Optimal Control of Partial Differential Equations

9.6. APPENDIX 105
9.6 Appendix
Lemma 9.6.1 For N = 2, we have
for all z ∈ H 1 0(Ω).
z L 4 (Ω) ≤ 2 1/4 ∇z 1/2
L 2 (Ω) z1/2
L 2 (Ω) ,
Proof. Let us prove the result for z ∈ D(Ω). We have
and
Thus
R 2
This completes the proof.
|z(x)| 2 ≤ 2
x1
−∞
|z(ξ1, x2)||∂1z(ξ1, x2)|dξ1,
|z(x)| 2 x2
≤ 2 |z(x1, ξ2)||∂1z(x1, ξ2)|dξ2.
−∞
|z(x)| 4 dx ≤ 4z 2
L 2 (Ω) ∂1z L 2 (Ω)∂2z L 2 (Ω) ≤ 2z 2
L 2 (Ω) ∇z2
L 2 (Ω) .
Theorem 9.6.1 For N = 2, the imbedding from C([0, T ]; L 2 (Ω)) ∩ L 2 (0, T ; H 1 0(Ω)) into
L 4 ((0, T ) × Ω) is continuous. Moreover we have
Proof. Due to Lemma 9.6.1 we have
T
|z| 4 T
≤ 2
0
Ω
The proof is complete.
zL4 (Q) ≤ 2 1/4 z 1/2
L2 (0,T ;H1 0 (Ω))z1/2 C([0,T ];L2 (Ω)) .
The other results are stated in dimension N ≥ 2.
0
z 2
L2 (Ω) ∇z2L
2 (Ω) ≤ 2z2C([0,T
];L2 (Ω) z2 L2 (0,T ;H1 0 (Ω)).
Theorem 9.6.2 Set D(Ap) = W 2,p (Ω)∩W 1,p
0 (Ω) and Apz = ∆z for z ∈ D(Ap), with 1 < p <
∞. The operator (Ap, D(Ap)) is the infinitesimal generator of a strongly continuous analytic
semigroup on L p (Ω).
See for example [5, Theorem 7.6.1]. This theorem together with properties of fractional powers
of (−Ap) can be used to prove the theorem below.
Theorem 9.6.3 Let h ∈ D(Q), and zh be the solution to equation
∂z
∂t
− ∆z = ∂xi h in Q, z = 0 on Σ, z(x, 0) = 0 in Ω, (9.6.11)
where i ∈ {1, . . . , N}. Suppose that 1 < s < ∞ and 1 < p < ∞. The mapping h ↦→ zh is
continuous from L s (0, T ; L p (Ω)) into C([0, T ]; L r (Ω)) ∩ L s (0, T ; W 1,p (Ω)) if
N 1
+
2p s
1 1
< +
r 2 .