Optimal Control of Partial Differential Equations

8.4. THE HEAT EQUATION 89
For every z0 ∈ Z, f ∈ L 2 (Q), and u ∈ L 2 (0, T ; L 2 (Γ)) equation (8.4.16) admits a unique
solution in H1 (0, T ; D(A∗ ) ′ ) which is
t
z(t) = e t(A∗ ) ∗
z0 +
0
(I − (A ∗ ) ∗ )e (t−s)(A∗ ) ∗
Nu(s) ds +
t
0
e (t−s)A f(s) ds. (8.4.17)
When u ∈ C 1 ([0, T ]; H 1
2 (Γ)), the solutions given by Theorem 5.3.4 and by the formula (8.4.17)
coincide. Henceforth, by density arguments it follows that the solution defined by (8.4.17) and
the one of Theorem 5.3.4 also coincide when u ∈ L 2 (0, T ; L 2 (Γ)). In this case to simplify the
writing, we often write
z ′ = Az + f + (I − A)Nu, z(0) = z0, (8.4.18)
in place of (8.4.16). Since N ∈ L(L2 (Γ); D((I −A) α )) for all α ∈]0, 3[,
the operator (I −A)N =
4
(I − A) 1−α (I − A) αN can be decomposed in the form (I − A)N = (I − A) 1−αB1, where
B1 = (I − A) αN belongs to L(L2 (Γ); L2 (Ω)). This decomposition will be very useful to study
the Riccati equation corresponding to problem (P3) (see [26]).
8.4.2 Dirichlet boundary control
We set Z = L 2 (Ω) and we define the unbounded operator A in Z by
D(A) = H 2 (Ω) ∩ H 1 0(Ω), Az = ∆z.
We know that A is the infinitesimal generator of an analytic semigroup on Z and that (see
[2])
D((−A) α ) = H 2α (Ω) if α ∈]0, 1
4 [
and
D((−A) α ) = {z ∈ H 2α (Ω) | z = 0 on Γ} if α ∈] 1
, 1[.
4
We define the Dirichlet operator G from L2 (Γ) into L2 (Ω) by G :
solution to
u ↦→ w , where w is the
∆w = 0 in Ω, w = u on Γ.
From [12] we deduce that G ∈ L(L 2 (Γ); H 1
2 (Ω)). This implies that G ∈ L(L 2 (Γ); D((−A) α ))
for all α ∈]0, 1
4
[. We also have G ∈ L(H 3
2 (Γ); H 2 (Ω)).
Suppose that u ∈ C 1 ([0, T ]; H 3
2 (Γ)) and denote by z the solution to equation (5.4.20). Set
y(x, t) = z(x, t)−(Gu(t))(x). Since u ∈ C 1 ([0, T ]; H 3
2 (Γ)), Gu(·) belongs to C 1 ([0, T ]; H 2 (Ω)).
Thus we have
∂y
∂t
− ∆y = f − ∂Gu
∂t
in Q, y = 0 on Σ, y(x, 0) = (z0 − Gu(0))(x) in Ω.
As for Neumann controls we can check that, when u is regular enough, equation (5.4.20) may
be written in the form
z ′ = Az + f + (−A)Gu, z(0) = z0.
And in the case when u ∈ L 2 (0, T ; L 2 (Γ)) we still continue to use the above formulation even
if for a correct writing A should be replaced by its extension (A ∗ ) ∗ .