Chapter 25
Lecture 29 More Wedge
Problems
Example 25.1 A wedge with Inhomogeneous BC
1
1
urr + ur + 2 uθθ = 0
r
r
u(r, 0) = u0
u(r, α) = u1
0 < r < a,
0<θ<α
(25.1)
u(r, θ) < ∞ as r → 0 u(a, θ) = f (θ). (25.2)
Let us look for the simplest function of θ only that satisfies the inhomogeθ
neous BC of the from: w(θ) = (u1 − u0 ) + u0 . Note that wθθ = 0 and that
α
w(0) = u0 and w(α) = u1 . Then let u(r, θ) = w(θ) + v(r, θ).
⎫
1
1
1
1
urr + ur + 2 uθθ = vrr + vr + 2 vθθ = 0 ⎪
⎬
Essentially the problem
r
r
r
r
(25.3)
v(r, 0) = 0 v(r, α) = 0
⎪
⎭ solved in Example 24.2
v(a, θ) = f (θ) − w(θ)
The solution is
∞
u(r, θ) = (u1 − u0 )
nπ
θ
+ u0 +
cn r( α ) sin
α
n=1
nπθ
α
(25.4)
147
Lecture 29 More Wedge Problems
where
nπ
2
cn = a−( α )
α
∞
f (θ) − w(θ) sin
0
nπθ
α
dθ.
(25.5)
Example 25.2 A wedge with insulating BC on θ = 0 and θ = α < 2π.
1
1
urr + ur + 2 uθθ = 0
r
r
uθ (r, 0) = 0 uθ (r, α) = 0
u(a, θ) = f (θ).
(25.6)
Let
u(r, θ) = R(r)Θ(θ) ⇒ r
2
1 R + R /R(r) = −Θ /Θ = λ2
r
(25.7)
Θ equation
Θ + λ2 Θ = 0
Θ (0) = 0 = Θ (α)
Θ(θ) = A cos λθ + B sin(λθ)
Θ (0) = Bλ = 0 λ = 0 or B = 0;
(25.8)
Θ (θ) = −Aλ sin(λθ) + Bλ cos(λθ)
(25.9)
Θ (α) = −Aλ sin(λα) = 0 λn = nπ
α ; n = 0, 1, . . .
R equation r2 Rn + rRn − λ2n Rn = 0.
n = 0: rR0 + R0 = (rR0 ) = 0 ⇒ rR0 = a?0 ⇒ R0 (r) = a?0 ln r + c0 .
n ≥ 1: r2 Rn + rRn − λ2 Rn = 0 ⇒ Rn = cn rλn + Dn r−λn .
Since u(r, θ) < ∞ (i.e. must be bounded) as r → 0 we require d0 = 0 =
148
Dn . Therefore
∞
u(r, θ) =
f (θ) = u(a, θ) =
c0 =
u(r, θ) =
nπ
c0 +
cn r( α ) cos
2
c0
+
2
2
α
α
n=1
∞
nπ
cn a( α ) cos
c0
+
2
n=1
f (θ)d?θ
0
∞
cn r(
nπθ
α
nπθ
α
(25.10)
(25.11)
nπ
2
cn = a−( α )
α
nπ
α
) cos
n=1
nπθ
α
α
f (θ) cos
0
.
nπθ
α
dθ (25.12)
(25.13)
Example 25.3 Mixed BC - a ‘crack like’ problem.
1
1
Δu = urr + ur + 2 uθθ = 0
r
r
(25.14)
subject to
∂u
(r, π) = 0
∂θ
u(a, θ) = f (θ).
u(r, 0) = 0
Let u(r, θ) = R(r)Θ(θ).
r
2
R + 1r R
Θ (θ)
=−
= λ2
R
Θ(θ)
(25.15)
(25.16)
(25.17)
Θ equation
Θ + λ2 Θ = 0
Θ(0) = 0
Θ (π) = 0
or λn = (2n + 1)
Θ = −Aλ sin λθ + Bλ cos λθ
π 3π (25.18)
Θ (π) = Bλ cos(λπ) = 0 ⇒ πλ1 = , , . . .
2 2
Θ = A cos λθ + B sin λθ
Θ(0) = A = 0
1
n = 0, 1, . . . λ = 0 as this would be trivial.
2
149
Lecture 29 More Wedge Problems
R equation r2 R + rR − λ2 R = 0 R(r) = rγ ⇒ γ 2 − λ2 = 0 γ = ±λ.
Therefore
un (r, θ) = cn rλn + dn r−λn sin λn θ.
(25.19)
Since u should be bounded as r → 0 we conclude that dn = 0. The general
solution is thus
∞
(2n + 1)
(2n+1)/2
θ
(25.20)
cn r
sin
u(r, θ) =
2
n=0
∞
2n + 1
(2n+1)/2
cn a
sin
f (θ) = u(a, θ) =
θ . (25.21)
2
n=0
Check orthogonality
π
sin
0
2m + 1
2
2n + 1
0
m = n
θ sin
θ dθ =
. (25.22)
π/2 m = n
2
Therefore
1
cn =
u(r, θ) =
2a−(n+ 2 )
π
∞
n=0
150
π
f (θ) sin
0
1
cn r(n+ 2 ) sin
1
n+
2
θ dθ
1
θ .
n+
2
(25.23)
(25.24)