30
4.2
4.2.1
CHAPTER 4. HARMONIC FUNCTIONS
The Poisson integral formula
Properties
The Poisson integral formula produces a function which is harmonic on the disk
of radius a whose value on the boundary is prescribed, u(a, θ) = h(θ). The
formula is
Z 2π
dφ
a2 − r 2
.
(4.1)
u(r, θ) =
h(φ) 2
2
a − 2ar cos(θ − φ) + r 2π
0
The function
K(r, θ, φ) =
a2 − r 2
a2 − 2ar cos(θ − φ) + r 2
appearing in (4.1) is called Poisson’s kernel. In establishing this formula we also
established the identity
K(r, θ, φ) = 1 + 2
∞ X
r n
n=1
a
cos(n[θ − φ]).
(4.2)
The next result shows that the harmonic function given by the Poisson integral
formula (4.1) is an average of the values of the function h(φ).
Theorem 4.2.1. For 0 ≤ r < a we have K(r, θ, φ) > 0, and
Z 2π
dφ
= 1.
K(r, θ, φ)
2π
0
In addition, for any ǫ > 0 and any δ > 0 there is an R1 such that R1 ≤ r < R
implies K(r, θ, φ) < ǫ when the distance from θ to φ on the circle is at least δ.
Proof. First, (a − r)2 > 0, so a2 + r 2 > 2ar ≥ 0. Since −1 ≤ cos(θ − φ) ≤ 1 it
follows that
a2 + r 2 > 2ar ≥ 2ar cos(θ − φ).
Thus a2 − 2ar cos(θ − φ) + r 2 > 0 and K(r, θ, φ) > 0.
Second, the 2π periodicity of sin(ψ − φ) gives
Z 2π
sin(n[θ − φ]) 2π
= 0.
cos(n[θ − φ]) dφ =
n
φ=0
0
The integral in the theorem may then be computed by termwise integration of
the uniformly convergent series in (4.2).
Z
0
2π
dφ
K(r, θ, φ)
=
2π
Z
0
2π
1+2
∞ X
r n
n=1
a
cos(n[θ − φ])
dφ
= 1.
2π
4.2. THE POISSON INTEGRAL FORMULA
31
Finally, it turns out that for values of r near a, the contributions to u(r, θ)
come mostly from θ near φ. As long as 1 − cos(θ − φ) ≥ σ > 0 we have
a2 − r 2
a2 − r 2
a2 − r 2
=
≤
,
a2 − 2ar cos(θ − φ) + r 2
(a − r)2 + 2ar(1 − cos(θ − φ))
2arσ
which has a limit 0 as r → a.
4.2.2
Applications
Suppose Ω ⊂ R2 is a bounded connected open set, and u(x, y) is continuous
on Ω, the closure of Ω, and harmonic on Ω. The weak maximum principle for
harmonic functions has already established (p. 155-6) that there is at most one
such function which agrees with u on ∂Ω. It is not hard to show that if h(θ) is
continuous on the circle of radius a, then the function produced by the Poisson
integral formula is continuous on the closed disk D, and harmonic inside. Thus
for the disk we have an explicit formula for solving our boundary problem
uxx + uyy = 0, u = h(θ).
∂D
Here are some applications of the Poisson formula to more general domains.
Proposition 4.2.2. Suppose u(x, y) is continuous on a closed disk D ⊂ R2 of
radius a > 0 and center (x0 , y0 ). If u is harmonic in D, then u(x0 , y0 ) is the
average of u on ∂D.
Proof. Use the translation X = x − x0 , Y = y − y0 to shift the problem to the
disk centered at 0. The shifted function is still harmonic in the shifted disk.
Now put r = 0 in the Poisson formula (4.1).
Theorem 4.2.3. Suppose u(x, y) is harmonic in a connected open set Ω ⊂ R2 .
If there is a point (x0 , y0 ) ∈ Ω such that u(x, y) ≤ u(x0 , y0 ) for all (x, y) ∈ Ω,
then u(x, y) = u(x0 , y0 ) for all (x, y) ∈ Ω.
Proof. Pick any point (x1 , y1 ) ∈ Ω, and connect (x0 , y0) to (x1 , y1 ) by a path
γ : [0, 1] → Ω. For a sufficiently small the disk of radius a centered at (x0 , y0 )
is contained in Ω. Since u(x0 , y0) is the average of the values of (the continuous
function) u over the circle of radius a with center (x0 , y0), but u(x, y) ≤ u(x0 , y0 ),
it follows that u(x, y) = u(x0 , y0) for all (x, y) whose distance from (x0 , y0 ) is at
most a.
Let t1 be the least upper bound of numbers t ∈ [0, 1] such that u(γ(s)) =
u(γ(0)) for all s < t. By the first part of the argument, t1 > 0. Since u is
32
CHAPTER 4. HARMONIC FUNCTIONS
continuous, u(γ(t1 )) = u(γ(0)). By repeating the original argument we see that
t1 cannot be smaller than 1, since we can always push further along the path.
Thus u(x1 , y1 ) = u(x0 , y0 ). Since (x1 , y1 ) was any point in Ω, the proof is
finished.
Theorem 4.2.4. Suppose u(x, y) is harmonic in a connected open set Ω ⊂ R2 .
Then u has partial derivatives of all orders at every (x, y) ∈ Ω.
Pick any point (x0 , y0 ) ∈ Ω. Pick an a > 0 sufficiently small so that the disk
of radius a centered at (x0 , y0 ) is contained in Ω. Use the Poisson formula (4.1)
to represent u(x, y) for (x, y) near (x0 , y0 ). The main point is that the Poisson
kernel has derivatives of all orders. To compute derivatives of u we can simply
differentiate the Poisson kernel in (4.1). More details are in the text, page 170.