is a weak solution of Neumann's problem

6.6 #3
Assume U is connected. A function u ∈ H1 (U) is a weak solution of Neumann’s
problem
�
−∆u = f in U
∂u
= 0
∂ν
on ∂U
(1)
if �
�
Du · Dv dx = fv dx (2)
U
U
for all v ∈ H1 (U). Suppose f ∈ L2 (U). Prove (1) has a weak solution if and only if
�
U
f dx = 0. (3)
Solution First we show that if (1) has a weak solution, then (3) holds. Indeed, using
v ≡ 1 in the weak form (2), which is possible since (2) holds for any v ∈ H1 (U), we
get � �
f dx = DuD(1)dx = 0,
U
U
which is (3).
Now let (3) holds. We show existence of weak solution of (1). Note that if a weak
solution u of (1) exists, then u + c is also a weak solution for any constant c, this
follows from (2). We can fix a constant by requiring
�
u dx = 0. (4)
U
Thus we seek a weak solution of (1) satisfying (4).
Note that the set ˆ L2 (U) := {f ∈ L2 (U) | �
f dx = 0} is a closed subspace of
U
L2 (U). To see that, we note that constant functions are in L2 (U), and thus
�
G(f) :=
U
1 · f dx
is a linear bounded functional on L 2 (U). From its definition,
ˆL 2 (U) = {f ∈ L 2 (U) | G(f) = 0}. Thus ˆ L 2 (U) is a closed subspace of L 2 (U). It
follows that ˆ L2 (U) with norm � · �L2 (U) is a Hilbert space.
This also implies that the set ˆ H1 (U) := {u ∈ H1 (U) | �
u dx = 0} is a closed
U
subspace of H1 (U). Thus ˆ H1 (U) with norm � · �H1 (U) is a Hilbert space.
Define the bilinear form on ˆ H1 (U) by
�
B[u, v] = Du · Dv dx for any u, v ∈ ˆ H 1 (U).
Then we prove energy estimates for B[u, v]:
U

By Holder inequality for any u, v ∈ ˆ H 1 (U)
|B[u, v]| ≤ �Du� L 2 (U)�Dv� L 2 (U) ≤ �u� ˆ H 1 (U) �v� ˆ H 1 (U) . (5)
From definition of B[u, v], for any u ∈ ˆ H 1 (U)
�Du� 2
L2 (U) = B[u, u]. (6)
Also, since (4) holds, then denoting (u)U := 1
�
|U| U u(x) dx, we get (u)U
using Poincare inequality, we estimate
= 0. Thus,
�u� 2
L2 (U) =
�
u 2 �
dx = (u − (u)U) 2 �
dx ≤ C |Du| 2 dx = CB[u, u],
U
where C depends only on U. Combining this and (6), get
1
C + 1 �u�2 Hˆ 1 = (U) 1
�
C + 1
Also, if f ∈ ˆ L2 , then the functional
�
F (v) =
U
U
u
U
2 + |Du| 2 dx ≤ B[u, u]. (7)
U
fv dx
is linear bounded on ˆ H 1 (U). Thus (5) and (7) allow to apply Lax-Milgram theorem
to obtain a unique u ∈ ˆ H 1 (U) such that
B[u, v] = F (v) for all v ∈ ˆ H 1 (U).
This implies that (2) holds for all v ∈ H1 (U) satisfying �
v dx = 0.
U
Recall that u satisfies (4) and f satisfies (3). Thus if (2) holds for some v ∈ H1 (U),
then (2) also holds for v replaced by v + c, where c
�
is any constant. Then (2) holds
for all v ∈ H 1 (U): indeed, if v ∈ H 1 (U) and c = 1
|U|
then we know that (2) holds for v − c, thus for v.
Thus u is a weak solution of (1).
U v dx, then v − c ∈ ˆ H 1 (U), and