Eigenvalue and Resonance Asymptotics in Perturbed Periodically Twisted Tubes: Twisting Versus Bending

Abstract

We consider a three-dimensional waveguide that is a small deformation of a periodically twisted tube (including in particular the case of a straight tube). The deformation is given by a bending and an additional twisting of the tube, both parametrized by a coupling constant $\delta$. In this deformed waveguide, we consider the Dirichlet Laplacian. We expand its resolvent near the bottom of its essential spectrum, and we show the existence of exactly one resonance, in the asymptotic regime of $\delta$ small. We are able to perform the asymptotic expansion of the resonance in $\delta$, which in particular permits us to give a quantitative geometric criterion for the existence of a discrete eigenvalue below the essential spectrum. In the case of perturbations of straight tubes, we are able to show the existence of resonances not only near the bottom of the essential spectrum but near each threshold in the spectrum, showing in particular what are the spectral effects of the bending for higher energies. We also obtain the asymptotic behavior of the resonances in this situation, which is generically different from the first case.

Appendix A: Some Explicit Expansions

1.1 A.1: The Perturbation as a Second-Order Differential Operator

In this appendix, we give some explicit computations that are straightforward. For simplicity, we set $\xi (s):=\tau (s)-\epsilon (s)-\beta$. Also, it is easy to see that $h(s,t)=1-\kappa (s)(t_2\cos (\theta (s))+t_3\sin (\theta (s)))$ satisfies

$$\begin{aligned} (\partial _\varphi h)(s,t)=\kappa (s)(t_3\cos (\theta (s))-t_2\sin (\theta (s)))=:{\tilde{h}}(s,t)\ \text { and } \partial _\varphi {\tilde{h}}=1-h. \end{aligned}$$

With this, we can start computing an expression for W. Here we systematically use the notation ${\dot{f}}=\partial _s f$ and ${\tilde{f}}=\partial _\varphi f$. By definition, we have

$$\begin{aligned} W=-\frac{\kappa ^2}{4h^2}-(h^{-\frac{1}{2}}(\partial _s+\xi \partial _\varphi )h^{-\frac{1}{2}})^2+\partial _s^2-2\beta \partial _s\partial _\varphi +\beta ^2\partial _\varphi ^2\ . \end{aligned}$$

Noticing that

$$\begin{aligned} h^{-\frac{1}{2}}(\partial _s+\xi \partial _\varphi )h^{-\frac{1}{2}}=\frac{1}{h}\partial _s+\frac{\xi }{h}\partial _\varphi -\frac{{\dot{h}}+\xi {\tilde{h}}}{2h^2}, \end{aligned}$$

we get

$$\begin{aligned}&\left( h^{-\frac{1}{2}}(\partial _s+\xi \partial _\varphi )h^{-\frac{1}{2}}\right) ^2\\&\quad = \frac{1}{h^2}\partial _s^2-\frac{{\dot{h}}}{h^3}\partial _s+\frac{\xi }{h^2}\partial _s\partial _\varphi +\left( \frac{{\dot{\xi }}}{h^2}-\frac{\xi {\dot{h}}}{h^3}\right) \partial _\varphi -\frac{{\dot{h}}+\xi {\tilde{h}}}{2h^3}\partial _s\\&\qquad -\frac{\ddot{h}+{\dot{\xi }}{\tilde{h}}+\xi \dot{{\tilde{h}}}}{2h^3}+\frac{{\dot{h}}^2+\xi {\tilde{h}}{\dot{h}}}{h^4} \\&\qquad +\frac{\xi }{h^2}\partial _s\partial _\varphi -\frac{\xi {\tilde{h}}}{h^3}\partial _s +\frac{\xi ^2}{h^2}\partial _\varphi ^2-\frac{\xi ^2{\tilde{h}}}{h^3}\partial _\varphi -\frac{\xi {\dot{h}}+\xi ^2{\tilde{h}}}{2h^3}\partial _\varphi \\&\qquad -\frac{\xi \dot{{\tilde{h}}}+\xi ^2(1-h)}{2h^3}+\frac{\xi {\dot{h}}{\tilde{h}}+\xi ^2{\tilde{h}}^2}{h^4}\\&\qquad -\frac{{\dot{h}}+\xi {\tilde{h}}}{2h^3}\partial _s-\frac{\xi {\dot{h}}+\xi ^2{\tilde{h}}}{2h^3}\partial _\varphi +\frac{{\dot{h}}^2}{4h^4}+\frac{\xi {\dot{h}}{\tilde{h}}}{2h^4}+\frac{\xi ^2{\tilde{h}}^2}{4h^4}. \end{aligned}$$

Then, writing

$$\begin{aligned} W=f_{0,0}+f_{1,0}\partial _s+f_{0,1}\partial _\varphi +f_{1,1}\partial _s\partial _\varphi +f_{2,0}\partial _s^2+f_{0,2}\partial _\varphi ^2 \end{aligned}$$

(48)

we obtain

$$\begin{aligned} f_{0,0}&=-\frac{\kappa ^2}{4h^2}+\frac{\ddot{h}}{2h^3}+\frac{{{\dot{\xi }}}{\tilde{h}}}{2h^3}+\frac{\xi \dot{{\tilde{h}}}}{h^3}-\frac{5{\dot{h}}^2}{4h^4}-\frac{5\xi {\tilde{h}}{\dot{h}}}{2h^4}+\frac{\xi ^2(1-h)}{2h^3}-\frac{5\xi ^2{\tilde{h}}^2}{4h^4};\\ f_{1,0}&=\frac{2{\dot{h}}}{h^3}+\frac{2\xi {\tilde{h}}}{h^3};\\ f_{0,1}&=\frac{2\xi {\dot{h}}}{h^3}+\frac{2\xi ^2{\tilde{h}}}{h^3}-\frac{{\dot{\xi }}}{h^2};\\ f_{1,1}&=-2\beta -\frac{2\xi }{h^2};\\ f_{2,0}&=1-\frac{1}{h^2};\\ f_{0,2}&=\beta ^2-\frac{\xi ^2}{h^2}. \end{aligned}$$

1.2 A.2: Asymptotics of the Coefficient in the Perturbative Regime

We are now interested in the asymptotic behavior of $\langle \eta \otimes \psi _n|\eta ^{-1}W_\delta \psi _n\rangle =\langle \psi _n|W_\delta \psi _n\rangle$ as $\delta \rightarrow 0$. For this, we need to make appear the $\delta$-dependence in the previous expressions. We set $\xi _\delta (s)=\delta \tau (s)-\delta \epsilon (s)-\beta$ and introduce the auxiliary function $\Xi _\delta (s):=\xi _\delta (s)+\beta =\delta \tau (s)-\delta \epsilon (s)=:\delta \Xi (s)$. Furthermore, setting $E(s)=\int _{-\infty }^s\epsilon (s)\mathrm{d}s$ we get the expression $\theta _\delta (s)=\beta s+\delta E(s)$. Before studying $\langle \psi _n|W_\delta \psi _n\rangle$, we need the asymptotic behavior of $h_\delta (s,t)=1-\delta \kappa (s)(t_2\cos (\theta _\delta (s))+t_3\sin (\theta _\delta (s)))$ given by

$$\begin{aligned} h_\delta =1+\delta g_1+\delta ^2 g_2+O(\delta ^3) \end{aligned}$$

(49)

where

$$\begin{aligned} \begin{aligned} g_1(s,t)&=-\kappa (s)(t_2\cos (\beta s)+t_3\sin (\beta s)),\\ g_2(s,t)&=-\kappa (s)E(s)(-t_2\sin (\beta s)+t_3\cos (\beta s))\ .\end{aligned} \end{aligned}$$

(50)

We will also need the fact that:

$$\begin{aligned} h_\delta ^{-1}&=1-g_1\delta +(g_1^2-g_2)\delta ^2+{\mathcal {O}}(\delta ^3);\\ h_\delta ^{-2}&=1-2g_1\delta +(3g_1^2-2g_2)\delta ^2+{\mathcal {O}}(\delta ^3);\\ h_\delta ^{-n}&=1-ng_1\delta +O(\delta ^2). \end{aligned}$$

The relation $\partial _\varphi h={\tilde{h}}$ gives the corresponding asymptotic for ${\tilde{h}}_\delta$

$$\begin{aligned} {\tilde{h}}_\delta ={\tilde{g}}_1\delta +{\tilde{g}}_2\delta ^2+O(\delta ^3), \end{aligned}$$

where ${\tilde{g}}_i=\partial _\varphi g_i$.

Lemma 16

Under the assumptions of decay of $\kappa ,\tau ,\varepsilon$ and its derivatives (see (9)), we have the following asymptotic:

$$\begin{aligned} \langle \psi _n|W_\delta \psi _n\rangle =\breve{\mu }_{1,n}\delta +\breve{\mu }_{2,n}\delta ^2+O(\delta ^3) . \end{aligned}$$

(51)

The constant $\breve{\mu }_{1,n}$ is given by

$$\begin{aligned} \breve{\mu }_{1,n}=2\beta \Vert \partial _\varphi \psi _n\Vert ^2 \int _{\mathbb {R}}\varepsilon -\tau +\beta ^2\int _{{\mathbb {R}}\times \omega } \kappa (\tfrac{1}{2}|\psi _n|^2+2|\partial _\varphi \psi _n|^2)\vartheta , \end{aligned}$$

(52)

where $\vartheta (s,t)$ is given by

$$\begin{aligned} \vartheta (s,t):=(t_2\cos (\beta s)+t_3\sin (\beta s)). \end{aligned}$$

(53)

Assume, moreover, that $\beta =0$. Then, $\breve{\mu }_{1,n}=0$ and $\breve{\mu }_{2,n}$ is given by

$$\begin{aligned} \breve{\mu }_{2,n}=-\left\| \frac{\kappa }{2}\right\| ^2+\left\| \frac{{\dot{\kappa }}}{2}t_2\psi _n+(\tau -\varepsilon )\partial _\varphi \psi _n\right\| ^2. \end{aligned}$$

(54)

Proof

In order to compute the differential operator in the r.h.s. of (48) applied to $\psi _n$, since $\partial _s \psi _n=0$, we only need the expansions of $f_{0,j}$ for $j\in \{0,1,2\}$:

$$\begin{aligned} f_{0,0}(\delta )=&\,\left( -\frac{\kappa ^2}{4}\delta ^2 +O(\delta ^3)\right) +\left( \frac{\ddot{g}_1}{2}\delta +\frac{\ddot{g}_2-3g_1{\ddot{g}_1}}{2}\delta ^2+O(\delta ^3) \right) +\left( \frac{{\dot{\Xi }}{\tilde{g}}_1}{2}\delta ^2+O(\delta ^3) \right) \\&+\,\left( -\beta \dot{{\tilde{g}}}_1\delta +(3\beta g_1\dot{{\tilde{g}}}_1-\beta \dot{{\tilde{g}}}_2+\dot{{\tilde{g}}}_1\Xi )\delta ^2+O(\delta ^3) \right) +\left( \frac{-5{\dot{g}}_1^2}{4}\delta ^2+O(\delta ^3) \right) \\&+\,\left( \frac{5\beta {\dot{g}}_1{\tilde{g}}_1}{2}\delta ^2+O(\delta ^3) \right) +\left( \frac{-\beta ^2g_1}{2}\delta +\frac{\beta ^2(3g_1^2-g_2)+2\beta \Xi g_1}{2}\delta ^2+O(\delta ^3) \right) \\&+\,\left( \frac{-5\beta ^2{\tilde{g}}_1^2}{4}\delta ^2+O(\delta ^3) \right) \ . \end{aligned}$$

This in turn gives

$$\begin{aligned} f_{0,0}(\delta )=&\,\left( \frac{\ddot{g}_1}{2}-\frac{\beta ^2g_1}{2}-\beta \dot{{\tilde{g}}}_1\right) \delta +\Bigg [\frac{1}{2}\left( \ddot{g}_2-3g_1{\ddot{g}_1}+{\dot{\Xi }}{\tilde{g}}_1+5\beta {\dot{g}}_1{\tilde{g}}_1+\beta ^2(3g_1^2-g_2)\right) \\&-\,\frac{5}{4}\left( \beta ^2{\tilde{g}}_1^2+{\dot{g}}_1^2\right) +\beta \Xi g_1+3\beta g_1\dot{{\tilde{g}}}_1-\beta \dot{{\tilde{g}}}_2+\Xi \dot{{\tilde{g}}}_1-\frac{\kappa ^2}{4}\bigg ]\delta ^2+O(\delta ^3). \end{aligned}$$

The higher-order terms are

$$\begin{aligned} f_{0,1}(\delta )=&\,\left( 2\beta ^2{\tilde{g}}_1-2\beta {\dot{g}}_1-{\dot{\Xi }}\right) \delta \\&+\,\left( 2\Xi {\dot{g}}_1+2{\dot{\Xi }}g_1-4\beta \Xi {\tilde{g}}_1-2\beta {\dot{g}}_2+6\beta g_1{\dot{g}}_1+2\beta ^2{\tilde{g}}_2-6\beta ^2g_1{\tilde{g}}_1)\right) \delta ^2+O(\delta ^3) \end{aligned}$$

and

$$\begin{aligned} f_{0,2}(\delta )=(2\beta \Xi +2\beta ^2g_1)\delta -\left( \Xi ^2+4\beta \Xi g_1+\beta ^2(3g_1^2-2g_2)\right) \delta ^2+O(\delta ^3) \end{aligned}$$

Let us now compute $\breve{\mu }_{1,n}$ when $\beta \ne 0$. Notice that for any function ${\mathbb {R}}\times \omega \ni (s,t)\rightarrow F(s,t)$ such that ${\dot{F}}(\cdot ,t)$ is integrable for every $t\in \omega$, we have $\langle {\dot{F}}\psi _n,\psi _n\rangle =0$ for every $1\le n\le \infty$. Therefore, we have

$$\begin{aligned} \langle \psi _n|f_{0,0}\psi _n\rangle= & {} -\,\frac{1}{2}\beta ^2\delta \int _{{\mathbb {R}}\times \omega } g_1 \psi _n^2+O(\delta ^2) \ \text{ and } \ \langle \psi _n|f_{0,1}\partial _\varphi \psi _n\rangle \\= & {} 2\beta ^2\delta \int _{{\mathbb {R}}\times \omega } {\tilde{g}}_1(\partial _\varphi \psi _n) \psi _n+O(\delta ^2). \end{aligned}$$

An integration by parts provides

$$\begin{aligned} \langle \psi _n|f_{0,2}\partial ^2_\varphi \psi _n\rangle =&\,\int _{{\mathbb {R}}\times \omega }2\delta \beta \Xi (\partial ^2_\varphi \psi _n)\psi _n+2\delta \beta ^2g_1\partial ^2_\varphi \psi _n \psi _n+O(\delta ^2)\\ =&\,-\int _{{\mathbb {R}}\times \omega }2\delta \beta \Xi (\partial _\varphi \psi _n)^2+2\delta \beta ^2\left( {\tilde{g}}_1(\partial _\varphi \psi _n) \psi _n+g_1(\partial _\varphi \psi _n)^2\right) +O(\delta ^2). \end{aligned}$$

Combining these results, we get

$$\begin{aligned} \langle \psi _n,W_\delta \psi _n\rangle =\left( \frac{-\beta ^2}{2}\int _{{\mathbb {R}}\times \omega } g_1 |\psi _n|^2-2\int _{{\mathbb {R}}\times \omega }(\beta \Xi +\beta ^2g_1)|\partial _\varphi \psi _n|^2\right) \delta +O(\delta ^2).\nonumber \\ \end{aligned}$$

(55)

from where one easily deduces the expression for $\breve{\mu }_{1,n}$.

Assume now that $\beta =0$. It is clear from (55) that $\mu _{1,n}=0$. Let us now compute the expression of $\mu _{2,n}$. We have

$$\begin{aligned} \langle \psi _n |f_{0,0} \psi _n \rangle =&\,\delta ^2\int _{{\mathbb {R}}\times \omega } \left( -\frac{\kappa ^2}{4}-\frac{3}{2}g_1 \ddot{g_1}+\left( \frac{1}{2}{\dot{\Xi }}\tilde{g_1}+\Xi \dot{\tilde{g_1}}\right) -\frac{5}{4}\dot{g_1}^2\right) \psi _n ^2+O(\delta ^3)\nonumber \\ =&\,\delta ^2\int _{{\mathbb {R}}\times \omega } \left( -\frac{\kappa ^2}{4}+\frac{1}{4}\dot{g_1}^2+\frac{1}{2}{\Xi } \dot{\tilde{g_1}}\right) \psi _n ^2+O(\delta ^3) \end{aligned}$$

(56)

where we have used an integration by parts in s and that $\int _{{\mathbb {R}}} {\dot{\Xi }}\tilde{g_1}+\dot{\tilde{g_1}}\Xi =0$. Moreover,

$$\begin{aligned} \langle \psi _n | f_{0,1}\partial _\varphi \psi _n \rangle ={\mathcal {O}}(\delta ^3) \ \text{ and } \ \langle f_{0,2}\partial ^2_\varphi \psi _n |\psi _n \rangle =\delta ^2\int _{{\mathbb {R}}\times \omega } \Xi ^2 (\partial _\varphi \psi _n)^2+O(\delta ^3).\nonumber \\ \end{aligned}$$

(57)

Combining (56) and (57) and integrating by parts, we deduce $\breve{\mu }_{2,n}$ by

$$\begin{aligned} \langle \psi _n |W_\delta \psi _n \rangle =&\,\delta ^2\int _{{\mathbb {R}}\times \omega } -\frac{\kappa ^2}{4}\psi _n ^2+\frac{1}{4}\dot{g_1}^2\psi _n ^2-{\Xi } \dot{g_1}(\partial _\varphi \psi _n)\psi _n +\Xi ^2 (\partial _\varphi \psi _n)^2+O(\delta ^3)\\ =&\,\delta ^2\left( -\left\| \frac{\kappa }{2}\right\| ^2+\int _{{\mathbb {R}}\times \omega }\left( \frac{{\dot{g}}_1\psi _n}{2}\right) ^2-{\Xi } \dot{g_1}(\partial _\varphi \psi _n)\psi _n +(\Xi \partial _\varphi \psi _n)^2\right) +O(\delta ^3)\\ =&\,\delta ^2\left( -\left\| \frac{\kappa }{2}\right\| ^2+\int _{{\mathbb {R}}\times \omega }\left( \frac{{\dot{g}}_1\psi _n}{2}-\Xi \partial _\varphi \psi _n\right) ^2\right) +O(\delta ^3). \end{aligned}$$

$\square$