Chapter 24
Lecture 28 - Neumann
Problem - only flux BC and
Circular domains
Example 24.1 Eg. 5 Neumann Problem:
uxx + uyy = 0,
0<x<a 0<y<b
(24.1)
ux (0, y) = 0 ux (a, y) = f (y)
(24.2)
uy (x, 0) = 0 = uy (x, b).
(24.3)
Let u(x, y) = X(x)Y (y).
X (x)
Y (y)
=−
= λ2
X(x)
Y (y)
Y (y) + λ2 Y (y) = 0
Y (0) = 0 = Y (b)
Y = A cos λy + B sin λy
Y = −Aλ sin λy + Bλ cos λy
(24.4)
(24.5)
141
Lecture 28 - Neumann Problem - only flux BC and Circular domains
Y (0) = λB = 0 λ = 0 or B = 0.
(24.6)
λn = (nπ/b)
nπyn = 0, 1, . . .
, Y0 = 1
Yn = cos
b
Y (b) = −Aλ sin λb = 0
(24.7)
Xn − λ2 Xn = 0
(24.8)
= 0
(24.9)
Xn (0)
n = 0: X0 = 0, X0 = c0 x + D0 ⇒ X0 = c0 ⇒ X0 (0) = c0 = 0.
Choose D0 = 1: X0 = 1
n ≥ 1 Xn = cn cosh(λn x) + Dn sinh(λn x)
Xn = cn λ sinh(λn x) + Dn λ cosh(λn x)
Xn (0) = λn Dn = 0
(24.10)
Choose cn = 1: Xn = cosh(λn x).
Thus
un (x, y) = Xn Yn = cosh(λn x) cos(λn y)
u0 (x, y) = X0 Y0 = 1
satisfy homog. BC.(24.11)
Therefore
u(x, y) = A0 +
∞
An cosh
nπx b
n=1
cos
nπy b
.
(24.12)
Now f (y) = ux (a, y).
ux (x, y) =
ux (a, y) =
∞
An
n=1
∞ nπ An
n=1
b
sinh
nπ b
nπx sinh
b
cos
nπa b
nπy (24.13)
b
cos
nπy b
= f (y) . .(24.14)
.
This is like a Fourier Cosine Series for f (y) but without the constant
term a0 .
Recall
nπy a0 2
f (y) =
+
, an =
an cos
2
b
b
∞
n=1
142
b
f (y) cos
0
nπy b
dy.(24.15)
24.1. NOTE
Thus the expansion (24.14) is consistent only if a0 = 0. For this to be
true we require that
b
f (y) dy = 0
(24.16)
0
b
f (y) dy = 0 then there is no solution to the boundary value problem 1.
if
0
24.1
Note
b
f (y) dy = 0 there is a net flux into the domain through the right
1. If
0
hand boundary and, since the other boundaries are insulated, there
can be no steady solution – the temperature will continually change
with time.
b
f (y) dy = 0 there is no net flux through the boundary and a
2. If
0
steady state can exist. i.e. It is possible that uxx + uyy = ut = 0. If
b
f (y) dy = 0 then
0
An
nπ b
sinh
nπa 2
=
b
b
b
f (y) cos
nπy dy.
(24.17)
n≥1
(24.18)
b
0
Therefore
2
An =
nπ sinh nπa
b
b
f (y) cos
0
nπy b
dy
and
u∞ (x, y) = A0 +
∞
n=1
An cosh
nπx L
cos
nπy b
(24.19)
where A0 is undetermined. u(x, y) is said to be known up to an arbitrary constant.
143
Lecture 28 - Neumann Problem - only flux BC and Circular domains
3. If u∞ (x, y) is the steady state of a 2D Heat Equation ut = uxx + uyy
with u(x, y, 0) = u0 (x, y) then
∂u
ds = 0.
(24.20)
ut dx dy = ∇ · ∇u dx dy =
∂n
D
D
∂D
Therefore
⎛
⎞
∂ ⎝
u dx dy ⎠ = 0 ⇒ u dx dy = const for all time = u0 (x, y) dx dy.(24.21)
∂t
D
Now
D
D
u∞ (x, y) = A0 × area(D) =
∂D
u0 (x, y) dx
(24.22)
∂D
Which the condition that determines A0 .
24.2
Circular domains:
Recall Laplacian in polar coordinates:
1
1
0 = Δu = uxx + uyy = urr + ur + 2 uθθ
r
r
r = (x2 + y 2 )1/2
. (24.23)
θ = tan−1 (y/x)
Example 24.2 Wedge with homogeneous BC on θ = 0, θ = α < 2π
1
1
urr + ur + 2 uθθ = 0
r
r
u(r, 0) = 0 u(r, α) = 0
u(r, θ) bounded as r → 0
0 < r < a,
u(a, θ) = f (θ).
0 < θ < α (24.24)
(24.25)
(24.26)
Let u(r, θ) = R(r) · Θ(θ).
r2
(R + 1r R )
Θ (θ)
r2 R + rR − λ2 R = 0 Euler Eq.
=−
= λ2 ⇒
Θ + λ2 Θ = 0
R
Θ(θ)
144
24.2. CIRCULAR DOMAINS:
u(r, 0) = R(r)Θ(0) = 0 ⇒ Θ(0) = 0; u(r, α) = R(r)Θ(α) = 0 ⇒ Θ(α) = 0
Θ = A cos λθ + B sin(λθ)
Eigenvalue
Θ + λ2 Θ = 0
(24.27)
Θ(0) = 0 = Θ(α)
Θ(0) = A = 0 Θ(α) = B sin(λα) = 0
Problem
Therefore
λn = (nπ/α) n = 1, 2, . . .
Θn = sin
nπθ
α
.
(24.28)
To solve the Euler Eq. let R(r) = rγ , R = γrγ−1 , R = γ(γ − 1)rγ−2 .
Therefore
γ(γ − 1) + γ − λ2 = γ 2 − λ2 = 0 ⇒ γ = ±λ.
(24.29)
R(r) = c1 rλ + c2 r−λ .
(24.30)
Therefore
Now since u(r, θ) < ∞ as r → 0 we require c2 = 0. Therefore
nπθ
u(r, θ) =
α
n=1
∞
nπ
nπθ
(
)
α
.
u(a, θ) = f (θ) =
sin
cn a
α
∞
nπ
cn r( α ) sin
(24.31)
(24.32)
n=1
This is just a Fourier Sine Series for f (θ): Therefore
nπ
cn a( α ) =
cn =
2
α
α
f (θ) sin
0
2 −( nπ )
a α
α
nπθ
α
dθ
α
f (θ) sin
0
nπθ
α
(24.33)
dθ.
(24.34)
Therefore
u(x, θ) =
∞
n=1
nπ
cn r( α ) sin
nπθ
α
.
(24.35)
145