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Partial Differential Equations in Two
dimensions
W. A. Day
February 11, 2008
Example 12. A conductor occupies the region r ≤ a and the temperature satisfies the boundary condition T (a, θ) = sin^3 θ. Find T (r, θ) in r < a. Note that sin^3 θ = 34 sin θ − 14 sin 3θ. (This is already the Fourier series ex- pansion of sin^3 θ). Hence
T =
r a
sin θ −
( (^) r a
sin 3θ.
Example 13. Find T (r, θ) so as to satisfy Laplace’s equation in the disc r < a and the boundary condition
T (a, θ) = | sin θ|, 0 ≤ θ ≤ 2 π.
the solution is
T = A +
∑^ ∞
n=
rn(Cn cos nθ + Dn sin nθ),
where
A =
π
∫ (^2) π
0
| sin θ|dθ =
2 π
[∫ (^) π
0
sin θdθ −
∫ (^2) π
π
sin θdθ
]
π
Cn =
πan
∫ (^2) π
0
| sin θ| cos nθdo =
0 , n odd − (^) πa^4 n · (^) n (^21) − 1 , n even
DN =
πan
∫ (^2) π
0
| sin θ| sin nθdθ = 0,
and so
T (r, θ) =
π
π
∑^ ∞
n=
( (^) r
a
) 2 n (^) cos 2nθ
4 n^2 − 1
Poisson’s Formula. The solution of the BVP of Example 11 is
T (r, θ) =
(a^2 + 2) 2 π
∫ (^2) π
0
f (φ) a^2 + r^2 − 2 ar cos(θ − φ)
dφ.
The proof depends upon the
Lemma. If λ and α are real and |λ| < 1 then
1 2
∑^ ∞
n=
λn^ cos nα =
1 − λ^2 2(1 + λ^2 − 2 λ cos α)
Proof. (^) ∞ ∑
n=
λn^ cos nα = Re
∑^ ∞
n=
λneinα
= Re
∑^ ∞
n=
(λeiα)n^ (geometric series)
= Re
[
λeiα 1 − λeiα
]
λ cos α − λ^2 1 + λ^2 − 2 λ cos α
Hence
∑^ ∞
n=
λn^ cos nα =
1 − λ^2 2(1 + λ^2 − 2 λ cos α)
Proof of Poisson’s Formula. Taking care over the dummy variable of integration we get
T (r, θ) =
2 π
∫ (^2) π
0
f (φ)dφ
π
∑^ ∞
n=
r a
)n
[
cos nθ
∫ (^2) π
0
f (φ) cos nφdφ + sin nθ
∫ (^2) π
0
f (φ) sin nφdφ
]
[In this case ∂T ∂n
is a genuine derivative.] Let g(θ) have the Fourrier expansion
g(θ) =
p 0 +
∑^ ∞
n=
(pn cos nθ + qn sin nθ|,
where
p 0 =
π
∫ (^2) π
0
g(θ)dθ
pn =
π
∫ (^2) π
0
g(θ) cos nθdθ
qn =
π
∫ (^2) π
0
g(θ) sin nθdθ.
Look for solutions in the form
T (r, θ) = A +
∑^ ∞
n=
rn(Cn cos nθ + Dn sin nθ).
Then ∂T ∂r
∑^ ∞
n=
nrn−^1 (Cn cos nθ + Dn sin nθ)
and the boundary condition gives
∑^ ∞
n=
nan−^1 (Cn cos nθ + Dn sin nθ) = g(θ), 0 ≤ θ ≤ 2.
We conclude immediately that the condition ∫ (^2) π
0
g(θ)dθ = 0
is necessary for a solution to exist.
If this condition is satisfied then there are solutions
T (r, θ) = A +
∑^ ∞
n=
rn(Cn cos nθ + Dn sin nθ),
where
Cn =
nπan−^1
∫ (^2) π
0
g(θ) cos nθdθ
Dn =
nπan−^1
∫ (^2) π
0
g(θ) sin nθdθ
and A is an arbitrary instant, i.e. soutions are non-unique if they exist.
Example 15. Find T so as to satisfy Laplace’s equation in the disc 0 ≤ r < a and the boundary condition
∂T ∂n
(a, θ) = sin^3 θ, 0 ≤ θ ≤ 2.
Here
sin^3 θ =
sin θ −
sin 3θ, ∫ (^2) π
0
sin^3 θdθ = 0,
and the solutions are
T = A +
r sin θ −
r^3 a^2
sin 3θ,
where A is arbitrary.
