Three Dimensional Case

We distinguish here two cases, the purely subsonic case and the general case which may include transonic regimes.

Purely Subsonic Case. In this case $1 - M^2 > 0$ and hence $\vert (1-M^2) k_1^2 + k_2^2 \vert = (1-M^2) k_1^2 + k_2^2$. The preconditioner symbol is

$$\hat {\cal R} ({\bf k}) = \left[ (1-M^2) k_1^2 + k_2^2 \right] / (\rho_0^2u_0 ^4 k_1^4).$$ (54)

The preconditioned iteration will have in the Fourier space the direction

$$\hat{\tilde \alpha} ({\bf k}) = \hat {\cal R} ({\bf k}) \hat g ({\bf k})$$ (55)

which after some rearrangements reads as

$$\rho_0^2 u_0^4 k_1^4 \hat{\tilde\alpha} ({\bf k}) = ( (1-M^2) k_1^2 + k_2^2 ) \hat g ({\bf k}).$$ (56)

This equation in the Fourier space can be translated into the following differential equation for the change in the design variable,

$$\rho_0^2 u_0^4 \frac{\partial ^4 \tilde\alpha}{\partial x^4} = -(1 - M^2) \frac{\partial ^2 g}{\partial x^2} - \frac{\partial ^2 g}{\partial y^2}.$$ (57)

Actually, since our analysis was accurate only for the high frequency changes, we may not want to use that preconditioning for the very smooth components in the solution. This suggest a combination of the standard gradient descent method and this preconditioning, which for example, can be employed as

$$\mu \tilde \alpha + \rho_0^2 u_0^4 \frac{\partial ^4 \tilde\alpha... ...\frac{\partial ^2 g}{\partial x^2} - \frac{\partial ^2 g}{\partial y^2} + \mu g$$ (58)

The addition of the $\mu$ term is so that the low frequency range will not be affected by this preconditioner and would just use the gradient direction. High frequency on the other hand, are accurately analyzed by our method and should use the above preconditioner. It is also possible to use BFGS method in conjunction with the infinite dimensional preconditioner developed here.

Supersonic and Transonic Cases. In this case the term $\vert(1-M^2) k_1^2 + k_2^2 \vert$ cannot be simplified and we have to treat a certain pseudo differential operator. To approximate $\vert(1-M^2) k_1^2 + k_2^2 \vert$ we use the relation

$$\bar\sigma \sigma = \vert (1-M^2) k_1^2 + k_2^2 \vert$$ (59)

which was derived before, using the interior equations. We want to derive an implementation in real space of the equation whose form in the Fourier space is

$$\begin{eqnarray*} \mu \hat \alpha ({\bf k}) + \rho_0 ^2 u_0^4 k_1^4 \alpha ({\bf... ...^2) k_1^2 + k_2^2 \vert \hat g ({\bf k}) + \mu \hat g ({\bf k}). \end{eqnarray*}$$

The symbol $\bar\sigma \sigma$ represent two normal derivatives to solution of the small disturbance equation we started with, one with the $\phi$ equation and the other with the $\lambda$ equation. The difference between the two is at the far field boundary condition which is responsible for the proper choice in $\sigma, \bar\sigma$. The operator whose symbol is $\vert(1-M^2) k_1^2 + k_2^2 \vert$ is therefore constructed in two step.

$$\begin{array}{rr} (1-M^2) \psi _{xx} + \psi _{yy} + \psi _{zz} = 0 & \Omega \\ \psi = g & \partial \Omega \end{array}$$ (60)

$$\begin{array}{rr} (1-M^2) \bar\psi _{xx} + \bar\psi _{yy} + \bar\... ...a \\ \bar\psi = \frac{\partial \psi}{\partial z} & \partial \Omega \end{array}$$ (61)

and the full preconditioned direction for $\tilde\alpha$ is therefore

$$\mu \tilde \alpha + \rho _0^2 u_0^4 \frac{\partial ^4 \tilde \alpha} { \partial x ^4} = \frac{\partial \bar\psi}{\partial z} + \mu g$$ (62)

where $\bar\psi, \psi$ satisfy the above equations.