Fourier Series and Partial Differential Equations Lecture Notes

Chapter 5. Laplace’s equation in the plane 49
The choices λ = n 2 π 2 /a 2 , F(x) = sin(nπx/a) (n = 1,2,3...) give the solutions
T(x,y) = sin
nπx
a
G(y), (5.8)
which satisfy the boundary conditions (5.3). Furthermore G(y) is a solution of the ODE
G ′′ (y) = n2 y 2
G(y), (5.9)
a2 and so
nπy
nπy
G(y) = Acosh +Bsinh , (5.10)
a a
in which hyperbolic functions, rather than trigonometric functions, occur. The choice
A = 0 ensures that the boundary condition T(x,0) = 0, (0 < x < a) holds and thus we
are led to consider solutions of the BVP of the form
T(x,y) =
∞
nπx
nπy
Bnsin sinh . (5.11)
a a
n=1
On setting y = b we see that the coefficients Bn are determined by the condition that
Hence
∞
nπx
nπb
Bnsin sinh = f(x), 0 < x < a. (5.12)
a a
n=1
and the BVP has the solution
T(x,y) = 2
a
∞
n=1
nπb
Bnsinh
a
0
= 2
a
nπs
f(s)sin ds, (5.13)
a a
0
a 1
nπs
nπx
nπy
f(s)sin ds sin sinh . (5.14)
sinh(nπb/a) a a a
5.2 BVP in polar coordinates
Next, let us consider separable solutions of Laplace’s equation of the form T(r,θ) =
F(r)G(θ). Substitution into (5.3) gives
Hence
F ′′
and there is a constant λ such that
(r)G(θ)+ 1 ′
F (r)G(θ)+
r 1
r2F(r)G′′ (θ) = 0. (5.15)
r 2 F ′′
(r)+rF ′
(r)
F(r)
+ G′′ (θ)
= 0, (5.16)
G(θ)
r 2 F ′′ (r)+rF ′
(r) = λF(r), (5.17)
G ′′
(θ) = −λG(θ). (5.18)
The function G(θ) must be periodic with period 2π so that G(θ + 2π) = G(θ), and this
is possible only if λ = n 2 , where n is an integer. If n = 0, the only solutions of (5.18)
which are periodicare G(θ) ≡ constant. If n = 0 the periodic solutions are arbitrary linear