c Anthony Peirce.
Introductory lecture notes on Partial Differential Equations - °
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1
Lecture 26: Circular domains
(Compiled 3 March 2014)
In this lecture we consider the solution of Laplace’s equations on domains that have a boundary that has at least one
boundary segment that comprises a circular arc.
Key Concepts: Laplace’s equation; Circular domains; Pizza Slice-shaped regions, Dirichlet and Mixed BC.
Reference Section: Boyce and Di Prima Section 10.8
26 General Analysis of Laplace’s Equation on Circular Domains:
26.1 Laplacian in Polar Coordinates
For domains whose boundary comprises part of a circle, it is convenient to transform to polar coordinates. For this
purpose the Laplacian is transformed from cartesian coordinates (x, y) to polar coordinates (r, θ) as follows:
r2 = x2 + y 2
³y´
θ = tan−1
x
Differentiating with respect to x and y we obtain:
x
y
2rrx = 2x rx =
ry =
r
r
(− xy2 )
y
y
¢ =− 2
θx = ¡
=− 2
x + y2
r
1 + ( xy )2
θy =
x
r2
u(x, y) = U (r, θ)
³ y´
¡x¢
+ Uθ − 2
³ry ´
³ xr ´
uy = Ur ry + Uθ θy = Ur
+ Uθ 2
r
r
uxx = (Ur )x rx + Ur rxx + (Uθ )x θx + Uθ θxx
ux = Ur rx + Uθ θx = Ur
rxx
uxx
= Urr rx2 + Urθ θx rx + Ur rxx + Urθ rx θx + Uθθ θx2 + Uθ θxx
³ 2´
r − xr
y2
2yx
=
=
θxx = 4
2
3
rµ ¶ r
r
µ 2¶
µ
¶
³ y ´ ³x´
x2
y2
y
2xy
= Urr
+
2U
−
+
U
+
U
+
U
rθ
θθ 4
r
θ
r2
r2
r
r
r3
r4
uyy = Urr ry2 + Urθ ry θy + Ur ryy + Uθr ry θy + Uθθ θy2 + Uθ θyy
µ 2¶
µ 2¶
µ
¶
µ 2¶
³ x ´ ³y´
x
x
2xy
y
+
2U
+
U
+
U
+
U
−
= Urr
rθ
θθ
r
θ
r2
r2
r
r4
r3
r4
1
1
uxx + uyy = Urr + Ur + 2 Uθθ
r
r
2
26.2 Introductory remarks about circular domains
Recall the Laplacian in polar coordinates:
1
1
0 = ∆u = uxx + uyy = urr + ur + 2 uθθ
r
r
r = (x2 + y 2 )1/2
.
θ = tan−1 (y/x)
(26.1)
Let
u(r, θ) = R(r)Θ(θ)
Θ00
r2 R00 + rR0
=−
= λ2
R(r)
Θ(θ)
(26.2)
(26.3)
which leads to r2 R00 + rR0 − λ2 R = 0 and Θ00 + λ2 Θ = 0.
The R Equation: r2 R00 + rR0 − λ2 R = 0:
λ = 0: r2 R00 + rR0 = 0, R = rγ ⇒ γ(γ − 1) + γ = γ 2 = 0 ⇒ R(r) = C + D ln r
λ 6= 0: r2 R00 + rR0 − λ2 = 0, R = rγ ⇒ γ(γ − 1) + γ − λ2 = γ 2 − λ2 = 0 ⇒ R(r) = Crλ + Dr−λ .
The Θ Equation Θ00 + λ2 Θ = 0:
Θ00 + λ2 Θ = 0, Θ = A cos λθ + B sin λθ, Θ0 = −Aλ sin λθ + Bλ cos λθ
Different Boundary Conditions and corresponding eigenfunctions:
(I) Θ(0) = 0 = Θ(α) λn = nπ/α, n = 1, 2, . . . , Θn (θ) = sin λn θ
(II) Θ0 (0) = 0 = Θ0 (α), λn = nπ/α n = 0, 1, 2, . . . , Θn (θ) ∈ {1, cos λn θ}
(III) Θ(0) = 0 = Θ0 (α), λn = (2n − 1)π/2α n = 1, 2, . . . , Θn (θ) = sin λn θ
(IV) Θ0 (0) = 0 = Θ(α), λn = (2n − 1)π/2α, n = 1, 2, . . . , Θn (θ) = cos λn θ
¾
Θ(−π) = Θ(π)
λn = n, n = 0, 1, 2, . . . , Θn (θ) ∈ {1, cos λn θ, sin λn θ}.
(V)
Θ0 (−π) = Θ0 (π)
The most general solution is thus of the form
u(r, θ) = {A0 + α0 ln r} · 1 +
+
∞
X
©
n=1
∞
X
©
ª
An rλn + αn r−λn cos λn θ
(26.4)
ª
Bn rλn + βn r−λn sin λn θ.
(26.5)
n=1
Observations:
• For problems that include the origin, the condition |u| < ∞ as r → 0 dictates that α0 = 0, αn = 0 and βn = 0.
• For problems that involve infinite domains the condition |u| < ∞ as r → ∞ dictates that An = 0 and Bn = 0.
• The values of λn and the corresponding eigenfunctions depend on the boundary conditions (I)–(V) that apply.
26.3 Wedge Problems
Example 26.1 Wedge with homogeneous BC on θ = 0, θ = α < 2π
1
1
urr + ur + 2 uθθ = 0
0 < r < a, 0 < θ < α
r
r
u(r, 0) = 0 u(r, α) = 0, u(r, θ) bounded as r → 0, u(a, θ) = f (θ)
(26.6)
(26.7)
Laplace’s Equation
3
Figure 1. Homogeneous Dirichlet Boundary conditions on a wedge shaped domain (26.7)
Let u(r, θ) = R(r) · Θ(θ).
r2
(R00 + 1r R0 )
Θ00 (θ)
r2 R00 + rR0 − λ2 R
=−
= λ2 ⇒
Θ00 + λ2 Θ
R
Θ(θ)
= 0 Euler Eq.
= 0
u(r, 0) = R(r)Θ(0) = 0 ⇒ Θ(0) = 0; u(r, α) = R(r)Θ(α) = 0 ⇒ Θ(α) = 0
½ 00
Eigenvalue
Θ + λ2 Θ = 0
Θ = A cos λθ + B sin(λθ)
Problem
Θ(0) = 0 = Θ(α)
Θ(0) = A = 0 Θ(α) = B sin(λα) = 0
Therefore
µ
λn = (nπ/α) n = 1, 2, . . .
Θn = sin
nπθ
α
(26.8)
¶
.
(26.9)
To solve the Euler Eq. let R(r) = rγ , R0 = γrγ−1 , R00 = γ(γ − 1)rγ−2 . Therefore
γ(γ − 1) + γ − λ2 = γ 2 − λ2 = 0 ⇒ γ = ±λ.
(26.10)
R(r) = c1 rλ + c2 r−λ .
(26.11)
Therefore
Now since u(r, θ) < ∞ as r → 0 we require c2 = 0. Therefore
¶
nπθ
α
n=1
µ
¶
∞
o
Xn
nπ
nπθ
u(a, θ) = f (θ) =
cn a( α ) sin
.
α
n=1
u(r, θ) =
∞
X
µ
cn r(
nπ
α
) sin
(26.12)
(26.13)
This is just a Fourier Sine Series for f (θ): Therefore
cn a(
nπ
α
)= 2
α
µ
Zα
f (θ) sin
0
nπ
2
cn = a−( α )
α
nπθ
α
¶
µ
Zα
f (θ) sin
0
dθ
nπθ
α
(26.14)
¶
dθ.
(26.15)
4
Therefore
u(x, θ) =
µ
∞
X
nπ
cn r( α ) sin
n=1
nπθ
α
¶
.
(26.16)
Example 26.2 A wedge with Inhomogeneous BC
Figure 2. Inhomogeneous Dirichlet Boundary conditions on a wedge shaped domain (26.18)
1
1
urr + ur + 2 uθθ = 0
r
r
u(r, 0) = u0 ,
u(r, α) = u1 ,
0 < r < a,
0<θ<α
u(r, θ) < ∞ as r → 0,
u(a, θ) = f (θ)
(26.17)
(26.18)
θ
Let us look for the simplest function of θ only that satisfies the inhomogeneous BC of the from: w(θ) = (u1 −u0 ) +u0 .
α
Note that wθθ = 0 and that 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
(26.19)
v(r, 0) = 0 v(r, α) = 0

 solved in Example 26.1
v(a, θ) = f (θ) − w(θ)
The solution is
u(r, θ) = (u1 − u0 )
µ
¶
∞
X
nπ
θ
nπθ
+ u0 +
cn r( α ) sin
α
α
n=1
(26.20)
where
nπ
2
cn = a−( α )
α
Z∞
0
£
¤
f (θ) − w(θ) sin
µ
nπθ
α
¶
dθ.
(26.21)
Example 26.3 A wedge with insulating BC on θ = 0 and θ = α < 2π.
Let
1
1
urr + ur + 2 uθθ = 0
r
r
uθ (r, 0) = 0, uθ (r, α) = 0, u(a, θ) = f (θ)
(26.22)
µ
¶
1 0
00
u(r, θ) = R(r)Θ(θ) ⇒ r R + R /R(r) = −Θ00 /Θ = λ2
r
(26.23)
2
Laplace’s Equation
5
Figure 3. Mixed Boundary conditions on a wedge shaped domain (26.22)
Θ equationi
¾
Θ00 + λ2 Θ = 0
Θ0 (0) = 0 = Θ0 (α)
Θ(θ) = A cos λθ + B sin(λθ)
Θ0 (0) = Bλ = 0 λ = 0 or B = 0;
(26.24)
Θ0 (θ) = −Aλ sin(λθ) + Bλ cos(λθ)
Θ0 (α) = −Aλ sin(λα) = 0 λn = nπ
α ; n = 0, 1, . . .
(26.25)
R equationi r2 Rn00 + rRn0 − λ2n Rn = 0.
0
n = 0: rR000 + R00 = (rR00 ) = 0 ⇒ rR00 = d0 ⇒ R0 (r) = c0 + d0 ln r.
n ≥ 1: r2 Rn00 + rRn0 − 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 = Dn . Therefore
∞
nπ
c0 X
u(r, θ) =
+
cn r( α ) cos
2
n=1
∞
f (θ) = u(a, θ) =
nπ
c0 X
+
cn a( α ) cos
2
n=1
2
c0 =
α
u(r, θ) =
Zα
f (θ)dθ
0
µ
µ
nπθ
α
nπθ
α
¶
(26.26)
¶
(26.27)
nπ
2
cn = a−( α )
α
∞
X
nπ
c0
+
cn r( α ) cos
2
n=1
µ
nπθ
α
¶
µ
Zα
f (θ) cos
0
.
nπθ
α
¶
dθ
(26.28)
(26.29)
Example 26.4 Mixed BC - a ‘crack like’ problem.
1
1
∆u = urr + ur + 2 uθθ = 0
r
r
(26.30)
subject to
∂u
(r, π) = 0
∂θ
u(a, θ) = f (θ).
u(r, 0) = 0
(26.31)
(26.32)
6
Figure 4. Inhomogeneous Neumann Boundary conditions on a rectangular domain as prescribed in (??)
Let u(r, θ) = R(r)Θ(θ).
¡ 00 1 0 ¢
R + rR
Θ00 (θ)
r
=−
= λ2
R
Θ(θ)
2
(26.33)
Θ equationi
Θ00 + λ2 Θ = 0
Θ = A cos λθ + B sin λθ
Θ(0) = 0 Θ0 (π) = 0
Θ(0) = A = 0
Θ0 = −Aλ sin λθ + Bλ cos λθ
π 3π
Θ0 (π) = Bλ cos(λπ) = 0 ⇒ πλ1 = ,
,...
2 2
1
n = 0, 1, . . . λ 6= 0 as this would be trivial.
2
2 00
R equationi r R + rR0 − λ2 R = 0 R(r) = rγ ⇒ γ 2 − λ2 = 0 γ = ±λ. Therefore
¡
¢
un (r, θ) = cn rλn + dn r−λn sin λn θ.
(26.34)
or λn = (2n + 1)
Since u should be bounded as r → 0 we conclude that dn = 0. The general solution is thus
µ
¶
∞
X
(2n + 1)
u(r, θ) =
cn r(2n+1)/2 sin
θ
2
n=0
µµ
¶ ¶
∞
X
2n + 1
f (θ) = u(a, θ) =
cn a(2n+1)/2 sin
θ .
2
n=0
(26.35)
(26.36)
(26.37)
Check orthogonality
µµ
Zπ
sin
0
2m + 1
2
¶ ¶
µµ
¶ ¶
½
2n + 1
0
θ sin
θ dθ =
π/2
2
m 6= n
.
m=n
(26.38)
Therefore
2a−(n+ 2 )
cn =
π
1
u(r, θ) =
∞
X
n=0
Zπ
0
µµ
¶ ¶
1
f (θ) sin
n+
θ dθ
2
1
cn r(n+ 2 ) sin
µµ
¶ ¶
1
n+
θ
2
(26.39)
(26.40)