Sparse Dirichlet optimal control problems

Abstract

In this paper, we analyze optimal control problems governed by an elliptic partial differential equation, in which the control acts as the Dirichlet data. Box constraints for the controls are imposed and the cost functional involves the state and possibly a sparsity-promoting term, but not a Tikhonov regularization term. Two different discretizations are investigated: the variational approach and a full discrete approach. For the latter, we use continuous piecewise linear elements to discretize the control space and numerical integration of the sparsity-promoting term. It turns out that the best way to discretize the state equation is to use the Carstensen quasi-interpolant of the boundary data, and a new discrete normal derivative of the adjoint state must be introduced to deal with this. Error estimates, optimization procedures and examples are provided.

Appendix A: On the regulairty of the conormal derivative

Let $\phi _g\in H^2(\varOmega )\cap H^1_0(\varOmega )$ be the solution of (2). Let us prove in detail that $\partial _{\nu _{{A^*}}}\phi _g\in H^{1/2}(\varGamma )$.

Let us denote $(S_j)_{0\le j\le N_S}$ the vertices of $\varOmega$, numbered counterclockwise and with the convention that $S_0=S_{N_S}$. Each side of $\varGamma$ is denoted $\varGamma _j = [S_{j-1},S_j]$, where $1\le j\le N_S$. The trace of $\nabla \phi _g$ on $\varGamma$ belongs to $(H^{1/2}(\varGamma ))^2$. Thus the trace of $\nabla \phi _g$ on $\varGamma _j$ belongs to $(H^{1/2}(\varGamma _j))^2$ and the Lipschitz regularity of the coefficients $a_{ij}$ implies that $\partial _{\nu _{{A^*}}} \phi _g\in H^{1/2}(\varGamma _j)$. Moreover we have the estimate

$$\begin{aligned} \Vert \partial _{\nu _{{A^*}}}\phi _g\Vert _{H^{1/2}(\varGamma _j)}\le C\Vert g\Vert _{L^2(\varOmega )}. \end{aligned}$$

To show that $\partial _{\nu _{{A^*}}}\phi _g$ belongs to $H^{1/2}(\varGamma )$ we have to analyze the behaviour at the corners $S_j=\varGamma _j\cap \varGamma _{j+1}$. Following [16], we parametrize $\varGamma _{j+1}$ by setting $x_j(\sigma )= S_j + \frac{\sigma }{|\varGamma _{j+1}|}(S_{j+1}- S_j )$ with $0\le \sigma \le |\varGamma _{j+1}|$, and $\varGamma _{j}$ by $x_j(-\sigma )= S_j - \frac{\sigma }{|\varGamma _{j}|}(S_{j}- S_{j-1} )$ with $0\le \sigma \le |\varGamma _j|$. For $0\le \sigma \le \delta _j=\min \{|\varGamma _j|,|\varGamma _{j+1}|\}$, $x_j(\sigma )\in \varGamma _{j+1}$, $x_j(-\sigma )\in \varGamma _{j}$ and $|x_j(\sigma )-S_j|=|x_j(-\sigma )-S_j|=\sigma$. According to Theorem 1.5.2.3.c in [16], to prove that $\partial _{\nu _{{A^*}}}\phi _g\in H^{1/2}(\varGamma _j\cup \varGamma _{j+1})$, we have to show that

$$\begin{aligned} \int \limits _0^{\delta _j}\frac{|\partial _{\nu _{{A^*}}}\phi _g(x_j(\sigma )) -\partial _{\nu _{{A^*}}}\phi _g(x_j(-\sigma ))|^2}{\sigma }d\sigma <+\infty . \end{aligned}$$

(41)

First notice that $\Vert \phi _g\Vert _{H^2(\varOmega )}\le C\Vert g\Vert _{L^2(\varOmega )}$. Since $\phi _g\in H^2(\varOmega )$, then $\nabla \phi _g\in (H^1(\varOmega ))^2$, and the usual trace theorem says that $\partial _i \phi _g\in H^{1/2}(\varGamma )$ for $i=1,2$. As is shown in the first part of the proof of Theorem 1.5.2.3.c) in [16], this implies that

$$\begin{aligned} \int \limits _0^{\delta _j}\frac{|\partial _i \phi _g(x_j(\sigma ))-\partial _i \phi _g(x_j(-\sigma ))|^2}{\sigma }d\sigma <+\infty , \end{aligned}$$

(42)

for $i=1,2$. We are going to transform the integral in (41) into a combinations of integrals involving the partial derivatives. To do that, without loss of generality, we can suppose that $\varGamma _j$ is on the negative part of the x axis, $S_j$ is at the origin and $\varGamma _{j+1}\subset \{ (-\sigma \,n_2,\sigma \,n_1)\mid \ 0\le \sigma \}$, so that the normal and tangent vectors to $\varGamma _j$ and $\varGamma _{j+1}$ respectively are $\nu _j=(0,-1)^T$, $\tau _j=(1,0)^T$ and $\nu _{j+1}=(n_1,n_2)^T$, $\tau _{j+1}=(-n_2,n_1)^T$ where $n_1>0$ and $n_1^2+n_2^2=1$. The functions $r,s,\gamma _1,\gamma _2\in C^{0,1}(\varGamma _j\cup \varGamma _{j+1})$ defined as

$$\begin{aligned} \gamma _2 =&-a_{22},&s =&{a_{12}}+\frac{n_2+1}{n_1}a_{22},\\ \gamma _1 =&n_1 a_{11}+ n_2 {a_{21}}{+n_2 s},&r =&-\gamma _1 - {a_{21}}, \\ \end{aligned}$$

satisfy that

$$\begin{aligned} {\mathcal {A}}^{{T}}\nu _j&= r\tau _{j}+\gamma _1 e_1+\gamma _2 e_2 \text{ on } \varGamma _j\\ {\mathcal {A}}^{{T}}\nu _{j+1}&= s\tau _{j+1}+\gamma _1 e_1+\gamma _2 e_2 \text{ on } \varGamma _{j+1}, \end{aligned}$$

where ${\mathcal {A}}^T$ is the transpose matrix of

$$\begin{aligned}{\mathcal {A}}(x) = \left( \begin{array}{cc} a_{11}(x) &{} a_{12}(x) \\ {a_{21}}(x) &{} a_{22}(x) \end{array}\right) ,\ e_1 = \left( \begin{array}{c} 1 \\ 0 \end{array}\right) ,\text{ and } e_2 = \left( \begin{array}{c} 0 \\ 1 \end{array}\right) . \end{aligned}$$

Therefore, we have

$$\begin{aligned} \partial _{\nu _{{A^*}}}\phi _g(x_j(\sigma )) -\partial _{\nu _{{A^*}}}&\phi _g(x_j(-\sigma )) = s(x_j(\sigma ))\partial _{\tau }\phi _g(x_j(\sigma ))-r(x_j(-\sigma )) \partial _{\tau }\phi _g(x_j(-\sigma )) \\&+ \gamma _1 (x_j(\sigma )) \partial _{1}\phi _g(x_j(\sigma )) - \gamma _1(x_j(-\sigma )) \partial _{1}\phi _g(x_j(-\sigma ))\\&+ \gamma _2 (x_j(\sigma )) \partial _{2}\phi _g(x_j(\sigma ))- \gamma _2(x_j(-\sigma )) \partial _{2}\phi _g(x_j(-\sigma ))). \end{aligned}$$

Since $\phi _g=0$ on $\varGamma$, the tangential derivatives in that expression are zero. Therefore, we only have to prove for $i=1,2$ that

$$\begin{aligned} \int \limits _0^{\delta _j} \frac{|\gamma _i (x_j(\sigma )) \partial _{i}\phi _g(x_j(\sigma ))- \gamma _i(x_j(-\sigma )) \partial _{i}\phi _g(x_j(-\sigma ))|^2}{\sigma }d\sigma < +\infty . \end{aligned}$$

To prove this, we first insert the term $\gamma _i (x_j(\sigma )) \partial _{i}\phi _g(x_j(-\sigma ))$ and apply Young’s inequality. Next we apply the fundamental theorem of Calculus and take advantage of the Lipschitz regularity of $\gamma _i$ to obtain

$$\begin{aligned} \int \limits _0^{\delta _j}&\frac{|\gamma _i (x_j(\sigma )) \partial _{i} \phi _g(x_j(\sigma ))- \gamma _i(x_j(-\sigma )) \partial _{i} \phi _g(x_j(-\sigma ))|^2}{\sigma }d\sigma \\ \le&2 \int \limits _0^{\delta _j} \frac{|\gamma _i (x_j(\sigma )) \partial _{i}\phi _g(x_j(\sigma ))- \gamma _i (x_j(\sigma )) \partial _{i} \phi _g(x_j(-\sigma )) |^2}{\sigma }d\sigma \\&+ 2 \int \limits _0^{\delta _j} \frac{|\gamma _i (x_j(\sigma )) \partial _{i} \phi _g(x_j(-\sigma )) -\gamma _i(x_j(-\sigma )) \partial _{i} \phi _g(x_j(-\sigma ))|^2}{\sigma }d\sigma \\ \le&2 \int \limits _0^{\delta _j} \gamma _i (x_j(\sigma ))^2 \frac{| \partial _{i}\phi _g(x_j(\sigma ))- \partial _{i}\phi _g(x_j(-\sigma )) |^2}{\sigma }d\sigma \\&+ 2 \int \limits _0^{\delta _j} \frac{|\gamma _i (x_j(\sigma )) -\gamma _i(x_j(-\sigma )) |^2}{\sigma } |\partial _{i}\phi _g(x_j(-\sigma ))|^2d\sigma \\ \le&2 \Vert \gamma _i\Vert ^2_{L^\infty (\varGamma _{j+1})} \int \limits _0^{\delta _j} \frac{| \partial _{i}\phi _g(x_j(\sigma ))- \partial _{i}\phi _g(x_j(-\sigma )) |^2}{\sigma }d\sigma \\&+ 2\int \limits _0^{\delta _j}\frac{|\int \limits _{-\sigma }^\sigma \gamma '(x_j(s))ds |^2}{\sigma }|\partial _{i}\phi _g(x_j(-\sigma ))|^2d\sigma \\ \le&2 \Vert \gamma _i\Vert ^2_{L^\infty (\varGamma _{j+1})} \int \limits _0^{\delta _j} \frac{| \partial _{i}\phi _g(x_j(\sigma ))- \partial _{i}\phi _g(x_j(-\sigma )) |^2}{\sigma }d\sigma \\&+ 8\Vert \gamma _i\Vert ^2_{C^{0,1}(\varGamma _{j}\cup \varGamma _{j+1})}\delta _j \Vert \partial _{i}\phi _g\Vert ^2_{L^2(\varGamma _j)} \end{aligned}$$

which is finite thanks to (42) and the fact that the trace of $\partial _{i}\phi _g$ is in $H^{1/2}(\varGamma )\hookrightarrow L^2(\varGamma ).$

Making the same analysis for each corner, we have proved the claimed regularity.