52H HW9 solutions
1. Let S = ∂U , for U ⊂ Rn a compact subset. By the divergence theorem
Z
Z
n · vdS =
div(v)dV = 0.
S
U
2. A. Let Γ = ∂S. We have
Z
Z
n · ∇u =
∆u.
Γ
S
If ∆u = 0 then trivially the LHS vanishes for every Γ. If ∆u 6= 0, then WLOG
we can suppose ∆u > 0 at a point. By continuity ∆u ≥ δ > 0 on some small
ball B (p) ⊂ U . Taking Γ = ∂B (p) we have
Z
Z
n · ∇u =
∆u ≥ π2 δ > 0.
∂B (p)
B (p)
B. By the product rule we have
div(u∇u) = |∇u|2 + u∆u.
Integrating this relation on S gives, using the divergence theorem,
Z
Z
Z
u(n · ∇u) =
|∇u|2 +
u∆u
Γ
S
S
which is the required relation.
3. Recall the following homotopy formula. Let F (t, x) : [0, 1] × Rn → Rn a
homotopy between f ≡ F (0, ·) and g ≡ F (1, ·). Then for any k-form α we have
g ∗ α − f ∗ α = dKF ∗ α + KF ∗ dα.
Here K : Ωk ([0, 1] × Rn ) → Ωk−1 (Rn ) is the map
Z 1
Kβ =
(∂t yβ)dt.
0
For a proof of the homotopy formula see the end of this question.
We consider the case F (t, x) = tx, and dα = 0. Then f ≡ 0, g ≡ id, and
hence
Z
1
α = dKF ∗ α = d
(∂t yF ∗ α)dt.
0
This is the dual notion to taking the cone over a manifold without boundary.
I claim that
∂t yF ∗ α = tk−1 xyα(tx).
Here’s an easy way to see why: observe that
DF |(t,x) ∂t = x,
DF |(t,x) ∂i = t∂i .
1
and let Ft ≡ F (t, ·) be F restricted to time-slice t. Then
∂t y(F ∗ α)|x = (Ft |x )∗ (DF |(t,x) ∂t yα|tx )
= tk−1 (xyα|tx ).
So we’ve shown: if α ∈ Ωk (Rn ) is closed, then α = dβ where
Z
β=
1
(tk−1 xyα(tx))dt.
0
For example, if α =
P
i
αi dxi is a closed 1-form, then
β=
P
Z
xi αi (tx)dt =
α.
path 0 → x
0
i
If α =
1
XZ
αij dxi ∧ dxj is a closed 2-form, then
i<j
β=
X Z
i<j
1
tαij (tx)dt (xi dxj − xj dxi ).
0
A. We stipulate div(v) = 0 on R3 , and seek a w so the curl(w) = v. Recall
that
div(v)dV = d(vydV )
and
curl(w) = D−1 ? dD.
Here D : T Rn → T ∗ Rn is the isomorphism
D(v)(w) = v · w.
So we know that vydV is closed in R3 , and we ask for a form ω = Dw such
that
dω = vydV.
P
Write v = i Vi ei , so
vydV = V1 dx2 ∧ dx3 − V2 dx1 ∧ dx3 + V3 dx1 ∧ dx2 .
By the previous discussion, we have that
Z 1
Z
w=
tx3 V2 (tx) − tx2 V3 (tx)dt e1 +
0
0
Z
+
1
tx2 V1 (tx) − tx1 V2 (tx)dt e3 .
0
2
1
tx1 V3 (tx) − tx3 V1 (tx)dt e2
If we abuse notation, like one often does in calculating the cross-product, we
can write w as the integral of the following determinant (interpreted in a purely
symbolic sense):
Z 1 e1
e2
e3 V1 (tx) V2 (tx) V3 (tx) dt.
w=
0 tx
tx2
tx3 1
B. We apply the previous theory to the case
v = x2 yze1 + 2xy 2 ze2 − 3xyz 2 e3 .
A trivial check verifies that div(v) = 0. We want w such that curl(w) = v. So
the above formula says
Z 1
Z 1
Z 1
2 2 5
2
2 5
w=
5xy z t dte1 −
4x yz t dte2 −
x2 y 2 zt5 dte3
0
0
0
5
4
1
= xy 2 z 2 e1 − x2 yz 2 e2 − x2 y 2 ze3 .
6
6
6
Appendix: proof of homotopy formula. Since F ∗ d = dF ∗ it suffices to prove:
given α ∈ Ωk ([0, 1] × Rn ), and K as above, then
j1∗ α − j0∗ α = dKα + Kdα.
Here jt : Rn → [0, 1] × Rn are the inclusions jt (x) = (t, x).
Write α = dt ∧ β + γ, for ∂t yβ = ∂t yγ = 0. We have
dβ =
X ∂βJ
J
∂t
˜
dt ∧ dxJ + β,
∂t yβ˜ = 0
where we use multi-index notation J = (j1 , . . . , jk−1 ). We have a similar formula
for γ:
X ∂γI
dt ∧ dxI + γ˜ , ∂t y˜
γ = 0.
dγ =
∂t
I
So dα = −dt ∧ β˜ + dγ, and therefore
Z 1
Kdα =
(∂t ydα)dt
0
Z
1
d
˜
γ − β)dt
dt
0
Z 1
∗
∗
˜
= j1 γ − j0 γ −
βdt
=
(
0
= j1∗ α − j0∗ α −
Z
0
3
1
˜
βdt.
We also have
1
Z
(∂t yα)dt
dKα = d
0
1
Z
=d
βdt
0
1
Z
˜
βdt,
=
0
R1
since 0 βdt is independant of t. The formula follows directly.
4. We observe that div(v) = 0 (c.f. HW 4 Q. 2):
X 1
3x2i
div(v) =
−
= 0.
r3
r5
i
Write ρ2 = x2 + y 2 . The boundary of S is determined by the formula
2
ρ2 + e2ρ −1 = 3/2
√
which has the unique solution ρ = 1/ 2. So ∂S is the circle
∂S = {x2 + y 2 = 1/2, z = 1} ⊂ {r2 = 3/2}
oriented appropriately.
Let S˜ be the section of the sphere
S˜ = {r2 = 3/2, z ≥ 1},
oriented so that (0, 0, 1) is the normal at (0, 0, 1). Then ∂ S˜ = ∂S as oriented
submanifolds, and therefore since v is divergence-free we have
F luxS (v) = F luxS˜ (v)
Z
=
n · vdA
˜
S
2 ˜
= |S|.
3
The last inequality follows because v = rn2 along the sphere of radius r.
Parameterize S˜ by
p
p
√
(ρ, θ) 7→ 3/2(ρ cos θ, ρ sin θ, 1 − ρ2 ), (ρ, θ) ∈ [0, 1/ 3] × [0, 2π].
The Jacbian of this map is J =
F luxS (v) =
3
2
√ρ
1−ρ2
, so we have
2 ˜
|S|
3
Z 1/√3 Z
2π
=
0
0
√
3− 6
= 2π
3
4
ρ
p
1 − ρ2
dθdρ