Optimal Control of Partial Differential Equations

6.7. EXERCISES 73
6.7 Exercises
Exercise 6.7.1
The notation are the ones of section 6.3. Let (un)n be a sequence in L 2 (ω), converging to u for
the weak topology of L 2 (ω). Let zn be the solution to equation (6.3.6) corresponding to un,
and zu be the solution to equation (6.3.6) corresponding to u. Prove that (zn(T ))n converges
to zu(T ) for the weak topology of H1 0(Ω), and that ( ∂zn
∂t (T ))n converges to ∂zu (T ) for the weak
∂t
topology of L2 (Ω). Prove that the control problem (P2) admits a unique solution. Prove that
problem (P3) admits a unique solution.
Exercise 6.7.2
We study a control problem for the system of the Timoshenko beam (see section 1.4). We
consider the following set of equations:
ρ ∂2u ∂t2
− K ∂2u ∂x2 − ∂φ
= 0, ∂x
in (0, L),
Iρ ∂2φ ∂t2 − EI ∂2φ ∂x2
+ K φ − ∂u
(6.7.20)
∂x = 0, in (0, L),
with the boundary conditions
u(0, t) = 0 and φ(0, t) = 0 for t ≥ 0,
K(φ(L, t) − ux(L, t)) = f1(t) and − EIφx(L, t) = f2(t) for t ≥ 0.
and the initial conditions
u(x, 0) = u0 pour ∂u
∂t (x, 0) = u1 pour x ∈ (0, L),
φ(x, 0) = φ0 and ∂φ
∂t (x, 0) = φ1 in (0, L).
(6.7.21)
(6.7.22)
We recall that u is the deflection of the beam, φ is the angle of rotation of the beam crosssections
due to bending. The coefficient ρ is the mass density per unit length, EI is the flexural
rigidity of the beam, Iρ is the mass moment of inertia of the beam cross section, and K is the
shear modulus. We suppose that u0 ∈ H 1 0(0, L), u1 ∈ L 2 (0, L), φ0 ∈ H 1 0(0, L), φ1 ∈ L 2 (0, L).
The control functions f1 and f2 are taken in L 2 (0, T ).
To study the system (6.7.20)-(6.7.22), we use a fixed point method as in exercise 5.5.4. The
Hille-Yosida theorem could also be used to directly study the system. Denote by H1 {0} (0, L) the
space of functions ψ in H1 (0, L) such that ψ(0) = 0. Let τ > 0, for ψ ∈ L2 (0, τ; H1 {0} (0, L)),
we denote by (uψ, φψ) the solution to
ρ ∂2u ∂t2
− K ∂2u ∂x2 − ∂ψ
= 0, ∂x
in (0, L),
Iρ ∂2φ ∂t2 − EI ∂2φ ∂x2
+ K φ − ∂u
(6.7.23)
∂x = 0, in (0, L),
with the boundary conditions
u(0, t) = 0 and φ(0, t) = 0 for t ≥ 0,
K(ψ(L, t) − ux(L, t)) = f1(t) and − EIφx(L, t) = f2(t) for t ≥ 0.
(6.7.24)