Fourier Series and Partial Differential Equations Lecture Notes

Chapter 5. Laplace’s equation in the plane 51
Thus
A = 1
2π
2π
Cn =
Dn =
1
πa n
1
πa n
0
2π
0
2π
0
f(θ)dθ, (5.29)
f(θ)cos(nθ) dθ, (5.30)
f(θ)sin(nθ) dθ. (5.31)
Example 5.3 Find T(r,θ) so as to satisfy Laplace’s equation in the disc r < a and the
boundary condition
T(a,θ) = |sinθ|, 0 ≤ θ ≤ 2π. (5.32)
The solution is
where
and so
A = 1
2 π
2π
Cn =
Dn =
1
πa n
1
πa n
T(r,θ) = A+
0
2π
0
2π
5.2.2 Poisson’s formula
0
∞
r n [Cncos(nθ)+Dnsin(nθ)], (5.33)
n=1
|sinθ|dθ = 1
π
2π
|sinθ|cos(nθ) dθ =
0
2π
sinθdθ−
π
sinθdθ
0 n odd,
−4/(πa n (n 2 −1)) n even,
= 2
, (5.34)
π
(5.35)
|sinθ|sin(nθ) dθ = 0, (5.36)
T(r,θ) = 2 4
−
π π
∞
n=1
r
2n cos(2nθ)
a 4n2 . (5.37)
−1
Consider again the problem from Section 5.2.1: find T so as to satisfy Laplace’s equation
in the disc 0 ≤ r < a and the boundary condition T = f(θ) on r = a, (0 ≤ θ ≤ 2π), where
f is a prescribed function. Poisson’s formula states that the solution to this problem can
be written
T(r,θ) = (a2 +2)
2π
2π
Lemma 5.1 If λ and α are real and |λ| < 1 then
1
2 +
0
∞
λ n cosnα =
n=1
f(φ)
a2 +r2 dφ. (5.38)
−2arcos(θ −φ)
1−λ 2
2(1+λ 2 . (5.39)
−2λcosα)