Departmentof Mathematics

Introduction

Here, we have a distributed optimal control problem of the Poisson equation with pointwise box constraints on the control and a one-sided pointwise state constraint. The present problem is given on the unit ball in $\Omega =B_1\left(0\right)\subset {ℝ}^3$. The control acts in a distributed way on the entire domain $\Omega$ and the state constraint is enforced on the entire domain, too. This problem and the analytical solution appear in [Rösch and Steinig, 2012, Section 8]. The problem is designed to have a vanishing Lagrange multiplier for the state constraint. It is thus potentially a good test case for a posteriori error estimation.

Variables & Notation

Unknowns

$$\begin{array}{ccc}\hfill u\in L^2\left(\Omega \right)& \text{control variable}\hfill & \hfill \\ \hfill y\in H^1\left(\Omega \right)& \text{state variable}\hfill \end{array}$$

Given Data

The given data is chosen in a way which admits an analytic solution.

$$\begin{array}{llllllll}\hfill \Omega & =B_1\left(0\right)\subset {ℝ}^3& \hfill & \text{computational domain}& \hfill & & \hfill & \\ \hfill \partial \Omega & & \hfill & \text{its boundary}& \hfill & & \hfill & \\ \hfill y_d\left(x\right)& =-4{\pi }^2|x{|}^2\sin <!--FUNCTION\; APPLICATION-->\left(\pi |x{|}^2\right)+6\pi \cos <!--FUNCTION\; APPLICATION-->\left(\pi |x{|}^2\right)+\cos <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{2}|x{|}^2\right)& \hfill & \text{desired state}& \hfill & & \hfill & \\ \hfill f\left(x\right)& =3\pi \sin <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{2}|x{|}^2\right)+{\pi }^2|x{|}^2\cos <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{2}|x{|}^2\right)+\sin <!--FUNCTION\; APPLICATION-->\left(\pi |x{|}^2\right)& \hfill & \text{source shift}& \hfill & & \hfill & \\ \hfill y_c\left(x\right)& =\left\{\begin{array}{cc}\cos <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{2}|x{|}^2\right)\hfill & \text{if }\left|x\right|\le 0.5\hfill \\ \left(-\frac{4}{3}\cos <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{8}\right)-\frac{40}{3}\right)|x{|}^2+\frac{4}{3}\cos <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{8}\right)+\frac{10}{3}\hfill & \text{else}\hfill \end{array}\right.& \hfill & \text{state constraint}& \hfill & & \hfill & \end{array}$$

Problem Description

$$\begin{array}{llll}\hfill \text{Minimize}& \frac{1}{2}\parallel y-y_d{\parallel }_{L^2\left(\Omega \right)}^2+\frac{1}{2}\parallel u{\parallel }_{L^2\left(\Omega \right)}^2& \hfill & \\ \hfill \text{s.t.}& \left\{\begin{array}{ccccc}\hfill -△y& =u+f\hfill & \hfill & \text{in }\Omega \hfill & \hfill \\ \hfill y& =0\hfill & \hfill & \text{on }\partial \Omega \hfill \\ \hfill \end{array}\right.& \hfill & \\ \hfill \text{and}& -1\le u\left(x\right)\le 1\text{in }\Omega .& \hfill & \\ \hfill \text{and}& y_c\left(x\right)\le y\left(x\right)\text{in }\Omega & \hfill & \end{array}$$

Optimality System

The following optimality system, given in the strong form, for the state $y\in H_0^1\left(\Omega \right)$, the control $u\in L^2\left(\Omega \right)$, the adjoint state $p\in H_0^1\left(\Omega \right)$ and the Lagrange multiplier $\mu \in \mathcal{ℳ}\left(\Omega \right)$ characterizes the unique minimizer.

$$\begin{array}{llllllll}\hfill -△y& =u+f& \hfill & \text{in }\Omega ,& \hfill & & \hfill & \\ \hfill y& =0& \hfill & \text{on }\partial \Omega ,& \hfill & & \hfill & \\ \hfill -△p& =y-y_d-\mu & \hfill & \text{in }\Omega ,& \hfill & & \hfill & \\ \hfill p& =0& \hfill & \text{on }\partial \Omega ,& \hfill & & \hfill & \\ \hfill u& ={\mathrm{proj}<!--FUNCTION\; APPLICATION-->}_{\left[-1,1\right]}\left(-p\right)& \hfill & \text{in }\Omega ,& \hfill & & \hfill & \\ \hfill {⟨\mu ,y-y_c⟩}_{C{\left(\bar{\Omega }\right)}^{\ast },C\left(\bar{\Omega }\right)}& =0,& \hfill & & \hfill & \\ \hfill \mu & \ge 0,& \hfill & & \hfill & \\ \hfill y_c& \le y.& \hfill & & \hfill & \end{array}$$

Supplementary Material

The optimal state and control are known analytically:

$$\begin{array}{ccc}\hfill y& =\cos <!--FUNCTION\; APPLICATION-->\left(\frac{\pi }{2}|x{|}^2\right),\hfill & \hfill \\ \hfill p& =\sin <!--FUNCTION\; APPLICATION-->\left(\pi |x{|}^2\right),\hfill \\ \hfill u& =-\sin <!--FUNCTION\; APPLICATION-->\left(\pi |x{|}^2\right),\hfill \\ \hfill \mu & =0.\hfill \end{array}$$

Note that the state constraint is active on the ball $\left|x\right|\le \frac{1}{2}$. This means that strict complementarity fails on the active set, a fact which makes the problem numerically challenging.

References