Optimization of the lowest eigenvalue of a soft quantum ring

Abstract

We consider the self-adjoint two-dimensional Schrödinger operator ${{\mathsf {H}}}_{\mu }$ associated with the differential expression $-\Delta -\mu$ describing a particle exposed to an attractive interaction given by a measure $\mu$ supported in a closed curvilinear strip and having fixed transversal one-dimensional profile measure $\mu _\bot$. This operator has nonempty negative discrete spectrum, and we obtain two optimization results for its lowest eigenvalue. For the first one, we fix $\mu _\bot$ and maximize the lowest eigenvalue with respect to shape of the curvilinear strip; the optimizer in the first problem turns out to be the annulus. We also generalize this result to the situation which involves an additional perturbation of ${{\mathsf {H}}}_{\mu }$ in the form of a positive multiple of the characteristic function of the domain surrounded by the curvilinear strip. Secondly, we fix the shape of the curvilinear strip and minimize the lowest eigenvalue with respect to variation of $\mu _\bot$, under the constraint that the total profile measure $\AA >0$ is fixed. The optimizer in this problem is $\mu _\bot$ given by the product of $\alpha$ and the Dirac $\delta$-function supported at an optimal position.

Appendix A: Traces on \(\Sigma _t\) and the lowest eigenvalue of \({{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}}\)

The first auxiliary result concerns a t-independent upper bound on $\Vert u|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)}$ for an $H^1$-function u in terms of the $L^2$-norms of u itself and its gradient.

Lemma A.1

Let the curve $\Sigma _t$ be as in (2.6). For any $\varepsilon > 0$, there exists a constant $C(\varepsilon ) > 0$ such that for any $t\in {{\mathcal {I}}}$ the following inequality

$$\begin{aligned} \Vert u|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)} \le \varepsilon \Vert \nabla u\Vert ^2_{L^2({{\mathbb {R}}}^2;{{\mathbb {C}}}^2)} + C(\varepsilon )\Vert u\Vert ^2_{L^2({{\mathbb {R}}}^2)} \end{aligned}$$

holds for all $u\in H^1({{\mathbb {R}}}^2)$. As a consequence, there is a constant $c > 0$ such that the inequality $\Vert u|_{\Sigma _t}\Vert _{L^2(\Sigma _t)}^2\le c\Vert u\Vert _{H^1({{\mathbb {R}}}^2)}^2$ holds for any $u\in H^1({{\mathbb {R}}}^2)$ and all $t\in {{\mathcal {I}}}$.

Proof

In view of the density of $C^\infty _0({{\mathbb {R}}}^2)$ in $H^1({{\mathbb {R}}}^2)$, it suffices to check the inequality for $C^\infty _0$-functions. For any $u\in C^\infty _0({{\mathbb {R}}}^2)$, $s\in [0,L)$, and $t\in {{\mathcal {I}}}$ we get using the fundamental theorem of calculus the following estimate:

$$\begin{aligned} \begin{aligned} {{\mathcal {D}}}_u(s,t)&:= \left| |u(\sigma (s)+t\nu (s))|^2 - |u(\sigma (s))|^2\right| \\&= \left| \int _0^1 \frac{{{\mathsf {d}}}}{{{\mathsf {d}}}r} (|u|^2(\sigma (s)+rt\nu (s))\,{{\mathsf {d}}}r\right| \\&\le \int _0^1 |\langle \nabla (|u|^2)(\sigma (s)+rt\nu (s)),t\nu (s)\rangle |\,{{\mathsf {d}}}r \\&\le 2|t| \int _0^1 \big |(|u|\cdot \nabla |u|) (\sigma (s)+rt\nu (s))\big |\,{{\mathsf {d}}}r.\\ \end{aligned} \end{aligned}$$

(A.1)

Let ${{\hat{\varepsilon }}} > 0$ be arbitrary. Applying the inequality ${{\hat{\varepsilon }}} a^2+{\hat{\varepsilon }}^{-1}b^2 \ge 2ab$ with $a,b > 0$, we can further estimate ${{\mathcal {D}}}_u(s,t)$ as follows:

$$\begin{aligned} \begin{aligned} {{\mathcal {D}}}_u(s,t)&\le |t| \int _0^1 \left( {{\hat{\varepsilon }}}\big |\nabla |u| (\sigma (s)+rt\nu (s))\big |^2 + {{\hat{\varepsilon }}}^{-1} \big |u (\sigma (s)+rt\nu (s))\big |^2\right) {{\mathsf {d}}}r\\&\le \int _{-d_-}^{d_+} \left( {{\hat{\varepsilon }}}\big |\nabla u (\sigma (s)+q\nu (s))\big |^2 + {{\hat{\varepsilon }}}^{-1} \big |u (\sigma (s)+q\nu (s))\big |^2\right) {{\mathsf {d}}}q, \end{aligned} \end{aligned}$$

(A.2)

where the substitution $q = rt$ was used, the interval of integration was enlarged and the diamagnetic inequality [29, Thm. 7.21] was applied.

By the trace theorem [34, Thm. 3.38] (see also [5, Lem. 2.6]) for any ${\tilde{\varepsilon }} > 0$, there exists ${\tilde{C}}({\tilde{\varepsilon }}) > 0$ such that

$$\begin{aligned} \Vert u|_\Sigma \Vert ^2_{L^2(\Sigma )} = \int _0^L |u(\sigma (s))|^2{{\mathsf {d}}}s \le {\tilde{\varepsilon }}\Vert \nabla u\Vert ^2_{L^2({{\mathbb {R}}}^2;{{\mathbb {C}}}^2)} + {\tilde{C}}({\tilde{\varepsilon }})\Vert u\Vert ^2_{L^2({{\mathbb {R}}}^2)} \end{aligned}$$

(A.3)

for any $u\in C^\infty _0({{\mathbb {R}}}^2)$. In view of (2.3), there exist constants $c_+> c_- > 0$ such that

$$\begin{aligned} 1+\kappa (s)t\in [c_-, c_+]\qquad \text {for all}\, s\in [0,L),\, t\in {{\mathcal {I}}}. \end{aligned}$$

In this way, we obtain the following simple estimate:

$$\begin{aligned} \Vert u|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)}= & {} \int _0^L |u(\sigma (s)+t\nu (s))|^2(1+\kappa (s)t)\,{{\mathsf {d}}}s \nonumber \\\le & {} c_+ \int _0^L |u(\sigma (s)+t\nu (s))|^2\,{{\mathsf {d}}}s. \end{aligned}$$

(A.4)

Furthermore, combining estimates (A.2), (A.3), and (A.4), we end up with

$$\begin{aligned} \begin{aligned} \Vert u|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)}&\le c_+ \int _0^L \left( |u(\sigma (s))|^2+ {{\mathcal {D}}}_u(s,t)\right) {{\mathsf {d}}}s\\&\le c_+{{\tilde{\varepsilon }}}\Vert \nabla u\Vert ^2_{L^2({{\mathbb {R}}}^2;{{\mathbb {C}}}^2)} + c_+ {\tilde{C}}({{\tilde{\varepsilon }}})\Vert u\Vert ^2_{L^2({{\mathbb {R}}}^2)}\\&\quad + \frac{{{\hat{\varepsilon }}} c_+}{c_-}\int _0^L\int _{-d_-}^{d_+} \big |\nabla u (\sigma (s)+q\nu (s))\big |^2(1+\kappa (s)q)\,{{\mathsf {d}}}q{{\mathsf {d}}}s\\&\quad + \frac{c_+}{{{\hat{\varepsilon }}} c_-} \int _0^L\int _{-d_-}^{d_+} \big |u (\sigma (s)+q\nu (s))\big |^2(1+\kappa (s)q)\,{{\mathsf {d}}}q{{\mathsf {d}}}s\\&\le \left( c_+{{\tilde{\varepsilon }}} + \frac{c_+{{\hat{\varepsilon }}}}{c_-}\right) \Vert \nabla u\Vert ^2_{L^2({{\mathbb {R}}}^2;{{\mathbb {C}}}^2)} + \left( c_+ {\tilde{C}}({{\tilde{\varepsilon }}}) + \frac{c_+}{c_-{{\hat{\varepsilon }}}}\right) \Vert u\Vert ^2_{L^2({{\mathbb {R}}}^2)}. \end{aligned} \end{aligned}$$

By choosing ${{\tilde{\varepsilon }}} >0$ and ${{\hat{\varepsilon }}}>0$ such that $\varepsilon =c_+{{\tilde{\varepsilon }}} + \frac{c_+{{\hat{\varepsilon }}}}{c_-}$, we get the sought claim. $\square$

In the next lemma, we prove that the function ${{\mathcal {I}}}\ni t\mapsto \Vert u|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)}$ is Hölder continuous with exponent $\frac{1}{2}$ for any $u\in H^1({{\mathbb {R}}}^2)$.

Lemma A.2

There exists a constant $c >0$ such that

$$\begin{aligned} \left| \Vert u|_{\Sigma _{t_2}}\Vert _{L^2(\Sigma _{t_2})}^2 - \Vert u|_{\Sigma _{t_1}}\Vert ^2_{L^2(\Sigma _{t_1})}\right| \le c|t_1 - t_2|^{1/2}\Vert u\Vert ^2_{H^1({{\mathbb {R}}}^2)} \end{aligned}$$

holds for all $u\in H^1({{\mathbb {R}}}^2)$ and for any $t_1,t_2\in {{\mathcal {I}}}$.

Proof

Throughout the proof $c > 0$ denotes a generic positive constant, which varies from line to line. In view of the density of $C^\infty _0({{\mathbb {R}}}^2)$ in $H^1({{\mathbb {R}}}^2)$, it suffices to check the inequality for $C^\infty _0$-functions. Without loss of generality, we may prove the claim only for the case that $t_1 = 0$ and that $t_2 = t \in (0,d_+)$. In this case we need to show that

$$\begin{aligned} {{\mathcal {S}}}_u(t) := \left| \Vert u|_{\Sigma _{t}}\Vert _{L^2(\Sigma _{t})}^2 - \Vert u|_{\Sigma }\Vert ^2_{L^2(\Sigma )}\right| \le ct^{1/2}\Vert u\Vert ^2_{H^1({{\mathbb {R}}}^2)}. \end{aligned}$$

By elementary means we obtain the bound

$$\begin{aligned} \begin{aligned} {{\mathcal {S}}}_u(t)&= \left| \int _0^L|u(\sigma (s)+t\nu (s))|^2(1+\kappa (s)t)\,{{\mathsf {d}}}s - \int _0^L|u(\sigma (s))|^2\,{{\mathsf {d}}}s \right| \\&\le ct\int _0^L|u(\sigma (s)+t\nu (s))|^2\,{{\mathsf {d}}}s + \int _0^L\left| |u(\sigma (s)) + t\nu (s))|^2 - |u(\sigma (s))|^2\right| {{\mathsf {d}}}s \end{aligned} \end{aligned}$$

Using Lemma A.1 and the estimate of (A.1), and taking (2.3) into account, we find

$$\begin{aligned} \begin{aligned} {{\mathcal {S}}}_u(t)&\le ct \Vert u|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)} +2\int _0^L\int _0^t|(|u|\cdot \nabla |u|)(\sigma (s)+q\nu (s))|\,{{\mathsf {d}}}q\,{{\mathsf {d}}}s\\&\le ct^{1/2}\Vert u\Vert ^2_{H^1({{\mathbb {R}}}^2)} + 2\int _0^L\int _0^t|(|u|\cdot \nabla |u|)(\sigma (s)+q\nu (s))|\,{{\mathsf {d}}}q\,{{\mathsf {d}}}s. \end{aligned} \end{aligned}$$

(A.5)

Furthermore, applying Cauchy–Schwarz inequality, diamagnetic inequality, and Lemma A.1 again, we get

$$\begin{aligned} \begin{aligned}&\int _0^L\int _0^t|(|u|\cdot \nabla |u|)(\sigma (s)+q\nu (s))|\,{{\mathsf {d}}}q\,{{\mathsf {d}}}s\\&\quad \le \left( \int _0^L\int _0^t|\nabla u(\sigma (s)+q\nu (s))|^2\,{{\mathsf {d}}}q{{\mathsf {d}}}s\right) ^{1/2} \left( \int _0^L\int _0^t|u(\sigma (s)+q\nu (s))|^2\,{{\mathsf {d}}}q{{\mathsf {d}}}s\right) ^{1/2}\\&\quad \le c\left( \int _0^L\int _0^t|\nabla u(\sigma (s)+q\nu (s))|^2(1+\kappa (s)q)\,{{\mathsf {d}}}q{{\mathsf {d}}}s\right) ^{1/2} \left( \int _0^t\Vert u|_{\Sigma _q}\Vert _{L^2(\Sigma _q)}^2\,{{\mathsf {d}}}q\right) ^{1/2}\\&\quad \le ct^{1/2}\Vert \nabla u\Vert _{L^2({{\mathbb {R}}}^2;{{\mathbb {C}}}^2)}\Vert u\Vert _{H^1({{\mathbb {R}}}^2)} \le ct^{1/2}\Vert u\Vert ^2_{H^1({{\mathbb {R}}}^2)}. \end{aligned} \end{aligned}$$

Combining the last estimate with (A.5), we arrive at the claim. $\square$

The purpose of the last lemma of the appendix is to establish the continuity of the lowest eigenvalue of ${{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}}$ with respect to t.

Lemma A.3

Let the curves $\Sigma _t$ be as in (2.6) and the number $\AA > 0$ be fixed. The operator-valued function

$$\begin{aligned} {{\mathcal {I}}}\ni t\mapsto {{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}} \end{aligned}$$

(A.6)

is continuous in the norm-resolvent topology and uniformly lower-semibounded, and as a consequence, the function ${{\mathcal {I}}}\ni t\mapsto \lambda _1(\alpha \delta _{\Sigma _t})$ is continuous.

Proof

Throughout the proof $c > 0$ again denotes a generic positive constant, which varies from line to line. Lemma A.1 in combination with the expression for the form (2.10) referring to $\mu = \alpha \delta _{\Sigma _t}$ shows that operators ${{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}}$, $t\in {{\mathcal {I}}}$, are uniformly bounded from below by some constant $\lambda _1< 0$. Without loss of generality, it suffices to prove that ${{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}}$ converges in the norm resolvent sense to ${{\mathsf {H}}}_{\alpha \delta _\Sigma }$ as $t\rightarrow 0$. The norm resolvent continuity of ${{\mathcal {I}}}\ni t\mapsto {{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}}$ at the other points of the interval ${{\mathcal {I}}}$ can be proven analogously.

We fix $\lambda _0 < \lambda _1$, use the notation ${{\mathsf {R}}}_t := ({{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}}-\lambda _0)^{-1}$, $t\in {{\mathcal {I}}}$ and claim that there is a constant $c > 0$ such that

$$\begin{aligned} \Vert {{\mathsf {R}}}_t - {{\mathsf {R}}}_0\Vert \le c|t|^{1/2}. \end{aligned}$$

(A.7)

In fact, first we note that

$$\begin{aligned} \begin{aligned} \Vert {{\mathsf {R}}}_t - {{\mathsf {R}}}_0\Vert&= \sup _{\Vert u\Vert , \Vert v\Vert = 1} \big |(({{\mathsf {R}}}_t - {{\mathsf {R}}}_0)u,v)_{L^2({{\mathbb {R}}}^2)}\big |\\&= \sup _{\Vert u\Vert , \Vert v\Vert = 1} \big |({{\mathsf {R}}}_t u,({{\mathsf {H}}}_{\alpha \delta _\Sigma } - \lambda _0) {{\mathsf {R}}}_0v)_{L^2({{\mathbb {R}}}^2)}- (({{\mathsf {H}}}_{\alpha \delta _{\Sigma _t}} - \lambda _0){{\mathsf {R}}}_tu, {{\mathsf {R}}}_0v)_{L^2({{\mathbb {R}}}^2)} \big |\\&= \sup _{\Vert u\Vert , \Vert v\Vert = 1} \big | {\mathfrak {h}}_{\alpha \delta _{\Sigma }}[{{\mathsf {R}}}_t u,{{\mathsf {R}}}_0v]- {\mathfrak {h}}_{\alpha \delta _{\Sigma _t}}[{{\mathsf {R}}}_t u,{{\mathsf {R}}}_0v] \big |. \end{aligned} \end{aligned}$$

The estimate (A.7) would follow if we prove that

$$\begin{aligned} \big | {\mathfrak {h}}_{\alpha \delta _{\Sigma }}[f,g]- {\mathfrak {h}}_{\alpha \delta _{\Sigma _t}}[f,g] \big |\le c|t|^{1/2}\left( \Vert f\Vert ^2_{H^1({{\mathbb {R}}}^2)} +\Vert g\Vert ^2_{H^1({{\mathbb {R}}}^2)} \right) ,\quad f,g\in H^1({{\mathbb {R}}}^2),\nonumber \\ \end{aligned}$$

(A.8)

since with the choice $f = {{\mathsf {R}}}_t u$ and $g = {{\mathsf {R}}}_0v$ the inequality (A.8) together with Lemma A.1 yields the existence of constants $c_1 >0$ and $c_2 >-c_1\lambda _0$ such that

$$\begin{aligned} \begin{aligned}&\big | {\mathfrak {h}}_{\alpha \delta _{\Sigma }}[{{\mathsf {R}}}_t u,{{\mathsf {R}}}_0v]- {\mathfrak {h}}_{\alpha \delta _{\Sigma _t}}[{{\mathsf {R}}}_t u,{{\mathsf {R}}}_0v] \big |\\&\quad \le c|t|^{1/2} \left( c_1{\mathfrak {h}}_{\alpha \delta _{\Sigma }}[{{\mathsf {R}}}_0 v] + c_2\Vert {{\mathsf {R}}}_0 v\Vert ^2_{L^2({{\mathbb {R}}}^2)} + c_1{\mathfrak {h}}_{\alpha \delta _{\Sigma _t}}[{{\mathsf {R}}}_t u] + c_2\Vert {{\mathsf {R}}}_t u\Vert ^2_{L^2({{\mathbb {R}}}^2)} \right) \\&\quad = c|t|^{1/2}\left[ c_1({{\mathsf {R}}}_0v,v)_{L^2({{\mathbb {R}}}^2)} + c_1({{\mathsf {R}}}_tu,u)_{L^2({{\mathbb {R}}}^2)} \right. \\&\qquad \left. +\, (c_1\lambda _0 + c_2)\left( \Vert {{\mathsf {R}}}_0v\Vert ^2_{L^2({{\mathbb {R}}}^2)} +\Vert {{\mathsf {R}}}_tu\Vert ^2_{L^2({{\mathbb {R}}}^2)}\right) \right] \\&\quad \le c|t|^{1/2}\left( \Vert u\Vert ^2_{L^2({{\mathbb {R}}}^2)} + \Vert v\Vert ^2_{L^2({{\mathbb {R}}}^2)}\right) , \end{aligned} \end{aligned}$$

where we used that $\Vert {{\mathsf {R}}}_t\Vert \le \frac{1}{\lambda _1-\lambda _0}$ holds for all $t\in {{\mathcal {I}}}$. In view of the polarization identity, it is enough to check (A.8) for $f = g$. By definition of the form ${\mathfrak {h}}_{\alpha \delta _{\Sigma _t}}$ in (2.10) with $\mu =\alpha \delta _{\Sigma _t}$, we get

$$\begin{aligned} \begin{aligned} \big |{\mathfrak {h}}_{\alpha \delta _{\Sigma _t}}[f] - {\mathfrak {h}}_{\alpha \delta _\Sigma }[f]\big | = \alpha \left| \Vert f|_{\Sigma _t}\Vert ^2_{L^2(\Sigma _t)}- \Vert f|_{\Sigma }\Vert ^2_{L^2(\Sigma )}\right| \le c|t|^{1/2}\Vert f\Vert _{H^1({{\mathbb {R}}}^2)}^2, \end{aligned} \end{aligned}$$

where Lemma A.2 is applied in the last step. The continuity of the lowest eigenvalue ${{\mathcal {I}}}\ni t\mapsto \lambda _1(\alpha \delta _{\Sigma _t})$ follows from the norm resolvent continuity of the operator-valued function in (A.6) in combination with the spectral convergence result from [40, Satz 9.24]. $\square$