Optimal Control of Partial Differential Equations
Optimal Control of Partial Differential Equations
Optimal Control of Partial Differential Equations
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70 CHAPTER 6. CONTROL OF THE WAVE EQUATION<br />
Theorem 6.4.5 Assume that f ∈ L 2 (0, T ; L 2 (Ω)), z0 ∈ L 2 (Ω), z1 ∈ (H 1 (Ω)) ′ , and zd ∈<br />
C([0, T ]; L 2 (Ω)). Problem (P2) admits a unique solution (¯z, ū). Moreover the optimal control<br />
ū is defined by ū = − 1<br />
β p|Σ, where p is the solution to the equation<br />
∂ 2 p<br />
∂t 2 − ∆p = ¯z − zd in Q,<br />
p(T ) = 0,<br />
∂p<br />
∂n<br />
= 0 on Σ = Γ×]0, T [,<br />
∂p<br />
∂t (T ) = ¯z(T ) − zd(T ) in Ω.<br />
Pro<strong>of</strong>. We leave the reader adapt the pro<strong>of</strong> <strong>of</strong> Theorem 6.3.1.<br />
6.5 Trace regularity<br />
(6.4.12)<br />
To study the wave equation with a control in a Dirichlet boundary condition, we have to<br />
establish a sharp regularity result stated below.<br />
Theorem 6.5.1 Let y be the solution to the equation<br />
∂2y ∂t2 − ∆y = f in Q, y = 0 on Σ, y(x, 0) = y0 and ∂y<br />
∂t (x, 0) = y1 in Ω. (6.5.13)<br />
We have<br />
∂y<br />
∂n L2 <br />
<br />
(Σ) ≤ C fL2 (Q) + y0H 1<br />
0 (Ω) + y1L2 (Ω) . (6.5.14)<br />
The pro<strong>of</strong> can be found in [14, Theorem 2.2].<br />
6.6 Dirichlet boundary control<br />
We consider the wave equation with a control in a Dirichlet boundary condition<br />
∂2z ∂t2 − ∆z = f in Q, z = u on Σ, z(x, 0) = z0 and ∂z<br />
∂t (x, 0) = z1 in Ω. (6.6.15)<br />
As for the heat equation with a Dirichlet boundary control, the solution to equation (6.6.15)<br />
is defined by the transposition method.<br />
Definition 6.6.1 A function z ∈ C([0, T ]; L2 (Ω)) ∩ C1 ([0, T ]; H−1 (Ω)) is called a weak solution<br />
to equation (6.6.15) if, and only if,<br />
<br />
<br />
fy dxdt =<br />
−〈 ∂z<br />
∂t (0), y(0)〉 H −1 (Ω)×H 1 0 (Ω) −<br />
Q<br />
<br />
Q<br />
Ω<br />
zϕ dxdt + 〈 ∂z<br />
∂t (T ), yT 〉 H −1 (Ω)×H 1 0 (Ω)<br />
<br />
z(T )νT dx +<br />
Ω<br />
z(0) ∂y<br />
<br />
∂y<br />
(0) dx + u dsdt (6.6.16)<br />
∂t Σ ∂n<br />
for all (ϕ, yT , νT ) ∈ L 1 (0, T ; L 2 (Ω)) × H 1 0(Ω) × L 2 (Ω), where y is the solution to<br />
∂ 2 y<br />
∂t 2 − ∆y = ϕ in Q, y = 0 on Σ, y(x, T ) = yT and ∂y<br />
∂t (x, T ) = νT in Ω. (6.6.17)