Page 130 -

Exercises 113

2
2
where Ω is a bounded Lipschitz domain, f ∈ L (Ω) and g ∈ L (∂Ω).
1
(a) Let u ∈ H (Ω) be a solution of this problem. Show that −∆u exists
2
in the weak sense in L (Ω) and
−∆u = f.
2
(b) If additionally u ∈ H (Ω), then ∂ ν u| ∂Ω exists in the sense of trace in
2
L (∂Ω) and
∂ ν u = g
where this equality is to be understood as

1
(∂ ν u − g)vdσ =0 for all v ∈ H (Ω) .
∂Ω
3.8 Consider the variational equation (3.37) for the Neumann problem
for the Poisson equation as in Exercise 3.7.
1
(a) If a solution u ∈ H (Ω) exists, then the compatibility condition

fdx + gdσ =0 (3.38)
Ω ∂Ω
has to be fulfilled.
1
(b) Consider the following bilinear form on H (Ω) :

˜ a(u, v):= ∇u ·∇vdx + udx vdx .
Ω Ω Ω
1
Show that ˜a is V -elliptic on H (Ω).
Hint: Do it by contradiction using the fact that a bounded sequence in
2
1
H (Ω) possesses a subsequence converging in L (Ω) (see, e.g., [37]).
1
(c) Consider the unique solution ˜u ∈ H (Ω) of

1
˜ a(u, v)= fv dx + gv dσ for all v ∈ H (Ω) .
Ω ∂Ω
Then:

|Ω| ˜ udx = fdx + gdσ .
Ω Ω ∂Ω
Furthermore, if (3.38) is valid, then ˜u is a solution of (3.37) (with

˜ udx =0).
Ω
1
3.9 Show analogously to Exercise 3.7: A weak solution u ∈ V ⊂ H (Ω)
of (3.31), where V is defined in (3.30), with data satisfying (3.14) and
2
(3.22), fulfills a differential equation in L (Ω). The boundary conditions
are fulfilled in the following sense:

uv dσ+ u+αu)vdσ = g 1 vdσ+ g 2 vdσ for all v ∈ V.
∂ ν K (∂ ν K
Γ 1 Γ 2 Γ 1 Γ 2