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0
x : polar coordinates (a, φ).
0
0
The origin and the points x and x form a triangle with sides r = |x|, a = |x |, and
0
|x − x |. By the law of cosines
2
2
0 2
|x − x | = a + r − 2ar cos(θ − φ).
0
The arc length element on the circumference is ds = adφ. Therefore, Poisson’s formula
takes the alternative form
− 2 Z 0
a |x| u(x ) 0
u(x) = ds (10.35)
0 2
2πa |x |=a |x − x |
0
0
for x ∈ D, where we write u(x ) = h(φ). This is a line integral with respect to arc length
0
0
ds = adφ, since s = aφ for a circle. For instance, in electrostatics this formula (10.35)
expresses the value of the electric potential due to a given distribution of charges on a
cylinder that are uniform along the length of the cylinder.
For the time being, we limit ourselves to a mathematically precise statement of it, as

follows.

Theorem 10.5.1 Let h(φ) = u(x ) be any continuous function on the
0
circle C = boundary D. Then the Poisson formula (10.34), or (10.35),

provides the only harmonic function in D for which

lim u(x) = h(x ) for all x ∈ C.
0
0
x→x 0
This means that u(x) is a continuous function on D = D ∪ C. It is also
differentiable to all orders inside D.

10.6 Separation of Variables in Polar Coordinates

Among the most important and ubiquitous of all partial differential equations is Laplace’s
Equation:
∆u = 0,
n
where the Laplacian operator ∆ acts on the function u : U → R (U is open in R ) by
taking the sum of the unmixed partial derivatives. For example:

00
n = 1 : ∆u = u xx = u = 0
In this simple case, the solution u = ax + b is found by integrating twice.

n = 2 : ∆u = u xx + u yy = 0

Here u = u(x, y) and the solution is much more diffcult to obtain.

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