The Linearized Euler Equations

Consider the linearized Euler equation around a mean flow $(\rho_0, \rho_0 u_0, 0, 0, p_0)$,

$$\begin{array}{cccc} \left( \begin{array}{ccccc} Q & {\it\rho _0}\... ...gin{array}{c} \rho \\ u \\ v \\ w \\ p \end{array}\right)& =& 0 \end{array}$$ (63)

in a domain $\Omega = \{ (x,y,z) \vert z \geq 0 \}$, where $Q=\vec{U_0}\cdot \vec{\nabla}$ ( $\vec{U_0}\equiv(u_0,0,0)$ denotes the velocity vector), with the solid wall boundary condition

$$w = u_0 \alpha _x \qquad \partial\Omega.$$ (64)

It is assumed that the problem was obtained by a linearization in a vicinity of a boundary point, and that the far field boundary conditions were given in terms of characteristic variables, which are not used explicitly in the derivation of the approximate Hessian. The minimization problem is

$$\min _{\alpha} \frac{1}{2} \int _{\partial \Omega} ( p - p^*) ^2 dx.$$ (65)

If a change $\tilde\alpha$ produces a change $\tilde p$ in the pressure then, the variation in this functional can be written as

$$\delta J = \int _{\partial \Omega} ( p - p^*) \tilde p dx + \frac{1}{2} \int _{\partial \Omega} \tilde p ^2 dx + O( \Vert \tilde p\Vert^3 )$$ (66)

We calculate the Hessian in a slightly different way than before to illustrate another approach. If one can express the quadratic term in $\tilde p$ in terms of $\tilde\alpha$ one can identify the Hessian. That is,

$$\int \tilde p ^2 dx = \int ({\cal H} \tilde \alpha) \tilde \alpha dx.$$ (67)

This means that we can calculate the Hessian without going through the adjoint variable. We need to express $\tilde p$ in terms of $\tilde\alpha$, and we do it in the Fourier space. From the boundary condition at the wall

$$\hat{ \tilde w} ( {\bf k} ) = i k_1 u_0 \hat {\tilde \alpha} ({\bf k} ).$$ (68)

The calculation of $\hat{\tilde p} ( {\bf k} )$ in terms of $\hat{\tilde\alpha} ({\bf k})$ is done by solving the system of the linearized Euler equation with the above boundary condition for $\tilde w$. We look for solution of the form

$$\tilde U = \left( \begin{array}{c} A_1 \\ A_2 \\ A_3 \\ A_4 \\ A_5 \end{array}\right) \exp ( i {\bf k} \cdot {\bf x} ) \exp ( i k_3 z ).$$ (69)

The term $i k_3$ here is the analog of our $\sigma$ in the previous example. It is more convenient here due to the form of the symbol of the full equation. The following relation follows by substituting the above expression for $\tilde U$ into the Linearized Euler equations (63),

$$\hat{L} ({\bf k}) \left( \begin{array}{c} A_1 \\ A_2 \\ A_3 \\ ... ...}\right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right)$$ (70)

This is a linear system for $(A_1, \dots, A_5)$ and it has a nontrivial solution when the determinant is zero,

$$det \hat{L} ({\bf k}) = k_1^3 ( u_0^2 k_1 ^2 - c_0^2 ( k_1^2 + k_2^2 + k_3^2) ) = 0$$ (71)

Note that there are five solutions for this equations. Each of them has a corresponding solution for the vector $(A_1, \dots, A_5)$,

$$\begin{array}{lr} V_1 = ( \rho _0 u_0 c_0^2 k_1, -k_1, -k_2, -k_3... ..., 1, 0,0, 0, ) & k_1 = 0 \\ V_5 = ( 0, 0, -k_3, k_2, 0 ) & k_1 = 0 \end{array}$$ (72)

where

$$\begin{array}{lr} \sigma _1 = - \sigma _2 = \sqrt{(1-M^2) k_1 ^2 ... ...i \sqrt{ - (1-M^2) k_1 ^2 - k_2^2 } & (1-M^2) k_1 ^2 + k_2^2 \leq 0 \end{array}$$ (73)

Note that for the subsonic case $\sigma _1$ correspond to the bounded solution, while $\sigma _2$ to the unbounded one. When $(1-M^2) k_1 ^2 + k_2^2 < 0$ we have two bounded solution. In that case $\sigma _2$ correspond to the incident wave and therefore its amplitude is zero for the perturbation variables. Thus, we are left with $\sigma _1$ for both subsonic and supersonic cases. The three solution corresponding to $k_1=0$ are not important for our analysis since they do not affect the changes in pressure (see the corresponding eigenvectors).

To summarize, only $V_1$ contributes to the pressure changes as a result of changes to the design variables by $\tilde\alpha$. The solution for $\tilde U$ is given by $A V_1$ for some scalar $A$. The $w$ component in this solution is $- A k_3$ and this must equal to $i u_0 k_1 \hat{ \tilde \alpha } ({\bf k})$ form the boundary condition which in the Fourier space is given by (68). From that we find $A = - i u_0 \frac{k_1}{k_3} \hat{\tilde \alpha}( {\bf k})$. Thus the solution is,

$$\tilde U = - i u_0 \frac{k_1}{k_3} \hat{\tilde \alpha}({\bf k}) \... ...ho _0 u_0 c_0^2 k_1\\ -k_1\\ -k_2\\ -k_3\\ \rho_0 u_0 k_1\end{array}\right)$$ (74)

The last component in this vector gives us the change in the pressure

$$\hat{\tilde p}({\bf k}) = -i \rho_0 u_0^2 \frac{k_1^2}{k_3} \hat{\tilde \alpha}( {\bf k})$$ (75)

and from this we get

$$\vert \hat{\tilde p}({\bf k}) \vert^2 = \rho_0^2 u_0^4 \frac{k_1^4}{k_3\bar k_3} \vert \hat{\tilde \alpha}( {\bf k}) \vert ^2.$$ (76)

Notice that we have taken the complex conjugate of $\hat{\tilde p} ( {\bf k} )$, and since $k_3$ is a complex number its conjugate was taken as well. Since $k_3 \bar k_3 = \vert (1-M^2) k_1^2 + k_2^2 \vert$ we obtain the symbol of the Hessian in the form,

$$\hat{\cal H} ({\bf k}) = \rho_0^2 u_0^4\frac{k_1^4}{\vert (1-M^2) k_1^2 + k_2^2 \vert }.$$ (77)

A preconditioner for this problem is done exactly as in the small disturbance equations using (60)-(62). It is also possible to construct the preconditioner based on solution of the linearized Euler equations, but is more complicated and unnecessary. The gradient $g$ appearing in (60)-(62) has to be changed to the gradient for this problem, using the adjoint formulation.