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Next: Two Dimensional Case Up: Application to Shape Design: Previous: Application to Shape Design:

Small Disturbance Potential Equation

Let $\Omega = \{ (x,y,z) \vert z > 0 \}$ and let $\phi$ satisfy
$\displaystyle \begin{array}{lr}
(1 - M^2) \phi _{xx} + \phi _{yy} + \phi _{zz} ...
... \frac{\partial \phi} {\partial z} = u_0 \alpha _x & \partial\Omega
\end{array}$     (37)

and consider the minimization problem
$\displaystyle \min _{\alpha} \frac{1}{2} \int _{\partial \Omega} ( \rho_0 u_0 \phi _x - f^*)^2.$     (38)

This problem is related to a shape design problem governed by the Full Potential equation in a general domain, using the cost functional $\frac{1}{2} \int _{\partial\Omega} ( p - p^* ) ^2 ds$.

It can be shown using standard computation, as explained in a previous lecture, that if $\lambda$ satisfies the equation

$\displaystyle \begin{array}{lr}
(1-M^2) \lambda _{xx} + \lambda _{yy} + \lambda... z} - \rho_0 u_0(\rho_0 u_0\phi _x - f^*) _x = 0 & \partial\Omega
\end{array}$     (39)

then the gradient of the functional is given by
$\displaystyle \nabla _\alpha J = - u_0 \frac{\partial \lambda }{ \partial x}.$     (40)

We would like to compute the Hessian for this problem and to construct an infinite dimensional preconditioner for it. We assume a perturbation in $\alpha$ of the form
$\displaystyle \tilde\alpha ({\bf x}) = \hat{\alpha} ({\bf k}) \exp {( i {\bf k} \cdot {\bf x} ) }$     (41)

where ${\bf x} = (x,y)$ and ${\bf k} = (k_1, k_2 )$, and then the corresponding change in $\phi$ is
$\displaystyle \tilde\phi ({\bf x},z) = \hat{\phi} ({\bf k}) \exp {( i {\bf k} \cdot {\bf x} )} \exp { (\sigma z )},$     (42)

and the change in the adjoint variable is
$\displaystyle \tilde\lambda ({\bf x},z) = \hat{\lambda} ({\bf k}) \exp {( i {\bf k} \cdot {\bf x} )} \exp {( \bar\sigma z )}$     (43)

where $\sigma, \bar\sigma$ are solutions for the following algebraic equation,
$\displaystyle \begin{array}{l}
- (1-M^2) k_1^2 - k_2^2 + \sigma ^2 = 0 \\
- (1-M^2) k_1^2 - k_2^2 + \bar\sigma ^2 = 0.
\end{array}$     (44)

In these expressions $k_1, k_2$ are given, as well as $\hat\alpha ({\bf k})$ which amount to perturbing the shape by one frequency with a given amplitude.

The Choice of $\sigma, \bar\sigma$. There is a nontrivial point with respect to the choice of the roots that needs some explanation. In the subsonic case, $M<1$, we have two real roots for $\sigma$. One is negative and correspond to a bounded solution in $\Omega$, the other is positive and correspond to an unbounded solution for $\phi, \lambda$ and it is discarded in our analysis. In the supersonic case, $M>1$, and the expression $(1-M^2) k_1^2 + k_2^2 $ may be either positive or negative. If it is positive we take for $\sigma$ the root mentioned above. If it is negative then $\sigma$ has two imaginary roots. One of these correspond to an incident wave and the other to a reflected wave. The perturbation in the incident wave is zero, since this wave comes from infinity and there was no change there. The reflected wave arise from the change in shape. In the supersonic regime, at the outflow there are no boundary conditions for $\phi$, while the adjoint variable $\lambda$ has two boundary conditions. Therefore, for the perturbation in the adjoint variable no waves are going toward the outflow. This means that the sign of $\bar\sigma$ is opposite to that of $\sigma$ in the supersonic case when wave solutions exist. Therefore, the roots are

$\displaystyle \begin{array}{lr}
\sigma = \bar \sigma = - \sqrt{ (1-M^2) k_1^2 +...
...= i \sqrt{ -(1-M^2) k_1^2 - k_2 ^2 }& (1-M^2) k_1^2 + k_2^2 \leq 0.
\end{array}$     (45)

From the boundary condition for $\phi$ and $\lambda$ we see that the following relations hold,
$\displaystyle \begin{array}{l}
- \sigma \hat\phi ({\bf k}) = i k_1 u_0 \hat\alp... \hat\lambda ({\bf k}) = \rho_0^2 u_0^2 k_1 ^2 \hat\phi ({\bf k})
\end{array}$     (46)

and from these the change in the gradient as a result of a change in $\alpha$ is given by $- \tilde\lambda _x$ and therefore,
$\displaystyle \hat {\cal H} ({\bf k}) = \rho_0^2 u_0 ^4 \frac{k_1^4}{\bar\sigma \sigma}.$     (47)

Now notice that
$\displaystyle \begin{array}{llr}
\bar\sigma\sigma & = (1-M^2) k_1^2 + k_2^2 & \...
...-(1-M^2) k_1^2 - k_2^2 & \qquad \qquad (1-M^2) k_1^2 + k_2^2 \leq 0
\end{array}$     (48)

and the two can be combined as
$\displaystyle \bar\sigma\sigma = \vert (1-M^2) k_1^2 + k_2^2 \vert.$     (49)

In summary, the symbol of the Hessian is
$\displaystyle \hat{\cal H} ({\bf k}) = \rho_0^2 u_0^4\frac{k_1^4}{\vert (1-M^2) k_1^2 + k_2^2 \vert }.$     (50)

next up previous
Next: Two Dimensional Case Up: Application to Shape Design: Previous: Application to Shape Design:
Shlomo Ta'asan 2001-08-22