ 2 ) 1/2 ∂ r v μ , η ( 1 , ⋅ ) . (22) Consequently,
N l i n : I × O K ⟶ C D 1 , α ( S 1 ) , G l i n : I × O K ⟶ Y K N_{\mathrm{lin}}:I\times\mathcal{O}_K\longrightarrow C_D^{1,\alpha}(\mathbb{S}^1),\qquad G_{\mathrm{lin}}:I\times\mathcal{O}_K\longrightarrow Y_K N lin : I × O K ⟶ C D 1 , α ( S 1 ) , G lin : I × O K ⟶ Y K are C ∞ C^\infty C ∞ N l i n ( μ , 0 ) = d μ N_{\mathrm{lin}}(\mu,0)=d_\mu N lin ( μ , 0 ) = d μ G l i n ( μ , 0 ) = 0 G_{\mathrm{lin}}(\mu,0)=0 G lin ( μ , 0 ) = 0 N l i n N_{\mathrm{lin}} N lin d μ ∗ ≠ 0 d_{\mu_*}\ne0 d μ ∗ = 0 u μ , η u_{\mu,\eta} u μ , η
G l i n ( μ , η ) = 0 , Q = N l i n ( μ , η ) 2 ‾ > 0. (23) G_{\mathrm{lin}}(\mu,\eta)=0,\qquad Q=\overline{N_{\mathrm{lin}}(\mu,\eta)^2}>0. \tag*{(23)} G lin ( μ , η ) = 0 , Q = N lin ( μ , η ) 2 > 0. (23) The mean is taken in the angular coordinate; its role is only to remove constants, so no arclength weight is needed.
Linearization, kernel, and range
Fix μ ∈ I \mu\in I μ ∈ I ξ ∈ X K \xi\in X_K ξ ∈ X K v μ , ε ξ v_{\mu,\varepsilon\xi} v μ , ε ξ
u ξ ′ = ∂ ε v μ , ε ξ ∣ ε = 0 − χ ( r ) U μ ′ ( r ) ξ ( θ ) . u'_\xi=\left.\partial_\varepsilon v_{\mu,\varepsilon\xi}\right|_{\varepsilon=0}-\chi(r)U_\mu'(r)\xi(\theta). u ξ ′ = ∂ ε v μ , ε ξ ∣ ε = 0 − χ ( r ) U μ ′ ( r ) ξ ( θ ) . It follows from radial ODE that U μ ′ ′ ( 1 ) = − U μ ′ ( 1 ) = − d μ U_\mu''(1)=-U_\mu'(1)=-d_\mu U μ ′′ ( 1 ) = − U μ ′ ( 1 ) = − d μ [9 Lemma 3.2] , we have
{ ( − Δ − μ ) u ξ ′ = 0 in A λ , u ξ ′ = 0 on r = λ , u ξ ′ = − U μ ′ ( 1 ) ξ = − d μ ξ on r = 1. (24) \begin{cases}
(-\Delta-\mu)u'_\xi=0 & \text{in } A_\lambda,\\
u'_\xi=0 & \text{on } r=\lambda,\\
u'_\xi=-U_\mu'(1)\xi=-d_\mu\xi& \text{on } r=1. \tag*{(24)}
\end{cases} ⎩ ⎨ ⎧ ( − Δ − μ ) u ξ ′ = 0 u ξ ′ = 0 u ξ ′ = − U μ ′ ( 1 ) ξ = − d μ ξ in A λ , on r = λ , on r = 1. (24) Either differentiating (2.13) or differentiating the physical normal trace therefore yields
D η N l i n ( μ , 0 ) ξ = ∂ r u ξ ′ ( 1 , ⋅ ) + U μ ′ ′ ( 1 ) ξ = ∂ r u ξ ′ ( 1 , ⋅ ) − d μ ξ . (25) D_\eta N_{\mathrm{lin}}(\mu,0)\xi=\partial_r u'_\xi(1,\cdot)+U_\mu''(1)\xi=\partial_r u'_\xi(1,\cdot)-d_\mu\xi. \tag*{(25)} D η N lin ( μ , 0 ) ξ = ∂ r u ξ ′ ( 1 , ⋅ ) + U μ ′′ ( 1 ) ξ = ∂ r u ξ ′ ( 1 , ⋅ ) − d μ ξ . (25) The normal-vector variation is tangential and has zero scalar product with the radial gradient of U μ U_\mu U μ
For every permitted frequency j = n K j=nK j = n K H j , μ H_{j,\mu} H j , μ
− H ′ ′ − r − 1 H ′ + j 2 r − 2 H − μ H = 0 , H ( λ ) = 0 , H ( 1 ) = 1. -H''-r^{-1}H'+j^2r^{-2}H-\mu H=0,\qquad H(\lambda)=0,\qquad H(1)=1. − H ′′ − r − 1 H ′ + j 2 r − 2 H − μ H = 0 , H ( λ ) = 0 , H ( 1 ) = 1. The Dirichlet invertibility established above makes this solution well defined. In Bessel notation it is
H j , μ ( r ) = J j ( t r ) Y j ( t λ ) − Y j ( t r ) J j ( t λ ) J j ( t ) Y j ( t λ ) − Y j ( t ) J j ( t λ ) , t = μ . (26) H_{j,\mu}(r)=\frac{J_j(tr)Y_j(t\lambda)-Y_j(tr)J_j(t\lambda)}{J_j(t)Y_j(t\lambda)-Y_j(t)J_j(t\lambda)},\qquad t=\sqrt{\mu}. \tag*{(26)} H j , μ ( r ) = J j ( t ) Y j ( t λ ) − Y j ( t ) J j ( t λ ) J j ( t r ) Y j ( t λ ) − Y j ( t r ) J j ( t λ ) , t = μ . (26) For ξ = cos ( j θ ) \xi=\cos(j\theta) ξ = cos ( j θ ) u ξ ′ = − d μ H j , μ ( r ) cos ( j θ ) u'_\xi=-d_\mu H_{j,\mu}(r)\cos(j\theta) u ξ ′ = − d μ H j , μ ( r ) cos ( j θ )
T j , λ ( μ ) = H j , μ ′ ( 1 ) + 1 , m j l i n ( μ ) = − 2 d μ 2 T j , λ ( μ ) , (27) T_{j,\lambda}(\mu)=H'_{j,\mu}(1)+1,\qquad m_j^{\mathrm{lin}}(\mu)=-2d_\mu^2T_{j,\lambda}(\mu), \tag*{(27)} T j , λ ( μ ) = H j , μ ′ ( 1 ) + 1 , m j lin ( μ ) = − 2 d μ 2 T j , λ ( μ ) , (27) we have
L μ [ cos ( j θ ) ] : = D η G l i n ( μ , 0 ) [ cos ( j θ ) ] = m j l i n ( μ ) cos ( j θ ) . (28) L_\mu[\cos(j\theta)]:=D_\eta G_{\mathrm{lin}}(\mu,0)[\cos(j\theta)]=m_j^{\mathrm{lin}}(\mu)\cos(j\theta). \tag*{(28)} L μ [ cos ( j θ )] := D η G lin ( μ , 0 ) [ cos ( j θ )] = m j lin ( μ ) cos ( j θ ) . (28) The symbol T j , λ T_{j,\lambda} T j , λ
− h ′ ′ − r − 1 h ′ + j 2 r − 2 h = μ h , h ( λ ) = 0 , h ′ ( 1 ) + h ( 1 ) = 0 -h''-r^{-1}h'+j^2r^{-2}h=\mu h,\qquad h(\lambda)=0,\qquad h'(1)+h(1)=0 − h ′′ − r − 1 h ′ + j 2 r − 2 h = μ h , h ( λ ) = 0 , h ′ ( 1 ) + h ( 1 ) = 0 has a nonzero solution. In particular, at μ = μ ∗ \mu=\mu_* μ = μ ∗ K K K j = n K j=nK j = n K n ≥ 2 n\ge2 n ≥ 2 μ ∗ \mu_* μ ∗ T n K , λ ( μ ∗ ) ≠ 0 T_{nK,\lambda}(\mu_*) \ne0 T n K , λ ( μ ∗ ) = 0 n ≥ 2 n \ge2 n ≥ 2
Lemma 2.3. For μ ∈ I \mu\in I μ ∈ I L μ : X K → Y K L_\mu: X_K \to Y_K L μ : X K → Y K μ = μ ∗ \mu= \mu_* μ = μ ∗
Ker L μ ∗ = span { cos ( K θ ) } , \operatorname{Ker} L_{\mu_*} = \operatorname{span}\{\cos(K\theta)\}, Ker L μ ∗ = span { cos ( K θ )} , Ran L μ ∗ = { g ∈ Y K : ∫ 0 2 π g ( θ ) cos ( K θ ) d θ = 0 } . (29) \operatorname{Ran} L_{\mu_*} = \left\{g \in Y_K : \int_0^{2\pi} g(\theta)\cos(K\theta)\,\mathrm{d}\theta= 0\right\}. \tag*{(29)} Ran L μ ∗ = { g ∈ Y K : ∫ 0 2 π g ( θ ) cos ( K θ ) d θ = 0 } . (29) Proof. For ξ ∈ X K \xi\in X_K ξ ∈ X K P μ ξ P_\mu\xi P μ ξ D K D_K D K
( − Δ − μ ) P μ ξ = 0 in A λ , P μ ξ ∣ r = λ = 0 , P μ ξ ∣ r = 1 = ξ . (-\Delta-\mu)P_\mu\xi= 0 \quad\text{in } A_\lambda,\qquad P_\mu\xi|_{r=\lambda}=0,\qquad P_\mu\xi|_{r=1}=\xi. ( − Δ − μ ) P μ ξ = 0 in A λ , P μ ξ ∣ r = λ = 0 , P μ ξ ∣ r = 1 = ξ . The Dirichlet invertibility on I I I P μ : X K → C 2 , α ( A λ ‾ ) P_\mu: X_K \to C^{2,\alpha}(\overline{A_\lambda}) P μ : X K → C 2 , α ( A λ )
Λ μ ξ = ∂ r ( P μ ξ ) ( 1 , ⋅ ) \Lambda_\mu\xi= \partial_r(P_\mu\xi)(1,\cdot) Λ μ ξ = ∂ r ( P μ ξ ) ( 1 , ⋅ ) defines a bounded map X K → Y K X_K \to Y_K X K → Y K
L μ = − 2 d μ 2 A μ , A μ = Λ μ + J , L_\mu= -2d_\mu^2 A_\mu,\qquad A_\mu= \Lambda_\mu+ J, L μ = − 2 d μ 2 A μ , A μ = Λ μ + J , where J : X K ↪ Y K J : X_K \hookrightarrow Y_K J : X K ↪ Y K [19 Proposition 3.2] , L μ L_\mu L μ
We use the corresponding Dirichlet–Robin problem to recall its Fredholm alternative in the precise spaces needed here:
{ ( − Δ − μ ) v = 0 in A λ , v = 0 on r = λ , ∂ r v + v = g on r = 1. \begin{cases}
(-\Delta-\mu)v = 0 & \text{in } A_\lambda,\\
v = 0 & \text{on } r=\lambda,\\
\partial_r v+v = g & \text{on } r=1.
\end{cases} ⎩ ⎨ ⎧ ( − Δ − μ ) v = 0 v = 0 ∂ r v + v = g in A λ , on r = λ , on r = 1. We restrict to D K D_K D K H 1 H^1 H 1 L 2 L^2 L 2 L 2 L^2 L 2
∫ 0 2 π g ( θ ) v 0 ( 1 , θ ) d θ = 0 \int_0^{2\pi} g(\theta)v_0(1,\theta)\,\mathrm{d}\theta= 0 ∫ 0 2 π g ( θ ) v 0 ( 1 , θ ) d θ = 0 for every homogeneous solution v 0 v_0 v 0 g ∈ Y K g \in Y_K g ∈ Y K v ∈ C 2 , α v \in C^{2,\alpha} v ∈ C 2 , α [7 ] . Dirichlet uniqueness identifies these homogeneous outer traces with Ker A μ \operatorname{Ker} A_\mu Ker A μ A μ [ v ( 1 , ⋅ ) ] = g A_\mu[v(1,\cdot)] = g A μ [ v ( 1 , ⋅ )] = g
Ran A μ = { g ∈ Y K : ∫ 0 2 π g ϕ d θ = 0 for every ϕ ∈ Ker A μ } . \operatorname{Ran} A_\mu= \left\{g \in Y_K : \int_0^{2\pi} g\phi\,\mathrm{d}\theta= 0 \text{ for every } \phi\in\operatorname{Ker} A_\mu\right\}. Ran A μ = { g ∈ Y K : ∫ 0 2 π g ϕ d θ = 0 for every ϕ ∈ Ker A μ } . The finite-dimensional kernel lies in X K ⊂ Y K X_K \subset Y_K X K ⊂ Y K dim Ker A μ \dim\operatorname{Ker} A_\mu dim Ker A μ A μ A_\mu A μ L μ L_\mu L μ d μ ≠ 0 d_\mu\ne0 d μ = 0 C 0 , α C^{0,\alpha} C 0 , α
At μ ∗ \mu_* μ ∗ (28) shows that the coefficient in mode j = n K j=nK j = n K L μ ∗ ξ L_{\mu_*}\xi L μ ∗ ξ m j l i n ( μ ∗ ) m^{\mathrm{lin}}_j(\mu_*) m j lin ( μ ∗ ) ξ \xi ξ n = 1 n=1 n = 1 n ≥ 2 n\ge2 n ≥ 2 Ker L μ ∗ = span { cos ( K θ ) } \operatorname{Ker} L_{\mu_*}=\operatorname{span}\{\cos(K\theta)\} Ker L μ ∗ = span { cos ( K θ )}
ℓ ( g ) = ∫ 0 2 π g ( θ ) cos ( K θ ) d θ \ell(g)=\int_0^{2\pi}g(\theta)\cos(K\theta)\,d\theta ℓ ( g ) = ∫ 0 2 π g ( θ ) cos ( K θ ) d θ vanishes on every range element and satisfies ℓ ( cos ( K θ ) ) = π ≠ 0 \ell(\cos(K\theta))=\pi\ne0 ℓ ( cos ( K θ )) = π = 0 Ker ℓ \operatorname{Ker}\ell Ker ℓ L μ ∗ L_{\mu_*} L μ ∗ (29) . This argument does not require Fourier partial sums to converge in the C 2 , α C^{2,\alpha} C 2 , α
Transversality and proof of Theorem 1.1
We verify the remaining Crandall–Rabinowitz condition explicitly. A dot will denote differentiation with respect to μ \mu μ H K , μ ( λ ) = 0 H_{K,\mu}(\lambda)=0 H K , μ ( λ ) = 0 H K , μ ( 1 ) = 1 H_{K,\mu}(1)=1 H K , μ ( 1 ) = 1 μ \mu μ
( − d 2 d r 2 − 1 r d d r + K 2 r 2 − μ ) H ˙ K , μ = H K , μ , H ˙ K , μ ( λ ) = H ˙ K , μ ( 1 ) = 0. \left(-\frac{d^2}{dr^2}-\frac{1}{r}\frac{d}{dr}+\frac{K^2}{r^2}-\mu\right)\dot{H}_{K,\mu}=H_{K,\mu},\qquad\dot{H}_{K,\mu}(\lambda)=\dot{H}_{K,\mu}(1)=0. ( − d r 2 d 2 − r 1 d r d + r 2 K 2 − μ ) H ˙ K , μ = H K , μ , H ˙ K , μ ( λ ) = H ˙ K , μ ( 1 ) = 0. Multiplying the equation for H ˙ K , μ \dot{H}_{K,\mu} H ˙ K , μ r H K , μ rH_{K,\mu} r H K , μ H K , μ H_{K,\mu} H K , μ r H ˙ K , μ r\dot{H}_{K,\mu} r H ˙ K , μ
∫ λ 1 r H K , μ 2 d r = − H ˙ K , μ ′ ( 1 ) , \int_\lambda^1 rH_{K,\mu}^2\,dr=-\dot{H}'_{K,\mu}(1), ∫ λ 1 r H K , μ 2 d r = − H ˙ K , μ ′ ( 1 ) , which gives
d T K , λ ( μ ) d μ = − ∫ λ 1 r H K , μ ( r ) 2 d r < 0. (30) \frac{dT_{K,\lambda}(\mu)}{d\mu}=-\int_\lambda^1 rH_{K,\mu}(r)^2\,dr<0. \tag*{(30)} d μ d T K , λ ( μ ) = − ∫ λ 1 r H K , μ ( r ) 2 d r < 0. (30) Because T K , λ ( μ ∗ ) = 0 T_{K,\lambda}(\mu_*)=0 T K , λ ( μ ∗ ) = 0 (27) at the critical value yields
d m K l i n d μ ( μ ∗ ) = 2 μ ∗ 2 ∫ λ 1 r H K , μ ∗ ( r ) 2 d r > 0. (31) \frac{d m_K^{\mathrm{lin}}}{d\mu}(\mu_*)=2\mu_*^2\int_\lambda^1 rH_{K,\mu_*}(r)^2\,dr>0. \tag*{(31)} d μ d m K lin ( μ ∗ ) = 2 μ ∗ 2 ∫ λ 1 r H K , μ ∗ ( r ) 2 d r > 0. (31) In particular,
D μ D η G l i n ( μ ∗ , 0 ) cos ( K θ ) ∉ Ran D η G l i n ( μ ∗ , 0 ) . D_\mu D_\eta G_{\mathrm{lin}}(\mu_*,0)\cos(K\theta)\notin\operatorname{Ran}D_\eta G_{\mathrm{lin}}(\mu_*,0). D μ D η G lin ( μ ∗ , 0 ) cos ( K θ ) ∈ / Ran D η G lin ( μ ∗ , 0 ) . We have now checked all hypotheses on the concrete map G l i n : I × O K → Y K G_{\mathrm{lin}}:I\times O_K\to Y_K G lin : I × O K → Y K G l i n ( μ , 0 ) = 0 G_{\mathrm{lin}}(\mu,0)=0 G lin ( μ , 0 ) = 0 η \eta η [2 ] gives a smooth local curve of nontrivial solutions. Choose the parameter by
s = 1 π ∫ 0 2 π η ( s , θ ) cos ( K θ ) d θ . s=\frac{1}{\pi}\int_0^{2\pi}\eta(s,\theta)\cos(K\theta)\,d\theta. s = π 1 ∫ 0 2 π η ( s , θ ) cos ( K θ ) d θ . With the complement of the kernel defined by vanishing K K K
μ ( 0 ) = μ ∗ , η ( s ) = s cos ( K θ ) + O ( s 2 ) in X K . \mu(0)=\mu_*,\qquad\eta(s)=s\cos(K\theta)+O(s^2)\ \text{in }X_K. μ ( 0 ) = μ ∗ , η ( s ) = s cos ( K θ ) + O ( s 2 ) in X K . Every nearby nontrivial solution belongs to this curve.
Let R η ( θ ) = η ( θ + π / K ) R\eta(\theta)=\eta(\theta+\pi/K) R η ( θ ) = η ( θ + π / K ) X K X_K X K Y K Y_K Y K K K K G l i n ( μ , R η ) = G_{\mathrm{lin}}(\mu,R\eta)= G lin ( μ , R η ) = R G l i n ( μ , η ) RG_{\mathrm{lin}}(\mu,\eta) R G lin ( μ , η ) [8 ] and local uniqueness of the bifurcation curve therefore yield
μ ( − s ) = μ ( s ) , η ( − s ) = R η ( s ) . \mu(-s)=\mu(s),\qquad\eta(-s)=\mathcal{R}\eta(s). μ ( − s ) = μ ( s ) , η ( − s ) = R η ( s ) . Hence μ ′ ( 0 ) = 0 \mu'(0)=0 μ ′ ( 0 ) = 0 μ ( s ) = μ ∗ + O ( s 2 ) \mu(s)=\mu_*+O(s^2) μ ( s ) = μ ∗ + O ( s 2 ) (23) defines the corresponding smooth positive Q ( s ) Q(s) Q ( s )
The power-law overdetermined problem
We now prove Theorem 1.2 for the specific equation Δ u = u − u p \Delta u=u-u^p Δ u = u − u p λ > 0 \lambda>0 λ > 0 p > 1 p>1 p > 1 a = 2 / ( p − 1 ) a=2/(p-1) a = 2/ ( p − 1 ) V p = V V_p=V V p = V (43) . We use the mean outer radius R R R
The working spaces and the power map
Fix temporarily a reference radius R ∗ > λ R_*> \lambda R ∗ > λ A ∗ = B R ∗ ∖ B λ ‾ A_*=B_{R_*}\setminus\overline{B_\lambda} A ∗ = B R ∗ ∖ B λ K ≥ 2 K\ge2 K ≥ 2 D K D_K D K θ ↦ − θ \theta\mapsto-\theta θ ↦ − θ θ ↦ θ + 2 π / K \theta\mapsto\theta+2\pi/K θ ↦ θ + 2 π / K X K X_K X K Y K Y_K Y K
Z p , K = { w ∈ C D K 2 , α ( A ∗ ‾ ) : w ∣ r = λ = w ∣ r = R ∗ = 0 } , W p , K = C D K 0 , α ( A ∗ ‾ ) . Z_{p,K}=\left\{w\in C^{2,\alpha}_{D_K}(\overline{A_*}):w|_{r=\lambda}=w|_{r=R_*}=0\right\},\qquad W_{p,K}=C^{0,\alpha}_{D_K}(\overline{A_*}). Z p , K = { w ∈ C D K 2 , α ( A ∗ ) : w ∣ r = λ = w ∣ r = R ∗ = 0 } , W p , K = C D K 0 , α ( A ∗ ) . Every D K D_K D K v = b ∗ + w v=b_*+w v = b ∗ + w w ∈ Z p , K w\in Z_{p,K} w ∈ Z p , K
b ∗ ( r ) = R ∗ − r R ∗ − λ . b_*(r)=\frac{R_*-r}{R_*-\lambda}. b ∗ ( r ) = R ∗ − λ R ∗ − r . We work in a sufficiently small C 2 , α C^{2,\alpha} C 2 , α R R R
For nonintegral p p p ϱ ( r ) = R ∗ − r \varrho(r)=R_*-r ϱ ( r ) = R ∗ − r v ∈ C 2 , α ( A ∗ ‾ ) v\in C^{2,\alpha}(\overline{A_*}) v ∈ C 2 , α ( A ∗ ) R ∗ R_* R ∗ t = R ∗ − r t=R_*-r t = R ∗ − r
v ( t , θ ) t = ∫ 0 1 ∂ t v ( τ t , θ ) d τ . (32) \frac{v(t,\theta)}{t}=\int_0^1\partial_t v(\tau t,\theta)\,\mathrm{d}\tau. \tag*{(32)} t v ( t , θ ) = ∫ 0 1 ∂ t v ( τ t , θ ) d τ . (32) Thus the quotient extends across t = 0 t=0 t = 0
∥ v ϱ ∥ C 1 , α ( A ∗ ‾ ) ≤ C ∥ v ∥ C 2 , α ( A ∗ ‾ ) . \left\lVert\frac{v}{\varrho}\right\rVert_{C^{1,\alpha}(\overline{A_*})}\le C\lVert v\rVert_{C^{2,\alpha}(\overline{A_*})}. ϱ v C 1 , α ( A ∗ ) ≤ C ∥ v ∥ C 2 , α ( A ∗ ) . This estimate allows us to use ordinary small Hölder neighborhoods without introducing a separate space for the quotient. We first describe the power map in the Dirichlet space.
Lemma 3.1. Let U ∈ C D K 2 , α ( A ∗ ‾ ) U\in C^{2,\alpha}_{D_K}(\overline{A_*}) U ∈ C D K 2 , α ( A ∗ ) A ∗ A_* A ∗ U ∣ r = λ = 1 U|_{r=\lambda}=1 U ∣ r = λ = 1 U ∣ r = R ∗ = 0 U|_{r=R_*}=0 U ∣ r = R ∗ = 0 ∂ r U ( R ∗ , θ ) < 0 \partial_rU(R_*,\theta)<0 ∂ r U ( R ∗ , θ ) < 0 ϵ U > 0 \epsilon_U>0 ϵ U > 0
w ⟼ ( U + w ) p : { w ∈ Z p , K : ∥ w ∥ C 2 , α < ϵ U } ⟶ W p , K w\longmapsto(U+w)^p:\left\{w\in Z_{p,K}:\lVert w\rVert_{C^{2,\alpha}}<\epsilon_U\right\}\longrightarrow W_{p,K} w ⟼ ( U + w ) p : { w ∈ Z p , K : ∥ w ∥ C 2 , α < ϵ U } ⟶ W p , K is C ∞ C^\infty C ∞ U + w U+w U + w A ∗ A_* A ∗ D K D_K D K
Proof. The quotient U / ϱ U/\varrho U / ϱ R ∗ R_\ast R ∗ − ∂ r U ( R ∗ , θ ) > 0 -\partial_rU(R_\ast,\theta)>0 − ∂ r U ( R ∗ , θ ) > 0 1 / ( R ∗ − λ ) 1/(R_\ast-\lambda) 1/ ( R ∗ − λ ) U > 0 U>0 U > 0
c U : = min A ‾ U ϱ > 0. c_U:=\min_{\overline{A}}\frac{U}{\varrho}>0. c U := A min ϱ U > 0. The division estimate gives ∥ w / ϱ ∥ C 1 , α ≤ C ∥ w ∥ C 2 , α \|w/\varrho\|_{C^{1,\alpha}}\leq C\|w\|_{C^{2,\alpha}} ∥ w / ϱ ∥ C 1 , α ≤ C ∥ w ∥ C 2 , α C ϵ U < c U / 2 C\epsilon_U<c_U/2 C ϵ U < c U /2
z : = U + w ϱ ≥ c U / 2 > 0. z:=\frac{U+w}{\varrho}\geq c_U/2>0. z := ϱ U + w ≥ c U /2 > 0. This proves positivity in the annulus and, at the outer boundary, ∂ r ( U + w ) = − z < 0 \partial_r(U+w)=-z<0 ∂ r ( U + w ) = − z < 0 w ↦ z w\mapsto z w ↦ z C 1 , α C^{1,\alpha} C 1 , α z ↦ z p z\mapsto z^p z ↦ z p C ∞ C^\infty C ∞ ϱ p ∈ C 0 , α \varrho^p\in C^{0,\alpha} ϱ p ∈ C 0 , α p > 1 > α p>1>\alpha p > 1 > α ϱ p \varrho^p ϱ p C 0 , α C^{0,\alpha} C 0 , α ( U + w ) p = ϱ p z p (U+w)^p=\varrho^pz^p ( U + w ) p = ϱ p z p C ∞ C^\infty C ∞ w i ∈ Z p , K w_i\in Z_{p,K} w i ∈ Z p , K
D m ( v p ) [ w 1 , … , w m ] = p ( p − 1 ) ⋯ ( p − m + 1 ) ϱ p z p − m ∏ i = 1 m w i ϱ , v = U + w . D^m(v^p)[w_1,\ldots,w_m]=p(p-1)\cdots(p-m+1)\varrho^pz^{p-m}\prod_{i=1}^{m}\frac{w_i}{\varrho},\qquad v=U+w. D m ( v p ) [ w 1 , … , w m ] = p ( p − 1 ) ⋯ ( p − m + 1 ) ϱ p z p − m i = 1 ∏ m ϱ w i , v = U + w . Each quotient belongs to C 1 , α C^{1,\alpha} C 1 , α z z z C 0 , α C^{0,\alpha} C 0 , α v v v ϱ \varrho ϱ v p − m v^{p-m} v p − m
The trivial branch: large positive radial solutions
Fix λ > 0 \lambda>0 λ > 0 R = λ + δ R=\lambda+\delta R = λ + δ δ > 0 \delta>0 δ > 0
U ′ ′ + 1 r U ′ = U − U p , U ( λ ) = 1 , U ( λ + δ ) = 0. (33) U''+\frac{1}{r}U'=U-U^p,\qquad U(\lambda)=1,\qquad U(\lambda+\delta)=0. \tag*{(33)} U ′′ + r 1 U ′ = U − U p , U ( λ ) = 1 , U ( λ + δ ) = 0. (33) We shall construct a family of size δ − a \delta^{-a} δ − a a = 2 / ( p − 1 ) a=2/(p-1) a = 2/ ( p − 1 ) V p V_p V p
Rescaling and the implicit-function theorem
The radial second derivative has size δ − a − 2 \delta^{-a-2} δ − a − 2 δ − a \delta^{-a} δ − a δ \delta δ p p p δ − a p \delta^{-ap} δ − a p a + 2 = a p a+2=ap a + 2 = a p a = 2 / ( p − 1 ) a=2/(p-1) a = 2/ ( p − 1 )
t = r − λ δ , U ( r ) = δ − a V δ ( t ) . (34) t=\frac{r-\lambda}{\delta},\qquad U(r)=\delta^{-a}V_\delta(t). \tag*{(34)} t = δ r − λ , U ( r ) = δ − a V δ ( t ) . (34) The derivatives and the power are
U ′ ( r ) = δ − a − 1 V δ ′ ( t ) , U ′ ′ ( r ) = δ − a − 2 V δ ′ ′ ( t ) , U ( r ) p = δ − a p V δ ( t ) p . U'(r)=\delta^{-a-1}V_\delta'(t),\qquad U''(r)=\delta^{-a-2}V_\delta''(t),\qquad U(r)^p=\delta^{-ap}V_\delta(t)^p. U ′ ( r ) = δ − a − 1 V δ ′ ( t ) , U ′′ ( r ) = δ − a − 2 V δ ′′ ( t ) , U ( r ) p = δ − a p V δ ( t ) p . We substitute these expressions into (3.2) and multiply by δ a + 2 \delta^{a+2} δ a + 2 a + 2 − a p = 0 a+2-ap=0 a + 2 − a p = 0 1 1 1 δ / ( λ + δ t ) \delta/(\lambda+\delta t) δ / ( λ + δ t ) δ 2 \delta^2 δ 2 V δ ( 0 ) = δ a U ( λ ) = δ a V_\delta(0)=\delta^aU(\lambda)=\delta^a V δ ( 0 ) = δ a U ( λ ) = δ a V δ ( 1 ) = 0 V_\delta(1)=0 V δ ( 1 ) = 0
{ V δ ′ ′ + δ λ + δ t V δ ′ − δ 2 V δ + V δ p = 0 , 0 < t < 1 , V δ ( 0 ) = δ a , V δ ( 1 ) = 0. (35) \begin{cases}
V_\delta''+\dfrac{\delta}{\lambda+\delta t}V_\delta'-\delta^2V_\delta+V_\delta^p=0, & 0<t<1,\\
V_\delta(0)=\delta^a,\quad V_\delta(1)=0. \tag*{(35)}
\end{cases} ⎩ ⎨ ⎧ V δ ′′ + λ + δ t δ V δ ′ − δ 2 V δ + V δ p = 0 , V δ ( 0 ) = δ a , V δ ( 1 ) = 0. 0 < t < 1 , (35) As the width tends to zero, the curvature coefficient and the linear reaction term vanish. The rescaled inner trace also tends to zero, although the original inner value remains 1. These observations give the limiting equation (43) and its two zero endpoint values.
We cannot apply the smooth implicit function theorem directly at δ = 0 \delta=0 δ = 0 δ a \delta^a δ a a a a b b b V = v + b ( 1 − t ) V=v+b(1-t) V = v + b ( 1 − t ) v ( 0 ) = v ( 1 ) = 0 v(0)=v(1)=0 v ( 0 ) = v ( 1 ) = 0 C 0 2 , α ( [ 0 , 1 ] ) = { v ∈ C 2 , α ( [ 0 , 1 ] ) : v ( 0 ) = v ( 1 ) = 0 } C^{2,\alpha}_0([0,1])=\{v\in C^{2,\alpha}([0,1]):v(0)=v(1)=0\} C 0 2 , α ([ 0 , 1 ]) = { v ∈ C 2 , α ([ 0 , 1 ]) : v ( 0 ) = v ( 1 ) = 0 } C 0 , α ( [ 0 , 1 ] ) C^{0,\alpha}([0,1]) C 0 , α ([ 0 , 1 ])
E ( δ , b , v ) = v ′ ′ + δ λ + δ t ( v ′ − b ) − δ 2 ( v + b ( 1 − t ) ) + ( v + b ( 1 − t ) ) + p . (36) \mathcal{E}(\delta,b,v)=v''+\frac{\delta}{\lambda+\delta t}(v'-b)-\delta^2(v+b(1-t))+(v+b(1-t))_+^p. \tag*{(36)} E ( δ , b , v ) = v ′′ + λ + δ t δ ( v ′ − b ) − δ 2 ( v + b ( 1 − t )) + ( v + b ( 1 − t ) ) + p . (36) Here s + = max { s , 0 } s_+=\max\{s,0\} s + = max { s , 0 } ∣ δ ∣ < λ / 2 |\delta|<\lambda/2 ∣ δ ∣ < λ /2 [ 0 , 1 ] [0,1] [ 0 , 1 ] ( 0 , 0 , V p ) (0,0,V_p) ( 0 , 0 , V p ) b = δ a > 0 b=\delta^a>0 b = δ a > 0
Lemma 3.2. For every sufficiently small δ > 0 \delta>0 δ > 0 (35) admits a positive solution V δ V_\delta V δ
∥ V δ − V p ∥ C 2 , α ( [ 0 , 1 ] ) ≤ C ( δ + δ a ) . (37) \lVert V_\delta-V_p\rVert_{C^{2,\alpha}([0,1])}\leq C(\delta+\delta^a). \tag*{(37)} ∥ V δ − V p ∥ C 2 , α ([ 0 , 1 ]) ≤ C ( δ + δ a ) . (37) The family is locally unique among scaled profiles close to V p V_p V p C ∞ C^\infty C ∞ δ > 0 \delta>0 δ > 0
∥ δ ∂ δ V δ ∥ C 2 , α ( [ 0 , 1 ] ) ≤ C ( δ + δ a ) . (38) \lVert\delta\,\partial_\delta V_\delta\rVert_{C^{2,\alpha}([0,1])}\leq C(\delta+\delta^a). \tag*{(38)} ∥ δ ∂ δ V δ ∥ C 2 , α ([ 0 , 1 ]) ≤ C ( δ + δ a ) . (38) Proof. We first verify the regularity needed at ( 0 , 0 , V p ) (0,0,V_p) ( 0 , 0 , V p ) F ( v ) = ( v + ) p F(v)=(v_+)^p F ( v ) = ( v + ) p α < β < min { 1 , p − 1 } \alpha<\beta<\min\{1,p-1\} α < β < min { 1 , p − 1 } F ′ ( s ) = p ( s + ) p − 1 F'(s)=p(s_+)^{p-1} F ′ ( s ) = p ( s + ) p − 1 β \beta β C 1 C^1 C 1 p ( v + ) p − 1 p(v_+)^{p-1} p ( v + ) p − 1 C 0 , β C^{0,\beta} C 0 , β v n → v v_n\to v v n → v C 2 , α C^{2,\alpha} C 2 , α
∥ g ∥ C 0 , α ≤ C ∥ g ∥ C 0 1 − α / β ∥ g ∥ C 0 , β α / β \lVert g\rVert_{C^{0,\alpha}}\leq C\lVert g\rVert_{C^0}^{1-\alpha/\beta}\lVert g\rVert_{C^{0,\beta}}^{\alpha/\beta} ∥ g ∥ C 0 , α ≤ C ∥ g ∥ C 0 1 − α / β ∥ g ∥ C 0 , β α / β then gives their convergence in C 0 , α C^{0,\alpha} C 0 , α C 2 , α C^{2,\alpha} C 2 , α C 0 , α C^{0,\alpha} C 0 , α D F ( v ) w = p ( v + ) p − 1 w DF(v)w=p(v_+)^{p-1}w D F ( v ) w = p ( v + ) p − 1 w
F ( v + w ) − F ( v ) − p ( v + ) p − 1 w = ∫ 0 1 p ( ( v + s w ) + p − 1 − ( v + ) p − 1 ) w d s F(v+w)-F(v)-p(v_+)^{p-1}w=\int_0^1 p\left((v+sw)_+^{p-1}-(v_+)^{p-1}\right)w\,ds F ( v + w ) − F ( v ) − p ( v + ) p − 1 w = ∫ 0 1 p ( ( v + s w ) + p − 1 − ( v + ) p − 1 ) w d s shows that the remainder is o ( ∥ w ∥ C 2 , α ) o(\lVert w\rVert_{C^{2,\alpha}}) o (∥ w ∥ C 2 , α ) C 0 , α C^{0,\alpha} C 0 , α F F F C 1 C^1 C 1 E \mathcal{E} E ( 0 , 0 , V p ) (0,0,V_p) ( 0 , 0 , V p ) V p V_p V p (43) , we have E ( 0 , 0 , V p ) = 0 \mathcal{E}(0,0,V_p)=0 E ( 0 , 0 , V p ) = 0
D v E ( 0 , 0 , V p ) w = w ′ ′ + p V p p − 1 w = − A D w , w ( 0 ) = w ( 1 ) = 0. D_v\mathcal{E}(0,0,V_p)w=w''+pV_p^{p-1}w=-A_Dw,\qquad w(0)=w(1)=0. D v E ( 0 , 0 , V p ) w = w ′′ + p V p p − 1 w = − A D w , w ( 0 ) = w ( 1 ) = 0. Lemma 3.5 gives Ker A D = { 0 } \operatorname{Ker} A_D=\{0\} Ker A D = { 0 } V p p − 1 V_p^{p-1} V p p − 1 C 0 , α C^{0,\alpha} C 0 , α V p V_p V p α < p − 1 \alpha<p-1 α < p − 1 f ∈ C 0 , α f\in C^{0,\alpha} f ∈ C 0 , α − A D w = f -A_Dw=f − A D w = f
∥ w ∥ C 2 , α ≤ C ∥ f ∥ C 0 , α . \|w\|_{C^{2,\alpha}}\leq C\|f\|_{C^{0,\alpha}}. ∥ w ∥ C 2 , α ≤ C ∥ f ∥ C 0 , α . In particular, D v E ( 0 , 0 , V p ) D_v\mathcal{E}(0,0,V_p) D v E ( 0 , 0 , V p ) C 1 C^1 C 1 ( δ , b ) ↦ v ( δ , b ) (\delta,b)\mapsto v(\delta,b) ( δ , b ) ↦ v ( δ , b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) v ( 0 , 0 ) = V p v(0,0)=V_p v ( 0 , 0 ) = V p
V δ = v ( δ , δ a ) + δ a ( 1 − t ) . V_\delta=v(\delta,\delta^a)+\delta^a(1-t). V δ = v ( δ , δ a ) + δ a ( 1 − t ) . The first derivatives of v v v ( 0 , 0 ) (0,0) ( 0 , 0 ) ( δ , δ a ) (\delta,\delta^a) ( δ , δ a )
∥ v ( δ , δ a ) − V p ∥ C 2 , α ≤ C ( δ + δ a ) . \|v(\delta,\delta^a)-V_p\|_{C^{2,\alpha}}\leq C(\delta+\delta^a). ∥ v ( δ , δ a ) − V p ∥ C 2 , α ≤ C ( δ + δ a ) . Adding the boundary lift proves (37) . The local uniqueness supplied by the theorem applies to any solution of (35) sufficiently close to V p V_p V p δ a ( 1 − t ) \delta^a(1-t) δ a ( 1 − t )
We next check that the extension by the positive part has introduced no change in the desired equation. By Lemma 3.4 , V p ′ ( 0 ) > 0 V_p'(0)>0 V p ′ ( 0 ) > 0 V p ′ ( 1 ) < 0 V_p'(1)<0 V p ′ ( 1 ) < 0 ϵ ∈ ( 0 , 1 / 2 ) \epsilon\in(0,1/2) ϵ ∈ ( 0 , 1/2 ) c > 0 c>0 c > 0 V p ′ ≥ 2 c V_p'\geq2c V p ′ ≥ 2 c [ 0 , ϵ ] [0,\epsilon] [ 0 , ϵ ] V p ′ ≤ − 2 c V_p'\leq-2c V p ′ ≤ − 2 c [ 1 − ϵ , 1 ] [1-\epsilon,1] [ 1 − ϵ , 1 ] C 1 C^1 C 1 (37) gives the corresponding bounds V δ ′ ≥ c V_\delta'\geq c V δ ′ ≥ c V δ ′ ≤ − c V_\delta'\leq-c V δ ′ ≤ − c δ \delta δ b = δ a b=\delta^a b = δ a
V δ ( t ) ≥ b + c t ( 0 ≤ t ≤ ϵ ) , V δ ( t ) ≥ c ( 1 − t ) ( 1 − ϵ ≤ t ≤ 1 ) . V_\delta(t)\geq b+ct\quad(0\leq t\leq\epsilon),\qquad V_\delta(t)\geq c(1-t)\quad(1-\epsilon\leq t\leq1). V δ ( t ) ≥ b + c t ( 0 ≤ t ≤ ϵ ) , V δ ( t ) ≥ c ( 1 − t ) ( 1 − ϵ ≤ t ≤ 1 ) . On [ ϵ , 1 − ϵ ] [\epsilon,1-\epsilon] [ ϵ , 1 − ϵ ] V p V_p V p δ \delta δ V δ > 0 V_\delta>0 V δ > 0 [ 0 , 1 ) [0,1) [ 0 , 1 ) 1 1 1 ( V δ ) + p = V δ p (V_\delta)_+^p=V_\delta^p ( V δ ) + p = V δ p (35) with the original power.
Although the curve b = δ a b=\delta^a b = δ a δ > 0 \delta>0 δ > 0
δ ∂ δ V δ = δ ∂ δ v ( δ , δ a ) + a δ a ( ∂ b v ( δ , δ a ) + ( 1 − t ) ) . \delta\partial_\delta V_\delta=\delta\partial_\delta v(\delta,\delta^a)+a\delta^a\left(\partial_bv(\delta,\delta^a)+(1-t)\right). δ ∂ δ V δ = δ ∂ δ v ( δ , δ a ) + a δ a ( ∂ b v ( δ , δ a ) + ( 1 − t ) ) . The bounded parameter derivatives prove (38) .
To obtain smooth dependence for each fixed small δ 0 > 0 \delta_0>0 δ 0 > 0 V δ 0 / ( 1 − t ) V_{\delta_0}/(1-t) V δ 0 / ( 1 − t ) C 1 , α C^{1,\alpha} C 1 , α [ 0 , 1 ] [0,1] [ 0 , 1 ] 1 − t 1-t 1 − t C 1 , α C^{1,\alpha} C 1 , α (32) . Their quotients remain bounded away from zero. Writing the power as ( 1 − t ) p ( V / ( 1 − t ) ) p (1-t)^p(V/(1-t))^p ( 1 − t ) p ( V / ( 1 − t ) ) p C ∞ C^\infty C ∞ p p p V δ p − 1 V_\delta^{p-1} V δ p − 1 V p p − 1 V_p^{p-1} V p p − 1 C 0 , α C^{0,\alpha} C 0 , α C 0 , β C^{0,\beta} C 0 , β δ / ( λ + δ ) → 0 \delta/(\lambda+\delta)\to0 δ / ( λ + δ ) → 0 δ 2 → 0 \delta^2\to0 δ 2 → 0 D v E ( δ , δ a , v ( δ , δ a ) ) D_v\mathcal{E}(\delta,\delta^a,v(\delta,\delta^a)) D v E ( δ , δ a , v ( δ , δ a )) δ \delta δ ( δ 0 , δ 0 a ) (\delta_0,\delta_0^a) ( δ 0 , δ 0 a ) C ∞ C^\infty C ∞
Returning to the original variables, we define
U δ ( r ) = δ − a V δ ( r − λ δ ) . (39) U_{\delta}(r)=\delta^{-a}V_{\delta}\left(\frac{r-\lambda}{\delta}\right). \tag*{(39)} U δ ( r ) = δ − a V δ ( δ r − λ ) . (39) The scaling calculation and positivity above show that this function solves (33) , is positive in the annulus, and has inner value 1. Its outer normal derivative is
d δ = U δ ′ ( λ + δ ) = δ − a − 1 V δ ′ ( 1 ) < 0. (40) d_{\delta}=U_{\delta}'(\lambda+\delta)=\delta^{-a-1}V_{\delta}'(1)<0. \tag*{(40)} d δ = U δ ′ ( λ + δ ) = δ − a − 1 V δ ′ ( 1 ) < 0. (40) Thus the radial Bernoulli constant is d δ 2 > 0 d_{\delta}^{2}>0 d δ 2 > 0 C 1 C^{1} C 1 U δ ′ ( λ ) > 0 U_{\delta}'(\lambda)>0 U δ ′ ( λ ) > 0
max U δ = δ − a ( M p + o ( 1 ) ) . \max U_{\delta}=\delta^{-a}(M_{p}+o(1)). max U δ = δ − a ( M p + o ( 1 )) . In particular, these radial solutions lie outside the ordered class 0 < U δ < 1 0<U_{\delta}<1 0 < U δ < 1
3.2.2. A derivative estimate for the linearized potential. To verify transversality, we also need an estimate for the parameter derivative of the potential. When 1 < p < 2 1<p<2 1 < p < 2 V δ p − 2 V_{\delta}^{p-2} V δ p − 2
Lemma 3.3. As δ ↓ 0 \delta\downarrow0 δ ↓ 0
∥ V δ p − 1 − V p p − 1 ∥ L ∞ ⟶ 0 , ∥ δ ∂ δ ( V δ p − 1 ) ∥ L ∞ ⟶ 0. (41) \left\|V_{\delta}^{p-1}-V_{p}^{p-1}\right\|_{L^{\infty}}\longrightarrow0,\qquad
\left\|\delta\partial_{\delta}\left(V_{\delta}^{p-1}\right)\right\|_{L^{\infty}}\longrightarrow0. \tag*{(41)} V δ p − 1 − V p p − 1 L ∞ ⟶ 0 , δ ∂ δ ( V δ p − 1 ) L ∞ ⟶ 0. (41) Moreover, the second norm is at most C ( δ + δ a ) C(\delta+\delta^{a}) C ( δ + δ a )
Proof. The profiles are uniformly bounded and converge uniformly to V p V_{p} V p s ↦ s p − 1 s\mapsto s^{p-1} s ↦ s p − 1 Z δ = δ ∂ δ V δ Z_{\delta}=\delta\partial_{\delta}V_{\delta} Z δ = δ ∂ δ V δ b = δ a b=\delta^{a} b = δ a
∥ Z δ ∥ C 1 ≤ C ( δ + b ) , Z δ ( 0 ) = a b , Z δ ( 1 ) = 0. \|Z_{\delta}\|_{C^{1}}\leq C(\delta+b),\qquad Z_{\delta}(0)=ab,\qquad Z_{\delta}(1)=0. ∥ Z δ ∥ C 1 ≤ C ( δ + b ) , Z δ ( 0 ) = ab , Z δ ( 1 ) = 0. For 0 < t < 1 0<t<1 0 < t < 1
δ ∂ δ ( V δ p − 1 ) = ( p − 1 ) V δ p − 2 Z δ . \delta\partial_{\delta}\left(V_{\delta}^{p-1}\right)=(p-1)V_{\delta}^{p-2}Z_{\delta}. δ ∂ δ ( V δ p − 1 ) = ( p − 1 ) V δ p − 2 Z δ . If p ≥ 2 p\geq2 p ≥ 2 V δ p − 2 V_{\delta}^{p-2} V δ p − 2 (38) proves the required bound directly.
Suppose now that 1 < p < 2 1<p<2 1 < p < 2 c , ϵ > 0 c,\epsilon>0 c , ϵ > 0
V δ ( t ) ≥ c ( t + b ) ( 0 ≤ t ≤ ϵ ) , V δ ( t ) ≥ c ( 1 − t ) ( 1 − ϵ ≤ t ≤ 1 ) , V_{\delta}(t)\geq c(t+b)\quad(0\leq t\leq\epsilon),\qquad
V_{\delta}(t)\geq c(1-t)\quad(1-\epsilon\leq t\leq1), V δ ( t ) ≥ c ( t + b ) ( 0 ≤ t ≤ ϵ ) , V δ ( t ) ≥ c ( 1 − t ) ( 1 − ϵ ≤ t ≤ 1 ) , and gives a uniform positive lower bound on the middle interval. Since Z δ ( 1 ) = 0 Z_{\delta}(1)=0 Z δ ( 1 ) = 0 ∣ Z δ ( t ) ∣ ≤ C ( δ + b ) ( 1 − t ) |Z_{\delta}(t)|\leq C(\delta+b)(1-t) ∣ Z δ ( t ) ∣ ≤ C ( δ + b ) ( 1 − t )
V δ ( t ) p − 2 ∣ Z δ ( t ) ∣ ≤ C ( δ + b ) ( 1 − t ) p − 1 ≤ C ( δ + b ) ( 1 − ϵ ≤ t < 1 ) . V_{\delta}(t)^{p-2}|Z_{\delta}(t)|\leq C(\delta+b)(1-t)^{p-1}\leq C(\delta+b)\qquad(1-\epsilon\leq t<1). V δ ( t ) p − 2 ∣ Z δ ( t ) ∣ ≤ C ( δ + b ) ( 1 − t ) p − 1 ≤ C ( δ + b ) ( 1 − ϵ ≤ t < 1 ) . In particular, this product extends continuously by zero at t = 1 t=1 t = 1 Z δ ( 0 ) = a b Z_{\delta}(0)=ab Z δ ( 0 ) = ab ∣ Z δ ( t ) ∣ ≤ C b + C ( δ + b ) t |Z_{\delta}(t)|\leq Cb+C(\delta+b)t ∣ Z δ ( t ) ∣ ≤ C b + C ( δ + b ) t
V δ ( t ) p − 2 ∣ Z δ ( t ) ∣ ≤ C b + ( δ + b ) t ( t + b ) 2 − p ≤ C b p − 1 + C ( δ + b ) t p − 1 ≤ C ( b p − 1 + δ + b ) ( 0 ≤ t ≤ ϵ ) . V_{\delta}(t)^{p-2}|Z_{\delta}(t)|
\leq C\frac{b+(\delta+b)t}{(t+b)^{2-p}}
\leq Cb^{p-1}+C(\delta+b)t^{p-1}
\leq C\left(b^{p-1}+\delta+b\right)\qquad(0\leq t\leq\epsilon). V δ ( t ) p − 2 ∣ Z δ ( t ) ∣ ≤ C ( t + b ) 2 − p b + ( δ + b ) t ≤ C b p − 1 + C ( δ + b ) t p − 1 ≤ C ( b p − 1 + δ + b ) ( 0 ≤ t ≤ ϵ ) . The second inequality follows from t + b ≥ b t+b\ge b t + b ≥ b t + b ≥ t t+b\ge t t + b ≥ t b p − 1 = δ a ( p − 1 ) = δ 2 b^{p-1}=\delta^{a(p-1)}=\delta^2 b p − 1 = δ a ( p − 1 ) = δ 2 C ( δ + b ) C(\delta+b) C ( δ + b ) δ ≤ 1 \delta\le1 δ ≤ 1 V δ V_\delta V δ C 0 C^0 C 0 Z δ Z_\delta Z δ p − 1 p-1 p − 1
Radial Dirichlet invertibility
The radial Dirichlet linearization in the rescaled variables is
A δ , 0 h = − h ′ ′ − δ λ + δ t h ′ + ( δ 2 − p V δ p − 1 ) h , h ( 0 ) = h ( 1 ) = 0. (42) A_{\delta,0}h=-h''-\frac{\delta}{\lambda+\delta t}h'+\left(\delta^2-pV_\delta^{p-1}\right)h,\qquad h(0)=h(1)=0. \tag*{(42)} A δ , 0 h = − h ′′ − λ + δ t δ h ′ + ( δ 2 − p V δ p − 1 ) h , h ( 0 ) = h ( 1 ) = 0. (42) Put w δ ( t ) = 1 + δ t / λ w_\delta(t)=1+\delta t/\lambda w δ ( t ) = 1 + δ t / λ w δ ′ / w δ = δ / ( λ + δ t ) w_\delta'/w_\delta=\delta/(\lambda+\delta t) w δ ′ / w δ = δ / ( λ + δ t ) − w δ − 1 ( w δ h ′ ) ′ -w_\delta^{-1}(w_\delta h')' − w δ − 1 ( w δ h ′ ) ′
q δ ( h ) = ∫ 0 1 w δ ( ∣ h ′ ∣ 2 + ( δ 2 − p V δ p − 1 ) h 2 ) d t , h ∈ H 0 1 ( 0 , 1 ) . q_\delta(h)=\int_0^1w_\delta\left(|h'|^2+\left(\delta^2-pV_\delta^{p-1}\right)h^2\right)\,dt,\qquad h\in H_0^1(0,1). q δ ( h ) = ∫ 0 1 w δ ( ∣ h ′ ∣ 2 + ( δ 2 − p V δ p − 1 ) h 2 ) d t , h ∈ H 0 1 ( 0 , 1 ) . with squared norm ∫ 0 1 w δ h 2 d t \int_0^1w_\delta h^2\,dt ∫ 0 1 w δ h 2 d t A δ , 0 A_{\delta,0} A δ , 0 L 2 L^2 L 2
The weight tends uniformly to 1 1 1 − p V p p − 1 -pV_p^{p-1} − p V p p − 1 Lemma 3.3 . If q D ( h ) = ∫ 0 1 ( ∣ h ′ ∣ 2 − p V p p − 1 h 2 ) d t q_D(h)=\int_0^1(|h'|^2-pV_p^{p-1}h^2)\,dt q D ( h ) = ∫ 0 1 ( ∣ h ′ ∣ 2 − p V p p − 1 h 2 ) d t
∣ q δ ( h ) − q D ( h ) ∣ ≤ ϵ δ ( ∥ h ′ ∥ L 2 2 + ∥ h ∥ L 2 2 ) , ϵ δ ⟶ 0 , |q_\delta(h)-q_D(h)|\le\epsilon_\delta\left(\|h'\|_{L^2}^2+\|h\|_{L^2}^2\right),\qquad\epsilon_\delta\longrightarrow0, ∣ q δ ( h ) − q D ( h ) ∣ ≤ ϵ δ ( ∥ h ′ ∥ L 2 2 + ∥ h ∥ L 2 2 ) , ϵ δ ⟶ 0 , as well as ∫ 0 1 w δ h 2 d t = ( 1 + O ( δ ) ) ∥ h ∥ L 2 2 \int_0^1w_\delta h^2\,dt=(1+O(\delta))\|h\|_{L^2}^2 ∫ 0 1 w δ h 2 d t = ( 1 + O ( δ )) ∥ h ∥ L 2 2 h h h A δ , 0 A_{\delta,0} A δ , 0 A D A_D A D Lemma 3.5 , the first limiting eigenvalue is negative and the second is strictly positive. For sufficiently small δ \delta δ A δ , 0 A_{\delta,0} A δ , 0 A δ , 0 A_{\delta,0} A δ , 0
We can also see the absence of a zero eigenvalue directly. If A δ n , 0 h n = 0 A_{\delta_n,0}h_n=0 A δ n , 0 h n = 0 δ n ↓ 0 \delta_n\downarrow0 δ n ↓ 0 ∥ h n ∥ C 0 = 1 \|h_n\|_{C^0}=1 ∥ h n ∥ C 0 = 1 C 2 C^2 C 2 ∥ h n ∥ W 2 , q ≤ C ( ∥ A δ n , 0 h n ∥ L q + ∥ h n ∥ L q ) \|h_n\|_{W^{2,q}}\le C(\|A_{\delta_n,0}h_n\|_{L^q}+\|h_n\|_{L^q}) ∥ h n ∥ W 2 , q ≤ C ( ∥ A δ n , 0 h n ∥ L q + ∥ h n ∥ L q ) h n ′ h_n' h n ′ q > 1 q>1 q > 1 h n ′ ′ h_n'' h n ′′ C 1 C^1 C 1 h h h ∥ h ∥ C 0 = 1 \|h\|_{C^0}=1 ∥ h ∥ C 0 = 1 A D h = 0 A_Dh=0 A D h = 0 Lemma 3.5 . This also explains why radial Dirichlet invertibility persists along the large annular family.
The preceding construction reduces the thin-annulus problem to a fixed interval. We now prove directly the one-dimensional profile properties used above and the mixed–Dirichlet gap that will determine the critical angular scale.
The limiting profile for the radial branch
The rescaling in the preceding subsection explains the form of the limiting profile. In (35) , the coefficients of the curvature term and the linear reaction term tend to zero as δ → 0 \delta\to0 δ → 0 δ a → 0 \delta^a\to0 δ a → 0 1 1 1 a ( p − 1 ) = 2 a(p-1)=2 a ( p − 1 ) = 2
− V ′ ′ = V p in ( 0 , 1 ) , V ( 0 ) = V ( 1 ) = 0. (43) -V^{\prime\prime}=V^p \quad\text{in }(0,1), \qquad V(0)=V(1)=0. \tag*{(43)} − V ′′ = V p in ( 0 , 1 ) , V ( 0 ) = V ( 1 ) = 0. (43) We now prove the properties of this profile used in the radial construction. We also compare its Dirichlet and mixed spectra, since the state equation and the linearized outer boundary condition impose different conditions at t = 1 t=1 t = 1
Existence, uniqueness, and elementary identities
Lemma 3.4. For each p > 1 p>1 p > 1 (43) admits a unique positive solution V = V p V=V_p V = V p t = 1 / 2 t=1/2 t = 1/2 ( 0 , 1 / 2 ) (0,1/2) ( 0 , 1/2 ) ( 1 / 2 , 1 ) (1/2,1) ( 1/2 , 1 ) V ′ ( 0 ) > 0 V'(0)>0 V ′ ( 0 ) > 0 V ′ ( 1 ) = − V ′ ( 0 ) < 0 V'(1)=-V'(0)<0 V ′ ( 1 ) = − V ′ ( 0 ) < 0
Proof. We first construct a positive trajectory with prescribed maximum M > 0 M>0 M > 0 M M M ( v + ) p (v_+)^p ( v + ) p C 1 C^1 C 1 p > 1 p>1 p > 1 V ′ V' V ′
1 2 V ′ ( t ) 2 + 1 p + 1 V ( t ) p + 1 = 1 p + 1 M p + 1 . (44) \frac{1}{2}V'(t)^2+\frac{1}{p+1}V(t)^{p+1}=\frac{1}{p+1}M^{p+1}. \tag*{(44)} 2 1 V ′ ( t ) 2 + p + 1 1 V ( t ) p + 1 = p + 1 1 M p + 1 . (44) On its decreasing part, the derivative is therefore
V ′ = − 2 p + 1 M p + 1 − V p + 1 . V'=-\sqrt{\frac{2}{p+1}}\sqrt{M^{p+1}-V^{p+1}}. V ′ = − p + 1 2 M p + 1 − V p + 1 . Separation of variables shows that the time from the maximum to the first zero is
T ( M ) = p + 1 2 ∫ 0 M d v M p + 1 − v p + 1 = p + 1 2 M − ( p − 1 ) / 2 ∫ 0 1 d z 1 − z p + 1 . T(M)=\sqrt{\frac{p+1}{2}}\int_0^M\frac{dv}{\sqrt{M^{p+1}-v^{p+1}}}
=\sqrt{\frac{p+1}{2}}M^{-(p-1)/2}\int_0^1\frac{dz}{\sqrt{1-z^{p+1}}}. T ( M ) = 2 p + 1 ∫ 0 M M p + 1 − v p + 1 d v = 2 p + 1 M − ( p − 1 ) /2 ∫ 0 1 1 − z p + 1 d z . This integral is finite: its only singularity is at z = 1 z=1 z = 1 ( 1 − z ) − 1 / 2 (1-z)^{-1/2} ( 1 − z ) − 1/2 T ( M ) T(M) T ( M )
The total distance between the two zeros is 2 T ( M ) 2T(M) 2 T ( M ) M M M M ↓ 0 M\downarrow0 M ↓ 0 M → ∞ M\to\infty M → ∞ M M M 2 T ( M ) = 1 2T(M)=1 2 T ( M ) = 1 0 0 0 (43) , with its maximum at 1 / 2 1/2 1/2
Conversely, any positive solution of (43) has V ′ ′ = − V p < 0 V^{\prime\prime}=-V^p<0 V ′′ = − V p < 0 V ′ ( 0 ) > 0 V'(0)>0 V ′ ( 0 ) > 0 V ′ ( 1 ) < 0 V'(1)<0 V ′ ( 1 ) < 0 (44) to this solution shows that the distances from its maximum to both endpoints are T ( M ) T(M) T ( M ) 2 T ( M ) = 1 2T(M)=1 2 T ( M ) = 1 1 / 2 1/2 1/2 (44) gives V ′ ( 1 ) = − V ′ ( 0 ) V'(1)=-V'(0) V ′ ( 1 ) = − V ′ ( 0 )
Let I p = ∫ 0 1 ( 1 − z p + 1 ) − 1 / 2 d z I_p=\int_0^1(1-z^{p+1})^{-1/2}\,dz I p = ∫ 0 1 ( 1 − z p + 1 ) − 1/2 d z 2 T ( M p ) = 1 2T(M_p)=1 2 T ( M p ) = 1 t = 0 t=0 t = 0
M p = ( 2 ( p + 1 ) I p ) 2 / ( p − 1 ) , V p ′ ( 0 ) = 2 p + 1 M p ( p + 1 ) / 2 . (45) M_p=\left(\sqrt{2(p+1)}I_p\right)^{2/(p-1)},\qquad V_p'(0)=\sqrt{\frac{2}{p+1}}M_p^{(p+1)/2}. \tag*{(45)} M p = ( 2 ( p + 1 ) I p ) 2/ ( p − 1 ) , V p ′ ( 0 ) = p + 1 2 M p ( p + 1 ) /2 . (45) These formulas determine the leading size of the radial solution and its Bernoulli constant. The simple endpoint zeros also show that V p ( t ) V_p(t) V p ( t ) t t t 1 − t 1-t 1 − t κ p \kappa_p κ p
Dirichlet nondegeneracy
Consider the self-adjoint Dirichlet operator
A D = − d 2 d t 2 − p V p p − 1 on H 2 ( 0 , 1 ) ∩ H 0 1 ( 0 , 1 ) . A_D=-\frac{d^2}{dt^2}-pV_p^{p-1}\quad\text{on }H^2(0,1)\cap H_0^1(0,1). A D = − d t 2 d 2 − p V p p − 1 on H 2 ( 0 , 1 ) ∩ H 0 1 ( 0 , 1 ) . Its potential is continuous on [ 0 , 1 ] [0,1] [ 0 , 1 ] ν D \nu_D ν D
Lemma 3.5. The operator A D A_D A D ν D < 0 \nu_D<0 ν D < 0
Proof. We first test the Dirichlet quadratic form with V = V p V=V_p V = V p − V ′ ′ = V p -V''=V^p − V ′′ = V p V V V ∫ 0 1 V ′ 2 d t = ∫ 0 1 V p + 1 d t \int_0^1 V'^2\,dt=\int_0^1V^{p+1}\,dt ∫ 0 1 V ′2 d t = ∫ 0 1 V p + 1 d t
∫ 0 1 ( V ′ 2 − p V p + 1 ) d t = ( 1 − p ) ∫ 0 1 V p + 1 d t < 0. \int_0^1\left(V'^2-pV^{p+1}\right)\,dt=(1-p)\int_0^1V^{p+1}\,dt<0. ∫ 0 1 ( V ′2 − p V p + 1 ) d t = ( 1 − p ) ∫ 0 1 V p + 1 d t < 0. The variational characterization of ν D \nu_D ν D ν D < 0 \nu_D<0 ν D < 0
To exclude zero and count the negative eigenvalues, we use the scaling of the profile equation. Locally in the scaling parameter s > 0 s>0 s > 0 s a V ( s t ) s^aV(st) s a V ( s t ) ( 0 , 1 ) (0,1) ( 0 , 1 ) a + 2 = a p a+2=ap a + 2 = a p s = 1 s=1 s = 1
Z ( t ) = a V ( t ) + t V ′ ( t ) , a = 2 p − 1 . (46) Z(t)=aV(t)+tV'(t),\qquad a=\frac{2}{p-1}. \tag*{(46)} Z ( t ) = aV ( t ) + t V ′ ( t ) , a = p − 1 2 . (46) We can also verify its equation directly. In the interior,
Z ′ = ( a + 1 ) V ′ + t V ′ ′ , Z'=(a+1)V'+tV'', Z ′ = ( a + 1 ) V ′ + t V ′′ , Z ′ ′ = ( a + 2 ) V ′ ′ + t V ′ ′ ′ = − ( a + 2 ) V p − p t V p − 1 V ′ , Z''=(a+2)V''+tV'''=-(a+2)V^p-ptV^{p-1}V', Z ′′ = ( a + 2 ) V ′′ + t V ′′′ = − ( a + 2 ) V p − pt V p − 1 V ′ , Z ′ ′ + p V p − 1 Z = ( p a − a − 2 ) V p = 0. Z''+pV^{p-1}Z=(pa-a-2)V^p=0. Z ′′ + p V p − 1 Z = ( p a − a − 2 ) V p = 0. The equation extends to the endpoints by continuity. Moreover,
Z ( 0 ) = 0 , Z ′ ( 0 ) = ( a + 1 ) V ′ ( 0 ) > 0 , Z ( 1 ) = V ′ ( 1 ) < 0. Z(0)=0,\qquad Z'(0)=(a+1)V'(0)>0,\qquad Z(1)=V'(1)<0. Z ( 0 ) = 0 , Z ′ ( 0 ) = ( a + 1 ) V ′ ( 0 ) > 0 , Z ( 1 ) = V ′ ( 1 ) < 0. Every solution of the zero-energy equation with zero left endpoint value is determined by its left derivative. It is consequently a multiple of Z Z Z Z ( 1 ) ≠ 0 Z(1)\ne0 Z ( 1 ) = 0 A D A_D A D
It remains to count the zeros of Z Z Z ( 0 , 1 / 2 ] (0,1/2] ( 0 , 1/2 ] V > 0 V>0 V > 0 V ′ ≥ 0 V'\ge0 V ′ ≥ 0 Z > 0 Z>0 Z > 0 ( 1 / 2 , 1 ) (1/2,1) ( 1/2 , 1 )
Z ′ = ( a + 1 ) V ′ + t V ′ ′ < 0 , Z'=(a+1)V'+tV''<0, Z ′ = ( a + 1 ) V ′ + t V ′′ < 0 , because both V ′ V' V ′ V ′ ′ V'' V ′′ Z ( 1 / 2 ) = a V ( 1 / 2 ) > 0 Z(1/2)=aV(1/2)>0 Z ( 1/2 ) = aV ( 1/2 ) > 0 Z ( 1 ) < 0 Z(1)<0 Z ( 1 ) < 0 Z Z Z [22 ] , applied to the solution of A D Z = 0 A_DZ=0 A D Z = 0 A D A_D A D
In particular, the homogeneous Dirichlet linearization at V p V_p V p C 0 , α C^{0,\alpha} C 0 , α (6) , the Dirichlet Fredholm alternative and the one-dimensional Schauder estimate give the isomorphism used in the preceding implicit-function construction of the radial family.
The mixed eigenvalue and the strict gap
On the form domain E = { h ∈ H 1 ( 0 , 1 ) : h ( 0 ) = 0 } E=\{h\in H^1(0,1):h(0)=0\} E = { h ∈ H 1 ( 0 , 1 ) : h ( 0 ) = 0 }
q D N ( h ) = ∫ 0 1 ( ∣ h ′ ∣ 2 − p V p p − 1 h 2 ) d t . q_{\mathrm{DN}}(h)=\int_0^1\left(|h'|^2-pV_p^{p-1}h^2\right)\,dt. q DN ( h ) = ∫ 0 1 ( ∣ h ′ ∣ 2 − p V p p − 1 h 2 ) d t . The associated lowest eigenvalue is the infimum of q D N ( h ) / ∥ h ∥ L 2 2 q_{\mathrm{DN}}(h)/\|h\|_{L^2}^2 q DN ( h ) /∥ h ∥ L 2 2 h ∈ E h\in E h ∈ E A D N A_{\mathrm{DN}} A DN A D A_D A D
{ h ∈ H 2 ( 0 , 1 ) : h ( 0 ) = 0 , h ′ ( 1 ) = 0 } . \{h\in H^2(0,1):h(0)=0,\ h'(1)=0\}. { h ∈ H 2 ( 0 , 1 ) : h ( 0 ) = 0 , h ′ ( 1 ) = 0 } . Thus its eigenvalue problem is (7) . The following lemma establishes a strict separation between the principal mixed and Dirichlet eigenvalues.
Lemma 3.6. The principal mixed eigenvalue is simple, has a positive eigenfunction, and satisfies
ν D N < ν D < 0 , g p : = ν D − ν D N > 0. (47) \nu_{\mathrm{DN}}<\nu_D<0,\qquad g_p:=\nu_D-\nu_{\mathrm{DN}}>0. \tag*{(47)} ν DN < ν D < 0 , g p := ν D − ν DN > 0. (47) If ϕ p \phi_p ϕ p ϕ p ( 1 ) = 1 \phi_p(1)=1 ϕ p ( 1 ) = 1
κ p 2 = − ν D N , A D + κ p 2 ≥ g p Id \kappa_p^2=-\nu_{\mathrm{DN}},\qquad A_D+\kappa_p^2\geq g_p\operatorname{Id} κ p 2 = − ν DN , A D + κ p 2 ≥ g p Id in the Dirichlet quadratic-form sense.
Proof. Put q ( t ) = p V p ( t ) p − 1 q(t)=pV_p(t)^{p-1} q ( t ) = p V p ( t ) p − 1 ∥ h ∥ L 2 = 1 \|h\|_{L^2}=1 ∥ h ∥ L 2 = 1
q D N ( h ) ≥ ∥ h ′ ∥ L 2 2 − ∥ q ∥ L ∞ when ∥ h ∥ L 2 = 1 , q_{\mathrm{DN}}(h)\geq\|h'\|_{L^2}^2-\|q\|_{L^\infty}\qquad\text{when }\|h\|_{L^2}=1, q DN ( h ) ≥ ∥ h ′ ∥ L 2 2 − ∥ q ∥ L ∞ when ∥ h ∥ L 2 = 1 , a minimizing sequence is bounded in H 1 ( 0 , 1 ) H^1(0,1) H 1 ( 0 , 1 ) H 1 H^1 H 1 L 2 L^2 L 2 E E E L 2 L^2 L 2
We may replace this minimizer by its absolute value, which preserves both its norm and its energy. The resulting nonnegative minimizer ϕ \phi ϕ
∫ 0 1 ( ϕ ′ h ′ − q ϕ h ) d t = ν D N ∫ 0 1 ϕ h d t ( h ∈ E ) . \int_0^1(\phi'h'-q\phi h)\,dt=\nu_{\mathrm{DN}}\int_0^1\phi h\,dt\qquad(h\in E). ∫ 0 1 ( ϕ ′ h ′ − q ϕ h ) d t = ν DN ∫ 0 1 ϕ h d t ( h ∈ E ) . Interior variations give − ϕ ′ ′ − q ϕ = ν D N ϕ -\phi''-q\phi=\nu_{\mathrm{DN}}\phi − ϕ ′′ − q ϕ = ν DN ϕ ϕ ∈ C 2 ( [ 0 , 1 ] ) \phi\in C^2([0,1]) ϕ ∈ C 2 ([ 0 , 1 ]) h ( 1 ) h(1) h ( 1 ) ϕ ′ ( 1 ) = 0 \phi'(1)=0 ϕ ′ ( 1 ) = 0 ϕ \phi ϕ ϕ ′ = 0 \phi'=0 ϕ ′ = 0 ϕ ≡ 0 \phi\equiv0 ϕ ≡ 0 ϕ ( 1 ) = 0 \phi(1)=0 ϕ ( 1 ) = 0 ϕ ′ ( 1 ) = 0 \phi'(1)=0 ϕ ′ ( 1 ) = 0 ϕ > 0 \phi>0 ϕ > 0 ( 0 , 1 ] (0,1] ( 0 , 1 ] ϕ ( 1 ) = 1 \phi(1)=1 ϕ ( 1 ) = 1 0 0 0 ϕ \phi ϕ
The Dirichlet form domain H 0 1 ( 0 , 1 ) H^1_0(0,1) H 0 1 ( 0 , 1 ) E E E ν D N ≤ ν D \nu_{\mathrm{DN}} \le\nu_{\mathrm{D}} ν DN ≤ ν D ψ \psi ψ h ∈ E h \in E h ∈ E ψ ′ ( 1 ) = 0 \psi'(1)=0 ψ ′ ( 1 ) = 0 ψ ( 1 ) = 0 \psi(1)=0 ψ ( 1 ) = 0 ψ ≡ 0 \psi\equiv0 ψ ≡ 0 (47) and makes κ p \kappa_p κ p
Finally, for every h ∈ H 0 1 ( 0 , 1 ) h \in H^1_0(0,1) h ∈ H 0 1 ( 0 , 1 )
∫ 0 1 ( ∣ h ′ ∣ 2 − p V p p − 1 h 2 + κ p 2 h 2 ) d t ≥ ( ν D − ν D N ) ∥ h ∥ L 2 2 = g p ∥ h ∥ L 2 2 . \int_0^1 \left( |h'|^2-pV_p^{p-1}h^2+\kappa_p^2h^2 \right)\,dt \ge(\nu_{\mathrm{D}}-\nu_{\mathrm{DN}})\lVert h\rVert_{L^2}^2=g_p\lVert h\rVert_{L^2}^2. ∫ 0 1 ( ∣ h ′ ∣ 2 − p V p p − 1 h 2 + κ p 2 h 2 ) d t ≥ ( ν D − ν DN ) ∥ h ∥ L 2 2 = g p ∥ h ∥ L 2 2 . This is the asserted strictly positive Dirichlet form bound.
The gap g p g_p g p A D N + κ p 2 A_{\mathrm{DN}}+\kappa_p^2 A DN + κ p 2 A D + κ p 2 A_{\mathrm{D}}+\kappa_p^2 A D + κ p 2 g p Id g_p\operatorname{Id} g p Id
The state equation and the Bernoulli equation
For the remainder of the construction, the trivial state is U R − λ U_{R-\lambda} U R − λ (39) . We now set up the equations near one such state on A ∗ ; A_*; A ∗ ; R ∗ = R K R_*=R_K R ∗ = R K
Choose a smooth radial cutoff χ \chi χ λ \lambda λ R ∗ R_* R ∗ ( R , η ) (R,\eta) ( R , η ) ( R ∗ , 0 ) (R_*,0) ( R ∗ , 0 )
q R , η ( r , θ ) = r + χ ( r ) ( R − R ∗ + η ( θ ) ) , T R , η ( r , θ ) = ( q R , η ( r , θ ) , θ ) . (48) q_{R,\eta}(r,\theta)=r+\chi(r)(R-R_*+\eta(\theta)),\qquad T_{R,\eta}(r,\theta)=(q_{R,\eta}(r,\theta),\theta). \tag*{(48)} q R , η ( r , θ ) = r + χ ( r ) ( R − R ∗ + η ( θ )) , T R , η ( r , θ ) = ( q R , η ( r , θ ) , θ ) . (48) The smallness condition makes q r > 0 q_r>0 q r > 0 R + η > λ R+\eta>\lambda R + η > λ T R , η T_{R,\eta} T R , η A ∗ A_* A ∗ Ω R , η ∖ B λ ‾ \Omega_{R,\eta}\setminus\overline{B_\lambda} Ω R , η ∖ B λ q = q R , η q=q_{R,\eta} q = q R , η
D q = ∂ θ − q θ q r ∂ r . D_q=\partial_\theta-\frac{q_\theta}{q_r}\partial_r. D q = ∂ θ − q r q θ ∂ r . The chain rule in polar coordinates gives
Δ R , η p v = 1 q r ∂ r ( v r q r ) + v r q q r + 1 q 2 D q 2 v . (49) \Delta^p_{R,\eta}v=\frac{1}{q_r}\partial_r\left(\frac{v_r}{q_r}\right)+\frac{v_r}{q q_r}+\frac{1}{q^2}D_q^2v. \tag*{(49)} Δ R , η p v = q r 1 ∂ r ( q r v r ) + q q r v r + q 2 1 D q 2 v . (49) Thus the Dirichlet equation for this specific power law is
E p ( R , η , w ) = − Δ R , η p ( b ∗ + w ) + ( b ∗ + w ) − ( b ∗ + w ) p = 0 , (50) \mathcal{E}_p(R,\eta,w)=-\Delta^p_{R,\eta}(b_*+w)+(b_*+w)-(b_*+w)^p=0, \tag*{(50)} E p ( R , η , w ) = − Δ R , η p ( b ∗ + w ) + ( b ∗ + w ) − ( b ∗ + w ) p = 0 , (50) where w ∈ Z p , K w\in Z_{p,K} w ∈ Z p , K b ∗ + w b_*+w b ∗ + w C 2 , α C^{2,\alpha} C 2 , α U R ∗ − λ U_{R_*-\lambda} U R ∗ − λ W p , K W_{p,K} W p , K (49) , multiplication in Hölder spaces, and Lemma 3.1 show that E p \mathcal{E}_p E p C ∞ C^\infty C ∞ R R R η \eta η q r > 0 q_r>0 q r > 0
At the trivial state, w ∗ = U R ∗ − λ − b ∗ w_*=U_{R_*-\lambda}-b_* w ∗ = U R ∗ − λ − b ∗
D w E p ( R ∗ , 0 , w ∗ ) w = − Δ w + ( 1 − p U R ∗ − λ p − 1 ) w , w ∈ Z p , K . (51) D_w\mathcal{E}_p(R_*,0,w_*)w=-\Delta w+(1-pU_{R_*-\lambda}^{p-1})w,\qquad w\in Z_{p,K}. \tag*{(51)} D w E p ( R ∗ , 0 , w ∗ ) w = − Δ w + ( 1 − p U R ∗ − λ p − 1 ) w , w ∈ Z p , K . (51) The potential belongs to C 0 , α C^{0,\alpha} C 0 , α Z p , K → W p , K Z_{p,K} \to W_{p,K} Z p , K → W p , K [7 ] and uniqueness, and multiplication by this potential is a compact perturbation, using the compact embedding C 2 , α ↪ C 0 , α C^{2,\alpha} \hookrightarrow C^{0,\alpha} C 2 , α ↪ C 0 , α
Proposition 3.7. Suppose that the Dirichlet operator in (3.20) has trivial kernel in Z p , K Z_{p,K} Z p , K
( R , η ) ⟼ v R , η = b ∗ + w ( R , η ) (R,\eta) \longmapsto v_{R,\eta}=b_*+w(R,\eta) ( R , η ) ⟼ v R , η = b ∗ + w ( R , η ) which is positive in A ∗ A_* A ∗ C ∞ C^\infty C ∞ C 2 , α ( A ∗ ‾ ) C^{2,\alpha}(\overline{A_*}) C 2 , α ( A ∗ ) η = 0 \eta=0 η = 0 U R − λ ∘ T R , 0 U_{R-\lambda}\circ T_{R,0} U R − λ ∘ T R , 0
Proof. The Fredholm statement and the kernel assumption make (3.20) a bounded isomorphism. The implicit function theorem applied to (3.19) gives the state map and uniqueness. The neighborhood is chosen small enough that Lemma 3.1 guarantees positivity and a negative outer normal derivative. The constructed radial family solves the same equation when η = 0 \eta=0 η = 0 □ \square □
Set u R , η = v R , η ∘ T R , η − 1 u_{R,\eta}=v_{R,\eta}\circ T^{-1}_{R,\eta} u R , η = v R , η ∘ T R , η − 1 χ = 1 \chi=1 χ = 1 R ∗ R_* R ∗
N p ( R , η ) ( θ ) = 1 + η θ ( θ ) 2 ( R + η ( θ ) ) 2 ∂ r v R , η ( R ∗ , θ ) . (52) N_{p}(R,\eta)(\theta)=\sqrt{1+\frac{\eta_\theta(\theta)^2}{(R+\eta(\theta))^2}}\,\partial_r v_{R,\eta}(R_*,\theta). \tag*{(52)} N p ( R , η ) ( θ ) = 1 + ( R + η ( θ ) ) 2 η θ ( θ ) 2 ∂ r v R , η ( R ∗ , θ ) . (52) Indeed, q r = 1 q_r=1 q r = 1 u θ ( R + η , θ ) = − η θ u r ( R + η , θ ) u_\theta(R+\eta,\theta)=-\eta_\theta u_r(R+\eta,\theta) u θ ( R + η , θ ) = − η θ u r ( R + η , θ )
F p , K ( R , η ) = Π ( N p ( R , η ) 2 ) = 0 , Π g = g − g ‾ . (53) F_{p,K}(R,\eta)=\Pi\left(N_p(R,\eta)^2\right)=0,\qquad\Pi g=g-\overline{g}. \tag*{(53)} F p , K ( R , η ) = Π ( N p ( R , η ) 2 ) = 0 , Π g = g − g . (53) The trace estimate and (3.21) imply that F p , K F_{p,K} F p , K C ∞ C^\infty C ∞ ( R ∗ , 0 ) (R_*,0) ( R ∗ , 0 ) R × X K \mathbb{R}\times X_K R × X K Y K Y_K Y K F p , K ( R , 0 ) = 0 F_{p,K}(R,0)=0 F p , K ( R , 0 ) = 0
Q = N p ( R , η ) 2 ‾ > 0. (54) Q=\overline{N_p(R,\eta)^2}>0. \tag*{(54)} Q = N p ( R , η ) 2 > 0. (54) The strict inequality follows from the negative outer derivative at the trivial state and continuity. The map is equivariant under reflection and rotation, including the half-period rotation η ( θ ) ↦ η ( θ + π / K ) \eta(\theta)\mapsto\eta(\theta+\pi/K) η ( θ ) ↦ η ( θ + π / K ) X K X_K X K
Linearization, the boundary symbol, and the range
We set A λ , R = B R ∖ B λ ‾ A_{\lambda,R}=B_R\setminus\overline{B_\lambda} A λ , R = B R ∖ B λ U = U R − λ U=U_{R-\lambda} U = U R − λ d = U ′ ( R ) < 0 d=U'(R)<0 d = U ′ ( R ) < 0 D K D_K D K U ( R ) = 0 U(R)=0 U ( R ) = 0
U ′ ′ ( R ) = − d R . (55) U''(R)=-\frac{d}{R}. \tag*{(55)} U ′′ ( R ) = − R d . (55) For η = ϵ ξ \eta=\epsilon\xi η = ϵ ξ v ˙ \dot{v} v ˙ R R R u ′ = v ˙ − χ ( r ) U ′ ( r ) ξ ( θ ) u'=\dot{v}-\chi(r)U'(r)\xi(\theta) u ′ = v ˙ − χ ( r ) U ′ ( r ) ξ ( θ )
{ ( − Δ + 1 − p U p − 1 ) u ′ = 0 in A λ , R , u ′ = 0 on r = λ , u ′ = − d ξ on r = R . (56) \begin{cases}
(-\Delta+ 1 - pU^{p-1})u' = 0 & \text{in } A_{\lambda,R},\\
u' = 0 & \text{on } r = \lambda,\\
u' = -d\xi& \text{on } r = R. \tag*{(56)}
\end{cases} ⎩ ⎨ ⎧ ( − Δ + 1 − p U p − 1 ) u ′ = 0 u ′ = 0 u ′ = − d ξ in A λ , R , on r = λ , on r = R . (56) The material derivative has zero boundary traces; the nonzero outer trace of u ′ u' u ′ (52) , or differentiating the physical normal, yields
D η N p ( R , 0 ) [ ξ ] = ∂ r u ′ ( R , θ ) + U ′ ′ ( R ) ξ ( θ ) . (57) D_\eta N_p(R,0)[\xi] = \partial_r u'(R,\theta) + U''(R)\xi(\theta). \tag*{(57)} D η N p ( R , 0 ) [ ξ ] = ∂ r u ′ ( R , θ ) + U ′′ ( R ) ξ ( θ ) . (57) The normal-vector variation is tangential and has zero scalar product with the radial gradient at the trivial state.
For j = n K j=nK j = n K n ≥ 1 n \ge1 n ≥ 1 H j = H j ( R ; ⋅ ) H_j=H_j(R;\cdot) H j = H j ( R ; ⋅ )
− H j ′ ′ − 1 r H j ′ + ( j 2 r 2 + 1 − p U p − 1 ) H j = 0 , H j ( λ ) = 0 , H j ( R ) = 1. (58) -H_j''-\frac{1}{r}H_j' + \left(\frac{j^2}{r^2}+1-pU^{p-1}\right)H_j=0,\qquad H_j(\lambda)=0,\qquad H_j(R)=1. \tag*{(58)} − H j ′′ − r 1 H j ′ + ( r 2 j 2 + 1 − p U p − 1 ) H j = 0 , H j ( λ ) = 0 , H j ( R ) = 1. (58) Dirichlet invertibility gives existence and uniqueness. Substitution of ξ = cos ( j θ ) \xi=\cos(j\theta) ξ = cos ( j θ ) (56) gives u ′ = − d H j ( r ) cos ( j θ ) u'=-dH_j(r)\cos(j\theta) u ′ = − d H j ( r ) cos ( j θ )
T j ( R ) = H j ′ ( R ) + 1 R . (59) T_j(R)=H_j'(R)+\frac{1}{R}. \tag*{(59)} T j ( R ) = H j ′ ( R ) + R 1 . (59) Equations (55) –(57) then give
D η F p , K ( R , 0 ) [ cos ( j θ ) ] = m j ( R ) cos ( j θ ) , m j ( R ) = − 2 d 2 T j ( R ) . (60) D_\eta F_{p,K}(R,0)[\cos(j\theta)]=m_j(R)\cos(j\theta),\qquad m_j(R)=-2d^2T_j(R). \tag*{(60)} D η F p , K ( R , 0 ) [ cos ( j θ )] = m j ( R ) cos ( j θ ) , m j ( R ) = − 2 d 2 T j ( R ) . (60) In particular, T j ( R ) = 0 T_j(R)=0 T j ( R ) = 0
[ − d 2 d r 2 − 1 r d d r + j 2 r 2 + 1 − p U p − 1 ] h = 0 , h ( λ ) = 0 , h ′ ( R ) + R − 1 h ( R ) = 0 (61) \left[-\frac{d^2}{dr^2}-\frac{1}{r}\frac{d}{dr}+\frac{j^2}{r^2}+1-pU^{p-1}\right]h=0,\qquad h(\lambda)=0,\qquad h'(R)+R^{-1}h(R)=0 \tag*{(61)} [ − d r 2 d 2 − r 1 d r d + r 2 j 2 + 1 − p U p − 1 ] h = 0 , h ( λ ) = 0 , h ′ ( R ) + R − 1 h ( R ) = 0 (61) has a nonzero solution. The Robin coefficient is produced by the radial equation at the outer boundary.
Let Λ p \Lambda_p Λ p − Δ + 1 − p U p − 1 -\Delta+1-pU^{p-1} − Δ + 1 − p U p − 1
L p , K : = D η F p , K ( R , 0 ) = − 2 d 2 ( Λ p + R − 1 Id ) : X K ⟶ Y K . L_{p,K}:=D_\eta F_{p,K}(R,0)=-2d^2(\Lambda_p+R^{-1}\operatorname{Id}):X_K\longrightarrow Y_K. L p , K := D η F p , K ( R , 0 ) = − 2 d 2 ( Λ p + R − 1 Id ) : X K ⟶ Y K . The following lemma establishes the Fredholm property and characterizes the range for the power law.
Lemma 3.8. The operator L p , K : X K → Y K L_{p,K}:X_K\to Y_K L p , K : X K → Y K T K ( R ) = 0 T_K(R)=0 T K ( R ) = 0 T n K ( R ) ≠ 0 T_{nK}(R)\ne0 T n K ( R ) = 0 n ≥ 2 n\ge2 n ≥ 2
Ker L p , K = span { cos ( K θ ) } , \operatorname{Ker}L_{p,K}=\operatorname{span}\{\cos(K\theta)\}, Ker L p , K = span { cos ( K θ )} , Ran L p , K = { g ∈ Y K : ∫ 0 2 π g ( θ ) cos ( K θ ) d θ = 0 } . \operatorname{Ran}L_{p,K}=\left\{g\in Y_K:\int_0^{2\pi}g(\theta)\cos(K\theta)\,d\theta=0\right\}. Ran L p , K = { g ∈ Y K : ∫ 0 2 π g ( θ ) cos ( K θ ) d θ = 0 } . Proof. We write c ( r ) = 1 − p U ( r ) p − 1 c(r)=1-pU(r)^{p-1} c ( r ) = 1 − p U ( r ) p − 1 L p = − Δ + c \mathcal{L}_p=-\Delta+c L p = − Δ + c α < p − 1 \alpha<p-1 α < p − 1 c ∈ C 0 , α ( A λ , R ‾ ) c\in C^{0,\alpha}(\overline{A_{\lambda,R}}) c ∈ C 0 , α ( A λ , R ) P p ξ P_p\xi P p ξ
L p ( P p ξ ) = 0 in A λ , R , P p ξ ∣ r = λ = 0 , P p ξ ∣ r = R = ξ . \mathcal{L}_p(P_p\xi)=0\quad\text{in }A_{\lambda,R},\qquad P_p\xi|_{r=\lambda}=0,\qquad P_p\xi|_{r=R}=\xi. L p ( P p ξ ) = 0 in A λ , R , P p ξ ∣ r = λ = 0 , P p ξ ∣ r = R = ξ . Uniqueness preserves the D K D_K D K Λ p ξ = ∂ r ( P p ξ ) ( R , ⋅ ) \Lambda_p\xi=\partial_r(P_p\xi)(R,\cdot) Λ p ξ = ∂ r ( P p ξ ) ( R , ⋅ ) X K X_K X K J : X K ↪ Y K J:X_K\hookrightarrow Y_K J : X K ↪ Y K c c c ∂ r v + R − 1 v = g \partial_r v+R^{-1}v=g ∂ r v + R − 1 v = g c c c d S = R d θ dS=R\,d\theta d S = R d θ d θ d\theta d θ Λ p + R − 1 J \Lambda_p+R^{-1}J Λ p + R − 1 J L 2 ( d θ ) L^2(d\theta) L 2 ( d θ ) Y K Y_K Y K − 2 d 2 -2d^2 − 2 d 2 L p , K L_{p,K} L p , K
Taking angular Fourier coefficients in the extension equation and using Dirichlet uniqueness gives, for j = n K j=nK j = n K
1 π ∫ 0 2 π ( L p , K η ) ( θ ) cos ( j θ ) d θ = m j ( R ) 1 π ∫ 0 2 π η ( θ ) cos ( j θ ) d θ . \frac{1}{\pi}\int_0^{2\pi}(L_{p,K}\eta)(\theta)\cos(j\theta)\,d\theta=m_j(R)\frac{1}{\pi}\int_0^{2\pi}\eta(\theta)\cos(j\theta)\,d\theta. π 1 ∫ 0 2 π ( L p , K η ) ( θ ) cos ( j θ ) d θ = m j ( R ) π 1 ∫ 0 2 π η ( θ ) cos ( j θ ) d θ . Under the stated hypotheses, only the K K K Ker L p , K = span { cos ( K θ ) } \operatorname{Ker}L_{p,K}=\operatorname{span}\{\cos(K\theta)\} Ker L p , K = span { cos ( K θ )}
For later use, the principal eigenvalue ζ j ( R ) \zeta_j(R) ζ j ( R ) (61) has the Rayleigh quotient
ζ j ( R ) = min h ∈ H 1 ( λ , R ) ∖ { 0 } h ( λ ) = 0 ∫ λ R ( r ∣ h ′ ∣ 2 + j 2 r h 2 + r ( 1 − p U p − 1 ) h 2 ) d r + h ( R ) 2 ∫ λ R r h 2 d r (62) \zeta_j(R)=\min_{\substack{h\in H^1(\lambda,R)\setminus\{0\}\\h(\lambda)=0}}\frac{\int_\lambda^R\left(r|h'|^2+\frac{j^2}{r}h^2+r(1-pU^{p-1})h^2\right)\,dr+h(R)^2}{\int_\lambda^R rh^2\,dr} \tag*{(62)} ζ j ( R ) = h ∈ H 1 ( λ , R ) ∖ { 0 } h ( λ ) = 0 min ∫ λ R r h 2 d r ∫ λ R ( r ∣ h ′ ∣ 2 + r j 2 h 2 + r ( 1 − p U p − 1 ) h 2 ) d r + h ( R ) 2 (62) The positive endpoint term gives the Robin coefficient R − 1 R^{-1} R − 1
ζ j ( R ) ≥ ζ ℓ ( R ) + j 2 − ℓ 2 R 2 > ζ ℓ ( R ) , j > ℓ . (63) \zeta_j(R)\geq\zeta_\ell(R)+\frac{j^2-\ell^2}{R^2}>\zeta_\ell(R),\qquad j>\ell. \tag*{(63)} ζ j ( R ) ≥ ζ ℓ ( R ) + R 2 j 2 − ℓ 2 > ζ ℓ ( R ) , j > ℓ . (63) Thus a zero of the principal K K K D K D_K D K R R R
Verification of the spectral and transversality conditions
We now construct the critical radii for the power law. We fix the angular frequency K K K K δ / λ K\delta/\lambda K δ / λ U R − λ U_{R-\lambda} U R − λ
The scaled mixed operator
For the radial solution U δ U_\delta U δ j j j
L δ , j = − d 2 d r 2 − 1 r d d r + j 2 r 2 + 1 − p U δ p − 1 . \mathcal{L}_{\delta,j}=-\frac{d^2}{dr^2}-\frac{1}{r}\frac{d}{dr}+\frac{j^2}{r^2}+1-pU_\delta^{p-1}. L δ , j = − d r 2 d 2 − r 1 d r d + r 2 j 2 + 1 − p U δ p − 1 . Set
ε = 1 K , ρ = K δ λ , δ = λ ε ρ , w ε , ρ ( t ) = 1 + ε ρ t . (64) \varepsilon=\frac{1}{K},\qquad\rho=\frac{K\delta}{\lambda},\qquad\delta=\lambda\varepsilon\rho,\qquad w_{\varepsilon,\rho}(t)=1+\varepsilon\rho t. \tag*{(64)} ε = K 1 , ρ = λ K δ , δ = λ ερ , w ε , ρ ( t ) = 1 + ερt . (64) We use t = ( r − λ ) / δ t=(r-\lambda)/\delta t = ( r − λ ) / δ w = w ε , ρ w=w_{\varepsilon,\rho} w = w ε , ρ r = λ w r=\lambda w r = λ w d / d r = δ − 1 d / d t d/dr=\delta^{-1}d/dt d / d r = δ − 1 d / d t δ 2 U δ p − 1 = V δ p − 1 \delta^2U_\delta^{p-1}=V_\delta^{p-1} δ 2 U δ p − 1 = V δ p − 1 j = K j=K j = K δ 2 K 2 / r 2 = ρ 2 / w 2 \delta^2K^2/r^2=\rho^2/w^2 δ 2 K 2 / r 2 = ρ 2 / w 2 δ 2 \delta^2 δ 2
A ε , ρ h = − 1 w ( w h ′ ) ′ + ( ρ 2 w 2 + λ 2 ε 2 ρ 2 − p V λ ε ρ p − 1 ) h , (65) A_{\varepsilon,\rho}h=-\frac{1}{w}(wh')'+\left(\frac{\rho^2}{w^2}+\lambda^2\varepsilon^2\rho^2-pV_{\lambda\varepsilon\rho}^{p-1}\right)h, \tag*{(65)} A ε , ρ h = − w 1 ( w h ′ ) ′ + ( w 2 ρ 2 + λ 2 ε 2 ρ 2 − p V λ ερ p − 1 ) h , (65) with mixed boundary conditions
h ( 0 ) = 0 , h ′ ( 1 ) + ε ρ 1 + ε ρ h ( 1 ) = 0. (66) h(0)=0,\qquad h'(1)+\frac{\varepsilon\rho}{1+\varepsilon\rho}h(1)=0. \tag*{(66)} h ( 0 ) = 0 , h ′ ( 1 ) + 1 + ερ ερ h ( 1 ) = 0. (66) Indeed, the original Robin condition is δ − 1 h ′ ( 1 ) + R − 1 h ( 1 ) = 0 \delta^{-1}h'(1)+R^{-1}h(1)=0 δ − 1 h ′ ( 1 ) + R − 1 h ( 1 ) = 0 δ / R = ε ρ / ( 1 + ε ρ ) \delta/R=\varepsilon\rho/(1+\varepsilon\rho) δ / R = ερ / ( 1 + ερ ) (65) –(66) denote derivatives with respect to t t t
We use the fixed form domain E = { h ∈ H 1 ( 0 , 1 ) : h ( 0 ) = 0 } E=\{h\in H^1(0,1):h(0)=0\} E = { h ∈ H 1 ( 0 , 1 ) : h ( 0 ) = 0 }
n ε , ρ ( h ) = ∫ 0 1 w h 2 d t . n_{\varepsilon,\rho}(h)=\int_0^1 wh^2\,dt. n ε , ρ ( h ) = ∫ 0 1 w h 2 d t . The corresponding quadratic form is
q ε , ρ ( h ) = ∫ 0 1 w { ∣ h ′ ∣ 2 + ( ρ 2 w 2 + λ 2 ε 2 ρ 2 − p V λ ε ρ p − 1 ) h 2 } d t + ε ρ h ( 1 ) 2 . (67) q_{\varepsilon,\rho}(h)=\int_0^1w\left\{|h'|^2+\left(\frac{\rho^2}{w^2}+\lambda^2\varepsilon^2\rho^2-pV_{\lambda\varepsilon\rho}^{p-1}\right)h^2\right\}\,dt+\varepsilon\rho h(1)^2. \tag*{(67)} q ε , ρ ( h ) = ∫ 0 1 w { ∣ h ′ ∣ 2 + ( w 2 ρ 2 + λ 2 ε 2 ρ 2 − p V λ ερ p − 1 ) h 2 } d t + ερ h ( 1 ) 2 . (67) To check the endpoint term, integration by parts gives
∫ 0 1 w h A ε , ρ h d t = ∫ 0 1 w { ∣ h ′ ∣ 2 + ( ρ 2 w 2 + λ 2 ε 2 ρ 2 − p V λ ε ρ p − 1 ) h 2 } d t − w ( 1 ) h ( 1 ) h ′ ( 1 ) . \int_0^1whA_{\varepsilon,\rho}h\,dt=\int_0^1w\left\{|h'|^2+\left(\frac{\rho^2}{w^2}+\lambda^2\varepsilon^2\rho^2-pV_{\lambda\varepsilon\rho}^{p-1}\right)h^2\right\}\,dt-w(1)h(1)h'(1). ∫ 0 1 w h A ε , ρ h d t = ∫ 0 1 w { ∣ h ′ ∣ 2 + ( w 2 ρ 2 + λ 2 ε 2 ρ 2 − p V λ ερ p − 1 ) h 2 } d t − w ( 1 ) h ( 1 ) h ′ ( 1 ) . The left endpoint contributes zero because h ( 0 ) = 0 h(0)=0 h ( 0 ) = 0 ε ρ h ( 1 ) 2 \varepsilon\rho h(1)^2 ερ h ( 1 ) 2 w ( 1 ) w(1) w ( 1 ) ε ρ \varepsilon\rho ερ
Denote by e ε ( ρ ) e_\varepsilon(\rho) e ε ( ρ ) q ε , ρ / n ε , ρ q_{\varepsilon,\rho}/n_{\varepsilon,\rho} q ε , ρ / n ε , ρ E ∖ { 0 } E\setminus\{0\} E ∖ { 0 } E ↪ L 2 ( 0 , 1 ) E\hookrightarrow L^2(0,1) E ↪ L 2 ( 0 , 1 ) ε = 0 \varepsilon=0 ε = 0
q 0 , ρ ( h ) = ∫ 0 1 ( ∣ h ′ ∣ 2 + ( ρ 2 − p V p p − 1 ) h 2 ) d t . q_{0,\rho}(h)=\int_0^1\left(|h'|^2+(\rho^2-pV_p^{p-1})h^2\right)\,dt. q 0 , ρ ( h ) = ∫ 0 1 ( ∣ h ′ ∣ 2 + ( ρ 2 − p V p p − 1 ) h 2 ) d t . Its natural endpoint condition is h ′ ( 1 ) = 0 h'(1)=0 h ′ ( 1 ) = 0 ρ 2 \rho^2 ρ 2 ρ 2 \rho^2 ρ 2
e 0 ( ρ ) = ν D N + ρ 2 = ρ 2 − κ p 2 . (68) e_0(\rho)=\nu_{\mathrm{DN}}+\rho^2=\rho^2-\kappa_p^2. \tag*{(68)} e 0 ( ρ ) = ν DN + ρ 2 = ρ 2 − κ p 2 . (68) Fix a compact interval J ⋐ ( 0 , ∞ ) J\Subset(0,\infty) J ⋐ ( 0 , ∞ ) κ p \kappa_p κ p ρ ∈ J \rho\in J ρ ∈ J
Lemma 3.9. As ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0
e ε ( ρ ) ⟶ ρ 2 − κ p 2 , e ε ′ ( ρ ) ⟶ 2 ρ uniformly on J . (69) e_\varepsilon(\rho)\longrightarrow\rho^2-\kappa_p^2,\qquad e_\varepsilon'(\rho)\longrightarrow2\rho\quad\text{uniformly on }J. \tag*{(69)} e ε ( ρ ) ⟶ ρ 2 − κ p 2 , e ε ′ ( ρ ) ⟶ 2 ρ uniformly on J . (69) Proof. Since δ = λ ε ρ \delta=\lambda_{\varepsilon}\rho δ = λ ε ρ w → 1 w\to1 w → 1 ε ρ → 0 \varepsilon\rho\to0 ερ → 0 ∣ h ( 1 ) ∣ 2 ≤ C ∥ h ∥ H 1 2 |h(1)|^2\le C\|h\|_{H^1}^2 ∣ h ( 1 ) ∣ 2 ≤ C ∥ h ∥ H 1 2 ω ε ↓ 0 \omega_\varepsilon\downarrow0 ω ε ↓ 0
∣ q ε , ρ ( h ) − q 0 , ρ ( h ) ∣ ≤ ω ε ∥ h ∥ H 1 2 , ∣ n ε , ρ ( h ) − ∥ h ∥ 2 2 ∣ ≤ C ε ∥ h ∥ 2 2 . |q_{\varepsilon,\rho}(h)-q_{0,\rho}(h)|\le\omega_\varepsilon\|h\|_{H^1}^2,\qquad\left|n_{\varepsilon,\rho}(h)-\|h\|_2^2\right|\le C\varepsilon\|h\|_2^2. ∣ q ε , ρ ( h ) − q 0 , ρ ( h ) ∣ ≤ ω ε ∥ h ∥ H 1 2 , n ε , ρ ( h ) − ∥ h ∥ 2 2 ≤ C ε ∥ h ∥ 2 2 . The bounded potentials and the positive weights also imply
q ε , ρ ( h ) ≥ c ∥ h ′ ∥ 2 2 − C ∥ h ∥ 2 2 , c > 0 , q_{\varepsilon,\rho}(h)\ge c\|h'\|_2^2-C\|h\|_2^2,\qquad c>0, q ε , ρ ( h ) ≥ c ∥ h ′ ∥ 2 2 − C ∥ h ∥ 2 2 , c > 0 , with constants independent of small ε \varepsilon ε ρ \rho ρ e ε ( ρ ) ≤ e 0 ( ρ ) + o ( 1 ) e_\varepsilon(\rho)\le e_0(\rho)+o(1) e ε ( ρ ) ≤ e 0 ( ρ ) + o ( 1 ) ϕ ε , ρ \phi_{\varepsilon,\rho} ϕ ε , ρ n ε , ρ ( ϕ ε , ρ ) = 1 n_{\varepsilon,\rho}(\phi_{\varepsilon,\rho})=1 n ε , ρ ( ϕ ε , ρ ) = 1 H 1 H^1 H 1 q 0 , ρ ( h ) ≥ e 0 ( ρ ) ∥ h ∥ 2 2 q_{0,\rho}(h)\ge e_0(\rho)\|h\|_2^2 q 0 , ρ ( h ) ≥ e 0 ( ρ ) ∥ h ∥ 2 2 o ( 1 ) o(1) o ( 1 )
We next justify differentiation of the eigenvalue. For fixed ε > 0 \varepsilon>0 ε > 0 V λ ε ρ ( t ) = ( 1 − t ) z ρ ( t ) V_{\lambda_\varepsilon\rho}(t)=(1-t)z_\rho(t) V λ ε ρ ( t ) = ( 1 − t ) z ρ ( t ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ρ \rho ρ C 1 , α C^{1,\alpha} C 1 , α
V λ ε ρ p − 1 = ( 1 − t ) p − 1 z ρ p − 1 V_{\lambda_\varepsilon\rho}^{p-1}=(1-t)^{p-1}z_\rho^{p-1} V λ ε ρ p − 1 = ( 1 − t ) p − 1 z ρ p − 1 depends smoothly on ρ \rho ρ C 0 , α C^{0,\alpha} C 0 , α p p p 1 < p < 2 1<p<2 1 < p < 2 t = 1 t=1 t = 1 E E E
For clarity, let q ε , ρ ( h , k ) q_{\varepsilon,\rho}(h,k) q ε , ρ ( h , k ) n ε , ρ ( h , k ) n_{\varepsilon,\rho}(h,k) n ε , ρ ( h , k )
q ε , ρ ( ϕ , k ) = e ε ( ρ ) n ε , ρ ( ϕ , k ) , k ∈ E . q_{\varepsilon,\rho}(\phi,k)=e_\varepsilon(\rho)n_{\varepsilon,\rho}(\phi,k),\qquad k\in E. q ε , ρ ( ϕ , k ) = e ε ( ρ ) n ε , ρ ( ϕ , k ) , k ∈ E . Differentiating q ( ϕ , ϕ ) = e ε n ( ϕ , ϕ ) q(\phi,\phi)=e_\varepsilon n(\phi,\phi) q ( ϕ , ϕ ) = e ε n ( ϕ , ϕ ) k = ∂ ρ ϕ k=\partial_\rho\phi k = ∂ ρ ϕ ∂ ρ ϕ \partial_\rho\phi ∂ ρ ϕ n ( ϕ , ϕ ) = 1 n(\phi,\phi)=1 n ( ϕ , ϕ ) = 1
e ε ′ ( ρ ) = ∂ ρ q ε , ρ ( ϕ ε , ρ ) − e ε ( ρ ) ∫ 0 1 ( ∂ ρ w ) ϕ ε , ρ 2 d t . (70) e_\varepsilon'(\rho)=\partial_\rho q_{\varepsilon,\rho}(\phi_{\varepsilon,\rho})-e_\varepsilon(\rho)\int_0^1(\partial_\rho w)\phi_{\varepsilon,\rho}^2\,dt. \tag*{(70)} e ε ′ ( ρ ) = ∂ ρ q ε , ρ ( ϕ ε , ρ ) − e ε ( ρ ) ∫ 0 1 ( ∂ ρ w ) ϕ ε , ρ 2 d t . (70) In particular, this identity includes the change in the weighted normalization. It also includes the Robin variation through the endpoint term in q q q
We now estimate each coefficient derivative explicitly. Put P ε , ρ = V λ ε ρ p − 1 P_{\varepsilon,\rho}=V_{\lambda_\varepsilon\rho}^{p-1} P ε , ρ = V λ ε ρ p − 1 ∂ ρ w = ε t \partial_\rho w=\varepsilon t ∂ ρ w = εt h h h
∂ ρ q ε , ρ ( h ) = ∫ 0 1 ε t ∣ h ′ ∣ 2 d t + ε h ( 1 ) 2 + ∫ 0 1 { 2 ρ w − ε ρ 2 t w 2 + λ 2 ε 2 ( 2 ρ w + ε ρ 2 t ) − p ε t P ε , ρ − p w ∂ ρ P ε , ρ } h 2 d t . \begin{aligned}
\partial_\rho q_{\varepsilon,\rho}(h)
={}&\int_0^1\varepsilon t|h'|^2\,dt+\varepsilon h(1)^2\\
&+\int_0^1\left\{\frac{2\rho}{w}-\frac{\varepsilon\rho^2t}{w^2}+\lambda^2\varepsilon^2(2\rho w+\varepsilon\rho^2t)-p\varepsilon tP_{\varepsilon,\rho}-pw\partial_\rho P_{\varepsilon,\rho}\right\}h^2\,dt.
\end{aligned} ∂ ρ q ε , ρ ( h ) = ∫ 0 1 εt ∣ h ′ ∣ 2 d t + ε h ( 1 ) 2 + ∫ 0 1 { w 2 ρ − w 2 ε ρ 2 t + λ 2 ε 2 ( 2 ρw + ε ρ 2 t ) − pεt P ε , ρ − pw ∂ ρ P ε , ρ } h 2 d t . The derivative of the profile is taken along the radial branch:
∂ ρ P ε , ρ = λ ε ∂ δ ( V δ p − 1 ) = 1 ρ ∂ δ ( V δ p − 1 ) . \partial_\rho P_{\varepsilon,\rho}=\lambda_\varepsilon\partial_\delta(V_\delta^{p-1})=\frac{1}{\rho}\partial_\delta(V_\delta^{p-1}). ∂ ρ P ε , ρ = λ ε ∂ δ ( V δ p − 1 ) = ρ 1 ∂ δ ( V δ p − 1 ) . Because J J J L ∞ L^\infty L ∞ 2 ρ 2\rho 2 ρ o ( 1 ) ∥ h ∥ H 1 2 o(1)\lVert h\rVert_{H^1}^2 o ( 1 ) ∥ h ∥ H 1 2
∂ ρ q ε , ρ ( h ) = 2 ρ ∥ h ∥ 2 2 + o ( 1 ) ∥ h ∥ H 1 2 . \partial_\rho q_{\varepsilon,\rho}(h)=2\rho\lVert h\rVert_2^2+o(1)\lVert h\rVert_{H^1}^2. ∂ ρ q ε , ρ ( h ) = 2 ρ ∥ h ∥ 2 2 + o ( 1 ) ∥ h ∥ H 1 2 . The normalized eigenfunctions have uniformly bounded H 1 H^1 H 1 w → 1 w\to1 w → 1 ∥ φ ε , ρ ∥ 2 2 → 1 \lVert\varphi_{\varepsilon,\rho}\rVert_2^2\to1 ∥ φ ε , ρ ∥ 2 2 → 1 O ( ε ) O(\varepsilon) O ( ε )
Choose 0 < b < κ p / 2 0<b<\kappa_p/2 0 < b < κ p /2 J = [ κ p − b , κ p + b ] J=[\kappa_p-b,\kappa_p+b] J = [ κ p − b , κ p + b ] 2 ρ 2\rho 2 ρ J J J ε \varepsilon ε e ε e_\varepsilon e ε J J J ρ ε ∈ J \rho_\varepsilon\in J ρ ε ∈ J
ρ ε ⟶ κ p , e ε ′ ( ρ ε ) > 0. (71) \rho_\varepsilon\longrightarrow\kappa_p,\qquad e_\varepsilon'(\rho_\varepsilon)>0. \tag*{(71)} ρ ε ⟶ κ p , e ε ′ ( ρ ε ) > 0. (71) Taking ε = 1 / K \varepsilon=1/K ε = 1/ K
δ K = λ ρ 1 / K K , R K = λ + δ K . (72) \delta_K=\frac{\lambda\rho_{1/K}}{K},\qquad R_K=\lambda+\delta_K. \tag*{(72)} δ K = K λ ρ 1/ K , R K = λ + δ K . (72) The original mixed eigenvalue is ζ K ( R ) = δ − 2 e 1 / K ( ρ ) \zeta_K(R)=\delta^{-2}e_{1/K}(\rho) ζ K ( R ) = δ − 2 e 1/ K ( ρ ) (9) follows immediately. The uniqueness just proved concerns the chosen radial family and the stated interval of the thin-annulus scaling parameter.
Dirichlet invertibility in the symmetry class
Let e ε D ( ρ ) e_\varepsilon^D(\rho) e ε D ( ρ ) h ( 0 ) = h ( 1 ) = 0 h(0)=h(1)=0 h ( 0 ) = h ( 1 ) = 0 H 0 1 ( 0 , 1 ) H_0^1(0,1) H 0 1 ( 0 , 1 )
e ε D ( ρ ) ⟶ ν D + ρ 2 . (73) e_\varepsilon^D(\rho)\longrightarrow\nu_D+\rho^2. \tag*{(73)} e ε D ( ρ ) ⟶ ν D + ρ 2 . (73) In particular, at ρ = ρ ε \rho=\rho_\varepsilon ρ = ρ ε ν D + κ p 2 = g p > 0 \nu_D+\kappa_p^2=g_p>0 ν D + κ p 2 = g p > 0
Proposition 3.10. For every sufficiently large K K K − Δ + 1 − p U K p − 1 -\Delta+1-pU_K^{p-1} − Δ + 1 − p U K p − 1 A λ , R K A_{\lambda,R_K} A λ , R K D K D_K D K
Proof. For the radial mode, the scaled operator is (3.11). Its first two Dirichlet eigenvalues converge to the first two eigenvalues of A D A_D A D δ K \delta_K δ K
For j = K j=K j = K ρ 1 / K → κ p \rho_{1/K}\to\kappa_p ρ 1/ K → κ p e 1 / K D ( ρ 1 / K ) ≥ g p / 2 e_{1/K}^D(\rho_{1/K})\ge g_p/2 e 1/ K D ( ρ 1/ K ) ≥ g p /2 K K K j = n K j=nK j = n K n ≥ 2 n\ge2 n ≥ 2
( n 2 − 1 ) ρ 1 / K 2 ( 1 + ρ 1 / K / K ) 2 ≥ 0. \frac{(n^2-1)\rho_{1/K}^2}{(1+\rho_{1/K}/K)^2}\ge0. ( 1 + ρ 1/ K / K ) 2 ( n 2 − 1 ) ρ 1/ K 2 ≥ 0. Its Rayleigh quotient is therefore at least that of the K K K g p / 2 g_p/2 g p /2 δ K − 2 \delta_K^{-2} δ K − 2
A D K D_K D K n K nK n K n ≥ 1 n \ge1 n ≥ 1 L 2 L^2 L 2 1 − p U K p − 1 1-pU_K^{p-1} 1 − p U K p − 1 C 0 , α C^{0,\alpha} C 0 , α U K U_K U K α \alpha α
The proposition concerns the symmetry class used in the bifurcation argument. This class excludes the lower angular modes 1 , … , K − 1 1,\ldots,K-1 1 , … , K − 1
Isolation of the boundary kernel
At R K R_K R K K K K R K R_K R K 1 1 1 R K R_K R K H K H_K H K T K ( R K ) = 0 T_K(R_K)=0 T K ( R K ) = 0
For every n ≥ 2 n \ge2 n ≥ 2
ζ n K ( R K ) ≥ ( n 2 − 1 ) K 2 R K 2 > 0. \zeta_{nK}(R_K) \ge\frac{(n^2-1)K^2}{R_K^2} > 0. ζ n K ( R K ) ≥ R K 2 ( n 2 − 1 ) K 2 > 0. Since the principal eigenvalue is already positive in each of these sectors, zero cannot be any of its higher mixed eigenvalues. Equivalently, T n K ( R K ) ≠ 0 T_{nK}(R_K) \ne0 T n K ( R K ) = 0
Transversality of the boundary operator
Fix a sufficiently large K K K ε = 1 / K \varepsilon=1/K ε = 1/ K ρ \rho ρ ρ 1 / K \rho_{1/K} ρ 1/ K z z z h z , ρ h_{z,\rho} h z , ρ
( A 1 / K , ρ − z ) h z , ρ = 0 , h z , ρ ( 0 ) = 0 , h z , ρ ( 1 ) = 1. (A_{1/K,\rho}-z)h_{z,\rho}=0,\qquad h_{z,\rho}(0)=0,\qquad h_{z,\rho}(1)=1. ( A 1/ K , ρ − z ) h z , ρ = 0 , h z , ρ ( 0 ) = 0 , h z , ρ ( 1 ) = 1. The positive Dirichlet gap persists in a neighborhood of the critical parameters. Subtracting a fixed lift of the endpoint values reduces this problem to an invertible zero-Dirichlet problem, which gives existence, uniqueness, and smooth dependence on ( z , ρ ) (z,\rho) ( z , ρ )
B ( z , ρ ) = h z , ρ ′ ( 1 ) + ρ / K 1 + ρ / K . (74) B(z,\rho)=h'_{z,\rho}(1)+\frac{\rho/K}{1+\rho/K}. \tag*{(74)} B ( z , ρ ) = h z , ρ ′ ( 1 ) + 1 + ρ / K ρ / K . (74) The radial solution and the coefficients of A 1 / K , ρ A_{1/K,\rho} A 1/ K , ρ ρ \rho ρ
T K ( R ) = 1 δ B ( 0 , ρ ) , ρ = K ( R − λ ) λ . (75) T_K(R)=\frac{1}{\delta}B(0,\rho),\qquad\rho=\frac{K(R-\lambda)}{\lambda}. \tag*{(75)} T K ( R ) = δ 1 B ( 0 , ρ ) , ρ = λ K ( R − λ ) . (75) Indeed, H K ′ ( R ) = δ − 1 h 0 , ρ ′ ( 1 ) H'_K(R)=\delta^{-1}h'_{0,\rho}(1) H K ′ ( R ) = δ − 1 h 0 , ρ ′ ( 1 ) δ / R = ( ρ / K ) / ( 1 + ρ / K ) \delta/R=(\rho/K)/(1+\rho/K) δ / R = ( ρ / K ) / ( 1 + ρ / K ) 1 1 1 B ( e 1 / K ( ρ ) , ρ ) = 0 B(e_{1/K}(\rho),\rho)=0 B ( e 1/ K ( ρ ) , ρ ) = 0
We first compute the sign of B z B_z B z h = h z , ρ h=h_{z,\rho} h = h z , ρ h ˙ = ∂ z h z , ρ \dot{h}=\partial_z h_{z,\rho} h ˙ = ∂ z h z , ρ
( A 1 / K , ρ − z ) h ˙ = h , h ˙ ( 0 ) = h ˙ ( 1 ) = 0. \left(A_{1/K,\rho}-z\right)\dot{h}=h,\qquad\dot{h}(0)=\dot{h}(1)=0. ( A 1/ K , ρ − z ) h ˙ = h , h ˙ ( 0 ) = h ˙ ( 1 ) = 0. Multiply this equation by w h wh w h h h h w h ˙ w\dot{h} w h ˙
∫ 0 1 w h 2 d t = ∫ 0 1 w ( h ( A − z ) h ˙ − h ˙ ( A − z ) h ) d t = [ w ( h h ˙ ′ − h ˙ h ′ ) ] 0 1 = − w ( 1 ) h ˙ ′ ( 1 ) . \begin{aligned}
\int_0^1 wh^2\,dt=\int_0^1 w\left(h(A-z)\dot{h}-\dot{h}(A-z)h\right)\,dt \\
&=\left[w\left(h\dot{h}'-\dot{h}h'\right)\right]_0^1=-w(1)\dot{h}'(1).
\end{aligned} ∫ 0 1 w h 2 d t = ∫ 0 1 w ( h ( A − z ) h ˙ − h ˙ ( A − z ) h ) d t = [ w ( h h ˙ ′ − h ˙ h ′ ) ] 0 1 = − w ( 1 ) h ˙ ′ ( 1 ) . Both contributions at t = 0 t=0 t = 0 h ( 0 ) = h ˙ ( 0 ) = 0 h(0)=\dot{h}(0)=0 h ( 0 ) = h ˙ ( 0 ) = 0 t = 1 t=1 t = 1 h ˙ ( 1 ) \dot{h}(1) h ˙ ( 1 ) h ( 1 ) = 1 h(1)=1 h ( 1 ) = 1 z z z
B z ( z , ρ ) = − ∫ 0 1 w h 2 d t w ( 1 ) < 0. (76) B_z(z,\rho)=-\frac{\int_0^1 wh^2\,dt}{w(1)}<0. \tag*{(76)} B z ( z , ρ ) = − w ( 1 ) ∫ 0 1 w h 2 d t < 0. (76) This calculation in fact applies to every ( z , ρ ) (z,\rho) ( z , ρ )
We next differentiate B ( e 1 / K ( ρ ) , ρ ) = 0 B(e_{1/K}(\rho),\rho)=0 B ( e 1/ K ( ρ ) , ρ ) = 0 e 1 / K ( ρ 1 / K ) = 0 e_{1/K}(\rho_{1/K})=0 e 1/ K ( ρ 1/ K ) = 0
B ρ ( 0 , ρ 1 / K ) = − B z ( 0 , ρ 1 / K ) e 1 / K ′ ( ρ 1 / K ) > 0. (77) B_\rho(0,\rho_{1/K})=-B_z(0,\rho_{1/K})e'_{1/K}(\rho_{1/K})>0. \tag*{(77)} B ρ ( 0 , ρ 1/ K ) = − B z ( 0 , ρ 1/ K ) e 1/ K ′ ( ρ 1/ K ) > 0. (77) Here B ρ B_\rho B ρ z z z h z , ρ ′ ( 1 ) h'_{z,\rho}(1) h z , ρ ′ ( 1 ) K − 1 ( 1 + ρ / K ) − 2 K^{-1}(1+\rho/K)^{-2} K − 1 ( 1 + ρ / K ) − 2 β = ( ρ / K ) / ( 1 + ρ / K ) \beta=(\rho/K)/(1+\rho/K) β = ( ρ / K ) / ( 1 + ρ / K ) ρ \rho ρ z z z
( A − z ) h ρ = − A ρ h , h ρ ( 0 ) = h ρ ( 1 ) = 0 , w ( 1 ) h ρ ′ ( 1 ) = ∫ 0 1 w h A ρ h d t . (A-z)h_\rho=-A_\rho h,\qquad h_\rho(0)=h_\rho(1)=0,\qquad w(1)h'_\rho(1)=\int_0^1 whA_\rho h\,dt. ( A − z ) h ρ = − A ρ h , h ρ ( 0 ) = h ρ ( 1 ) = 0 , w ( 1 ) h ρ ′ ( 1 ) = ∫ 0 1 w h A ρ h d t . At a mixed eigenvalue, h ′ ( 1 ) = − β h'(1)=-\beta h ′ ( 1 ) = − β h h h
∫ 0 1 w h A ρ h d t = ∂ ρ q 1 / K , ρ ( h ) − z ∂ ρ n 1 / K , ρ ( h ) − w ( 1 ) β ρ . \int_0^1 whA_\rho h\,dt=\partial_\rho q_{1/K,\rho}(h)-z\partial_\rho n_{1/K,\rho}(h)-w(1)\beta_\rho. ∫ 0 1 w h A ρ h d t = ∂ ρ q 1/ K , ρ ( h ) − z ∂ ρ n 1/ K , ρ ( h ) − w ( 1 ) β ρ . The term − w ρ ( 1 ) h ( 1 ) h ′ ( 1 ) -w_\rho(1)h(1)h'(1) − w ρ ( 1 ) h ( 1 ) h ′ ( 1 ) w ρ ( 1 ) β w_\rho(1)\beta w ρ ( 1 ) β w ( 1 ) β ρ w(1)\beta_\rho w ( 1 ) β ρ
B ρ ( z , ρ ) = ∂ ρ q 1 / K , ρ ( h ) − z ∂ ρ n 1 / K , ρ ( h ) w ( 1 ) = n 1 / K , ρ ( h ) w ( 1 ) e 1 / K ′ ( ρ ) , z = e 1 / K ( ρ ) . B_\rho(z,\rho)=\frac{\partial_\rho q_{1/K,\rho}(h)-z\partial_\rho n_{1/K,\rho}(h)}{w(1)}
=\frac{n_{1/K,\rho}(h)}{w(1)}e'_{1/K}(\rho),\qquad z=e_{1/K}(\rho). B ρ ( z , ρ ) = w ( 1 ) ∂ ρ q 1/ K , ρ ( h ) − z ∂ ρ n 1/ K , ρ ( h ) = w ( 1 ) n 1/ K , ρ ( h ) e 1/ K ′ ( ρ ) , z = e 1/ K ( ρ ) . The last equality is (3.39) with an unnormalized eigenfunction. It agrees with (3.46) and explicitly accounts for the weight, profile, and Robin variations.
Since d ρ / d R = K / λ d\rho/dR=K/\lambda d ρ / d R = K / λ d δ / d R = 1 d\delta/dR=1 d δ / d R = 1
T K ′ ( R ) = − 1 δ 2 B ( 0 , ρ ) + K λ δ B ρ ( 0 , ρ ) . T'_K(R)=-\frac{1}{\delta^2}B(0,\rho)+\frac{K}{\lambda\delta}B_\rho(0,\rho). T K ′ ( R ) = − δ 2 1 B ( 0 , ρ ) + λ δ K B ρ ( 0 , ρ ) . The first term vanishes at R K R_K R K
T K ′ ( R K ) = K λ δ K ∫ 0 1 w h 0 , ρ 1 / K 2 d t w ( 1 ) e 1 / K ′ ( ρ 1 / K ) > 0. (78) T'_K(R_K)=\frac{K}{\lambda\delta_K}\frac{\int_0^1 wh_{0,\rho_{1/K}}^2\,dt}{w(1)}e'_{1/K}(\rho_{1/K})>0. \tag*{(78)} T K ′ ( R K ) = λ δ K K w ( 1 ) ∫ 0 1 w h 0 , ρ 1/ K 2 d t e 1/ K ′ ( ρ 1/ K ) > 0. (78) Finally, (3.29) gives m K ( R ) = − 2 d R − λ 2 T K ( R ) m_K(R)=-2d_{R-\lambda}^2T_K(R) m K ( R ) = − 2 d R − λ 2 T K ( R ) − 4 d R − λ ( ∂ R d R − λ ) T K -4d_{R-\lambda}(\partial_Rd_{R-\lambda})T_K − 4 d R − λ ( ∂ R d R − λ ) T K − 2 d R − λ 2 T K ′ -2d_{R-\lambda}^2T'_K − 2 d R − λ 2 T K ′ T K ( R K ) = 0 T_K(R_K)=0 T K ( R K ) = 0
m K ′ ( R K ) = − 2 d δ K 2 T K ′ ( R K ) < 0. (79) m'_K(R_K)=-2d_{\delta_K}^2T'_K(R_K)<0. \tag*{(79)} m K ′ ( R K ) = − 2 d δ K 2 T K ′ ( R K ) < 0. (79) To express the Crandall–Rabinowitz condition directly, put ϕ K ( θ ) = cos ( K θ ) \phi_K(\theta)=\cos(K\theta) ϕ K ( θ ) = cos ( K θ ) ℓ K ( g ) = ∫ 0 2 π g ( θ ) ϕ K ( θ ) d θ \ell_K(g)=\int_0^{2\pi}g(\theta)\phi_K(\theta)\,d\theta ℓ K ( g ) = ∫ 0 2 π g ( θ ) ϕ K ( θ ) d θ Ran D η F p , K ( R K , 0 ) = Ker ℓ K \operatorname{Ran}D_\eta F_{p,K}(R_K,0)=\operatorname{Ker}\ell_K Ran D η F p , K ( R K , 0 ) = Ker ℓ K
ℓ K ( D R D η F p , K ( R K , 0 ) [ ϕ K ] ) = m K ′ ( R K ) ∫ 0 2 π cos 2 ( K θ ) d θ = π m K ′ ( R K ) < 0. \ell_K\left(D_RD_\eta F_{p,K}(R_K,0)[\phi_K]\right)=m'_K(R_K)\int_0^{2\pi}\cos^2(K\theta)\,d\theta=\pi m'_K(R_K)<0. ℓ K ( D R D η F p , K ( R K , 0 ) [ ϕ K ] ) = m K ′ ( R K ) ∫ 0 2 π cos 2 ( K θ ) d θ = π m K ′ ( R K ) < 0. Consequently D R D η F p , K ( R K , 0 ) [ ϕ K ] D_RD_\eta F_{p,K}(R_K,0)[\phi_K] D R D η F p , K ( R K , 0 ) [ ϕ K ]
Application of the Crandall–Rabinowitz theorem
We first fix the threshold for the modes independently of the Hölder exponent. Carry out the radial construction and spectral estimates with α ∗ = 1 2 min { 1 , p − 1 } \alpha_*=\frac{1}{2}\min\{1,p-1\} α ∗ = 2 1 min { 1 , p − 1 } λ , p \lambda,p λ , p δ > 0 \delta>0 δ > 0 V δ ∈ C 3 ( [ 0 , 1 ] ) V_\delta\in C^3([0,1]) V δ ∈ C 3 ([ 0 , 1 ]) s ↦ s p s\mapsto s^p s ↦ s p C 1 C^1 C 1 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) V δ ∈ C 2 , α V_\delta\in C^{2,\alpha} V δ ∈ C 2 , α α < 1 \alpha<1 α < 1 C 2 , α C^{2,\alpha} C 2 , α K 0 = K 0 ( λ , p ) ≥ 2 K_0=K_0(\lambda,p)\ge2 K 0 = K 0 ( λ , p ) ≥ 2 α \alpha α
Fix K ≥ K 0 K\ge K_0 K ≥ K 0 R ∗ = R K R_*=R_K R ∗ = R K U K = U δ K U_K=U_{\delta_K} U K = U δ K Z p , K → W p , K Z_{p,K}\to W_{p,K} Z p , K → W p , K [2 ] can now be verified for this concrete map:
(1) F p , K F_{p,K} F p , K C ∞ C^\infty C ∞ ( R K , 0 ) (R_K,0) ( R K , 0 ) R × X K \mathbb{R}\times X_K R × X K Y K Y_K Y K F p , K ( R , 0 ) = 0 F_{p,K}(R,0)=0 F p , K ( R , 0 ) = 0
(2) By Lemma 3.8 and the isolation of the principal mixed mode, L K = D η F p , K ( R K , 0 ) L_K=D_\eta F_{p,K}(R_K,0) L K = D η F p , K ( R K , 0 ) Ker L K = span { ϕ K } \operatorname{Ker}L_K=\operatorname{span}\{\phi_K\} Ker L K = span { ϕ K } ϕ K ( θ ) = cos ( K θ ) \phi_K(\theta)=\cos(K\theta) ϕ K ( θ ) = cos ( K θ ) Ran L K = { g ∈ Y K : ∫ 0 2 π g ϕ K = 0 } \operatorname{Ran}L_K=\{g\in Y_K:\int_0^{2\pi}g\phi_K=0\} Ran L K = { g ∈ Y K : ∫ 0 2 π g ϕ K = 0 }
(3) By (3.48),
D R D η F p , K ( R K , 0 ) ϕ K = m K ′ ( R K ) ϕ K ∉ Ran L K . D_RD_\eta F_{p,K}(R_K,0)\phi_K=m'_K(R_K)\phi_K\notin\operatorname{Ran}L_K. D R D η F p , K ( R K , 0 ) ϕ K = m K ′ ( R K ) ϕ K ∈ / Ran L K . The theorem gives a C ∞ C^\infty C ∞
R = R ( s ) , η ( s ) = s ϕ K + s z ( s ) , z ( 0 ) = 0 , R=R(s),\qquad\eta(s)=s\phi_K+sz(s),\qquad z(0)=0, R = R ( s ) , η ( s ) = s ϕ K + sz ( s ) , z ( 0 ) = 0 , with z ( s ) z(s) z ( s )
X K ⊥ = { η ∈ X K : ∫ 0 2 π η ( θ ) ϕ K ( θ ) d θ = 0 } . X_K^\perp=\left\{\eta\in X_K:\int_0^{2\pi}\eta(\theta)\phi_K(\theta)\,d\theta=0\right\}. X K ⊥ = { η ∈ X K : ∫ 0 2 π η ( θ ) ϕ K ( θ ) d θ = 0 } . We use the Fourier coefficient s = π − 1 ∫ 0 2 π η ( s , θ ) ϕ K ( θ ) d θ s=\pi^{-1}\int_0^{2\pi}\eta(s,\theta)\phi_K(\theta)\,d\theta s = π − 1 ∫ 0 2 π η ( s , θ ) ϕ K ( θ ) d θ R η ( θ ) = η ( θ + π / K ) \mathcal{R}\eta(\theta)=\eta(\theta+\pi/K) R η ( θ ) = η ( θ + π / K )
R ( − s ) = R ( s ) , η ( − s ) = R η ( s ) , Q ( − s ) = Q ( s ) . (80) R(-s)=R(s),\qquad\eta(-s)=\mathcal{R}\eta(s),\qquad Q(-s)=Q(s). \tag*{(80)} R ( − s ) = R ( s ) , η ( − s ) = R η ( s ) , Q ( − s ) = Q ( s ) . (80) Here Q ( s ) Q(s) Q ( s ) (54) . Consequently,
R ( s ) = R K + O ( s 2 ) , η ( s ) = s cos ( K θ ) + O ( s 2 ) , Q ( s ) > 0 R(s)=R_K+O(s^2),\qquad\eta(s)=s\cos(K\theta)+O(s^2),\qquad Q(s)>0 R ( s ) = R K + O ( s 2 ) , η ( s ) = s cos ( K θ ) + O ( s 2 ) , Q ( s ) > 0 for sufficiently small ∣ s ∣ |s| ∣ s ∣
The positivity in Proposition 3.7 carries over to the physical solutions. Moreover, U K ′ ( λ ) > 0 U_K'(\lambda)>0 U K ′ ( λ ) > 0 C 1 C^1 C 1 ∂ r u s ( λ , θ ) > 0 \partial_r u_s(\lambda,\theta)>0 ∂ r u s ( λ , θ ) > 0 ∣ s ∣ |s| ∣ s ∣ s ≠ 0 s\ne0 s = 0 K K K η ( s ) \eta(s) η ( s ) 2 π / K 2\pi/K 2 π / K K ≥ 2 K\ge2 K ≥ 2 η = 0 \eta=0 η = 0
We next verify the local completeness statement, including the state symmetry. Suppose that a nearby positive solution of (1.4) has boundary perturbation η ∈ X K \eta\in X_K η ∈ X K g ∈ D K g\in D_K g ∈ D K u u u u ∘ g u\circ g u ∘ g ∂ ν u = − Q \partial_\nu u=-\sqrt{Q} ∂ ν u = − Q w w w
Δ w = c g ( x ) w , c g ( x ) = 1 − p ∫ 0 1 ( t u ( x ) + ( 1 − t ) u ( g x ) ) p − 1 d t , \Delta w=c_g(x)w,\qquad c_g(x)=1-p\int_0^1\bigl(tu(x)+(1-t)u(gx)\bigr)^{p-1}\,dt, Δ w = c g ( x ) w , c g ( x ) = 1 − p ∫ 0 1 ( t u ( x ) + ( 1 − t ) u ( g x ) ) p − 1 d t , where c g c_g c g w w w [1 ] implies w = 0 w=0 w = 0 D K D_K D K v R , η v_{R,\eta} v R , η F p , K ( R , η ) = 0 F_{p,K}(R,\eta)=0 F p , K ( R , η ) = 0 C 2 , α C^{2,\alpha} C 2 , α
Finally, (3.40)–(3.41) give
R K = λ + λ κ p K + o ( K − 1 ) . R_K=\lambda+\frac{\lambda\kappa_p}{K}+o(K^{-1}). R K = λ + K λ κ p + o ( K − 1 ) . This proves Theorem 1.2.
Further properties of the power-law solutions
Proof of Proposition 1.3. Set d = U ′ ( R ) d=U'(R) d = U ′ ( R ) U ( R ) = 0 U(R)=0 U ( R ) = 0 d ≤ 0 d\le0 d ≤ 0 d = 0 d=0 d = 0 r = R r=R r = R U ( λ ) = 1 U(\lambda)=1 U ( λ ) = 1 d < 0 d<0 d < 0
Differentiating the radial equation, we find that U ′ U' U ′
− ( U ′ ) ′ ′ − 1 r ( U ′ ) ′ + ( 1 r 2 + 1 − p U p − 1 ) U ′ = 0. (81) -(U')''-\frac{1}{r}(U')'+\left(\frac{1}{r^2}+1-pU^{p-1}\right)U'=0. \tag*{(81)} − ( U ′ ) ′′ − r 1 ( U ′ ) ′ + ( r 2 1 + 1 − p U p − 1 ) U ′ = 0. (81) The identity U ′ ′ ( R ) = − d / R U''(R)=-d/R U ′′ ( R ) = − d / R U ′ U' U ′ h h h h ( R ) = 1 h(R)=1 h ( R ) = 1 U ′ / d U'/d U ′ / d h = U ′ / d h=U'/d h = U ′ / d h ( λ ) = 0 h(\lambda)=0 h ( λ ) = 0 U ′ ( λ ) = 0 U'(\lambda)=0 U ′ ( λ ) = 0 1 − 1 p = 0 1-1^{p}=0 1 − 1 p = 0 u = 1 u=1 u = 1 U ( λ ) = 1 U(\lambda)=1 U ( λ ) = 1 U ′ ( λ ) = 0 U'(\lambda)=0 U ′ ( λ ) = 0 U ≡ 1 U\equiv1 U ≡ 1 U ( R ) = 0 U(R)=0 U ( R ) = 0
Acknowledgement
The research of Wang was supported by supported by National Key R&D Program of China 2022YFA1005601 and National Natural Science Foundation of China 12371114. The research of Xu was supported by National Natural Science Foundation of China (No. 12401134). The research of Zhang was supported by National Natural Science Foundation of China (No. 12301133) and the Postdoctoral Science Foundation of China (No. 2024T170353).
References
Jun Wang, School of Mathematical Sciences, Jiangsu University, Zhenjiang, Jiangsu 212013, P.R. China Email address: wangmath2011@126.com
Fei Xu, School of Mathematical Sciences, Jiangsu University, Zhenjiang, Jiangsu 212013, P.R. China Email address: xufeiujs@126.com
Yong Zhang, School of Mathematical Sciences, Jiangsu University, Zhenjiang, Jiangsu 212013, P.R. China Email address: zhangyong@ujs.edu.cn
