Eigenvalue Counting Function for Robin Laplacians on Conical Domains

Abstract

We study the discrete spectrum of the Robin Laplacian $Q^{\Omega }_\alpha$ in $L^2(\Omega )$, $u\mapsto -\Delta u, \quad D_n u=\alpha u \text { on }\partial \Omega$, where $D_n$ is the outer unit normal derivative and $\Omega \subset {\mathbb {R}}^{3}$ is a conical domain with a regular cross-section $\Theta \subset {\mathbb {S}}^2$, n is the outer unit normal, and $\alpha >0$ is a fixed constant. It is known from previous papers that the bottom of the essential spectrum of $Q^{\Omega }_\alpha$ is $-\alpha ^2$ and that the finiteness of the discrete spectrum depends on the geometry of the cross-section. We show that the accumulation of the discrete spectrum of $Q^\Omega _\alpha$ is determined by the discrete spectrum of an effective Hamiltonian defined on the boundary and far from the origin. By studying this model operator, we prove that the number of eigenvalues of $Q^{\Omega }_\alpha$ in $(-\infty ,-\alpha ^2-\lambda )$, with $\lambda >0$, behaves for $\lambda \rightarrow 0$ as

$$\begin{aligned} \dfrac{\alpha ^2}{8\pi \lambda } \int _{\partial \Theta } \kappa _+(s)^2\mathrm {d}s +o\left( \frac{1}{\lambda }\right) , \end{aligned}$$

where $\kappa _+$ is the positive part of the geodesic curvature of the cross-section boundary.

Appendix 1: Proof of Proposition 7

In this section, we study two 1D problems on $\big (0,\delta (r)\big )$ with Robin condition (with parameter r) at 0 and with Dirichlet (resp. Neumann) condition at $\delta (r)$ in order to prove Proposition 7. The only novelty lies on the $L^2$-estimate of the r-derivative of the ground state, but for the sake of completeness we provide the proofs of all the statements.

Let us look for eigenvalues of the form $E^{D/N}=-(k^{D/N}r)^2$, $k^{D/N}>0$, then the boundary condition $u(\delta )=0$ (respectively, $u'(\delta )=0$) gives the following forms for the positive normalized eigenfunctions:

$$\begin{aligned} \psi ^{D}(t)=C^{D}(r)\sinh \big (k^{D}r(\delta -t)\big ) \ \ \text{ and } \ \ \psi ^{N}(t)=C^{N}(r)\cosh \big (k^{N}r(\delta -t)\big ), \end{aligned}$$

(33)

where $C^{D/N}(r)>0$ are normalization factors. The second boundary condition gives then

$$\begin{aligned} k^{D}\coth (k^{D}r\delta )=1 \ \ \text{ and } \ \ k^{N}\tanh (k^{N}r\delta )=1 \end{aligned}$$

(34)

which can be rewritten as $F^{D/N}(kr\delta )=r\delta$ with $F^{D}(t)=t\coth t$ and $F^{N}(t)=t\tanh t$. The function $F^{D}$ (respectively, $F^{N}$) is a bijection between $(0,+\infty )$ and $(1,+\infty )$ (respectively, $(0,+\infty )$); hence, there exists a unique solution if $r\delta >1$ (respectively, $r\delta >0$), which holds, in particular, for large r. Furthermore, as both $\coth t$ and $\tanh t$ are bounded and tend to 1 at $+\infty$, it follows first that $k^{D/N}r\delta$ tends to $+\infty$ for large r, and then that $k^{D/N}r\delta =r\delta +o(r\delta )$ for large r, i.e., $k^{D/N}=1+o(1)$ and $k^{D/N}r\delta \rightarrow +\infty$, and (34) gives $k^{D/N}=1+{\mathcal {O}}(e^{-2r\delta })$ implying the estimates (8). Taking the derivative of (34) with respect to r, we obtain

$$\begin{aligned} (k^{D/N})'(r)=\frac{b(\rho -1)k^{D/N} r^{-\rho }}{\mp \cosh (k^{D/N}r\delta )\sinh (k^{D/N}r\delta )+r\delta } ={\mathcal {O}}(r^{-\infty }), \quad r\rightarrow +\infty . \end{aligned}$$

(35)

Recall that $C^{D/N}$ are normalization factors in (33), we get

$$\begin{aligned} C^{D/N}(r)^{-2}=G^{D/N}(2rk^{D/N}\delta ) \delta , \quad G^{D/N}(t):=\dfrac{1}{2}\Big (\dfrac{\sinh t}{t}\mp 1\Big ), \end{aligned}$$

which gives

$$\begin{aligned} \psi ^{D}(t)&=\delta ^{-1/2}G^{D}(2k^{D}r\delta )^{-1/2} \sinh (rkt),\\ \psi ^{N}(t)&=\delta ^{-1/2}G^{N}(2k^{N}r\delta )^{-1/2} \cosh (rkt). \end{aligned}$$

We have (we drop the indices D / N when the expressions are the same)

$$\begin{aligned} \partial _r(\delta ^{-1/2})= & {} \dfrac{\rho }{2r} \,\delta ^{-1/2},\quad \partial _r \big (G(2kr\delta )^{-1/2}\big )=-\dfrac{G'(2kr\delta )}{G(2kr\delta )}\,\partial _r(kr\delta )\cdot G(2kr\delta )^{-1/2},\\ \dfrac{(G^{D/N})'(t)}{G^{D/N}(t)}= & {} \dfrac{ t \cosh t - \sinh t}{t(\sinh t\mp t)}, \end{aligned}$$

hence, in both cases

$$\begin{aligned} \dfrac{G'(2kr\delta )}{G(2kr\delta )}={\mathcal {O}}(1), \quad r\rightarrow +\infty . \end{aligned}$$

Furthermore, using (35) (here again we drop the indices)

$$\begin{aligned} \partial _r(kr\delta )=k'\cdot (r\delta ) + k \cdot (r\delta )'={\mathcal {O}}(r^{-\rho }), \quad r\rightarrow +\infty . \end{aligned}$$

Therefore,

$$\begin{aligned} \left\{ \begin{aligned} \partial _r \psi ^{D}(t)={\mathcal {O}}(r^{-\rho }) \psi ^{D} - C^{D}(r)k^{D}t \cosh \big (rk^{D}(\delta -t)\big ) \\ \partial _r \psi ^{N}(t)={\mathcal {O}}(r^{-\rho }) \psi ^{N} - C^{N}(r)k^{N}t \sinh \big (rk^{N}(\delta -t)\big ) \end{aligned} \right. \end{aligned}$$

and

$$\begin{aligned} \Vert \partial _r \psi ^{D/N}\Vert ^2_{L^2(0,\delta )} \le {\mathcal {O}}(r^{-2\rho })+ {\mathcal {O}}(r^{-2\rho }) \dfrac{\sinh (2k^{D/N}r\delta )\pm 2k^{D/N}r\delta }{\sinh (2k^{D/N}r\delta )\mp 2k^{D/N}r\delta }={\mathcal {O}}(r^{-2\rho }). \end{aligned}$$

Finally,

$$\begin{aligned} \psi ^{D/N}(0)^2=2r k^{D/N} \dfrac{\cosh (2k^{D/N}r\delta )\mp 1}{\sinh (2k^{D/N}r\delta )\pm 2rk^{D/N}\delta } =2r\big (1+ {\mathcal {O}}(r\delta e^{-2r\delta })\big ) \end{aligned}$$

and

$$\begin{aligned} \psi ^{N}(\delta )^2=C^{N}(r)^{-2}= \dfrac{2k^{N}r}{\sinh (2k^{N}r\delta )+2k^{N}r\delta }={\mathcal {O}}(r^2\delta e^{-2r\delta }), \end{aligned}$$

and the proposition is proved.