EXERCISES PDE 23-25.01.13 1. Exercise Fourier analysis in ...

EX11 3therefore (2.1) is equivalent to prove that2(B(u N , u N ) − B(u, u N ) + B(u, v N ) − B(u N , v N )) = 0 .This can be easily proved to be true by observing that by definition of u and u N , for every v N ∈ H None hasB(u, v N ) = F (v N ) = B(u N , v N )and in particular, for v N = u N ,B(u, u N ) = B(u N , u N ) .Now, by (2.1), and the coercivity and continuity of B we get for every v N ∈ H Nc 0 ‖u − u N ‖ 2 H ≤ B(u − u N, u − u N ) ≤ B(u − u N , u − u N ) + B(v N − u N , v N − u N ) =B(u − v N , u − v N ) ≤ C 1 ‖u − v N ‖ 2 Hwhence√C‖u − u N ‖ H ≤ 1c 0‖u − v N ‖ Hfor every v N ∈ H N . Taking the infimum in the rigth-hand side gives (2.2).(d) To prove existence of u and u N when B is given by (2.3), it suffices to check that the Lax-Milgram theorem can be applied. The only relevant point to this end is to prove coercivity of B ,which follows from the Poincaré inequality. Namely, one hasB(u, u) ≥ ‖u ′ ‖ 2 2 ≥ 11 + π 2 ‖u‖2 H . 1It follows now from the previous exercise thatinf{‖u − v N ‖ H 10 ((0,π)) : v N ∈ H N } → 0as N goes to +∞. In particular, by (1.1), if u ∈ H 2 one has for every N ∈ Ninf{‖u − v N ‖ H 10 ((0,π)) : v N ∈ H N } ≤ C N ‖u‖ H 2 .Combining these two facts with (2.2) gives the conclusion.□3. ExerciseConsider a bounded smooth open subset Ω ⊂ R N , and Γ D ⊂ ∂Ω with H N−1 (Γ D ) > 0. LetH 1 Γ D(Ω) := {u ∈ H 1 (Ω), T u = 0 on Γ D } ,with T the trace operator. For λ > 0 defineˆB λ (u, v) = ∇u(x) · ∇v(x) dx − 1 ˆT u(ξ) T v(ξ) dH N−1 (ξ)Ω λ ∂Ωfor u and v ∈ HΓ 1 D(Ω). Prove that, when λ is sufficiently large, for every f ∈ L 2 (Ω) there exists aunique u f ∈ HΓ 1 D(Ω) such thatˆB λ (u f , v) = f(x) v(x) dxfor every v ∈ H 1 Γ D(Ω). Hint: does the Poincaré inequality apply in H 1 Γ D(Ω)?Solution: Following the hint, let us first prove that there exists a constant C such thatΩ‖u‖ L2 (Ω) ≤ C‖∇u‖ L2 (Ω) (3.1)for every u ∈ HΓ 1 D(Ω). To prove this, it suffices to argue as in the proof of the Poincaré-Wirtingerinequality (Theorem 1, Section 5.8.1 in Evans’book): contradicting (3.1) would produce the existenceof a sequence u n in HΓ 1 D(Ω) with ‖u n ‖| L2 (Ω) = 1 for every n which is converging to a constant c inH 1 (Ω). By continuity of the trace operator, HΓ 1 D(Ω) is a Hilbert subspace of H 1 (Ω), so it must bec = 0, since this one is the only constant in HΓ 1 D(Ω). On the other hand, by strong L 2 convergencewe must have ‖c‖ L 2 (Ω) = c|Ω| 1 2 = 1, and this gives a contradiction.