Optimal Control of Partial Differential Equations

2.7. OTHER FUNCTIONALS 23
where wu is the solution to equation (2.2.2):
−∆w + w = 0 in Ω,
Let us consider the variational equation
Ω
∂w
∂n
= u on Γ.
Find p ∈ H1 (Ω), such that
(∇p∇φ + pφ) = Ω (∇¯z − ∇zd)∇φ for all φ ∈ H1 (Ω).
(2.7.13)
With the Lax-Milgram theorem we can prove that equation (2.7.13) admits a unique solution
p ∈ H1 (Ω). Taking φ = wu in the variational equation (2.7.13), we obtain
F ′
4(ū)u = (∇¯z − ∇zd)∇wu + β ūu = (∇p∇wu + pwu) + β ūu.
Ω
Γ Ω
Γ
With the Green formula we finally have
F ′
4(ū)u = (p + βū)u.
Interpretation of equation (2.7.13). Observe that if p is the solution of (2.7.13), then
Γ
−∆p + p = −div(∇¯z − ∇zd)
in the sense of distributions in Ω. If ¯z and zd belongs to H 2 (Ω), then p ∈ H 2 (Ω). We can
verify that a function p is a solution of (2.7.13) in H 2 (Ω) if and only if
−∆p + p = −div(∇¯z − ∇zd) in Ω, and
∂p
∂n = (∇¯z − ∇zd) · n on Γ. (2.7.14)
In the case when ¯z and zd do not belong to H2 (Ω), (∇¯z−∇zd)·n|Γ = ∂¯z ∂zd − is not necessarily
∂n ∂n
defined. However, we still use the formulation (2.7.14) in place of (2.7.13), even if it is abusive.
We can state the following theorem.
Theorem 2.7.1 If (¯z, ū) is the solution to (P4) then ū = − 1
β p|Γ, where p is the solution to
equation (2.7.14).
Conversely, if a pair (˜z, ˜p) ∈ H 1 (Ω) × H 1 (Ω) obeys ˜z = z(f, − 1
β ˜p|Γ) and
−∆˜p + ˜p = −div(∇˜z − ∇zd) in Ω,
then the pair (˜z, − 1 ˜p) is the optimal solution to problem (P4).
β
2.7.2 Pointwise observation
We consider the control problem
∂p
∂n = (∇˜z − ∇zd) · n on Γ,
(P5) inf{J5(z, u) | (z, u) ∈ H 1 (Ω) × L 2 (Γ), (z, u) satisfies (2.2.1)}.