1 Properties of Laplace’s Equation in R2
Consider an open set Ω ⊆ R2 . Solutions to Laplace’s equation
∆u = uxx + uyy = 0 in Ω
2
are called harmonic functions. Harmonic functions in R are closely related to analytic functions in
complex analysis. We discuss several properties related to Harmonic functions from a PDE perspective.
We first state a fundamental consequence of the divergence theorem (also called the divergence form of
Green’s theorem in 2 dimensions) that will allow us to simplify the integrals throughout this section.
Definition 1. Let Ω be a bounded open subset in R2 with smooth boundary. For u, v ∈ C 2 (Ω̄), we
have ZZ ZZ I
∂u
∇v · ∇u dxdy + v∆u dxdy = v ds. (1)
Ω Ω ∂Ω ∂n
∂u
where n = n(x, y) is the outward pointing unit normal at (x, y) ∈ ∂Ω and ∂n := ∇u · n. This identity
is called Green’s first identity.
Remark 1. To simplify notation, we restricted the analysis to R2 but all results in this section
generalize easily to Rn . The domains Ω is always assumed to have smooth boundary.
1.1 Mean Value Property
We will show that the values of harmonic functions is equal to the average over balls of the form
p
Br (x0 , y0 ) = {(x, y) ∈ R2 : (x − x0 )2 + (y − y0 )2 ≤ r} ⊂ Ω.
Theorem 1 (Mean Value Property )
If u ∈ C 2 (Ω) is harmonic in Ω, then
I ZZ
1 1
u(x0 , y0 ) = u ds = u dxdy (2)
2πr ∂Br (x0 ,y0 ) πr2 Br (x0 ,y0 )
for any ball Br (x0 , y0 ) ⊂ Ω.
Proof. We fix (x0 , y0 ) ∈ Ω. We begin by showing the first equality in (2).
First Equality: Consider the function
I Z π
1 1
f (r) = u ds = u(x0 + r cos(θ), y0 + r sin(θ)) dθ
2πr ∂Br (x0 ,y0 ) 2π −π
using the counter clockwise parametrization of ∂Br (x0 , y0 ) with x(θ) = x0 + r cos(θ) and y(θ) =
x0 + r sin(θ) for −π ≤ θ ≤ π. Differentiating with respect to r implies that
Z π
1
f 0 (r) = cos(θ)ux (x0 + r cos(θ), y0 + r sin(θ)) + sin(θ)uy (x0 + r cos(θ), y0 + r sin(θ)) dθ
2π −π
x − x0 y − y0
I
1
= ux (x, y) + uy (x, y) ds
2πr ∂Br (x0 ,y0 ) r r
I
1 ∂u 1
= ds n = (x − x0 , y − y0 )
2πr ∂Br (x0 ,y0 ) ∂n r
ZZ
1
= ∆u dxdy Green’s first identity (1)
2πr Br (x0 ,y0 )
=0 (3)
Page 1 of 6
, July 31, 2020 APM346 – Week 11 Justin Ko
since ∆u = 0 on Br (x0 , y0 ) ⊂ Ω. Therefore f (r) is constant. To figure out the value of f (r), we take
the limit as r → 0 and apply continuity to see that
I Z π
1 1
lim f (r) = lim u ds = lim u(x0 + r cos(θ), y0 + r sin(θ)) dθ = u(x0 , y0 ).
r→0 r→0 2πr ∂B (x ,y ) r→0 2π −π
r 0 0
Second Equality: From the previous section, for all ρ ≤ r we have
Z π
1
u(x0 , y0 ) = u(x0 + ρ cos(θ), y0 + ρ sin(θ)) dθ.
2π −π
We can multiply both sides by ρ and integrate to conclude that
Z r Z π Z r
1
ρu dρ = u(x0 + ρ cos(θ), y0 + ρ sin(θ))ρ dρdθ.
0 2π −π 0
2
The term on the left simplifies to r2 u(x, y) and the term on the right is the integral over Br (x0 , y0 )
expressed in polar coodinates, so
Z π Z r ZZ
1 1
u(x, y) = 2 u(x0 + ρ cos(θ), y0 + ρ sin(θ))ρ dρdθ = 2 u dxdy.
πr −π 0 πr Br (x0 ,y0 )
Remark 2. The normalizations appearing in (2) are the circumference of a circle and the area of a
disc. This normalization means that the integrals can be interpreted as the expected value of u over
a uniform probability measure on the circle and disc.
The converse of Theorem 1 is also true, so the mean value property characterizes harmonic functions.
Theorem 2 (Converse of the Mean Value Property )
If u ∈ C 2 (Ω) satisfies (2) for every ball Br (x0 , y0 ) ⊂ Ω, then u is harmonic in Ω.
Proof. Suppose that ∆u 6≡ 0 in Ω. Without loss of generality, suppose there exists a ball Br (x0 , y0 )
such that ∆u > 0 within Br (x0 , y0 ). For ρ ≤ r, consider
I
1
f (ρ) = u ds.
2πρ ∂Bρ (x0 ,y0 )
If u satisfies (2), then clearly f must be constant since it must equal u(x0 , y0 ) for all ρ ≤ r. The
computations leading to (3) implies that
ZZ
0 1
f (r) = ∆u dxdy > 0,
2πr Br (x0 ,y0 )
which contradicts the fact that f (r) must be constant. If ∆u < 0 within Br (x0 , y0 ), then we arrive at
the same contradiction because the f will be strictly decreasing in that scenario.
Remark 3. Theorem 1 and Theorem 2 implies that u is harmonic if and only if it satisfies the mean
value property. This characterization of harmonic functions is also valid in Rn .
Page 2 of 6
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller 9kfhgia89h1. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.99. You're not tied to anything after your purchase.
