Optimal Control of Partial Differential Equations

50 CHAPTER 5. CONTROL OF THE HEAT EQUATION
with g ∈ L2 (Q) and pT ∈ L2 (Ω). It is well known that the backward heat equation is not well
posed. Due to the condition p(x, T ) = pT equation (5.2.4) is a terminal value problem, which
must be integrated backward in time. But equation (5.2.4) is not a backward heat equation
because we have − ∂p
∂p
− ∆p = g and not − ∆p = g (as in the case of the backward heat
∂t ∂t
equation). Let us explain why the equation is well posed. If p is a solution of (5.2.4) and if
we set w(t) = p(T − t), we can check, at least formally, that w is the solution of
∂w
∂t − ∆w = g(x, T − t) in Q, w = 0 on Σ, w(x, 0) = pT in Ω. (5.2.5)
Since equation (5.2.5) is well posed, equation (5.2.4) is also well posed even if (5.2.4) is
a terminal value problem. In particular equation (5.2.4) admits a unique weak solution in
L 2 (0, T ; L 2 (Ω)), and this solution belongs to W (0, T ; H 1 0(Ω), H −1 (Ω)). To obtain the expression
of the gradient of F1 we need a Green formula which is stated below.
Theorem 5.2.3 Suppose that φ ∈ L2 (Q), g ∈ L2 (Q), and pT ∈ L2 (Ω). Then the solution z
of equation
∂z
− ∆z = φ
∂t
in Q, z = 0 on Σ, z(x, 0) = 0 in Ω,
and the solution p of (5.2.4) satisfy the following formula
φ p = z g + z(T )pT . (5.2.6)
Q
Q
Proof. If pT ∈ H1 0(Ω), due to Theorem 5.2.2, z and p belong to L2 (0, T ; D(A)))∩H1 (0, T ; L2 (Ω)).
In that case, with the Green formula we have
−∆z(t)p(t) dx = −∆p(t)z(t) dx
for almost every t ∈ [0, T ], and
T
0
Ω
Ω
T
∂z
p = −
∂t 0 Ω
Ω
Ω
∂p
z + z(T )pT .
∂t Ω
Thus formula (5.2.6) is established in the case when pT ∈ H 1 0(Ω) (Theorem 5.2.2 (ii)). If (pT n)n
is a sequence in H 1 0(Ω) converging to pT in L 2 (Ω), due to Theorem 5.2.2, (pn)n, where pn is the
solution to equation (5.2.4) corresponding to pT n, converges to p in W (0, T ; H 1 0(Ω), H −1 (Ω))
when n tends to infinity. Thus, in the case when pT ∈ L 2 (Ω), formula (5.2.6) can be deduced
by passing to the limit in the formula satisfied by pn.
The gradient of F1. Let (z(f, ū, z0), ū) = (¯z, ū) be the solution to problem (P1). By a direct
calculation we obtain
F1(ū + λu) − F1(ū) = 1
(zλ − ¯z)(zλ + ¯z − 2zd)
2 Q
+ 1
(zλ(T ) − ¯z(T ))(zλ(T ) + ¯z(T ) − 2zd(T )) +
2 Ω
β
T
(2λuū + λ
2 0 ω
2 u 2 ),