Optimal Control of Partial Differential Equations

8 CHAPTER 1. EXAMPLES OF CONTROL PROBLEMS
than the previous one) can be placed in contact with the electrolyte. In this device the noble
metal plays the role of a cathode, and the other one the role of the anode. A current can be
prescribed to the anode to modify the electric field in the electrolyte. This process is known
as cathodic protection.
The system can be described by the elliptic equation
−div(σ∇φ) = 0 in Ω,
−σ ∂φ
∂n = i on Γa, −σ ∂φ
∂n = 0 on Γi, −σ ∂φ
∂n
= f(φ) on Γc,
(1.2.1)
where φ is the electrical potential, Ω is the domain occupied by the electrolyte, Γa is a part
of the boundary of Ω occupied by the anode, Γc is a part of the boundary of Ω occupied by
the cathode, Γi is the rest of the boundary Γ, Γi = Γ \ (Γa ∪ Γc). The control function is the
current density i, the constant σ is the conductivity of the electrolyte, the function f is known
as the cathodic polarization function, and in general it is a nonlinear function of φ.
The cathode is protected if the electrical potential is closed to a given potential ¯ φ on Γc.
Thus the cathodic protection can be achived by choosing the current i as the solution to the
minimization problem
(P1) inf{J1(φ) | (φ, i) ∈ H 1 (Ω) × L 2 (Γa), (φ, i) satisfies (1.2.1), a ≤ i ≤ b},
where a and b are some bounds on the current i, and
J1(φ) = (φ − ¯ φ) 2 .
Γc
A compromise between ’the cathodic protection’ and ’the consumed energy’ can be obtained
by looking for a solution to the problem
(P2) inf{J2(φ, i) | (φ, i) ∈ H 1 (Ω) × L 2 (Γa), (φ, i) satisfies (1.2.1), a ≤ i ≤ b},
where
and β is a positive constant.
J2(φ, i) =
Γc
(φ − ¯ φ) 2
+ β
1.2.2 Optimal control problem in radiation and scattering
Here the problem consists in determining the surface current of a radiating structure which
maximize the radiated far field in some given directions [22]. Let Ω ⊂ R N be the complementary
subset in R N of a regular bounded domain (Ω is called an exterior domain), and let Γ its
boundary. The radiated field y satisfies the Helmholtz equation
Γa
i 2 ,
∆y + k 2 y = 0, in Ω, (1.2.2)
where k ∈ C, Imk > 0, and the radiation condition
∂y
1
− iky = O
∂r |x| (N+1)/2
, when r = |x| → ∞. (1.2.3)