(37) |

and consider the minimization problem

(38) |

This problem is related to a shape design problem governed by the Full Potential equation in a general domain, using the cost functional .

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

(39) |

then the gradient of the functional is given by

(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 of the form

(41) |

where and , and then the corresponding change in is

(42) |

and the change in the adjoint variable is

(43) |

where are solutions for the following algebraic equation,

(44) |

In these expressions are given, as well as which amount to perturbing the shape by one frequency with a given amplitude.

**The Choice of
.** There is a nontrivial
point with respect to the choice of the roots that needs some explanation.
In the subsonic case, , we have two real roots for . One is
negative
and correspond to a bounded solution in , the other is
positive and correspond to an unbounded solution for and it
is discarded in our analysis.
In the supersonic case, , and the expression
may be
either positive or negative. If it is positive we take for
the root mentioned above. If it is negative
then 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 , while the adjoint
variable 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 is opposite to that
of in the supersonic case
when wave solutions exist.
Therefore, the roots are

(45) |

From the boundary condition for and we see that the following relations hold,

(46) |

and from these the change in the gradient as a result of a change in is given by and therefore,

(47) |

Now notice that

(48) |

and the two can be combined as

(49) |

In summary, the symbol of the Hessian is

(50) |