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Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes CA$11.49   Add to cart
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Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes
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Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes

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  • December 12, 2022
  • 6
  • 2019/2020
  • Class notes
  • Justin ko
  • All classes

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  • partial differential equation

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July 31, 2020 APM346 – Week 11 Justin Ko



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

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