%&amstex \input amstex \documentstyle{amsppt} \magnification=1200 \baselineskip=15 pt \TagsOnRight \loadbold \topmatter \title Weakly Coupled Bound States in Quantum Waveguides \endtitle \author W.~Bulla$^1$, F.~Gesztesy$^2$, W.~Renger$^2$, and B.~Simon$^3$ \endauthor \leftheadtext{W.~Bulla, F.~Gesztesy, W.~Renger, and B.~Simon} \thanks $^1$ Institute for Theoretical Physics, Technical University of Graz, A-8010 Graz, Austria. E-mail for W.B.: bulla\@itp.tu-graz.ac.at \endthanks \thanks $^2$ Department of Mathematics, University of Missouri, Columbia, MO 65211. E-mail for F.G.: mathfg\@ mizzou1.missouri.edu; e-mail for W.R.: walter\@mumathnx3.cs.missouri.edu \endthanks \thanks $^3$ Division of Physics, Mathematics, and Astronomy, California Institute of Technology, \linebreak Pasadena, CA 91125. This material is based upon work supported by the National Science Foundation under Grant No.~DMS-9401491. The Government has certain rights in this material. \endthanks \thanks To be submitted to {\it{Commun.~Math.~Phys.}} \endthanks \abstract We study the eigenvalue spectrum of Dirichlet Laplacians which model quantum waveguides associated with tubular regions outside of a bounded domain. Intuitively, our principal new result in two dimensions asserts that any domain $\Omega$ obtained by adding an arbitrarily small ``bump'' to the tube $\Omega_0=\Bbb R\times(0,1)$ (i.e., $\Omega\supsetneqq\Omega_0$, $\Omega\subset\Bbb R^2$ open and connected, $\Omega=\Omega_0$ outside a bounded region) produces at least one positive eigenvalue below the essential spectrum $[\pi^2,\infty)$ of the Dirichlet Laplacian $-\Delta ^{D}_\Omega$. For $|\Omega\backslash\Omega_0|$ sufficiently small ($|\,.\,|$ abbreviating Lebesgue measure), we prove uniqueness of the ground state $E_\Omega$ of $-\Delta ^{D}_\Omega$ and derive the ``weak coupling'' result $E_\Omega=\pi^2-\pi^4|\Omega\backslash\Omega_0|^2 +O(|\Omega\backslash\Omega_0|^3)$. As a corollary of these results we obtain the following surprising fact: Starting from the tube $\Omega_0$ with Dirichlet boundary conditions at $\partial\Omega_0$, replace the Dirichlet condition by a Neumann boundary condition on an arbitrarily small segment $(a,b)\times\{1\}$, $a0$ is. \endabstract \endtopmatter \document \flushpar {\bf {\S 1. Introduction}} \vskip 0.1in Our goal in this paper is to study the bound state spectra of the Dirichlet Laplacian $-\Delta ^{D}_\Omega$ for open regions $\Omega\subset\Bbb R^n$ which are tubes outside of a bounded region (quantum waveguides). (Following the traditional notation in quantum physics, we denote the Laplacian by $-\Delta$ as opposed to $\Delta$ in the following.) In particular, let $\Omega_0\subset\Bbb R^2$ be defined by $$ \Omega_0 = \Bbb R\times (0,1). $$ Consider open connected sets $\Omega$ such that: \roster \item"\rom{(i)}" For some $R>0$, $\Omega\cap\{x\in\Bbb R^2\mid |x|>R\} =\Omega_0\cap\{x\in\Bbb R^2\mid |x|>R\}$. \item"\rom{(ii)}" $\Omega_0 \subset\Omega$, $\Omega_0\neq \Omega$. \endroster Because of condition (i), $$ \sigma_{\text{\rom{ess}}}(-\Delta ^{D}_\Omega)=\sigma_{\text{\rom{ess}}} (-\Delta^D_{\Omega_0})=[\pi^2,\infty). \tag 1 $$ Then one of our main goals will be to prove \proclaim{Theorem 1.1} If $\Omega$ obeys {\rom{(i), (ii)}}, then $-\Delta ^{D}_\Omega$ has at least one eigenvalue in $(0,\pi^2)$. \endproclaim Actually, the eigenvalue lies in $[\frac{\pi^2}{4 R^2},\pi^2)$ since $\Omega\subset {\Bbb R}\times (-R,R)$ implies \linebreak $\inf\,\text{spec}(-\Delta ^{D}_\Omega)\geq \inf\,\text{spec}(-\Delta ^{D}_{{\Bbb R}\times (-R,R)}) =\frac{\pi^2}{4 R^2}$. We will focus especially on the particular case $$ \Omega =\Omega_\lambda, $$ where $$ \Omega_\lambda =\{(x,y)\in\Bbb R^2\mid 00$. \proclaim{Theorem 1.2} Let $\Omega_\lambda$ be given by {\rom{(2)}} where $f$ is a $C^\infty_0(\Bbb R)$ function with \linebreak $\int_{\Bbb R} f(x)\, dx >0$. Then for all small positive $\lambda$, $-\Delta^D_{\Omega_\lambda}$ has a unique eigenvalue $E(\lambda)$ in $(0,\pi^2)$, it is simple, $E(\lambda)$ is analytic at $\lambda=0$, and $$ E(\lambda)=\pi^2- \pi^4\lambda^2 \biggl(\int_{\Bbb R} f(x)\,dx\biggr)^2 + O(\lambda^3). \tag 3 $$ \endproclaim This is the main result of this paper, which we'll prove in Section 2 using a calculation in the appendix. The technique used in our proof is closely patterned after the theory of bound states of $-\frac{d^2} {dx^2} +\lambda V(x)$ for $\lambda$ small as developed in [2],[9],[10],[13]. The key idea there is that $(-\frac{d^2}{dx^2}+k^2)^{-1}$ has a well-behaved limit as $k\downarrow 0$ except for a divergent rank one piece. In exactly the same way, $(-\Delta^D_{\Omega_0}-\pi^2 +k^2)^{-1}$ has a nice limit as $k\downarrow 0$ except for a rank one piece. \vskip 0.2in Theorem 1.1 (or 1.2) leads to the following remarkable result which, roughly speaking, asserts that if on an arbitrarily small segment in the boundary $\partial\Omega_0$ of $\Omega_0$ the original Dirichlet boundary condition is replaced by a Neumann boundary condition, at least one additional eigenvalue is instantly created in the interval $(0,\pi^2)$. \proclaim{Corollary 1.3} Let $\Omega_0=\Bbb R\times (0,1)$ and denote by $H(a,b)$ in $L^2(\Omega_0)$ the Laplacian on $\Omega_0$ with a Neumann boundary condition on the segment $(a,b)\times\{1\},\, -\infty0$ is. \endproclaim \demo{Proof} Clearly $H(a,b)\geq 0$ and $\sigma_{\text{\rom{ess}}}(H(a,b))=[\pi^2,\infty)$. Enlarge $\Omega_0$ to $\Omega_\lambda$ of the type (2) with $\lambda>0$ sufficiently small and some $0\leq f\in C^\infty(\Bbb R)$ with ${\text{supp}}(f)=[a,b]$, $f>0$ on $(a,b)$. By Theorem 1.1, $-\Delta^D_{\Omega_\lambda}$ has at least one eigenvalue $E_\lambda\in(0,\pi^2)$. Next, decouple $\Omega_0$ and $\Omega_\lambda\backslash\Omega_0$ by a Neumann boundary condition along the segment $(a,b)\times\{1\}$. Denoting the resulting Laplace operator by $\hat{H}_{\Omega_\lambda}$, we obtain the direct sum decomposition $\hat{H}_{\Omega_\lambda}=H(a,b)\oplus-\widetilde{\Delta}(a,b)$ with respect to $L^2(\Omega_\lambda)=L^2(\Omega_0)\oplus L^2(\Omega_\lambda\backslash\Omega_0)$, where $-\widetilde{\Delta}(a,b)$ has Dirichlet (resp. Neumann) boundary conditions on $\partial\Omega_\lambda\backslash\partial\Omega_0$ (resp. $(a,b)\times\{1\}$). By Neumann decoupling (see, e.g., [11], p.270) $$ 0\leq\inf\,\text{spec}(\hat{H}_{\Omega_\lambda})\leq \inf\,\text{spec}(-\Delta^D_{\Omega_\lambda})\leq E_\lambda <\pi^2. $$ Choosing $f$ appropriately such that $\inf\,\text{spec}(-\widetilde{\Delta}(a,b))>\pi^2$ (e.g., choose $f$ such that $\Omega_\lambda\backslash\Omega_0$ is a smoothed out rectangle of the type $(a,b)\times(1,1+c)$ with $00$. Indeed, if $f$ is very close to $-1$ on a long region, we expect that $-\Delta^D_{\Omega_{\lambda=1}}$ has no bound states. (7) We owe to Mark Ashbaugh the following observation: \proclaim{Corollary 1.4} Let $\tilde{\Omega}=\{\Bbb R\times (0,2)\}\backslash \{\Bbb R\times\{1\}\} \cup\{(a,b)\times\{1\}\}$, $-\infty0$ of the slit $(a,b)\times\{1\}$. \endproclaim \demo{Proof} $\tilde\Omega$ has reflection symmetry under $y\to 2-y$. Thus, $-\Delta^D_{\tilde{\Omega}}$ is a direct sum of operators even and odd under this symmetry and so $-\Delta^D_{\tilde{\Omega}}\cong H(a,b)\oplus -\Delta^D_{\Omega_0}$, where $H(a,b)$ is the operator in Corollary 1.3 (since even is equivalent to Neumann and odd to Dirichlet boundary conditions) and $\cong$ abbreviates unitary equivalence. Since $\sigma_{\text{\rom{ess}}}(H(a,b)) =\sigma_{\text{\rom{ess}}}(-\Delta^D_{\Omega_0})=[\pi^2, \infty) =\sigma_{\text{\rom{ess}}}(-\Delta^D_{\tilde{\Omega}})$, it suffices to prove that $-\Delta^D_{\tilde{\Omega}}$ has spectrum in $[\frac{\pi^2}{4},\pi^2)$. Let $\widehat{\Omega}=\Omega_0 \cup \{(x,y)\in \Bbb R^2\mid a0$ such that the triangle spanned by the points $(-a,1)$, $(a,1)$ and $(0,1+b)$ is in $\Omega$. Define on $\Omega$ $$ \hat{\psi}_{\beta,\delta}(x,y)= \cases \sin(\pi y) e^{-\delta(|x|-a)}, &\qquad |x|>a,\quad 00$. This trial function certainly vanishes on $\partial\Omega$ and at $\infty$ and it is in the form domain $Q(-\Delta^D_{\Omega})$. By a straightforward calculation we obtain $$ E(\hat{\psi}_{\beta,\delta})= \frac{(\nabla\hat{\psi}_{\beta,\delta}, \nabla\hat{\psi}_{\beta,\delta})} {(\hat{\psi}_{\beta,\delta},\hat{\psi}_{\beta,\delta})} =\pi^2(1-2a\delta\beta)+ O(\beta^2\delta)+O(\delta^2). $$ If we first choose $\beta$ and then $\delta$ small enough we get $$ E(\hat{\psi}_{\beta,\delta})<\pi^2 =\inf \sigma_{\text{\rom{ess}}}(-\Delta^{D}_\Omega). $$ Since $\inf\,\text{spec}(-\Delta^{D}_\Omega) 0$ this proves Theorem 1.1. Note that $\sin(\pi y)$ in (4) represents the function $u(x,y)$ in (7) used prominently in Lemma 2.2 and in the proof of Theorem 1.2. \endremark \vskip0.1in Spectral properties of quantum waveguides received considerable attention recently. While a complete bibliography is beyond the scope of this paper, the interested reader is referred to [1],[3]--[7],[12] and the literature cited therein. In particular, a weak coupling mechanism different from the one discussed in the present paper, based on arbitrarily small bending of tubes, has been studied in detail in [4] and[12]. \vskip0.1in Without entering into further details we remark that Theorem 1.1 admits a variety of extensions. For instance, $\Omega$ and $\Omega_0$ need not coincide outside a sphere of radius $R$ as assumed in our condition (i), $\Omega$ only needs to approach $\Omega_0$ asymptotically (still assuming condition (ii)) since equality of the essential spectra of $-\Delta^{D}_{\Omega}$ and $-\Delta^{D}_{\Omega_0}$ as recorded in (1) is the crucial property in question. In addition, $\Omega$ could have various further branches running off to infinity as long as the asymptotic width of these branches is less than or equal to one in order to guarantee the validity of (1). Moreover, combining our results with the ones in [4] and [12] produces the same ground state effect for a bent tube of constant width one (and again additional bent branches running off to infinity of asymptotic widths not larger than one can be accommodated). \vskip 0.3in \flushpar {\bf {\S 2. Weak Coupling Analysis}} \vskip 0.1in We'll study $-\Delta^D_{\Omega_\lambda}$ by a perturbation method. Since $L^2 (\Omega_\lambda)$ is $\lambda$ dependent, it is difficult to use perturbation theory directly, so we'll map all the operators onto the same space. Let $U_\lambda:L^2(\Omega_\lambda)\to L^2(\Omega_0)$ by $$ (U_\lambda\psi)(x,y)=\sqrt{1+\lambda f(x)}\,\psi(x, (1+\lambda f(x))y). $$ Then $U_\lambda$ is unitary and $$ H_\lambda =U_\lambda(-\Delta^D_\lambda)U_\lambda^{-1}-\pi^2 $$ acts in $L^2(\Omega_0)$. We subtract $\pi^2$ so that $\sigma_{\text{\rom{ess}}}(H_\lambda)=[0,\infty)$. A straightforward calculation found in the appendix (cf. (A.6)) proves that $$ H_\lambda =H_0 +\lambda \sum^3_{i=1} A^*_i B_i +\lambda^2 \sum^8_{i=4} A^*_i B_i\,, \tag 5 $$ where each $A_i$ and $B_i$ is a first-order differential operator with coefficients which have compact support and ($g$ is a $C^\infty$ function chosen such that $g\equiv 1$ on $\text{supp}\,f$) $$\alignat3 &(\text{i}) && \qquad A^*_1 =2f(x)\frac{\partial}{\partial y}, && \qquad B_1 =g(x)\frac{\partial}{\partial y}, \\ &(\text{ii}) && \qquad A^*_2 =f''(x), && \qquad B_2 =g(x)\biggl( y\frac{\partial} {\partial y} +\frac{1}{2}\biggr ), \\ &(\text{iii}) && \qquad A^*_3 = \biggl(2y \frac{\partial}{\partial y} + 1\biggr) g(x), && \qquad B_3 =f'(x)\frac{\partial}{\partial x} \endalignat $$ (we'll see below that to leading order only $A^*_1 B_1$ matters). Rewrite (5) as follows. Define $C(\lambda)$, $D:L^2(\Omega_0) \to L^2(\Omega_0)\otimes\Bbb C^8$ by $$\align (C\varphi)_i &= \cases A_i \varphi & \qquad i=1,2,3 \\ \lambda A_i \varphi, & \qquad i=4,\dots, 8 \endcases \\ (D\varphi)_i &= \,\,\, B_i \varphi, \qquad\qquad i=1,\dots,8. \endalign $$ Then (5) becomes $$ H_\lambda = H_0 +\lambda C^*(\bar{\lambda}) D. $$ \proclaim{Lemma 2.1} Let $k\in\Bbb C$, $\text{\rom{Re}}\,k>0$ and $\lambda\in\Bbb R$. Then $-k^2$ is an eigenvalue of $H_\lambda$ if and only if $$ \lambda D(H_{0}+k^2)^{-1}C^* \equiv Q_\lambda $$ has $-1$ as an eigenvalue. \endproclaim \demo{Proof} If $Q_\lambda \psi\equiv-\psi$, then $-\lambda(H_0 +k^2)^{-1} C^* \psi\equiv\varphi$ is seen to satisfy $H_\lambda \varphi=-k^2 \varphi$. Conversely, if $H_\lambda \varphi=-k^2 \varphi$, then $\varphi\in Q(H_\lambda)\subset D(D)$ so $\psi=D\varphi$ is in $L^2(\Omega_0)$ and $Q_\lambda\psi=-\psi$. \qed \enddemo \proclaim{Lemma 2.2} Let $h$ be a $C^\infty$ function of compact support in $\Bbb R$. Then $$ h(H_{0}+k^2)^{-1}h =\frac{(hu,\,\cdot\,)hu}{2k} + A(k), \tag 6 $$ where $u$ is the function $$ u(x,y)=2^{\frac12}\sin(\pi y) \tag 7 $$ and $A(k)$ is a bounded operator--valued function of $k$, which can be analytically continued from $\{k\in\Bbb C\mid\text{\rom{Re}}\,k>0\}$ to a region that includes $k=0$. Indeed, even $(H_{0} +1)^{1/2} A(k) (H_{0}+1)^{1/2}$ has an analytic continuation into such a region. Moreover, $(H_{0}+1)^{1/2}A(k)(H_{0}+1)^{1/2}$ is bounded uniformly in $\{k\in\Bbb C\mid |\text{\rom{Arg}}\, k| <\pi/3\} \cup \{$a small disk about $k=0 \}$ \rom($\frac{\pi}{3}$ can be replaced by any number strictly less than $\frac{\pi}{2}$\rom). \endproclaim \demo{Proof} Let $\Cal H_0 \subset L^2 (\Omega_0)$ be the space of $L^2(\Omega_0)$ functions of the form $\varphi (x)\sin(\pi y)$, $\sin(\pi y)$ being chosen as the lowest eigenfunction of $(-\frac{d^2}{dy^2})^D$. Let $P_0$ be the projection onto $\Cal H_0$. Then $(H_{0}+k^2)^{-1}(1-P_0)$ has an analytic continuation into the region $\{k\in\Bbb C\mid -k^2\in\Bbb C\backslash [3\pi^2, \infty)\}$ since the lowest point in the spectrum of $H_{0}(1-P_0)\restriction (1-P_0) L^2(\Omega_0)$ is $3\pi^2$. On the other hand, $h(H_{0}+k^2)^{-1}P_0 h$ has the explicit integral kernel: $$ (2k)^{-1}h(x)h(x')u(y)u(y')e^{-k|x-x'|} =a_1(k) + a_2(k), $$ where $a_1(k)$ is obtained by replacing $e^{-k|x-x'|}$ by $1$ and $a_2(k)$ by using $e^{-k|x-x'|}-1$ in its place. The first term is the explicit rank one piece in (6) and the second term is analytic as a Hilbert-Schmidt kernel at $k=0$. It is easy to modify this argument to accommodate the extra factors of $(H_{0}+1)^{1/2}$ and prove the boundedness. \qed \enddemo \demo{Proof of Theorem 1.2} Consider first the operator on $$ L^2 (\Omega_0)\otimes\Bbb C^8 :L_0=(C(\lambda=0)u, \,\cdot\,)Du, $$ where $u$ is given by (7). Then $L_0$ is a rank one operator, so it has a single eigenvalue at $$ e_0 =\text{Tr}(L_0). \tag 8 $$ But $C_i(0)=0$ for $i=4,\dots, 8$, $B_3 u=0$, and $(A_2 u, B_2 u)=0$ since $\int_{\Bbb R} f''(x)\,dx=0$. It follows that $$\align e_0 =(A_1 u, B_1 u) &=-2 \biggl( \int_{\Bbb R} f(x)\, dx\biggr) \int_0^1 2\pi^2\cos^2(\pi y)\, dy \\ &=-2\pi^2 \int_{\Bbb R} f(x)\, dx. \endalign $$ Let $k=\lambda\ell$. Then by Lemma 2.2, $$ \lambda D(H_{0}+k^2)^{-1} C^*=Q(\lambda,\ell) $$ has the form: $$ Q(\lambda,\ell) =\frac{1}{2\ell}\,L_\lambda + \lambda M(\lambda,\ell), $$ where \roster \item"\rom{(i)}" $L_\lambda$ is rank one and $L_\lambda =L_0+\lambda \tilde{L}$. \item"\rom{(ii)}" $M(\lambda,\ell)= DA(\lambda,\ell)C^*$, where $A$ is given by Lemma 2.2 and $h$ is chosen such that $h\equiv 1$ in a neighborhood of $\text{supp}\,f$. \endroster By Lemma 2.2, we are interested in when $Q(\lambda, \ell)$ has $-1$ as an eigenvalue for $\lambda >0$ and $\ell >0$. Since $M$ is uniformly bounded in $\lambda$ on a sector about $(0,\infty)$, this can happen where $\lambda$ is small if $\frac{e_0}{2\ell}$ is near $-1$, that is, $\ell$ is near $\frac{-e_0}{2} >0$ since $\int_{\Bbb R} f(x)\, dx> 0$ by hypothesis. For such $\ell$ and $\lambda$ small, $Q(\lambda, \ell)$ has exactly one eigenvalue near $-1$, call it $E(\lambda,\ell)$, which by eigenvalue perturbation theory ([8], Ch.2, [11], Ch.XII) is jointly analytic in $\lambda,\ell$. Let $$ F(\lambda,\ell)=2\ell (E(\lambda,\ell)+1). $$ Since $\left. 2\ell Q(\lambda, \ell)\right|_{\lambda=0}$ is independent of $\ell$, $\left. \frac{\partial(2\ell E(\lambda,\ell))}{\partial\ell} \right|_{\lambda=0}=0$ and so \linebreak $\left. \frac{\partial F(\lambda,\ell)} {\partial\ell}\right|_{\lambda=0,\ell=-e_0 /2}=2 \neq 0$. It follows by the implicit function theorem that for $\lambda$ sufficiently small, there is an analytic function $\ell(\lambda)= -\frac{e_0}{2} + O(\lambda)$ so that for $\lambda > 0$ and $\ell$ in the sector $|\text{Arg}\,\ell| <\frac{\pi}{3}$, $-1$ is an eigenvalue of $Q(\lambda, \ell)$ if and only if $\ell =\ell(\lambda)$. Since $H_\lambda$ for $\lambda$ real has only real eigenvalues, $\ell(\lambda)$ must be real for $\lambda >0$. Thus $H_\lambda$ has a unique eigenvalue, $e(\lambda)$, in $(-\infty, 0)$ given by $e(\lambda)= -\lambda^2 (-\frac{e_0}{2})^2 + O(\lambda^3)$ as claimed. \qed \enddemo \vskip 0.3in \flushpar {\bf Acknowledgments.} It is a pleasure to thank Mark Ashbaugh and Thomas Hoffmann-Ostenhof for numerous discussions on this subject. W.~B.~is indebted to the Department of Mathematics of the University of Missouri, Columbia for the hospitality extended to him during a stay in the spring of 1994. Furthermore, he gratefully acknowledges the financial support for this stay by the Technische Universit\"at Graz, Austria. \pagebreak \vskip 0.3in \flushpar {\bf{Appendix: Calculating $\widetilde{H}_\lambda =U_\lambda(-\Delta^D_{\Omega_\lambda})U^{-1}_\lambda$.}} \vskip 0.1in We use coordinates $(x,y)$ on $\Omega_0$ and $(s,u)$ on $\Omega_\lambda$. Thus $U_\lambda$ becomes $$\alignat2 U_\lambda: \quad & L^2(\Omega_\lambda) && \quad \to L^2(\Omega_0) \\ & \tilde{\psi}(s,u) && \quad \mapsto \psi(x,y) = \sqrt{1+\lambda f(x)}\,\tilde{\psi}(x,(1+\lambda f(x))y). \tag A.1 \endalignat $$ The coordinate transformation is $$ x=s, \qquad y = (1+\lambda f(s))^{-1} u. \tag A.2 $$ The form associated with $-\Delta^D_{\Omega_\lambda}$ is given by $$\alignat2 q^D_{\Omega_\lambda}: \quad & Q(-\Delta^D_{\Omega_\lambda}) \times Q(-\Delta^D_{\Omega_\lambda}) && \quad \to \Bbb C \\ & (\varphi,\psi) && \quad \mapsto (\nabla\varphi,\nabla\psi), \tag A.3 \endalignat $$ where $Q(-\Delta^D_{\Omega_\lambda}) = H_0^{1,2}(\Omega_\lambda)$ is the local Sobolev space (i.e., the completion of $C^\infty_0$ under the norm $\|\cdot\|_{\nabla} = (\|\nabla\cdot\|^2 +\|\cdot\|^2)^{1/2}$). By unitary equivalence, the quadratic form associated with $\widetilde{H}_\lambda$ is $$\alignat2 q_{\widetilde{H}_\lambda}: \quad & U_\lambda Q(-\Delta^D_{\Omega_\lambda}) \times U_\lambda Q(-\Delta^D_{\Omega_\lambda}) && \quad \to \Bbb C \\ & (\varphi,\psi) && \quad \mapsto q^D_{\Omega_\lambda} (U_\lambda^{-1}\varphi, U_\lambda^{-1}\psi). \tag A.4 \endalignat $$ The form domain of $\widetilde{H}_\lambda$ is $U_\lambda H_0^{1,2}(\Omega_\lambda)= \overline{U_\lambda C^\infty_0 (\Omega_\lambda)}$, where the bar denotes completion under the norm $\|\cdot\|_q =(q_{\widetilde{H}_\lambda}(\,\cdot\, ,\, \cdot\,) + \|\cdot\|^2)^{1/2}$. Next we calculate the quadratic form $q_{\widetilde{H}_\lambda}(\varphi,\psi)$ for $\varphi,\psi \in C^\infty_0 (\Omega_0)$. We use the shorthand $c(x)=1+\lambda f(x)$ and use subscripts to denote partial derivatives. $$\align q_{\widetilde{H}_\lambda}(\varphi,\psi) &= q^D_{\Omega_\lambda} (U_\lambda^{-1}\varphi,U_\lambda^{-1}\psi) \\ &= \int\limits_{\Omega_\lambda} (\partial_s c(s)^{-1/2}\, \overline{\varphi(s,\tfrac{u}{c(s)})}\,) (\partial_s c(s)^{-1/2} \psi(s,\tfrac{u}{c(s)}))\,dsdu \\ & \qquad + \int\limits_{\Omega_\lambda} (\partial_u c(s)^{-1/2}\, \overline{\varphi(s,\tfrac{u}{c(s)})}\,)(\partial_u c(s)^{-1/2} \psi(s,\tfrac{u}{c(s)}))\,dsdu \\ &= \int\limits_{\Omega_0} \biggl\{\biggl[-\frac{1}{2}\, \frac{c'(x)}{c(x)}\, \overline{\varphi(x,y)} + \overline{\varphi_x(x,y)} -\frac{yc'(x)} {c(x)}\, \overline{\varphi_y(x,y)} \biggr] \\ & \qquad \biggl[-\frac{1}{2}\, \frac{c'(x)}{c(x)}\, \psi(x,y) + \psi_x(x,y) -\frac{yc'(x)}{c(x)}\, \psi_y(x,y) \biggr] \\ & \qquad+\frac{1}{c(x)^2}\, \overline{\varphi_y(x,y)}\, \psi_y(x,y) \biggr\}\, dxdy \endalign $$ $$ \align \hskip 6.4ex &= \int\limits_{\Omega_0} \biggl\{\overline{\varphi_x(x,y)}\, \psi_x(x,y) + \frac{1+y^2c'(x)^2}{c(x)^2}\, \overline{\varphi_y(x,y)} \, \psi_y(x,y) \\ & \qquad -\frac{yc'(x)}{c(x)}\, (\overline{\varphi_x(x,y)}\, \psi_y(x,y) +\overline{\varphi_y(x,y)}\, \psi_x(x,y)) \\ & \qquad-\frac{c'(x)}{2c(x)}\, (\overline{\varphi_x(x,y)}\, \psi(x,y) +\overline{\varphi(x,y)}\, \psi_x(x,y)) \\ & \qquad +\frac{yc'(x)^2}{2c(x)^2}\, (\overline{\varphi_y(x,y)}\, \psi(x,y) +\overline{\varphi(x,y)}\, \psi_y(x,y)) \\ & \qquad+\frac{c'(x)^2}{4c(x)^2}\, \overline{\varphi(x,y)}\, \psi(x,y)\biggr\}\, dxdy. \tag A.5 \endalign $$ By partial integration we get the operator $$\align \widetilde{H}_\lambda &= -\frac{\partial^2}{\partial x^2} -\frac{1+y^2\lambda^2 f'(x)^2} {c(x)^2}\,\frac{\partial^2}{\partial y^2} +\biggl(\frac{y\lambda f''(x)} {c(x)}-\frac{3y\lambda^2 f'(x)^2}{c(x)^2} \biggr) \frac{\partial} {\partial y} \\ & \qquad + \frac{2y\lambda f'(x)}{c(x)}\, \frac{\partial}{\partial x} \, \frac{\partial}{\partial y} + \frac{\lambda f'(x)}{c(x)}\, \frac{\partial}{\partial x} + \frac{\lambda f''(x)}{2 c(x)} - \frac{3 \lambda^2 f'(x)^2}{4 c(x)^2} \\ &= -\Delta^D_{\Omega_0} +\lambda \biggl[ 2 f(x)\frac{\partial^2}{\partial y^2} +y f''(x)\frac{\partial}{\partial y} +2y f'(x) \frac{\partial}{\partial x}\, \frac{\partial}{\partial y} +f'(x)\frac{\partial}{\partial x} +\frac{f''(x)}{2} \biggr] \\ & -\lambda^2\biggl[\frac{3f(x)^2 +2\lambda f(x)^3 +y^2 f'(x)^2} {(1+\lambda f(x))^2}\, \frac{\partial^2}{\partial y^2} +\biggl(\frac{y f(x)f''(x)} {(1+\lambda f(x))}+\frac{3y f'(x)^2}{(1+\lambda f(x))^2}\biggr) \frac{\partial}{\partial y} \\ & \quad +\frac{2y f(x)f'(x)}{(1+\lambda f(x))}\, \frac{\partial}{\partial x}\, \frac{\partial}{\partial y} +\frac{f(x)f'(x)}{(1+\lambda f(x))}\, \frac{\partial}{\partial x} +\frac{f(x)f''(x)}{2(1+\lambda f(x))} +\frac{3f'(x)^2}{4(1+\lambda f(x))^2} \biggr]. \\ \tag A.6 \endalign $$ Since we assumed $f\in C_0^\infty(\Bbb R)$, clearly $C^\infty_0(\Omega_0)\subset D(\widetilde{H}_\lambda)=U_\lambda D(-\Delta^D_{\Omega_\lambda})$. Actually, $U_\lambda C^\infty_0(\Omega_\lambda) = C^\infty_0(\Omega_0)$. 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