Control Problems Governed by PDE

Shape optimization problems which are one of our main topics, are related to control problems governed by PDE. An understanding of boundary control problems will help us to get the proper insight into shape optimization problems.

We consider the small disturbance equation, for zero Mach number, in two dimensions. The domain $\Omega$ is a rectangle whose bottom boundary has a control variable to be optimized. It is required to achieve a certain pressure distribution on that boundary. Denote by $\Gamma$ the bottom boundary and by $\Gamma _0$ the rest of the boundary $\Omega$. The potential $\phi$ satisfies the equation

$$\mbox{\tt State Equation: } \begin{array}{lr} \Delta \phi = 0 & \... ...artial n}= \alpha _x & \Gamma \\ \vspace{2mm} \phi = g & \Gamma _0 \end{array}$$ (30)

where $\alpha$ is the design variable. This problem is related to a shape design problem in which the bottom boundary is described by the function $\alpha$, and the boundary condition for $\phi$ on this boundary is $\frac{\partial \phi}{\partial n}= 0$. We will come to this relation later on.

We consider the cost functional

$$E( \alpha ) = \frac{1}{2} \int _{\Gamma } ( p - p^*)^2 dx$$ (31)

where $p = \phi _x$, and we would like to construct a formula for the gradient of this functional. We consider a perturbation of the design variable by $\epsilon \tilde \alpha$ and the corresponding change in $\phi$ by $\epsilon \tilde \phi$, which satisfies

$$\begin{array}{lr} \Delta \tilde\phi = 0 & \Omega \\ \vspace{2mm}... ...lde\alpha _x & \Gamma \\ \vspace{2mm} \tilde \phi = 0 & \Gamma _0. \end{array}$$ (32)

The variation in the functional is

$$\begin{array}{ll} \delta E \equiv E(\alpha + \epsilon \alpha ) - ... ...int _{\Gamma } (\phi _x - p^* ) _x \tilde \phi dx + O (\epsilon ^2) \end{array}$$ (33)

where integration by parts have been used in the last equality. As in the abstract constrained optimization problem we discussed in section (2.2), we see that the change in the functional depends on the the sensitivity derivatives $\tilde \phi$, and we would like to eliminate this dependence, in order to get an efficient computation of the gradient. We do it by adding a term to $\delta E$ which is the differential analog of the term $\tilde U ^T L_U^T \lambda + \tilde \alpha ^T L_\alpha ^T \lambda$ in the algebraic case. Then a proper choice for $\lambda$ will result in the desired form for the variation in the functional.

Let $\lambda$ be an arbitrary function defined in the same domain as $\phi$. From equation (32) for $\tilde \phi$ we have

$$0 = \int _\Omega \lambda \Delta \tilde \phi dx = \int _\Omega \ti... ...al \tilde\phi}{\partial n}- \tilde \phi \frac{\partial \lambda}{\partial n}) ds$$ (34)

where the second equality follows from integration by parts. This is the analog of equation (10) that we have in the algebraic case. Adding the right hand side of (34) (multiplied by $\epsilon$) to $\delta E$ we get

$$\delta E = - \epsilon \int _{\Gamma } (\phi _x - p^* )_x \tilde \... ...rtial n}- \tilde \phi \frac{\partial \lambda}{\partial n}) ds + O(\epsilon ^2).$$ (35)

Since $\partial \Omega = \Gamma + \Gamma _0$ we can break the integral $\int _{\partial \Omega}$ into $\int _{\Gamma} + \int _{\Gamma _0}$ and then combine the $\int_\Gamma$ terms, to obtain

$$\delta E = - \epsilon \int _\Gamma \tilde \phi [ (\phi _x - p^* )... ...l n}ds + \epsilon \int _\Omega \tilde \phi \Delta \lambda dx + O ( \epsilon ^2)$$ (36)

In order to eliminate the dependence of $\delta E$ on $\tilde \phi$ we make the following choice for $\lambda$

$$\mbox{\tt Adjoint Equation: } \begin{array}{ll} \Delta \lambda = ... ..._x - p^* )_x = 0 & \Gamma \\ \vspace{2mm} \lambda = 0 & \Gamma _0. \end{array}$$ (37)

Therefore,

$$\delta E = \epsilon \int _\Gamma \lambda \frac{\partial \tilde\ph... ...(\epsilon ^2) = -\epsilon \int _\Gamma \lambda _x \tilde \alpha+ O(\epsilon ^2)$$ (38)

where integration by parts was used in the last equality. The expression for the changes in the functional is given as as a function of the changes in the design variable, as well the adjoint variable $\lambda$ which satisfies the adjoint equation (37), (or costate equation in control terminology). We would like now to pick a direction of change $\tilde \alpha$ that will result in reduction of the cost functional. We distinguish two cases.

I. Finite Dimensional Control. In this case we assume that the design variable $\alpha (x)$ has a representation

$$\alpha (x) = \sum_{j=1}^q \alpha _j f_j(x)$$ (39)

and a similar expression for $\tilde \alpha$, where the functions $f_j, j=1,\dots, q$ are prescribed. This implies

$$\delta E = - \epsilon \sum _{j=1}^q \tilde \alpha _j \int _\Gamma \lambda _x f_j dx + O(\epsilon ^2),$$ (40)

and hence

$$\frac{\partial E}{\partial \alpha _j} = - \int _\Gamma \lambda _x f _j dx$$ (41)

and the choice

$$\tilde \alpha _j = \int _\Gamma \lambda _x f_j dx$$ (42)

will result in a reduction of the cost functional by

$$\delta E = -\epsilon \sum _{j=1}^q (\int _\Gamma \lambda _x f_j dx ) ^2 + O ( \epsilon ^2).$$ (43)

At a minimum the following conditions hold,

$$\frac{\partial E}{\partial \alpha _j} = - \int _\Gamma \lambda _x f_j dx = 0 \qquad \qquad j=1, \dots , q.$$ (44)

Note that the gradient given in (44) is in terms of $\lambda$.

II. Infinite Dimensional Control. In this case we regard the variable $\alpha$ as a function defined on the boundary $\Gamma$. A proper choice that will result in a reduction of the functional is given in terms of $\lambda$,

$$\tilde \alpha (x) = \lambda _x (x) \qquad x \in \Gamma$$ (45)

and the corresponding reduction in the functional is given this time by

$$\delta E = - \epsilon \int _\Gamma \vert \lambda _x \vert ^ 2 dx + O(\epsilon ^2).$$ (46)

An algorithm for solving this control problem consists of repeated application of the following three steps, until convergence (of the gradient to zero),

ALGORITHM
(1) Solve the state equation (30) for $\phi$
(2) Solve the adjoint equation (37) for $\lambda$
(3) Update $\alpha$ by $\alpha - \delta \tilde \alpha \qquad$, where $\delta$ is found by line search.
$\mbox{\hspace{1cm}} \tilde \alpha$ is given by (42) or (45).