On the Convergence of a Class of Nonlocal Elliptic Equations and Related Optimal Design Problems

Abstract

A convergence result for a nonlocal differential equation problem is proved. As a by-product, some results about the convergence for a type of nonlocal optimal design are given. Since these problems give rise to local design problems in the limit, different results on classical existence are obtained as well. Concerning the nonlocal formulation, the state equation is of nonlocal elliptic type and the cost functional we analyze includes, among other cases, an approximation of the square of the gradient.

Appendices

Appendix A: A Nonlocal Inequality for Measurable Sets

We analyze the limit

$$\begin{aligned} \lim _{\delta \rightarrow 0}\int _{G}\int _{G}k_{\delta }\left( \left| x^{\prime }-x\right| \right) \frac{\left( u_{\delta }\left( x^{\prime }\right) -u_{\delta }\left( x\right) \right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x \end{aligned}$$

where $G\subset \varOmega .$ We assume $u_{\delta }\rightarrow u^{*}$ strongly in $L^{2}\left( \varOmega \right)$ where $\left( u_{\delta }\right) \subset X_{0},$ $u^{*}\in H_{0}^{1}\left( \varOmega \right) ,\$ and we assume the sequence

$$\begin{aligned} \xi _{\delta }\left( x^{\prime },x\right) \doteq k_{\delta }\left( \left| x^{\prime }-x\right| \right) \frac{\left( u_{\delta }\left( x^{\prime }\right) -u_{\delta }\left( x\right) \right) ^{2}}{\left| x^{\prime }-x\right| ^{2}} \end{aligned}$$

is uniformly bounded in $L^{1}\left( \varOmega \times \varOmega \right)$.

It is well known that, at least for a subsequence of $\left( u_{\delta }\right) ,$ the inequality

$$\begin{aligned} \lim _{\delta \rightarrow 0}\int _{G}\int _{G}k_{\delta }\left( \left| x^{\prime }-x\right| \right) \frac{\left( u_{\delta }\left( x^{\prime }\right) -u_{\delta }\left( x\right) \right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x\ge \int _{G}\left| \nabla u^{*}\left( x\right) \right| ^{2}\hbox {d}x \end{aligned}$$

(39)

holds for every open and bounded $G\subset \varOmega$ (see [32, p. 12]).

We prove the above inequality for any measurable set $G\subset \varOmega .$ Since $\xi _{\delta }$ is uniformly bounded in $L^{1}\left( \varOmega \times \varOmega \right) ,$ we can apply Chacon’s biting Lemma (see [40]): there exists a subsequence, not relabeled, a nonincreasing sequence of measurable sets $G_{n}\subset \varOmega \times \varOmega ,$ $\left| G_{n}\right| \searrow 0$ and $\xi \in L^{1}\left( \varOmega \times \varOmega \right)$ such that

$$\begin{aligned} \xi _{\delta }\rightharpoonup \xi \text { in }L^{1}\left( \varOmega \times \varOmega \setminus G_{n}\right) \end{aligned}$$

(40)

for all n.

We take into account the identity

$$\begin{aligned} \int _{G}\left| \nabla u^{*}\left( x\right) \right| ^{2} \hbox {d}x=\iint _{G\times G}\frac{\left| \nabla u^{*}\left( x^{\prime }\right) \right| ^{2}+\left| \nabla u^{*}\left( x\right) \right| ^{2} }{2\left| G\right| }\hbox {d}x^{\prime }\hbox {d}x, \end{aligned}$$

and we shall prove for every measurable A

$$\begin{aligned} \iint _{A\times A}\xi \left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x\ge \iint _{A\times A}U^{*}\left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x=\int _{A}\left| \nabla u^{*}\left( x\right) \right| ^{2}\hbox {d}x \end{aligned}$$

(41)

where $U^{*}\left( x^{\prime }x\right) \doteq \frac{\left| \nabla u^{*}\left( x^{\prime }\right) \right| ^{2}+\left| \nabla u^{*}\left( x\right) \right| ^{2}}{2\left| A\right| }$. We prove (41) by a contradiction argument. We assume there is a measurable set A such that

$$\begin{aligned} \iint _{A\times A}\xi \hbox {d}x^{\prime }\hbox {d}x<\iint _{A\times A}U^{*}\hbox {d}x^{\prime }\hbox {d}x. \end{aligned}$$

Thereby, we can assume

$$\begin{aligned} \iint _{A\times A}\xi \hbox {d}x^{\prime }\hbox {d}x<\iint _{A\times A}U^{*}\hbox {d}x^{\prime }\hbox {d}x \end{aligned}$$

where $A\subset \varOmega$ is an open set such that $\left| A\right| >0.$ In view of the above estimate, the inequality

$$\begin{aligned} \iint _{A\times A\setminus G_{n}}\xi \hbox {d}x^{\prime }\hbox {d}x<\iint _{A\times A\setminus G_{n}}U^{*}\hbox {d}x^{\prime }\hbox {d}x \end{aligned}$$

is true for any n. Then the sequence defined by

$$\begin{aligned} r_{n}\doteq \iint _{A\times A\setminus G_{n}}\left( U^{*}-\xi \right) \hbox {d}x^{\prime }\hbox {d}x \end{aligned}$$

fulfills these requirements:

$$\begin{aligned} r_{n}>0\text { and }r_{n}\nearrow r_{0}\doteq \iint _{A\times A}\left( U^{*}-\xi \right) \hbox {d}x^{\prime }\hbox {d}x>0. \end{aligned}$$

We write

$$\begin{aligned} \iint _{A\times A\setminus G_{n}}\xi \hbox {d}x^{\prime }\hbox {d}x=\iint _{A\times A\setminus G_{n}}U^{*}\hbox {d}x^{\prime }\hbox {d}x-r_{n}, \end{aligned}$$

and we have

$$\begin{aligned} \iint _{A\times A\setminus G_{n}}\xi \hbox {d}x^{\prime }\hbox {d}x<\iint _{A\times A\setminus G_{n}}U^{*}\hbox {d}x^{\prime }\hbox {d}x-\frac{r_{n}}{2} \end{aligned}$$

for any n. Thus, thanks to the convergence (40), the above inequality allows us to write

$$\begin{aligned} \iint _{A\times A\setminus G_{n}}\xi _{\delta }\hbox {d}x^{\prime }\hbox {d}x<\iint _{A\times A\setminus G_{n}}U^{*}\hbox {d}x^{\prime }\hbox {d}x-\frac{r_{n}}{2} \end{aligned}$$

for any n and any $\delta \le \delta _{0}.$ If we pass to the limit when $n\rightarrow +\infty ,$ we get

$$\begin{aligned} \iint _{A\times A}\xi _{\delta }\hbox {d}x^{\prime }\hbox {d}x\le \iint _{A\times A}U^{*}\hbox {d}x^{\prime }\hbox {d}x-\frac{r_{0}}{2}. \end{aligned}$$

By taking now limits in $\delta \rightarrow 0$ and using (39), we have

$$\begin{aligned} \iint _{A\times A}U^{*}\hbox {d}x^{\prime }\hbox {d}x\le \iint _{A\times A}U^{*}\hbox {d}x^{\prime }\hbox {d}x-\frac{r_{0}}{2} \end{aligned}$$

which is a contradiction, and (41) has been proved.

Then, for each n and each measurable set $G\subset \varOmega ,$ the biting convergence (40) enables us to write

$$\begin{aligned} \lim _{\delta \rightarrow 0}\iint _{G\times G}\xi _{\delta }\left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x&\ge \lim _{\delta \rightarrow 0}\iint _{G\times G\setminus G_{n}}\xi _{\delta }\left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x\\&=\iint _{G\times G\setminus G_{n}}\xi \left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x. \end{aligned}$$

Finally, we take limits when $n\rightarrow +\infty$ and use (41) to obtain the desired estimate:

$$\begin{aligned} \lim _{\delta \rightarrow 0}\iint _{G\times G}\xi _{\delta }\left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x&\ge \lim _{n\rightarrow +\infty }\iint _{G\times G\setminus G_{n}}\xi \left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x\\&=\iint _{G\times G}\xi \left( x^{\prime },x\right) \hbox {d}x^{\prime }\hbox {d}x\\&\ge \iint _{G\times G}\frac{\left| \nabla u^{*}\left( x^{\prime }\right) \right| ^{2}+\left| \nabla u^{*}\left( x\right) \right| ^{2}}{2\left| G\right| }\hbox {d}x^{\prime }\hbox {d}x\\&=\int _{G}\left| \nabla u^{*}\left( x\right) \right| ^{2}\hbox {d}x. \end{aligned}$$

Appendix B: $H_{r}\left( \delta _{0}\right) \$ is Precompact in $L^{2}$

For this goal we shall use (43), an inequality due to Stein (see [9]), (44), an spectral result given in [17] (see also [33]), and a basic compactness result (Proposition B.1).

If $h\in H_{r}\left( \delta _{0}\right)$ then

$$\begin{aligned} B_{0}\left( h,h\right) =\int _{\varOmega _{\delta _{0}}}\int _{\varOmega _{\delta _{0}} }k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h^{\prime }-h\right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x<r, \end{aligned}$$

and $h=0$ en $\varOmega _{\delta _{0}}-\varOmega .$ We extend h by zero outside of $\varOmega _{\delta _{0}}$ and we compute the following integral:

$$\begin{aligned}&\int _{\varOmega _{\delta _{0}+\delta }}\int _{\varOmega _{\delta _{0}+\delta }} k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h^{\prime }-h\right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x\\&\quad =\int _{\varOmega _{\delta _{0}}}\int _{\varOmega _{\delta _{0}}}+\int _{\varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}}\int _{\varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}}+2\int _{\varOmega _{\delta _{0}}} \int _{\varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}}. \end{aligned}$$

We note the second integral vanishes because $h\left( x\right) =h\left( x^{\prime }\right) =0$ if $\left( x^{\prime },x\right) \in \left( \varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}\right) \times \left( \varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}\right) .$

The third one is zero too because h vanishes outside $\varOmega$

$$\begin{aligned}&\int _{\varOmega _{\delta _{0}}}\int _{\varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}}k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h\left( x^{\prime }\right) -h\left( x\right) \right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x\\&\quad =\int _{\varOmega }\int _{\varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}} }k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h\left( x\right) \right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x=0 \end{aligned}$$

where the last equality is true thanks to $k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) =0$ for $\left( x^{\prime },x\right) \in \left( \varOmega _{\delta _{0}+\delta }\setminus \varOmega _{\delta _{0}}\right) \times \left( \varOmega \right) .$

Therefore

$$\begin{aligned} B_{0}\left( h,h\right) =\int _{\varOmega _{\delta _{0}+\delta }}\int _{\varOmega _{\delta _{0}+\delta }}k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h^{\prime }-h\right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x<r. \end{aligned}$$

(42)

Let $\varphi \in C_{0}^{\infty }\left( B\left( 0,1\right) \right)$ be such that $\varphi \ge 0$ and $\int \varphi =1.$ For any $\delta >0$ we define $\varphi _{\delta }\left( x\right) \doteq \frac{1}{\delta ^{N}}\varphi \left( \frac{x}{\delta }\right)$ where $x\in R^{N}.$ At this point we can employ the following inequality

$$\begin{aligned} \int _{\varOmega _{\delta _{0}+\delta }}\int _{\varOmega _{\delta _{0}+\delta }} k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h^{\prime }-h\right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x\ge \int _{\varOmega _{\delta _{0}}}\int _{\varOmega _{\delta _{0}} }k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h_{\delta }^{\prime }-h_{\delta }\right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x \end{aligned}$$

(43)

where

$$\begin{aligned} h_{\delta }\left( x\right) =\left( \varphi _{\delta }*h\right) \left( x\right) \text { for any }x\in \varOmega _{\delta _{0}-\delta },\text { }\delta _{0}>\delta >0. \end{aligned}$$

Then $h_{\delta }$ is regular, ${\mathrm{supp}}\,h\subset \overline{\varOmega },\,{\mathrm{supp}}\, \varphi _{\delta }=B\left( 0,\delta \right) ,$ and thus

$$\begin{aligned} {\mathrm{supp}}\, h_{\delta }\subset \overline{\overline{\varOmega } +\overline{B\left( 0,\delta \right) }}=\overline{\varOmega }_{\delta } \subset \varOmega _{\delta _{0}-\delta }\subset \varOmega _{\delta _{0}}. \end{aligned}$$

We also know $h_{\delta }\rightarrow h$ in $L^{2}\left( \varOmega _{\delta _{0}-\delta }\right)$ if $\delta \rightarrow 0$ and, in particular, $h_{\delta }\rightarrow h$ strongly in $L^{2}\left( \varOmega \right)$ if $\delta \rightarrow 0.$ These facts clearly ensure $h_{\delta }\in X_{0}\left( \delta _{0}\right) .$ We use Theorem 1.1 from [17] to write

$$\begin{aligned} B_{0}\left( h_{\delta },h_{\delta }\right) =\int _{\varOmega _{\delta _{0}}} \int _{\varOmega _{\delta _{0}}}k_{\delta _{0}}\left( \left| x^{\prime }-x\right| \right) \frac{\left( h_{\delta }^{\prime }-h_{\delta }\right) ^{2}}{\left| x^{\prime }-x\right| ^{2}}\hbox {d}x^{\prime }\hbox {d}x=\sum _{k}h_{\delta k}^{2}\gamma _{k}^{nl}\left( \delta _{0}\right) \end{aligned}$$

(44)

where $h_{\delta k}=\left( h_{\delta },w_{\delta _{0}}^{\left( k\right) }\right) _{L^{2}\left( \varOmega \right) \times L^{2}\left( \varOmega \right) }.$ If we take into account the limit

$$\begin{aligned} \lim _{\delta \rightarrow 0}h_{\delta k}=\lim _{\delta \rightarrow 0}\left( h_{\delta },w_{\delta _{0}}^{\left( k\right) }\right) _{L^{2}\left( \varOmega \right) \times L^{2}\left( \varOmega \right) }=\left( h,w_{\delta _{0} }^{\left( k\right) }\right) _{L^{2}\left( \varOmega \right) \times L^{2}\left( \varOmega \right) }\doteq h_{k}, \end{aligned}$$

the discussion from above, (42)–(44), gives rise to the inequality

$$\begin{aligned} r\ge \sum _{k}h_{\delta k}^{2}\gamma _{k}^{nl}\left( \delta _{0}\right) . \end{aligned}$$

If we pass to the limit, we get

$$\begin{aligned} r\ge \lim _{\delta \rightarrow 0}\sum _{k}h_{\delta k}^{2}\gamma _{k}^{nl}\left( \delta _{0}\right) \ge \sum _{k}h_{k}^{2}\gamma _{k}^{nl}\left( \delta _{0}\right) , \end{aligned}$$

where $h_{k}=\left( h,w_{\delta _{0}}^{\left( k\right) }\right) _{L^{2}\left( \varOmega \right) \times L^{2}\left( \varOmega \right) }.$ We also recall the fact that $\left( w_{\delta _{0}}^{\left( k\right) }\right) _{k}$ is an orthonormal basis in $L^{2}\left( \varOmega \right)$ so that $\left\| h\right\| _{L^{2}\left( \varOmega \right) }^{2}=\sum _{k}h_{k} ^{2}.$

We shall need this result (see [41]):

Proposition B.1

Let $\ell ^{2}$ be the set of sequences $\left( u_{n}\right) _{n}$ such that $\sum u_{n}^{2}<\infty \$and consider $\left( \rho _{n}\right) _{n}$ to be a sequence of positive numbers such that $\lim _{n\rightarrow +\infty }\rho _{n}=+\infty .$ Let V be the space of sequences $\left( u_{n}\right) _{n}$ such that $\sum _{n}\rho _{n}\left| u_{n}\right| ^{2}<\infty \,$and assume the space V is equipped with the scalar product $\left( \left( u_{n}\right) ,\left( v_{n}\right) \right) =\sum \rho _{n}u_{n}v_{n}.$ Then, V is a Hilbert space and $V\subset \ell ^{2}$ with compact injection.

Proposition B.2

The set ${\mathcal {H}}_{r}\left( \delta _{0}\right)$ is precompact in $L^{2}\left( \varOmega \right) .$

Proof

Let $\left( h_{j}\right) _{j}\subset {\mathcal {H}}_{r}\left( \delta _{0}\right)$ be any sequence. Then, for any j

$$\begin{aligned} r\ge \sum _{k}h_{jk}^{2}\gamma _{k}^{nl}\left( \delta _{0}\right) . \end{aligned}$$

This estimate and Proposition B.1 guarantees $\left\{ \left( h_{jk}\right) _{k}\right\} _{j}$ converges in $\ell ^{2}$ if $j\rightarrow +\infty ,$ to a sequence $\left( \widetilde{h}_{k}\right) _{k};$ in this way, if $\widetilde{h}$ is the function from $L^{2}\left( \varOmega \right)$ defined by means of the sequence $\left( \widetilde{h}_{k}\right) _{k}$ (in the basis $\left( w_{\delta _{0}}^{\left( k\right) }\right) _{k},\,\ \widetilde{h} _{k}=\left( \widetilde{h},\left( w_{\delta _{0}}^{\left( k\right) }\right) _{k}\right) )$, then we can claim

$$\begin{aligned} h_{j}\rightarrow \widetilde{h}\text { strongly in }L^{2}\left( \varOmega \right) . \end{aligned}$$

Finally, since $\left( h_{j}\right) _{j}$ is uniformly bounded in $L^{\infty }$, we know there is a function $h\in L^{\infty }\left( \varOmega \right)$ such that $h_{j}\rightharpoonup h$ weak$-*$ in $L^{\infty }\left( \varOmega \right) .$ Hence $\widetilde{h}=h,$ and consequently $h\in {\mathcal {H}} _{r}\left( \delta _{0}\right) .$ $\square$