Optimal Control of Partial Differential Equations

18 CHAPTER 2. CONTROL OF ELLIPTIC EQUATIONS
Set
un+1 = un + ρndn
and gn+1 = gn + ρnˆgn.
Step 2. If gn+1 L 2 (Γ)/g0 L 2 (Γ) ≤ ε, stop the algorithm and take u = un+1, else compute
and
Replace n by n + 1 and go to step 1.
βn = gn+1 2
L2 2
(Γ) /gnL2 (Γ) ,
dn+1 = −gn+1 + βndn.
Observe that gn+1 = gn + ρnˆgn is the gradient of F1 at un+1.
2.4 Dirichlet boundary control
Now we want to control the Laplace equation by a Dirichlet boundary control, that is
−∆z = f in Ω, z = u on Γ. (2.4.6)
We say that a function z ∈ H 1 (Ω) is a solution to equation (2.4.6) if the equation −∆z = f is
satisfied in the sense of distributions in Ω and if the trace of z on Γ is equal to u. When
f ∈ H −1 (Ω) and u ∈ H 1/2 (Γ) equation (2.4.6) can be solved as follows. From a trace theorem
in H 1 (Ω), we know that there exists a linear continuous operator from H 1/2 (Γ) to H 1 (Ω):
u ↦−→ zu,
such that zu|Γ = u. Thus we look for a solution z to equation (2.4.6) of the form z = zu + y,
with y ∈ H 1 0(Ω). The equation − ∆z = f is satisfied in the sense of distributions in Ω and
z = u on Γ if and only if y is the solution to equation
−∆y = f + ∆zu in Ω, y = 0 on Γ.
If we identify the distribution ∆zu with the mapping
ϕ ↦−→ − ∇zu∇ϕ,
Ω
we can check that −∆zu belongs to H −1 (Ω). Hence the existence of y is a direct consequence
of the Lax-Milgram theorem. Therefore we have the following theorem.
Theorem 2.4.1 For every f ∈ H −1 (Ω) and every u ∈ H 1/2 (Γ), equation (2.4.6) admits a
unique weak solution z(f, u) in H 1 (Ω), moreover the operator
(f, u) ↦→ z(f, u)
is linear and continuous from H −1 (Ω) × H 1/2 (Γ) into H 1 (Ω).