Optimal Control of Partial Differential Equations

9.5. OPTIMAL CONTROL PROBLEM 103
9.5 Optimal control problem
We consider the control problem
(P1) inf{J1(z, u) | (z, u) ∈ W (0, T ; H 1 0(Ω), H −1 (Ω)) × Uad, (z, u) satisfies (9.2.1)},
where Uad is a closed convex subset of L 2 (0, T ; L 2 (ω)),
J1(z, u) = 1
(z − zd)
2 Q
2 + 1
(z(T ) − zd(T ))
2 Ω
2 + β
χωu
2 Q
2 ,
and β > 0. In this section, we assume that f ∈ L 2 (Q) and that zd ∈ C([0, T ]; L 2 (Ω)).
We set Z = W (0, T ; H 1 0(Ω), H −1 (Ω)) . We define the mapping G from Z × L 2 (0, T ; L 2 (ω))
into L 2 (0, T ; H −1 (Ω)) × L 2 (Ω) by
G(z, u) =
∂z
∂t − ∆z + ∂x1(z 2
) − f − χωu, z(0) − z0 .
Theorem 9.5.1 The mapping G is of class C 1 , and for every (z, u) ∈ Z × L 2 (0, T ; L 2 (ω)),
G ′ z(z, u) is an isomorphism from Z into L 2 (0, T ; H −1 (Ω)) × L 2 (Ω).
Proof.
(i) Differentiability of F . The mapping
(z, u) ↦→
∂z
∂t − ∆z − χωu,
z(0) ,
is linear and bounded from Z × L 2 (0, T ; L 2 (ω)) into L 2 (0, T ; H −1 (Ω)) × L 2 (Ω). Thus to prove
that G is differentiable, we have only to check that
φ(z + h) − φ(z) − 2∂x1(zh) L 2 (0,T ;H −1 (Ω))
hZ
→ 0 as hZ → 0.
Since φ(z + h) − φ(z) − 2∂x1(zh)L 2 (0,T ;H−1 (Ω)) = φ(h)L2 (0,T ;H−1 (Ω)) ≤ h2 Z , the result is
obvious. We can also verify that
z ↦−→ (h ↦→ ∂x1(zh))
is differentiable from Z into L(Z; L 2 (0, T ; H −1 (Ω))). This means that G is twice differentiable.
In fact G is of class C ∞ .
(ii) G ′ z(z, u) is an isomorphism from Z into L 2 (0, T ; H −1 (Ω))×L 2 (Ω). Observe that G ′ z(z, u)w =
( ∂w
∂t − ∆w + 2∂x1(zw), w(0)). Thus, to prove that G ′ z(z, u) is an isomorphism from Z into
L 2 (0, T ; H −1 (Ω)) × L 2 (Ω), we have only to verify that, for any (f, z0) ∈ L 2 (0, T ; H −1 (Ω)) ×
L 2 (Ω), equation
∂w
∂t − ∆w + 2∂x1(zw) = f in Q, w = 0 on Σ, w(0) = z0 in Ω,
admits a unique solution in Z. This clearly follows from Theorem 9.6.5.