8.3. HEAT EQ ON A CIRCULAR RING - FULL FOURIER SERIES
Lecture 12
8.3
Heat Eq on a Circular Ring - Full Fourier Series
Physical Interpretation: Consider a thin circular wire in which there is
∂u
= 0.
no radial temperature dependence.
∂r
ut = α2 uxx
BC:
IC:
u(−L, t)
= u(L, t)
∂u
∂u
(−L, t) =
(L, t)
∂x
∂x
(8.22)
Periodic BC
u(x, 0) = f (x)
Assume u(x, t) = X(x)T (t).
X (x)
Ṫ (t)
As before:
= 2
= −λ2 .
X(x)
α T (t)
Ṫ (t)
2
= −λ2 ⇒ T (t) = ce−λ t .
IVP: 2
α T (t)
L
2L
= = Constant.
r=
2π
π
The Laplacian becomes
Δu =
=
1 ∂2u
∂ 2 u 1 ∂u
+
+ 2 2
2
∂r
r ∂r
r ∂θ
∂2u
∂(rθ)2
(8.23)
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Separation of Variables
if we let x = rθ we obtain
X + λ2 X =
=
BVP: X(−L)
=
X (−L)
(1.1). ⎫
0
⎬
X(L)
⎭
X (L)
Eigenvalue Problem
look for λ such that
nontrivial x can be found.
X(x) = A cos(λx) + B sin(λx)
X(−L) = A cos(λL) − B sin(λL) = A cos(λL) + B sin(λL) = X(L)
therefore 2B sin(λL) = 0
X (x) = −Aλ sin λx + Bλ cos(λx)
(8.24)
X (−L) = +Aλ sin(λL) + Bλ cos(λL) = −Aλ sin(λL) + Bλ cos(λL) = X (L)
therefore 2Aλ sin(λL) = 0
Therefore λn L = (nπ) n = 0, 1, . . . .
Solutions to (1.1) that satisfy the BC are thus of the form
nπx nπx nπ 2 2
+ Bn sin
.
un (x, t) = e−( L ) α t An cos
L
L
(8.25)
Superposition of all these solutions yields the general solution
u(x, t) = A0 +
∞ nπx nπx nπ 2 2
An cos
+ Bn sin
e−( L ) α t .
L
L
(8.26)
n=1
In order to match the IC we have
f (x) = u(x, 0) = A0 +
∞
An cos
nπx n=1
L
+ Bn sin
nπx L
.
(8.27)
As before we obtain expressions
n and Bn by projecting f (x) onto
nπx for theA
nπx the basis functions sin
and cos
.
L
L
L
f (x)
−L
L sin mπx
sin mπx
L L
dx = A0
dx
cos mπx
cos mπx
L
L
−L
+
∞
n=1
+
∞
n=1
58
(8.28)
L
An
−L
L
Bn
−L
nπx sin mπx L cos
dx
cos mπx
L
L
nπx sin mπx L dx.
sin
cos mπx
L
L
8.3. HEAT EQ ON A CIRCULAR RING - FULL FOURIER SERIES
As in the previous example we use the orthogonality relations:
L
sin
−L
L
cos
−L
mπx L
mπx L
sin
cos
nπx L
nπx L
dx = Lδmn
dx = Lδmn
= 2L
L
sin
−L
mπx L
cos
nπx L
m and n = 0
(8.29)
m=n=0
dx = 0 ∀m, n.
Plugging these orthogonality conditions into (1.6) we obtain
⎫
⎪
⎪
⎪
1
⎪
⎪
f (x) dx = average value of f (x) on [−L, L]
⎪
⎪
2L
⎬
−L
(8.30)
L
L
⎪
nπx nπx ⎪
1
1
⎪
⎪
dx and Bn =
dx. ⎪
f (x) cos
f (x) sin
⎪
⎪
L
L
L
L
⎭
L
A0
=
An =
−L
−L
Note:
1. (1.6) and (1.9) [typist’s note: check re-numbering when equations are
re-labeled, these could be renamed to (9.6) and (9.9) respectively]
represent the full Fourier Series Expansion for f (x) on the interval
[−L, L].
⎧
L
⎨ 2A0
1
nπx
2. By defining an =
dx =
f (x) cos
and bn = Bn
⎩
L
L
An
−L
the Fourier Series (1.6) is often written in the form
nπx nπx a0 f (x) =
+
+ bn sin
.
an cos
2
L
L
∞
(8.31)
n=1
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