Optimal Control of Partial Differential Equations
Optimal Control of Partial Differential Equations Optimal Control of Partial Differential Equations
58 CHAPTER 5. CONTROL OF THE HEAT EQUATION Theorem 5.4.5 (i) Let z(f, u, z0) be the solution to equation (5.4.20). The operator (f, u, z0) ↦→ z(f, u, z0), is linear and continuous from L 2 (Q) × L 2 (Σ) × L 2 (Ω) into C([0, T ]; H −1 (Ω)). (ii) If u ∈ L 2 (Σ), and if pT ∈ H 1 0(Ω), then the solution z of equation (5.4.21) and the solution p of − ∂p ∂t − ∆p = 0 in Q, p = 0 on Σ, p(x, T ) = pT in Ω, satisfy the following formula 〈z(T ), pT 〉 H−1 (Ω)×H1 0 (Ω) = − Σ u ∂p . (5.4.26) ∂n Proof. (i) We only need to prove the regularity result for the solution z of equation (5.4.21). For every ϕ ∈ H 1 0(Ω) and every τ ∈]0, T ], consider the solution y to equation − ∂y − ∆y = 0 ∂t Due to Theorem 5.2.2, we have in Q, y = 0 on Σ, y(τ) = ϕ in Ω. y L 2 (0,τ;H 2 (Ω)∩H 1 0 (Ω)) ≤ cϕ H 1 0 (Ω), and the constant c is independent of τ. Let (un)n ⊂ L 2 (Σ) a sequence of regular functions satisfying the compatibility condition un(x, 0) = 0, and converging to u in L 2 (Σ). Denote by zn the solution to (5.4.21) corresponding to un. Since zn is regular, it satisfies the formula Thus we have Ω zn(τ)ϕ = − zn(τ) H −1 (Ω) = sup ϕH 1 0 (Ω) =1 Γ×(0,τ) ∂y un ∂n . un Γ×(0,τ) ∂y ≤ cunL2 (Σ), ∂n where the constant c is independent of τ. From this estimate it follows that zn − zm C([0,T ];H −1 (Ω)) = zn − zm L ∞ (0,T ;H −1 (Ω)) ≤ cun − um L 2 (Σ). Therefore the sequence (zn)n converges to some ˜z in C([0, T ]; H −1 (Ω)). Due to Theorem 5.4.1, the sequence (zn)n converges to the solution z of equation (5.4.21). We finally have z = ˜z ∈ C([0, T ]; H −1 (Ω)). (ii) Formula (5.4.26) can be established for regular data, and next deduced in the general case from density arguments. Now we are in position to study the control problem (P4) inf{J4(z, u) | (z, u) ∈ L 2 (0, T ; L 2 (Ω)) × L 2 (Σ), (z, u) satisfies (5.4.20)}, with J4(z, u) = 1 2 |z(T ) − zT | 2 H−1 β (Ω) + u 2 Σ 2 . The proof of existence and uniqueness of solution to problem (P4) is standard (see exercise 5.5.3).
5.5. EXERCISES 59 Theorem 5.4.6 Assume that f ∈ L2 (Q), z0 ∈ L2 (Ω), and zd ∈ L2 (0, T ; L2 (Ω)). Let (¯z, ū) be the unique solution to problem (P4). The optimal control u is defined by u = 1 ∂p , where p is β ∂n the solution to the equation − ∂p ∂t − ∆p = 0 in Q, p = 0 on Σ, p(x, T ) = (−∆)−1 (¯z(T ) − zT ) in Ω. (5.4.27) Proof. We set F4(u) = J4(z(f, z0, u), u). If wu is the solution to equation 5.4.21, and p the solution to equation 5.4.27, with the formula stated in Theorem 5.4.5(ii), we have ūu. The proof is complete. 5.5 Exercises Exercise 5.5.1 F4(ū)u = 〈wu(T ), (−∆) −1 (¯z(T ) − zT )〉 H −1 (Ω)×H 1 0 (Ω) + β = Σ − ∂p + βū u. ∂n The notation are the ones of section 5.2. Let (un)n be a sequence in L 2 (0, T ; L 2 (ω)), converging to u for the weak topology of L 2 (0, T ; L 2 (ω)). Let zn be the solution to equation (5.2.1) corresponding to un, and zu be the solution to equation (5.2.1) corresponding to u. Prove that (zn(T ))n converges to zu(T ) for the weak topology of L 2 (Ω). Prove that the control problem (P1) admits a unique solution. Exercise 5.5.2 Prove that the control problem (P2) of section 5.3 admits a unique solution. Exercise 5.5.3 The notation are the ones of section 5.4. Let (un)n be a sequence in L 2 (Σ), converging to u for the weak topology of L 2 (Σ). Let zn be the solution to equation (5.4.20) corresponding to un, and zu be the solution to equation (5.4.20) corresponding to u. Prove that (zn(T ))n converges to zu(T ) for the weak topology of H −1 (Ω). Prove that the control problem (P4) admits a unique solution. Exercise 5.5.4 Let Ω be a bounded domain in R N , with a boundary Γ of class C 2 . Let T > 0, set Q = Ω×(0, T ) and Σ = Γ × (0, T ). We consider a convection-diffusion equation with a distributed control ∂z ∂t − ∆z + V · ∇z = f + χωu in Q, z = 0 on Σ, z(x, 0) = z0 in Ω. (5.5.28) Σ
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5.5. EXERCISES 59<br />
Theorem 5.4.6 Assume that f ∈ L2 (Q), z0 ∈ L2 (Ω), and zd ∈ L2 (0, T ; L2 (Ω)). Let (¯z, ū) be<br />
the unique solution to problem (P4). The optimal control u is defined by u = 1 ∂p<br />
, where p is<br />
β ∂n<br />
the solution to the equation<br />
− ∂p<br />
∂t − ∆p = 0 in Q, p = 0 on Σ, p(x, T ) = (−∆)−1 (¯z(T ) − zT ) in Ω. (5.4.27)<br />
Pro<strong>of</strong>. We set F4(u) = J4(z(f, z0, u), u). If wu is the solution to equation 5.4.21, and p the<br />
solution to equation 5.4.27, with the formula stated in Theorem 5.4.5(ii), we have<br />
<br />
ūu.<br />
The pro<strong>of</strong> is complete.<br />
5.5 Exercises<br />
Exercise 5.5.1<br />
F4(ū)u = 〈wu(T ), (−∆) −1 (¯z(T ) − zT )〉 H −1 (Ω)×H 1 0 (Ω) + β<br />
<br />
=<br />
Σ<br />
<br />
− ∂p<br />
<br />
+ βū u.<br />
∂n<br />
The notation are the ones <strong>of</strong> section 5.2. Let (un)n be a sequence in L 2 (0, T ; L 2 (ω)), converging<br />
to u for the weak topology <strong>of</strong> L 2 (0, T ; L 2 (ω)). Let zn be the solution to equation (5.2.1)<br />
corresponding to un, and zu be the solution to equation (5.2.1) corresponding to u. Prove that<br />
(zn(T ))n converges to zu(T ) for the weak topology <strong>of</strong> L 2 (Ω). Prove that the control problem<br />
(P1) admits a unique solution.<br />
Exercise 5.5.2<br />
Prove that the control problem (P2) <strong>of</strong> section 5.3 admits a unique solution.<br />
Exercise 5.5.3<br />
The notation are the ones <strong>of</strong> section 5.4. Let (un)n be a sequence in L 2 (Σ), converging to<br />
u for the weak topology <strong>of</strong> L 2 (Σ). Let zn be the solution to equation (5.4.20) corresponding<br />
to un, and zu be the solution to equation (5.4.20) corresponding to u. Prove that (zn(T ))n<br />
converges to zu(T ) for the weak topology <strong>of</strong> H −1 (Ω). Prove that the control problem (P4)<br />
admits a unique solution.<br />
Exercise 5.5.4<br />
Let Ω be a bounded domain in R N , with a boundary Γ <strong>of</strong> class C 2 . Let T > 0, set Q = Ω×(0, T )<br />
and Σ = Γ × (0, T ). We consider a convection-diffusion equation with a distributed control<br />
∂z<br />
∂t − ∆z + V · ∇z = f + χωu in Q, z = 0 on Σ, z(x, 0) = z0 in Ω. (5.5.28)<br />
Σ