Spectral shape optimization for the Neumann traces of the Dirichlet-Laplacian eigenfunctions

Abstract

We consider a spectral optimal design problem involving the Neumann traces of the Dirichlet-Laplacian eigenfunctions on a smooth bounded open subset $\Omega$ of . The cost functional measures the amount of energy that Dirichlet eigenfunctions concentrate on the boundary and that can be recovered with a bounded density function. We first prove that, assuming a $L^1$ constraint on densities, the so-called Rellich functions maximize this functional. Motivated by several issues in shape optimization or observation theory where it is relevant to deal with bounded densities, and noticing that the $L^\infty$-norm of Rellich functions may be large, depending on the shape of $\Omega$, we analyze the effect of adding pointwise constraints when maximizing the same functional. We investigate the optimality of bang–bang functions and Rellich densities for this problem. We also deal with similar issues for a close problem, where the cost functional is replaced by a spectral approximation. Finally, this study is completed by the investigation of particular geometries and is illustrated by several numerical simulations.

Appendices

Appendix: Proofs of Propositions 4, 5 and 6

In what follows, we will assume that $M=1$ for the sake of readability. The general result will be easily inferred by an immediate adaptation of the reasonings.

Proof of Proposition 4

This proof is inspired by [40, Proposition 1 and Theorem 1]. For this reason, we only provide a short sketch of proof, underlining the main steps.

Let us first solve the convexified optimal design problem ($\mathcal {P}_\infty$). According to the expression of $J_\infty (a)$ given by (32), letting n and k going to $+\infty$ yields

$$\begin{aligned} \max _{a\in \overline{\mathcal {U}}_{L,M}}J_\infty (a) \leqslant \frac{2}{\pi ^2\alpha \beta } \max _{a\in \overline{\mathcal {U}}_{L,M}} \min \left( \int _{\Sigma _1\cup \Sigma _3}a(x,y)dy,2L\pi (\alpha +\beta )-\int _{\Sigma _1\cup \Sigma _3}a(x,y)dy\right) ,\nonumber \\ \end{aligned}$$

(45)

by using that $\int _{\partial \Omega }a(x,y)\, d\mathcal {H}^{n-1}=2L\pi (\alpha +\beta )$ and as a consequence

$$\begin{aligned} \max _{a\in \overline{\mathcal {U}}_{L,M}}J_\infty (a) \leqslant \frac{2}{\pi ^2\alpha \beta }\max _{t\in [0,L\pi (\alpha +\beta )]}\min \{t,2L\pi (\alpha +\beta )-t\}=\frac{2L (\alpha +\beta )}{\pi \alpha \beta }. \end{aligned}$$

Observe moreover that both components of the minimum above are equal at the optimum. To compute the optimal value for the convexified problem, we will use Theorem 3. Let us investigate the existence of Rellich-admissible functions. A simple computation shows that every Rellich-admissible function, whenever it exists, is necessarily constant on each side of $\Omega$, namely on $\Sigma _1$, $\Sigma _2$, $\Sigma _3$ and $\Sigma _4$. Moreover, for a given $x_0$ in $\Omega$ and when x runs over $\partial \Omega$, the quantity $\langle x-x_0,\nu \rangle$ is successively equal to the distance of $x_0$ to each side of $\Omega$. Therefore, it is enough to investigate the case where $x_0=(0,0)$ to determine the set of parameters L, $\alpha$ and $\beta$ for which there exists Rellich-admissible functions. In other words, this question comes to determine L, $\alpha$ and $\beta$ such that the function $a^*$ defined by

$$\begin{aligned} a^*|_{\Sigma _1\cup \Sigma _3} = L\frac{\alpha +\beta }{2\beta }\quad \text {and } \quad a^*|_{\Sigma _2\cup \Sigma _4}=L\frac{\alpha +\beta }{2\alpha }, \end{aligned}$$

belongs to $\overline{\mathcal {U}}_{L,1}$.

It follows that there exists a Rellich-admissible function if, and only if $|L|\leqslant L^c_{n}$ (defined in the statement of Proposition 4). Moreover, in this case, one has

$$\begin{aligned} \max _{a\in \overline{\mathcal {U}}_{L,M}}J_\infty (a)=\frac{2 L(\alpha +\beta )}{\pi \alpha \beta }. \end{aligned}$$

(46)

Let us now investigate the converse case. Assume without loss of generality that $\beta <\alpha$ and $L\in (\frac{2\beta }{\alpha +\beta },1]$, the case $L\in [-1,-\frac{2\beta }{\alpha +\beta })$ being treatable in a similar way. Then, one has

$$\begin{aligned} \int _{\Sigma _1\cup \Sigma _3}a(x,y)\, dy\leqslant & {} {\text {Per}}(\Sigma _1\cup \Sigma _3)=2\beta \pi \\ {\mathrm{and}}\quad \int _{\Sigma _2\cup \Sigma _4}a(x,y)\, dx= & {} 2L\pi (\alpha +\beta )-\int _{\Sigma _1\cup \Sigma _3}a(x,y)\, dy \geqslant 4\beta \pi -2\beta \pi =2\beta \pi , \end{aligned}$$

meaning that $J_\infty (a)\leqslant \frac{4}{\pi \alpha }$ for every $a\in \overline{\mathcal {U}}_{L,1}$, according to (45). The right-hand side in this inequality is in fact reached by every function a equal to 1 on $\Sigma _1\cup \Sigma _3$, constant on each side $\Sigma _2$ and $\Sigma _3$, where each constant is chosen in such a way that a belongs to $\overline{\mathcal {U}}_{L,1}$. Notice that such a choice is not unique, and easy computations yield that maximizers are given by

$$\begin{aligned} a_{\mid \Sigma _1\cup \Sigma _3}=1,\quad a_{\mid \Sigma _2}=u\frac{L(\alpha +\beta )-\beta }{\alpha }\quad \text {and}\quad a_{\mid \Sigma _4}=(2-u)\frac{L(\alpha +\beta )-\beta }{\alpha }. \end{aligned}$$

with $u\in \left( 1-\left( \frac{\alpha }{L(\alpha +\beta )-\beta }-1\right) ,1+\left( \frac{\alpha }{L(\alpha +\beta )-\beta }-1\right) \right)$.

The rest of the proof is a direct adaptation of the results of [40, Theorem 1] and in particular of the fact that

$$\begin{aligned} \sup _{\begin{array}{c} \omega \subset (-\alpha \pi /2,\alpha \pi /2)\\ |\omega |=V_0 \end{array}} \int _\omega \sin \left( \frac{n\pi x}{\alpha }\right) ^2\, dx = \frac{V_0}{2}. \end{aligned}$$

(47)

Finally, the necessary and sufficient condition on L guaranteeing the existence of solutions for the initial optimal design problem follows by using the same Fourier series method as in the proof of [40, Theorem 1].

Proof of Proposition 5

The proof of this proposition is inspired by [19, Theorem 3.2] and [40, Theorem 1]. First, notice that for every a in $\overline{\mathcal {U}}_{L,M}$, one has

$$\begin{aligned} J_\infty (a)\leqslant \lim _{n\rightarrow +\infty }\int _0^{2\pi }a(\theta )\cos (n\theta )^2\, d\theta =L\pi \end{aligned}$$

and that $J_\infty (L)=L\pi$, yielding that the optimal value for Problem ($\mathcal {P}_\infty$) is $\pi L$. Moreover, the constant function equal to L belongs to $\overline{\mathcal {U}}_{L,M}$ since $M\geqslant 1$ and $a\in \overline{\mathcal {U}}_{L,M}$ reaches $\pi L$ if, and only if

$$\begin{aligned} \inf \limits _{n\geqslant 1} \int _0^{2\pi }a(\theta )\cos (n\theta )^2 R d\theta = \inf \limits _{n\geqslant 1} \int _0^{2\pi }a(\theta )\sin (n\theta )^2 R d\theta = \pi L \end{aligned}$$

or similarly

(48)

Considering the Fourier expansion $a(\theta )=L+\sum _{j=1}^{+\infty } ( \alpha _j \cos (j\theta ) + \beta _j \sin (j\theta ) )$, the equality (48) holds if, and only if $\alpha _{2j} = 0$, $\forall j\geqslant 1$.

The no-gap property between the optimal values for Problem ($\mathcal {P}_\infty ^\mathrm{bb}$) and its convexified version ($\mathcal {P}_\infty$) is a consequence of Theorem 4.

Let us now investigate the no-gap property. Assume that $a=2\chi _{\Gamma }-1$ solves ($\mathcal {P}_\infty ^\mathrm{bb}$) and consider $a_e:\theta \mapsto \frac{a(\theta )+a(2\pi -\theta )}{2}$, its even part. First, $a_e$ is still a solution of ($\mathcal {P}_\infty$) and $a_e(\theta ) = L + \sum _{j=1}^{+\infty } \alpha _j \cos (j \theta )$. Setting now $\tilde{a}(\theta ) = \frac{a_e(\theta )+a_e(\pi -\theta )}{2}$, one has

$$\begin{aligned} \tilde{a}(\theta ) = L + \sum _{j=1}^{+\infty } \frac{\alpha _j}{2}(1+(-1)^j)\cos (j \theta ) . \end{aligned}$$

Since $\tilde{a}$ solves ($\mathcal {P}_\infty$), we have $\alpha _{2j}=0$, for all $j\geqslant 1$. Thus, $\tilde{a}$ is necessarily constant equal to L. Finally since $a\in \mathcal {U}_{L,M}$, hence the range of $a_e$ is contained in $\lbrace -1, 0, 1\rbrace$, and finally $\tilde{a}$ in $\lbrace -1,-1/2,0,1/2,1 \rbrace$. This yields that Problem ($\mathcal {P}_\infty ^\mathrm{bb}$) has no solution if $L\notin \lbrace -1,-1/2,0,1/2,1 \rbrace$.

Conversely, if $L\in \lbrace -1,-1/2,0,1/2,1 \rbrace$, a direct adaptation of the construction of solutions done in the proof of [40, Theorem 1] or [19, Lemma 3.1] yields the expected conclusion.

Proof of Proposition 6

Denote by $\Sigma _1$, $\Sigma _2$ and $\Sigma _3$ the subsets of $\partial \Omega$ defined by

$$\begin{aligned} \Sigma _1=\partial \Omega \cap \lbrace \theta =-\theta _1\rbrace , \quad \Sigma _2=\partial \Omega \cap \lbrace \theta =\theta _1\rbrace \quad \text { and }\quad \Sigma _3=\partial \Omega \cap \lbrace r=R\rbrace . \end{aligned}$$

Fixing and letting k tend to $+\infty$ shows that the sequence does not converge to a constant on $\partial \Omega$. Therefore, $\Omega$ does not satisfy the (QUEB) property.

We do not know if $\Omega$ satisfies the (WQEB) property. Anyway, we bypass this difficulty by showing a weaker version of this property. This is the purpose of the next two lemmas.

The results stated in the two next lemmas are inspired by [43, Proposition 4].

Lemma 4

For $s\in (0,1)$, let us denote by $F_s$ the function $[s,1]\ni x\mapsto \int _s^x \frac{du}{u^2\sqrt{u^2-s^2}}$. We have

$$\begin{aligned} \sup _{a\in \mathcal {V}} \; \inf \limits _{s\in (0,1)} \; \frac{1}{F_s(1)}\int _0^1 \frac{a(u) \chi _{(s,1)}(u)}{u^2 \sqrt{u^2-s^2}} du = 1 \end{aligned}$$

where $\mathcal {V}= \lbrace a\in L^\infty ([0,1])\ \mid \ \int _0^1 a(u)\, du = 1 \rbrace$.

Proof

Define the function $K: \overline{\mathcal {U}}_{L,M}\ni a \mapsto \inf \limits _{s\in (0,1)} \; \frac{1}{F_s(1)}\int _0^1 \frac{a(u) \chi _{(s,1)}(u)}{u^2 \sqrt{u^2-s^2}} \, du$. Note that $K(a=1) = 1$ by definition of $F_s$ and that the infimum in the definition of K is reached for every $s\in (0,1)$. As an infimum of linear functions, the function K is concave. Therefore, to prove the lemma, it is enough to show that the directional derivative of K at $a=1$ in every admissible direction h satisfies the first-order necessary optimality conditions, namely that $dK(a=1).h \leqslant 0$ for every function h defined on [0, 1] such that $\int _0^1 h(u)du =0$.

According to Danskin’s theorem,²² we have

$$\begin{aligned} dK(a=1).h = \inf \limits _{s\in (0,1)} \frac{1}{F_s(1)}\int _0^1 \frac{h(u) \chi _{(s,1)}(u)}{u^2 \sqrt{u^2-s^2}} du. \end{aligned}$$

By contradiction assume that there exists a function h such that

$$\begin{aligned} \frac{1}{F_s(1)}\int _0^1 \frac{h(u) \chi _{(s,1)}(u)}{u^2 \sqrt{u^2-s^2}} du > 0 , \end{aligned}$$

for every $s\in (0,1)$. One has

$$\begin{aligned} 0 < \int _{s=0}^1 \int _{u=s}^1 \frac{s \; h(u)}{u \sqrt{u^2-s^2}} \, du \, ds= \int _{u=0}^1 \int _{s=0}^u \frac{s \; h(u)}{u \sqrt{u^2-s^2}} \, ds \, du = - \int _0^1 h(x) dx = 0, \end{aligned}$$

leading to a contradiction. The conclusion follows. $\square$

Lemma 5

For every $a\in \overline{\mathcal {U}}_{L,M}$, one has $J_\infty (a) \leqslant \frac{L \mathcal {H}^{n-1}(\partial \Omega )}{|\Omega |}.$

Proof

Let $a\in \overline{\mathcal {U}}_{L,M}$. We will show the existence of a subsequence of converging to $\frac{1}{\theta _1 R^2} \int _{\partial \Omega } a(x)\, d\mathcal {H}^{n-1}=\frac{L \mathcal {H}^{n-1}(\partial \Omega )}{|\Omega |}$.

Since $\frac{1}{\lambda _{n,k}} \left( \frac{\partial \varphi _{n,k}}{\partial \nu } \right) ^2 = \frac{2}{R^2 \theta _1} \sin \left( \frac{n\pi }{2\theta _1} (\theta + \theta _1)\right)$ on $\Sigma _3$, the sequence converges to $\frac{1}{\theta _1 R^2} \int _{\Sigma _3} a(x)\, d\mathcal {H}^{n-1}$, according to the Riemann–Lebesgue Lemma.

To conclude the proof, it is enough to prove the existence of a positive number $\alpha$ such that, by keeping the ratio n / k constant equal to $\alpha$, there holds

$$\begin{aligned} \lim _{n \rightarrow +\infty } \int _{\Sigma _1 \cup \Sigma _2} \frac{a(x)}{\lambda _{n,k}}\left( \frac{\partial \varphi _{n,k}}{\partial \nu } \right) ^2 d\mathcal {H}^{n-1}\leqslant \frac{1}{\theta _1 R^2} \int _{\Sigma _1\cup \Sigma _2} a(x)\, d\mathcal {H}^{n-1}. \end{aligned}$$

Setting $\mu = \frac{n\pi }{2\theta _1}$ and introducing $\Phi _{n,k}: [0,1]\ni u\mapsto \frac{ n^2 \pi ^2}{2 z_{n,k}^2 \theta _1^2} \left( \frac{J_\mu \left( z_{n,k}\frac{r}{R}\right) }{r J'_\mu (z_{n,k})} \right) ^2$, one has

$$\begin{aligned} \frac{1}{\lambda _{n,k}}\left( \frac{\partial \varphi _{n,k}(r)}{\partial \nu } \right) ^2 = \frac{1}{\theta _1}\Phi _{n,k}\left( \frac{r}{R}\right) \quad \text { on }\quad \Sigma _1 \cup \Sigma _2. \end{aligned}$$

Hence, we have to prove that for every $\rho \in L^\infty ([-1,1],[-1,1])$, there exists $\alpha >0$ such that for n and k chosen as previously,

$$\begin{aligned} \lim _{n \rightarrow +\infty } \int _0^1 \rho (u) \Phi _{n,k}(u) \, du \leqslant \int _0^1 \rho (u) \, du . \end{aligned}$$

(49)

According to [31, p. 257], one has $\int _0^1 \Phi _{n,k}(u)\, du = \frac{n^2 \pi ^2}{n^2\pi ^2 - \theta _1}$. Then, taking the weak limit of the sequence of measures $\Phi _{n,k}(u)\, du$ with a fixed ratio n / k, and making this ratio vary, we obtain the family of probability measures

$$\begin{aligned} f_s(u)\, du = \frac{1}{F_s(1)}\frac{\chi _{(s,1)}(u)}{u^2 \sqrt{u^2-s^2}} \, du \end{aligned}$$

parametrized by $s\in (0,1)$.

To prove (49), let us argue by contradiction, by assuming that for every $s \in (0,1)$, one has

$$\begin{aligned} \frac{1}{F_s(1)}\int _0^1 \frac{\rho (u) \chi _{(s,1)}(u)}{u^2 \sqrt{u^2-s^2}} \, du > \int _0^1 \rho (u)\, du . \end{aligned}$$

We obtain a contradiction by using Lemma 4 and the expected conclusion follows. $\square$

In the next lemma, one computes the critical value $L^c_{n}$ introduced in Theorem 3.

Lemma 6

Denote by $L^c_{n}$ the critical value for the constraint parameter L, as introduced in Theorem 3. One has $L^c_{n} = \frac{\theta _1 (1+\tan \theta _1)}{(1+\theta _1)\tan \theta _1}$.

Proof

Following the proof of Theorem 3, the critical value $L^c_{n}$ is given by (17). Therefore, the issue comes to determine the quantity $\delta :=\min _{x_0\in \overline{\Omega }} \max _{x\in \partial \Omega } \langle x-x_0, \nu _x \rangle$. For the sake of simplicity, the notations we will use are summed-up on Fig. 10. Writing in polar coordinates $x_0=(r_0,\theta _0)$, the quantity $\langle x-x_0, \nu _x \rangle$ is equal to $r_0\sin (\theta _1+\theta _0)$ on $\Sigma _1$, $r_0\sin (\theta _1-\theta _0)$ on $\Sigma _2$ and $R-r_0 \cos (\theta _0-\theta )$ on $\Sigma _3$, where we denoted $x=(R,\theta )$ in polar coordinates on $\Sigma _3$.

Fig. 10

Notations

As a consequence, one computes using symmetry arguments

$$\begin{aligned} \delta= & {} \min _{\begin{array}{c} 0\leqslant r_0\leqslant R\\ \theta _0\in [-\theta _1,\theta _1] \end{array}} \max \left\{ R-r_0\cos (\theta _0+\theta _1), R-r_0\cos (\theta _0-\theta _1), r_0 \sin (\theta _1-\theta _0),r_0 \sin (\theta _1+\theta _0) \right\} \\= & {} \min _{0\leqslant r_0\leqslant R} \max \left\{ R-r_0\cos \theta _1 , r_0 \sin \theta _1 \right\} , \end{aligned}$$

and the maximum is reached provided that $R-r_0\cos \theta _1= r_0 \sin \theta _1$. It follows that $\delta = \frac{R\tan (\theta _1)}{1+\tan (\theta _1)}$, leading to the expected conclusion. $\square$

In the following lemma, we compute the optimal value of Problem ($\mathcal {P}_\infty$).

Lemma 7

Let $L^c_{n} = \frac{\theta _1 (1+\tan \theta _1)}{(1+\theta _1)\tan \theta _1}$. We have

$$\begin{aligned} \max _{a\in \overline{\mathcal {U}}_{L,M}} J_\infty (a)= \left\{ \begin{array}{ll} \frac{L\mathcal {H}^{n-1}(\partial \Omega )}{|\Omega |} &{} \text { if } |L|\leqslant L^c_{n} , \\ \frac{(L+L^c_{n})\mathcal {H}^{n-1}(\partial \Omega )}{2|\Omega |} &{} \text { if } L^c_{n} < |L|. \end{array}\right. \end{aligned}$$

Proof

According to Lemma 5, one has $J_\infty (a) \leqslant \frac{L \mathcal {H}^{n-1}(\partial \Omega )}{|\Omega |}$ and according to Lemma 6, the right-hand side in this inequality is reached by every Rellich-admissible function if $|L|\leqslant L^c_{n}$.

Assume now that $L>L^c_{n}$ (the case $L<-L^c_n$ being exactly similar). Let us introduce $a_c$, as the solution of the optimal design problem ($\mathcal {P}_\infty$) in the case where $L=L^c_{n}$. One verifies that $a_c=1$ on $\Sigma _1\cup \Sigma _2$ and $a_c = R-r_0\cos (\theta )$ on $\Sigma _3$. Let $a\in \overline{\mathcal {U}}_{L,M}$ and let us decompose a as $a=a_c+h$, so that $\int _{-\theta _1}^{\theta _1}a(R,\theta ) R d\theta = L\mathcal {H}^{n-1}(\partial \Omega ) - 2R$, $\int _{-\theta _1}^{\theta _1}h(R,\theta ) R d\theta +\int _{\Sigma _1\cup \Sigma _2}h = (L-L^c_{n})\mathcal {H}^{n-1}(\partial \Omega )$ and $h <0$ on $\Sigma _1\cup \Sigma _2$.

Let us apply (2) with $L=L^c_{n}$. One gets

$$\begin{aligned}&\frac{R^2n^2\pi ^2}{2 z_{n,k}^2\theta _1^2} \int _0^R \frac{J_{\pi n /2\theta _1}(z_{n,k}\frac{r}{R})^2}{|J_{\pi n/2\theta _1}'(z_{n,k})|^2} \frac{dr}{r^2} \nonumber \\&\quad = L^c_{n}\mathcal {H}^{n-1}(\partial \Omega ) - \int _{-\theta _1}^{\theta _1}a_c(R,\theta )\sin \left( \frac{n\pi }{2\theta _1}(\theta +\theta _1)\right) ^2R d\theta . \end{aligned}$$

Using this identity, we claim that , where

$$\begin{aligned} j_{n,k}(a)= & {} \frac{2}{R^2\theta _1}\int _{-\theta _1}^{\theta _1}a(R,\theta )\sin \left( \frac{n\pi }{2\theta _1}(\theta +\theta _1)\right) ^2\, R d\theta \\&+ \frac{n^2\pi ^2}{ z_{n,k}^2\theta _1^3} \int _0^R (1+h(r,\theta _1)+h(r,-\theta _1))\frac{J_{\pi n /2\theta _1}(z_{n,k}\frac{r}{R})^2}{|J_{\pi n/2\theta _1}'(z_{n,k})|^2} \frac{dr}{r^2}\\\leqslant & {} \frac{2}{R^2\theta _1} \left( \int _{-\theta _1}^{\theta _1}h(R,\theta ) \sin \left( \frac{n\pi }{2\theta _1}(\theta +\theta _1)\right) ^2R d\theta + L^c_{n}\mathcal {H}^{n-1}(\partial \Omega ) \right) \\= & {} \frac{(L+L^c_{n})\mathcal {H}^{n-1}(\partial \Omega )}{2|\Omega |} - \frac{1}{R^2\theta _1} \int _{-\theta _1}^{\theta _1}h(R,\theta )\cos \left( \frac{n\pi }{\theta _1}(\theta +\theta _1) \right) \, R d\theta . \end{aligned}$$

As a consequence and according to the Riemann–Lebesgue lemma, one has

Note that the right-hand side in the last inequality is reached by the function $a^*$ equal to 1 on $\Sigma _1\cup \Sigma _2$ and constant on $\Sigma _3$, where the constant is chosen so that $\int _{\partial \Omega }a^*\, d\mathcal {H}^{n-1}=L\mathcal {H}^{n-1}(\partial \Omega )$. We have then proved the expected result. $\square$

It remains now to prove the no-gap result between the optimal values for Problem ($\mathcal {P}_\infty ^\mathrm{bb}$) and its convexified version ($\mathcal {P}_\infty$) in the case where $L\geqslant L^c_{n}$. Since every solution for the convexified problem ($\mathcal {P}_\infty$) is equal to 1 on $\Sigma _1\cup \Sigma _2$, it is enough to exhibit a sequence of elements in $\mathcal {U}_{L,M}$ such that $a_m =1$ on $\Sigma _1\cup \Sigma _2$ and

One refers to the proof of [40, Theorem 1] where the construction of such a sequence is explained.