Spectral Asymptotics of the Dirichlet Laplacian on a Generalized Parabolic Layer

Abstract

We perform quantitative spectral analysis of the self-adjoint Dirichlet Laplacian ${\mathsf {H}}$ on an unbounded, radially symmetric (generalized) parabolic layer ${\mathcal {P}}\subset {\mathbb {R}}^3$. It was known before that ${\mathsf {H}}$ has an infinite number of eigenvalues below the threshold of its essential spectrum. In the present paper, we find the discrete spectrum asymptotics for ${\mathsf {H}}$ by means of a consecutive reduction to the analogous asymptotic problem for an effective one-dimensional Schrödinger operator on the half-line with the potential the behaviour of which far away from the origin is determined by the geometry of the layer ${\mathcal {P}}$ at infinity.

Appendix A. The Signed Curvature of \(\Gamma \)

In this appendix, we analyze properties of the signed curvature $\gamma$ of the curve $\Gamma$ defined in (2.13). This material can be seen as an exercise in the differential geometry, however, the claims we need are scattered and not easy to find in textbooks. Recall that the curve $\Gamma$ is parametrized by $\overline{{\mathbb {R}}}_+\ni s \rightarrow (\phi (s),f(\phi (s)))$, where the increasing function $\phi :\overline{{\mathbb {R}}}_+\rightarrow \overline{{\mathbb {R}}}_+$ fulfils $\phi (0) = 0$ and satisfies the ordinary differential equation (2.11). In the remaining part of this appendix, all the functions depend on s and their derivatives are taken with respect to that variable. For the sake of brevity, the indication of the dependence on s is occasionally dropped.

First, we formulate and prove an auxiliary lemma on the second principal curvature $\kappa _2$ of $\Sigma$, explicitly given in (2.19).

Lemma A.1

Let $\phi :\overline{{\mathbb {R}}}_+\rightarrow \overline{{\mathbb {R}}}_+$ be the solution of (2.11) with $\phi (0) = 0$. Then the function

$$\begin{aligned} \overline{{\mathbb {R}}}_+ \ni p\mapsto \sup _{s\in [p,\infty )}\frac{\dot{f}(\phi (s))\dot{\phi }(s)}{\phi (s)} \end{aligned}$$

is bounded and vanishes as $p\rightarrow \infty$.

Proof

First, we observe that the function ${\mathbb {R}}_+\ni s \mapsto \frac{\dot{f}(\phi (s))\dot{\phi }(s)}{\phi (s)}$ is ${\mathcal {C}}^\infty$-smooth on ${\mathbb {R}}_+$. Moreover, using that $\phi (0) = 0$, $\dot{\phi }(0) = 1$ and the Taylor expansion $\dot{f}(x) = \ddot{f}(0)x + o(x)$ as $x\rightarrow 0^+$ we get

$$\begin{aligned} \lim _{s\rightarrow 0^+}\frac{\dot{f}(\phi )\dot{\phi }}{\phi }=&{} \ddot{f}(0)\,, \quad \lim _{s\rightarrow \infty }\frac{\dot{f}(\phi )\dot{\phi }}{\phi } = \lim _{s\rightarrow \infty } \frac{1}{\phi }\left( \frac{\dot{f}^2(\phi )}{1+\dot{f}^2(\phi )}\right) ^{\frac{1}{2}}= {} \lim _{s\rightarrow \infty }\frac{1}{\phi } = 0. \end{aligned}$$

The above two limits and smoothness of $\frac{f(\phi )\dot{\phi }}{\phi }$ yield the claims. $\square$

Next, we prove a proposition, on the asymptotic behaviour of $\phi$ and its derivatives up to the third, in the limit $s\rightarrow \infty$.

Proposition A.2

The solution $\phi :\overline{{\mathbb {R}}}_+\rightarrow \overline{{\mathbb {R}}}_+$ of (2.11) with $\phi (0) = 0$ has the following properties.

• (i) $\lim _{s\rightarrow \infty }s^{-\frac{1}{\alpha }} \phi = k^{-\frac{1}{\alpha }}$.

• (ii) $\lim _{s\rightarrow \infty }s^{\frac{\alpha -1}{\alpha }} \dot{\phi }= \frac{ k^{-\frac{1}{\alpha }}}{\alpha }$.

• (iii) $\lim _{s\rightarrow \infty }s^{\frac{2\alpha -1}{\alpha }} \ddot{\phi }= -\frac{\alpha -1}{\alpha ^2}k^{-\frac{1}{\alpha }}$.

• (iv) $\lim _{s\rightarrow \infty }s^{\frac{3\alpha -1}{\alpha }}\dddot{\phi }= \frac{(\alpha -1)(2\alpha -1)}{\alpha ^3} k^{-\frac{1}{\alpha }}$.

Proof

Notice that we obviously have $\lim _{s\rightarrow \infty }\phi (s) = \infty$, because the map $\overline{{\mathbb {R}}}_+\ni s\mapsto (\phi (s), f(\phi (s))$ is a re-parametrization of the curve $\overline{{\mathbb {R}}}_+\ni x\mapsto (x,f(x))$. The differential equation (2.11) implies that on the interval $[R,\infty )$ the following estimates hold:

$$\begin{aligned} \alpha k\phi ^{\alpha -1}\dot{\phi }\le 1 , \big (1 + \alpha k\phi ^{\alpha -1}\big )\dot{\phi }\ge 1. \end{aligned}$$

Integrating the above inequalities on the interval [R, s] we get

$$\begin{aligned} k\phi ^\alpha (s) \le s + {\mathcal {O}}(1), \phi (s) + k\phi ^\alpha (s) \ge s + {\mathcal {O}}(1), \qquad \forall \, s \ge R.\nonumber \\ \end{aligned}$$

(A.1)

Plugging the first inequality in (A.1) into the second, we obtain

$$\begin{aligned} k\phi ^\alpha (s) \ge s - \left( \frac{s}{k}\right) ^{\frac{1}{\alpha }} + {\mathcal {O}}(1), \qquad \forall \, s \ge R. \end{aligned}$$

(A.2)

Combining the first inequality in (A.1) with (A.2) and using the assumption $\AA > 1$ we get the limit in (i). Furthermore, the limit in (ii) can be shown as follows,

$$\begin{aligned} \begin{aligned} \lim _{s\rightarrow \infty }s^{\frac{\alpha -1}{\alpha }}{\dot{\phi }}&= \lim _{s\rightarrow \infty }s^{\frac{\alpha -1}{\alpha }}\big (1+\alpha ^2 k^2 \phi ^{2\alpha -2}\big )^{-\frac{1}{2}}\\&= \lim _{s\rightarrow \infty }\left( s^{-\frac{2(\alpha -1)}{\alpha }} + \alpha ^2 k^2 s^{-\frac{2(\alpha -1)}{\alpha }} \phi ^{2\alpha -2}\right) ^{-\frac{1}{2}} \\&= \lim _{s\rightarrow \infty }\frac{1}{\alpha k s^{-\frac{\alpha -1}{\alpha }}\phi ^{\alpha -1}} = \frac{ k^{-\frac{1}{\alpha }}}{\alpha }. \end{aligned} \end{aligned}$$

The differential equation (2.11) can be alternatively written as

$$\begin{aligned} \dot{\phi }(s) = \frac{1}{\big (1+\dot{f}^2(\phi (s))\big )^{\frac{1}{2}}}. \end{aligned}$$

(A.3)

Differentiating the left and right hand sides of Eq. (A.3), we express $\ddot{\phi }$ as follows,

$$\begin{aligned} \ddot{\phi }= -\frac{\dot{f}(\phi ) \ddot{f}(\phi ) \dot{\phi }}{(1+\dot{f}^2(\phi ))^{\frac{3}{2}}} = - \frac{\dot{f}(\phi ) \ddot{f}(\phi )}{(1+\dot{f}^2(\phi ))^2}. \end{aligned}$$

(A.4)

Hence, on the interval $[R,\infty )$, we have

$$\begin{aligned} \ddot{\phi }= - \frac{\alpha ^2k^2(\alpha -1)\phi ^{2\alpha -3}}{ \big (1+\alpha ^2 k^2 \phi ^{2\alpha -2}\big )^2 }. \end{aligned}$$

Eventually, using (i) we get

$$\begin{aligned} \lim _{s\rightarrow \infty }s^{\frac{2\alpha -1}{\alpha }}\ddot{\phi }= - \lim _{s\rightarrow \infty }\frac{\alpha -1}{\alpha ^2 k^2 s^{-\frac{2\alpha -1}{\alpha }} \phi ^{2\alpha -1}} = \frac{\alpha -1}{\alpha ^2 k^2 k^{-\frac{2\alpha -1}{\alpha }}} = -\frac{\alpha -1}{\alpha ^2}k^{-\frac{1}{\alpha }}, \end{aligned}$$

and in this way the limit in (iii) is also obtained.

Differentiating the left and the right hand sides of (A.4), we express $\dddot{\phi }$ as follows,

$$\begin{aligned} \begin{aligned} \dddot{\phi }&= - \frac{(\ddot{f}^2(\phi ) + \dot{f}(\phi )\dddot{f}(\phi )) (1+\dot{f}^2(\phi ))\dot{\phi }- 4\dot{f}^2(\phi )\ddot{f}^2(\phi )\dot{\phi }}{(1+\dot{f}^2(\phi ))^3} \\&= \frac{3\dot{f}^2(\phi )\ddot{f}^2(\phi ) -\ddot{f}^2(\phi ) - \dot{f}(\phi )\dddot{f}(\phi ) - \dot{f}^3(\phi )\dddot{f}(\phi )}{(1+\dot{f}^2(\phi ))^{\frac{7}{2}}}. \end{aligned} \end{aligned}$$

The latter yields that on the interval $[R,\infty )$

$$\begin{aligned} \begin{aligned} \dddot{\phi }&= \frac{ 3\alpha ^4(\alpha -1)^2 k^4\phi ^{4\alpha -6} - \alpha ^2(\alpha -1)(2\alpha -3)k^2\phi ^{2\alpha -4} - \alpha ^4(\alpha -1)(\alpha -2)k^4\phi ^{4\alpha -6} }{\big (1+ \alpha ^2 k^2\phi ^{2\alpha -2}\big )^{\frac{7}{2}}}\\&= \frac{ \alpha ^4(\alpha -1) (2\alpha -1)k^4\phi ^{4\alpha -6} - \alpha ^2(\alpha -1)(2\alpha -3)k^2\phi ^{2\alpha -4} }{\big (1+ \alpha ^2 k^2\phi ^{2\alpha -2}\big )^{\frac{7}{2}}}. \end{aligned} \end{aligned}$$

Again using (i) we obtain

$$\begin{aligned} \begin{aligned}&\lim _{s\rightarrow \infty }s^{\frac{3\alpha -1}{\alpha }}\dddot{\phi }\\&= \lim _{s\rightarrow \infty }s^{\frac{3\alpha -1}{\alpha }} \frac{ \alpha ^4(\alpha -1)(2\alpha -1)k^4\phi ^{4\alpha -6} - \alpha ^2(\alpha -1)(2\alpha -3)k^2\phi ^{2\alpha -4} }{\big (1+ \alpha ^2 k^2\phi ^{2\alpha -2}\big )^{\frac{7}{2}}}\\&= \lim _{s\rightarrow \infty }s^{\frac{3\alpha -1}{\alpha }} \frac{\alpha ^4(\alpha -1)(2\alpha -1)k^4 \phi ^{4\alpha -6}}{\alpha ^7k^7\phi ^{7\alpha -7}} = \lim _{s\rightarrow \infty }\frac{(\alpha -1)(2\alpha -1)}{\alpha ^3 k^3 s^{-\frac{3\alpha -1}{\alpha }}\phi ^{3\alpha -1}}\\&= \frac{(\alpha -1)(2\alpha -1)}{\alpha ^3 k^3 k^{-\frac{3\alpha -1}{\alpha }}} = \frac{(\alpha -1)(2\alpha -1)}{\alpha ^3} k^{-\frac{1}{\alpha }}, \end{aligned} \end{aligned}$$

by which the limit in (iv) is also shown. $\square$

Recall that the signed curvature of the curve $\Gamma$ is given by the formula

$$\begin{aligned} \gamma = \ddot{f}(\phi )\dot{\phi }^3. \end{aligned}$$

(A.5)

Finally, we prove a claim about the asymptotic behaviour of $\gamma$ and its derivatives up to the second order, in the limit $s\rightarrow \infty$.

Proposition A.3

Let the signed curvature $\gamma :\overline{{\mathbb {R}}}_+\rightarrow {\mathbb {R}}$ be as in (A.5). Then there exist $g_j\in {\mathbb {R}}$, $j=0,1,2$, such that:

• (i) $\lim _{s\rightarrow \infty }s^{\frac{2\alpha -1}{\alpha }} \gamma = g_0$,

• (ii) $\lim _{s\rightarrow \infty }s^{\frac{3\alpha -1}{\alpha }} {\dot{\gamma }} = g_1$,

• (iii) $\lim _{s\rightarrow \infty }s^{\frac{4\alpha -1}{\alpha }} \ddot{\gamma } = g_2$.

Proof

The first and the second derivatives of the signed curvature $\gamma$ are given by

$$\begin{aligned} \begin{aligned} {\dot{\gamma }}&= 3{\dot{\phi }}^2\ddot{\phi }\ddot{f}(\phi ) +{\dot{\phi }}^4 \dddot{f}(\phi ),\\ \ddot{\gamma }&= 6\dot{\phi }\ddot{\phi }^2\ddot{f}(\phi ) +3\dot{\phi }^2\dddot{\phi }\ddot{f}(\phi ) +7\dot{\phi }^3\ddot{\phi }\dddot{f}(\phi ) +\dot{\phi }^5 f^{(4)}(\phi ).\\ \end{aligned} \end{aligned}$$

Hence, using the notation $\kappa := \alpha (\alpha -1)k$ we infer that on the interval $[R,\infty )$ the following relations hold:

$$\begin{aligned} \begin{aligned} \gamma&= \kappa \phi ^{\alpha -2}\dot{\phi }^3,\\ {\dot{\gamma }}&= \kappa \left[ 3\phi ^{\alpha -2}\dot{\phi }^2\ddot{\phi }+(\alpha -2)\phi ^{\alpha -3}\dot{\phi }^4\right] ,\\ \ddot{\gamma }&= \kappa \Big [ \phi ^{\alpha -2} \big (6\dot{\phi }\ddot{\phi }^2 +3\dot{\phi }^2\dddot{\phi }\big ) + 7(\alpha -2)\phi ^{\alpha -3} \dot{\phi }^3\ddot{\phi }+(\alpha -2)(\alpha -3)\phi ^{\alpha -4} \dot{\phi }^5\Big ]. \end{aligned} \end{aligned}$$

Eventually, existence of finite limits in (i)-(iii) directly follows from Proposition A.2 (i)-(iv). $\square$