Optimal Control of Partial Differential Equations

88 CHAPTER 8. EQUATIONS WITH UNBOUNDED CONTROL OPERATORS
8.4.1 Neumann boundary control
We want to write equation (5.3.10) in the form (8.1.1). For this, we set Z = L 2 (Ω) and we
define the unbounded operator A in Z by
D(A) = {z ∈ H 2 (Ω) | ∂z
∂n
= 0}, Az = ∆z.
We know that A generates an analytic semigroup on Z (see [2]),
and
D((I − A) α ) = H 2α (Ω) if α ∈]0, 3
4 [,
D((I − A) α ) = {z ∈ H 2α (Ω) | ∂z
∂n
= 0} if α ∈]3 , 1[.
4
Consider now the Neumann operator N from L2 (Γ) into L2 (Ω) defined by N : u ↦→ w , where
w is the solution to
∆w − w = 0 in Ω,
∂w
= u
∂n
on Γ.
From [12] we deduce that N ∈ L(L 2 (Γ); H 3
2 (Ω)). This implies that N ∈ L(L 2 (Γ); D((I −A) α ))
for all α ∈]0, 3
4
[. We also have N ∈ L(H 1
2 (Γ); H 2 (Ω)).
Suppose that u ∈ C 1 ([0, T ]; H 1
2 (Γ)) and denote by z the solution to equation (5.3.10). Set
y(x, t) = z(x, t)−(Nu(t))(x). Since u ∈ C 1 ([0, T ]; H 3
2 (Γ)), Nu(·) belongs to C 1 ([0, T ]; H 2 (Ω)),
and we have
∂y
∂t
− ∆y = f − ∂Nu
∂t
Thus y is defined by
+ Nu in Q,
y(t) = e tA (z0 − Nu(0)) −
t
With an integration by parts we check that
(y + Nu)(t) = z(t) = e tA z0 +
0
∂y
∂n = 0 on Σ, y(x, 0) = (z0 − Nu(0))(x) in Ω.
(t−s)A d
e (Nu(s))ds +
dt
t
0
e (t−s)A f(s)ds +
t
0
t
0
e (t−s)A (Nu(s) + f(s))ds.
(I − A)e (t−s)A Nu(s)ds.
This means that, when u is regular enough, equation (5.3.10) may be written in the form
z ′ = Az + f + (I − A)Nu, z(0) = z0.
It is known that A is selfadjoint in L 2 (Ω), that is D(A) = D(A ∗ ) and Az = A ∗ z for all
z ∈ D(A). Since A ∗ ∈ L(D(A ∗ ); L 2 (Ω)), (A ∗ ) ∗ ∈ L(L 2 (Ω); D(A ∗ ) ′ ). Observing that N ∈
L(L 2 (Γ); H 3
2 (Ω)) and (I−(A ∗ ) ∗ ) ∈ L(L 2 (Ω); D(A ∗ ) ′ ), we have (I−(A ∗ ) ∗ )N ∈ L(L 2 (Γ); D(A ∗ ) ′ ).
In the case when u ∈ L 2 (0, T ; L 2 (Γ)) we consider the equation
z ′ = (A ∗ ) ∗ z + f + (I − (A ∗ ) ∗ )Nu, z(0) = z0. (8.4.16)