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Shape Design Using The Euler Equations

Our next example is a similar minimization problem but this time subject to the Euler equation. Namely,
$\displaystyle \min_{\Gamma} \frac{1}{2} \int_{\Gamma} (p - p^*) ^2 ds$     (82)

where $p = p(U)$, and $U$ is the solution of the Euler equation. Here $U$ stands for the variables $(\rho, \rho {\bf u} , E )$ and ${\bf u} = (u,v,w)$. The Euler equations in conservation form are written as
$\displaystyle f_x + g_y + h_z = 0$     (83)

$\displaystyle f = A U \qquad g = B U \qquad h = C U,$     (84)

and where the matrices $A,B,C$ can be found, for example, in Hirsch [12]. An important property of the equation that we use here is
$\displaystyle f_x = A U_x \qquad g_y = B U_y \qquad h = C u_z.$     (85)

The change $\tilde f$ in the flux vector $f$ satisfies,
$\displaystyle f( U + \epsilon \tilde U) = f(U) + \epsilon \frac{\partial f}{\pa...
...silon ^2) =
f(U) + \epsilon A \tilde U + O(\epsilon ^2) = f + \epsilon \tilde f$     (86)

and similar expressions for $\tilde g, \tilde h$. The equation for the perturbation quantities reads
$\displaystyle \tilde f _x + \tilde g _y + \tilde h _z = 0$     (87)

or equivalently,
$\displaystyle (A \tilde U )_x + (B \tilde U) _y + (C \tilde U) _z = 0.$     (88)

Now consider the following identity which follows from integration by parts,
$\displaystyle \begin{array}{ll}
\int _\Omega \Lambda \cdot (A \tilde U)_x dV & ...
...\int _{\partial\Omega} A^T \Lambda \cdot \tilde U {\bf n\cdot i} ds
\end{array}$     (89)

and similar integrals for the $g$ and $h$ terms. Combining these identities we arrive at
$\displaystyle \begin{array}{ll}
-\int _\Omega ( A^T \Lambda _x + B^T \Lambda _y...
...n \cdot j} + C^T {\bf n \cdot k} ) \Lambda \cdot \tilde U ds & = 0,
\end{array}$     (90)

for an arbitrary $\Lambda$. We will use the notation
$\displaystyle D = A {\bf n \cdot i } + B {\bf n \cdot j} + C {\bf n \cdot k}$     (91)

and note that $DU $ is the normal flux at the boundary which has the form, see Hirsch [12],
$\displaystyle D U = \left( \begin{array}{c} \rho {\bf u} \cdot {\bf n} \\  \rho...
... n}) {\bf u} + p {\bf n} \\  (E + p ) {\bf u} \cdot {\bf n} \end{array}\right),$     (92)

and at a wall where ${\bf u} \cdot {\bf n} = 0$, it reduces to
$\displaystyle D U _{\vert _\Gamma} = \left( \begin{array}{c} 0 \\  p {\bf n} \\  0 \end{array}\right).$     (93)

We have $D \tilde U = \tilde{DU} $ following (84),(86) and its analog for the $\tilde g, \tilde h$ terms, and
$\displaystyle \tilde {( p {\bf n} )} = \tilde p {\bf n} + p \tilde {\bf n}.$     (94)

Combining the last equalities and $\tilde{\bf n} = - \sum _{j=1}^{d-1} \frac{\partial \alpha}{\partial t_j} {\bf t}_j$ from (75), we get
$\displaystyle \int _\Gamma \Lambda \cdot D \tilde U ds = \int _\Gamma \Lambda \...
..._{j=1}^{d-1} \frac{\partial \alpha }{\partial t_j} (\lambda \cdot {\bf t}_j )ds$     (95)

where we used the notation $\Lambda = ( \lambda _1, \lambda , \lambda _5 )$, and $\lambda = (\lambda _2, \lambda _3, \lambda _4 )$. The wall boundary condition
$\displaystyle {\bf u \cdot n}_{\vert _\Gamma} = 0$     (96)

becomes upon perturbation
$\displaystyle {\bf u}^\epsilon \cdot {\bf n^\epsilon} _{\vert _\Gamma ^\epsilon} = 0,$     (97)

and as before we transfer this boundary condition to the original boundary $\Gamma$,
$\displaystyle ({\bf u + \epsilon \tilde u}) \cdot {\bf n}^\epsilon _{\vert _{\G...
...partial \alpha}{\partial t_j} {\bf t_j} ) _{\vert _{\Gamma}} + O (\epsilon ^2).$     (98)

Collecting only the $\epsilon$ terms we get
$\displaystyle \tilde {\bf u} \cdot {\bf n} - \sum_{j=1}^{d-1} \frac{\partial \a...
...j} + \alpha \frac{\partial {\bf u}}{\partial n}\cdot {\bf n} = 0 \qquad \Gamma.$     (99)

The variation of the functional
$\displaystyle \delta J = \int _{\Gamma} [ (p - p^*) \tilde p + \alpha (p - p^*) \frac{\partial p}{\partial n}- \alpha \frac{(p-p^*)^2}{2R} ] ds$     (100)

will be simplified by adding (90) to it, but with a choice of $\Lambda$ which makes the volume integral vanish. Thus, we assume that
$\displaystyle A^T \Lambda _x + B^T \Lambda _y + C^T \Lambda _z = 0 \qquad \qquad \Omega$     (101)

Using (90),(95) it leads to
$\displaystyle \begin{array}{ll}
\delta J & = \int _{\Gamma} [ p - p^* + {\bf\la...
...}& + \int _{\partial \Omega - \Gamma} D^T \Lambda \cdot \tilde U ds
\end{array}$     (102)

Now we come to use the boundary conditions for $\tilde U$. We begin with the far field $\partial \Omega - \Gamma$. We assume that the boundary conditions there are given in terms of characteristic variables and assume that $T$ is the matrix such that $T \tilde U$ are the characteristic variables. We write the far field term as
$\displaystyle \int _{\partial \Omega - \Gamma } D^T \Lambda \cdot \tilde U ds =...
...U ds =
\int _{\partial \Omega - \Gamma } T^{-T} D^T \Lambda \cdot T \tilde U ds$     (103)

We distinguish the following cases. Supersonic inflow: all variables are specified at inflow, and thus $\tilde U = 0$. Thus, no boundary conditions are imposed on $\Lambda$. Supersonic outflow: No boundary conditions are specified for $U$, hence $\tilde U$ is arbitrary there and therefore we are led to the choice $\Lambda = 0$ at supersonic outflow. Subsonic inflow: 4 conditions are specified (3 in 2D), and those are $(T \tilde U)_{1,2,3,4} = 0$, thus $(T \tilde U)_5$ is arbitrary, leading to $(T^{-T} D^T \Lambda )_5 = 0$. Subsonic outflow: one condition is given for $U$ which implies $(T \tilde U)_5 = 0 $ and therefore $(T^{-T} D \Lambda )_{1,2,3,4} = 0$. On the wall $\Gamma$ we choose
$\displaystyle {\bf\lambda } \cdot {\bf n} + p - p^* = 0.$     (104)

In summary, the boundary conditions for $\Lambda$ are
$\displaystyle \begin{array}{lr}
\mbox{\tt supersonic inflow} & \mbox{\rm none }...
...,3,4} = 0 \\
\mbox{\tt Wall} & \lambda \cdot {\bf n} + p - p^* = 0
\end{array}$     (105)

With this choice for $\Lambda$ together with the interior equation (101) we get that $\delta J$ involves integrals depending on $\alpha$ and $\Lambda$ and not on $\tilde U$ terms. Rearrangement by using integration by parts gives,

$\displaystyle \delta J =
\int _\Gamma \alpha [- \frac{(p-p^*)^2}{2R} + (p-p^*) \frac{\partial p}{\partial n}-div (p \lambda )]ds.$     (106)

The gradient of the functional in this case is therefore given by

$\displaystyle \nabla _\Gamma J = - \frac{(p-p^*)^2}{2R} + (p-p^*) \frac{\partial p}{\partial n}-div (p \lambda ).$     (107)

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Next: Bibliography Up: Applications to Fluid Dynamics Previous: Shape Design Using The
Shlomo Ta'asan 2001-08-22