Optimal Control of Partial Differential Equations

6.6. DIRICHLET BOUNDARY CONTROL 71
Theorem 6.6.1 For every (f, u, z0, z1) ∈ L 1 (0, T ; H −1 (Ω)) × L 2 (Σ) × L 2 (Ω) × H −1 (Ω), equation
(6.6.15) admits a unique weak solution z(f, u, z0, z1) in C([0, T ]; L 2 (Ω))∩C 1 ([0, T ]; H −1 (Ω)).
The mapping (f, u, y0, y1) ↦→ z(f, u, z0, z1) is linear and continuous from L 1 (0, T ; H −1 (Ω)) ×
L 2 (Σ) × L 2 (Ω) × H −1 (Ω) into C([0, T ]; L 2 (Ω)) ∩ C 1 ([0, T ]; H −1 (Ω)).
Proof. This existence and regularity result can be proved by the transposition method with
Theorem 6.5.1.
(i) Due to Theorem 6.2.3, the mapping
(f, z0, z1) ↦−→ z(f, 0, z0, z1)
is linear and continuous from L 1 (0, T ; H −1 (Ω)) × L 2 (Ω) × H −1 (Ω) into C([0, T ]; L 2 (Ω)) ∩
C 1 ([0, T ]; H −1 (Ω)). Thus we have only to consider the case where (f, z0, z1) = (0, 0, 0).
(ii) Denote by Λ the mapping ϕ ↦→ ∂y
, where y is the solution to equation (6.6.17) cor-
∂n
responding to (yT , νT ) = (0, 0). Due to Theorem 6.5.1, Λ is a linear operator from L2 (Q)
into L2 (Σ). If we set z = Λ∗u, with u ∈ L2 (Σ), we observe that z ∈ L2 (Q), and z is a
solution to equation (6.6.15) with (f, z0, z1) = (0, 0, 0), in the sense of definition 6.6.1. This
solution is unique in L2 (Q). Indeed if z1 and z2 are two solutions to equation (6.6.15) with
(f, z0, z1) = (0, 0, 0), in the sense of definition 6.6.1, we have
(z1 − z2)ϕ = 0 for all ϕ ∈ L
Ω
2 (Q).
To prove that z belongs to C([0, T ]; L 2 (Ω))∩C 1 ([0, T ]; H −1 (Ω)), we proceed by approximation.
Let (un)n be a sequence of regular functions such that zn = z(0, un, 0, 0) be regular. For
τ ∈]0, T ] and (yτ, ντ) ∈ H 1 0(Ω) × L 2 (Ω), we denote by y(yτ, ντ) the solution to the equation
∂2y ∂t2 − ∆y = 0 in Q, y = 0 on Σ, y(x, τ) = yτ and ∂y
∂t (x, τ) = ντ in Ω.
With Theorem 6.5.1, we have
∂y(yτ, ντ)
L2 (Σ) ≤ C yτH 1
0 (Ω) + ντL2 (Ω) , (6.6.18)
∂n
where the constant C depends on T , but is independent of τ.
equation (6.6.15), according to definition 6.6.1, we have
Since zn is the solution to
〈 ∂zn
∂t (τ), yτ〉
H−1 (Ω)×H1 0 (Ω) =
∂zn
∂t (τ)yτ
dx = −
∂y(yτ, 0)
u dsdt
∂n
and
From which we deduce:
Ω
zn(τ)νT dx =
Ω
zn − zm C([0,T ];L 2 (Ω)) = sup τ∈]0,T ]sup ντ L 2 (Ω) =1
Σ
≤ Cun − um L 2 (Σ)
Σ
∂y(0, ντ)
un dsdt.
∂n
Σ
∂y(0, ντ)
∂n
(un − um) dsdt