1. Let U ⊂ Rn be a bounded domain with smooth boundary. Using Green’s formulas and
properties of harmonic functions show that if u ∈ C ∞ (Ū ) satisfies
∆(∆u) = 0 on U, u|∂U = 0 and
∂u
= 0 on ∂U,
∂ν
then u ≡ 0.
Recall that by Green’s formula
Z
Z
(u∆v − v∆u)dx =
U
(u
∂U
∂v
∂u
− v )dS.
∂ν
∂ν
∂u
∂ν
Taking v = ∆u and using that ∆v = 0 and u|∂U =
= 0 by assumption, we get
Z
(∆u)2 dx = 0.
U
Therefore ∆u ≡ 0, so u is harmonic. Since u vanishes on the boundary, by the maximum principle
we conclude that u ≡ 0.
2. Let U = (−1, 1) ⊂ R. For α, β ∈ R define a function u on U by
(
α,
for x ≤ 0,
u(x) =
β
x , for x > 0.
Find under what assumptions on α and β we have u ∈ W 1,1 (U ).
If the weak derivative u0 exists then u0 (x) = 0 for (almost all) x ∈ (−1, 0) and u0 (x) = βxβ−1
1
) and u0 ∈ L1 (U ) we therefore need β ≥ 0. Now to find when
Rfor x0 ∈ (0, 1).R To 0have u ∈ L (U
∞
U u vdx = − U uv dx for v ∈ Cc (U ), we compute
Z
Z
Z 0
Z 1
0
0
0
u vdx +
uv dx = α
v dx + lim
(u0 v + uv 0 )dx
U
ε→0+
−1
U
ε
= αv(0) − lim u(ε)v(ε).
ε→0+
R
If β > 0 then the limit above equals zero, so U u0 vdx = − U uv 0 dx holds for all v only when α = 0.
If β = 0, the limit is v(0), so α = 1. Therefore the answer is that either α = 0 and β > 0 or α = 1
and β = 0.
R
d2
on (0, a) ⊂ R (with Dirichlet boundary
dx2
2
conditions) and an orthonormal basis in L (0, a) consisting of eigenvectors.
3. Let a > 0. Find the spectrum of −∆ = −
We know that all eigenvalues of −∆ are strictly positive. A number λ > 0 is an eigenvalue if there
exists a nonzero u ∈ C ∞ ([0, a]) with u(0) = u(a) = 0 such that
u00 + λu = 0.
√
√
The general solution of the above equation has the form u(x) = A sin( λx) + B cos( λx). As
u(0) = 0, we get B = 0. Since u(a) = 0, we conclude (assuming u is a nonzero solution) that
√
π 2 n2
λa = πn for some n ∈ N. Therefore the spectrum consists of the numbers λn =
, n ∈ N.
a2
Each eigenvalue has multiplicity one, and since
Z a
πnx
a
dx = ,
sin2
a
2
0
1
2
as an orthonormal basis consisting of eigenvectors we can take
r
2
πnx
un (x) =
sin
.
a
a
4. Consider the open unit disc D in R2 with boundary T. Define functions en , n ≥ 0, on T in the
polar coordinates by
e0 = 1, e2m (ϕ) = cos mϕ, e2m−1 (ϕ) = sin mϕ.
We easily check that the functions en form an orthogonal system in L2 (T), so that (en , em )L2 (T) = 0
for n 6= m, and compute that
ke0 k2L2 (T) = 2π, ken k2L2 (T) = π for n ≥ 1.
You don’t have to do this. Consider now the functions un , n ≥ 0, on D defined by
u0 = 1, u2m (r, ϕ) = rm cos mϕ, u2m−1 (r, ϕ) = rm sin mϕ.
Show that {un }n≥0 is an orthogonal system in H 1 (D) and compute kun kH 1 (D) . You may use that
∂
∂
sin ϕ ∂
∂
cos ϕ ∂
∂
= cos ϕ
= sin ϕ
−
,
+
.
∂x1
∂r
r ∂ϕ ∂x2
∂r
r ∂ϕ
Recall that by definition
Z
(u, v)H 1 (D) =
Z
D
We have
Z
Z
2π
Z
uk ul dx =
D
Du · Dv dx.
uv dx +
D
1
uk ul r dϕ dr.
0
0
As the functions eRn are mutually orthogonal, we see that the functions un are mutually orthogonal
in L2 (D), that is, D uk ul dx = 0 for k 6= l. Furthermore,
Z 2π Z 1
Z 1
π
2
2
2
.
ku2m kL2 (D) =
u2m r dϕ dr = ke2m kL2 (T)
r2m+1 dr =
2m + 2
0
0
0
Similarly
π
ku2m−1 k2L2 (D) =
and ku0 k2L2 (D) = π.
2m + 2
Next we compute
∂u2m
= mrm−1 (cos ϕ cos mϕ + sin ϕ sin mϕ) = mrm−1 cos(m − 1)ϕ,
∂x1
∂u2m
= mrm−1 (sin ϕ cos mϕ − cos ϕ sin mϕ) = −mrm−1 sin(m − 1)ϕ,
∂x2
∂u2m−1
= mrm−1 (cos ϕ sin mϕ − sin ϕ cos mϕ) = mrm−1 sin(m − 1)ϕ,
∂x1
∂u2m−1
= mrm−1 (sin ϕ sin mϕ + cos ϕ cos mϕ) = mrm−1 cos(m − 1)ϕ.
∂x2
R
Using again orthogonality of the functions en we conclude that D Duk · Dul dx = 0 for k 6= l (note
that for (k, l) = (2m, 2m − 1) or (k, l) = (2m − 1, 2m) we even have Duk · Dul ≡ 0). We thus see
that the functions un are mutually orthogonal in H 1 (D).
Furthermore, we get
2 2 !
Z
Z
Z
Z 1
∂u
∂u
2m
2m
2
2 2m−2
2
|Du2m | dx =
+
dx =
m r
dx = 2πm
r2m−1 dr = πm.
∂x1
∂x2
D
D
D
0
3
Similarly
R
2
D |Du2m−1 | dx = πm. Thus
ku0 k2H 1 (D) = ku0 k2L2 (D)
ku2m k2H 1 (D)
ku2m−1 k2H 1 (D)
= π,
Z
1
|Du2m |2 dx = π(
= ku2m k2L2 (D) +
+ m),
2m + 2
D
Z
1
2
|Du2m−1 |2 dx = π(
= ku2m−1 kL2 (D) +
+ m).
2m
+2
D
5. Let U ⊂ Rn be a bounded domain and H the space of harmonic functions in H 1 (U ). Consider
the bilinear symmetric form B on H 1 (U ) defined by −∆. Explain why for each u ∈ H 1 (U ) there
exists a unique w ∈ H01 (U ) such that
B(u, v) = B(w, v) for all v ∈ H01 (U ).
Conclude that for each u ∈ H 1 (U ) there exist unique v ∈ H and w ∈ H01 (U ) such that u = v + w.
Recall that B is defined by
B(u, v) =
XZ
i
U
∂u ∂v
dx.
∂xi ∂xi
H 1 (U ),
We know that B is a bounded form on
so for each u ∈ H 1 (U ) the functional B(u, ·) on H 1 (U )
is bounded. Furthermore, we know that on H01 (U ) the bilinear form B satisfies the assumptions of
the Lax-Milgram theorem, that is, B(v, v) ≥ βkvk2H 1 (U ) . Hence there exists a unique w ∈ H01 (U )
0
such that
B(u, v) = B(w, v)
for all v ∈ H01 (U ). Then B(u − w, v) = 0 for all v ∈ H01 (U ), so u − w is a weak solution of the
equation −∆f = 0, that is, u − w ∈ H.
If we have a decomposition u = ṽ + w̃ with ṽ ∈ H and w̃ ∈ H01 (U ), then B(ṽ, v) = 0 for all
v ∈ H01 (U ) as ṽ is harmonic, whence
B(u, v) = B(w̃, v)
for all v ∈
H01 (U ).
Hence w̃ = w by uniqueness.