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Optimal Control of Partial Differential Equations

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54 CHAPTER 5. CONTROL OF THE HEAT EQUATION<br />

Since (z0 − wn(0)) ∈ L2 (Ω) and f − ∂wn<br />

∂t − ∆wn belongs to L2 (Q), yn<br />

W (0, T ; H<br />

and zn belong to<br />

1 (Ω), (H1 (Ω)) ′ ). Thus, for every t ∈]0, T ], we have<br />

<br />

<br />

<br />

<br />

We first get<br />

Ω<br />

|zn(t)| 2 + 2<br />

t<br />

0<br />

Ω<br />

|∇zn| 2 = 2<br />

t<br />

0<br />

fzn + 2<br />

Ω<br />

t<br />

0<br />

unzn +<br />

Γ<br />

Ω<br />

|z0| 2 .<br />

y 2<br />

C([0,T ];L2 (Ω)) + 2∇y2L<br />

2 (0,T ;L2 (Ω)) ≤ 2fL2 (Q)yL2 (Q) + 2uL2 (Σ)yL2 (Σ) + y0 2<br />

L2 (Ω) .<br />

Thus with Young’s inequality, we obtain<br />

y C([0,T ];L 2 (Ω)) + y L 2 (0,T ;H 1 (Ω)) ≤ C<br />

In the same way, we can prove<br />

<br />

<br />

fL2 (Q) + uL2 (Σ) + y0L2 (Ω) .<br />

zn − zm C([0,T ];L 2 (Ω)) + zn − zm L 2 (0,T ;H 1 (Ω)) ≤ Cun − um L 2 (Σ).<br />

Hence the sequence (zn)n converges to some z in C([0, T ]; L 2 (Ω)) ∩ L 2 (0, T ; H 1 (Ω)). Due to<br />

Theorem 5.3.3, we can also prove that the sequence (zn)n converges to the solution <strong>of</strong> equation<br />

(5.3.10) in C([0, T ]; L 2 (Ω)). By using the same arguments as for Theorem 5.2.2, we can next<br />

prove an estimate in W (0, T ; H 1 (Ω), (H 1 (Ω)) ′ ). Therefore we have established the following<br />

theorem.<br />

Theorem 5.3.4 For every f ∈ L 2 (Q), every u ∈ L 2 (Σ), and every z0 ∈ L 2 (Ω), equation<br />

(5.3.10) admits a unique weak solution z(f, u, z0) in L 2 (0, T ; L 2 (Ω)), moreover the operator<br />

(f, u, z0) ↦→ z(f, u, z0)<br />

is linear and continuous from L 2 (Q) × L 2 (Σ) × L 2 (Ω) into W (0, T ; H 1 (Ω), (H 1 (Ω)) ′ ).<br />

We now consider the control problem<br />

(P2) inf{J2(z, u) | (z, u) ∈ C([0, T ]; L 2 (Ω)) × L 2 (0, T ; L 2 (Γ)), (z, u) satisfies (5.3.10)},<br />

where<br />

J2(z, u) = 1<br />

<br />

(z − zd)<br />

2 Q<br />

2 + 1<br />

<br />

(z(T ) − zd(T ))<br />

2 Ω<br />

2 + β<br />

<br />

u<br />

2 Σ<br />

2 .<br />

We assume that f ∈ L 2 (Q), z0 ∈ L 2 (Ω), and zd ∈ C([0, T ]; L 2 (Ω)). Problem (P2) admits a<br />

unique solution (¯z, ū) (see exercise 5.5.2). The adjoint equation for (P2) is <strong>of</strong> the form<br />

− ∂p<br />

∂t<br />

− ∆p = g in Q,<br />

∂p<br />

∂n = 0 on Σ, p(x, T ) = pT in Ω. (5.3.16)<br />

Theorem 5.3.5 Suppose that u ∈ L2 (Σ), g ∈ L2 (Q), pT ∈ L2 equation<br />

(Ω). Then the solution z <strong>of</strong><br />

∂z<br />

− ∆z = 0<br />

∂t<br />

in Q,<br />

∂z<br />

= u<br />

∂n<br />

on Σ, z(0) = 0 in Ω,<br />

and the solution p <strong>of</strong> (5.3.16) satisfy the following formula<br />

<br />

up = z g + z(T )pT . (5.3.17)<br />

Σ<br />

Q<br />

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