MTH 424 - PARTIAL DIFFERENTIAL EQUSTION
IIT KANPUR
Odd Semester, 2023
Assignment 4
1. Consider the following function
ϕ(x) =
1
.
|x − x0 |n−2
(a) By direct calculation show that ϕ is harmonic in Rn \ {x0 }.
1
, show that △ϕ(x) ̸= 0 for x ̸= 0.
(b) For n = 2 and ϕ(x) = |x|
2. For n = 2, show that the Laplace equation in polar coordinates for function v(rθ) be
given by
∂ 2 v 1 ∂v
1 ∂ 2v
+
+
=0
∂r2 r ∂r r2 ∂θ2
3. Let ϕ : R → R be smooth and convex function. Assume u is harmonic. Show that
v := ϕ(u) is subharmonic (i.e. −△v ≤ 0).
4. Assume u ∈ C(Ω). Let B(x0 , r) ⊂ Ω, show that
Z
Z
u(y)dy −−→ u(x) and –
–
B(x0 ,ϵ)
ϵ→0
u(y)dy −−→ u(x).
∂B(x0 ,ϵ)
ϵ→0
5. Suppose Ω is a bounded domain and u, v ∈ C 2 (Ω) ∩ C(Ω) satisfy
−△u ≤ 0 and
Prove that u ≤ v
− △v ≤ 0 in Ω and u ≤ v
in ∂Ω.
in Ω.
6. Let Ω ⊂ R3 be a bounded domain. Show that the following problem
(
△u = u
in Ω,
u=0
on ∂Ω
has only solution u ≡ 0. [HINT: Multiply both side of the equation by u and integrate
over Ω.]
7. let Ω be a domain in R2 symetric about the x-axis and let Ω+ = {(x, y) ∈ Ω : y > 0} be
the upper part of Ω. Assume u ∈ C(Ω+ ) is harmonic in Ω+ with u = 0 on ∂Ω+ ∩{y = 0}.
Define for (x, y) ∈ Ω
(
u(x, y)
if y ≥ 0,
v(x, y) =
−u(x, −y)
if y < 0.
Show that v is harmonic.
8. Let u ∈ C 2 (Ω) ∩ C(Ω) satisfies
n
X
n
X
∂ 2u
∂u
Lu =
ai,j (x)
(x) +
bi (x)
(x) + c(x)u(x) ≥ 0,
∂xi ∂xj
∂xi
i,j=1
i=1
where aij = aji , bi , c are real valued smooth functions on Ω and c < 0 and (ai,j )i,j is a
non-negative definite matrix. Then show that u can not take interior positive maximum
in Ω.
n
X
∂ 2u
(x0 ) ≤ 0]
[HINT: If x0 ∈ Ω is point of maximum, then prove first that
ai,j (x0 )
∂xi ∂xj
i,j=1
9. Let u ∈ C 2 (Ω) ∩ C(Ω) be a solution of
△u +
n
X
i=1
bi (x)
∂u
(x) + c(x)u(x) = 0 in Ω
∂xi
with c(x) < 0 in Ω, u = 0 on ∂Ω. Show that u ≡ 0. [HINT: Use previous result with u
and −u]
8*. If we do not assume positive maximum assumption, we can give counter example.
Take Lu = △u − u in B(0, 1), then for ϕ = −|x|2 − 2 we can see Lϕ ≥ 0 and ϕ having
maximum at x = 0.