Chapter 17
Lecture 21 Distributed, Time
Dependent Heat Sources eigenfunction expansions
Example 17.1 A Bar with a Time-Varying External Heat Source:
2πx
2
−t
0 < x < L, t > 0 (17.1)
ut = α uxx + e sin
L
BC: u(0, t) = 0; u(L, t) = L
(17.2)
IC: u(x, 0) = x.
(17.3)
Consider the function w(x) = x which satisfies the BC as well
homogeneous version of the PDE.
Now let u(x, t) = w(x) + v(x, t).
2πx
2
−t
ut = (w
+v)t = α (w
+v)xx + e sin
L
2πx
⇒ vt = α2 vxx + e−t sin
L
u(0, t) = w(0)
+v(0, t) = 0 ⇒ v(0, t) = 0
as the
u(L, t) = w(L)
+v(L, t) = ⇒
L v(L, t) = 0
(17.7)
(17.4)
(17.5)
(17.6)
x = u(x, 0) = w(x) + v(x, 0) = x + v(x, 0) ⇒ v(x, 0) = 0. (17.8)
105
Lecture 21 Distributed, Time Dependent Heat Sources - eigenfunction
expansions
Now assume that v(x, t) =
∞
v̂n (t) sin
n=1
nπx .
L
∞
∞
nπx nπx ∂ 2 v
∂v dv̂n
nπ 2
=
v̂n (t) −
sin
=
(t) sin
.(17.9)
∂t
dt
L
∂x2
L
L
n=1
n=1
Therefore
2
vt − α vxx =
∞ dv̂n
dt
n=1
+α
2
nπ 2
L
−t
v̂n − e δ2n sin
nπx L
= 0. (17.10)
Therefore
nπ 2
dv̂n
+ α2
v̂n = e−t δ2n
dt
L
2
d α2 ( nπ )2 t −1 t
α2 ( nπ
)
L
L
v̂n = e
δ2n .
e
dt
(17.11)
(17.12)
Therefore
e
2
−α2 nπ
t
L
(
) v̂ =
n
v̂n (t) =
e
2
α2 ( nπ
−1 t
L )
α2
α2
nπ 2
L
e−t δ
−1
2n
nπ 2
L
−1
δ2n + cn
cn arbitrary
(17.13)
2
2 nπ
+ e−α ( L ) t cn
v(x, 0) = 0 ⇒ v̂n (0) = 0 =
α2
δ2n
nπ 2
L
(17.14)
−1
cn =
+ cn ⇒
0
−
1
(17.15)
2
α2 ( 2π
−1
L )
n = 2
n=2
2πx
v(x, t) =
e −e
sin
(17.16)
L
α2 L − 1
2 2π 2
2πx
e−t − e−α ( L ) t
sin
.
u(x, t) = x + v(x, t) = x +
2
L
α2 2π − 1
1
2π 2
−t
2
−α2 ( 2π
t
L )
L
106
Example 17.2 A bar with a general external heat source s(x, t)
ut = α2 uxx + s(x, t)
BC: u(0, t) = A
(17.17)
u(L, t) = B
(17.18)
IC: u(x, 0) = f (x, t).
(17.19)
We look for a particular solution: w(x, t) by expanding s(x, t) as a Sine
Series. Note that the sine functions are the eigenfunctions that correspond
to the homogeneous form of the BC in (17.18). Thus if we add w(x, t) to a
solution of (17.17)-(17.18) without the source (i.e. with s(x, t) = 0) we will
not affect the BC.
1. Eigenfunction Expansion:
Let
s(x, t) =
∞
ŝn (t) sin
nπx where
2
ŝn (t) =
L
L
s(x, t) sin
nπx L
0
If we assume
w(x, t) =
∞
(17.20)
L
n=1
ŵn (t) sin
dx.
(17.21)
nπx n=1
(17.22)
L
then
wt =
∞
ŵn (t) sin
n=1
∞
wxx = −
n=1
ŵn
nπx nπ 2
L
(17.23)
L
sin
nπx L
.
(17.24)
Therefore substituting these expansions into wt = α2 wxx + s(x, t) we
obtain:
∞ nπ 2
nπx = 0.
(17.25)
ŵn − ŝn (t) sin
ŵn + α2
L
L
n=1
107
Lecture 21 Distributed, Time Dependent Heat Sources - eigenfunction
expansions
Therefore
ŵn (t) = −α2
nπ 2
L
ŵn (t) + ŝn (t).
(17.26)
α2
nπ 2
L
This is a linear 1st order ODE with integrating factor e
t
.
Therefore
t
wn (t) =
2
2
2 nπ
2 nπ
e−α ( L ) (t − τ )ŝn (τ ) dτ + cn e−α ( L )
t
(17.27)
0
where the cn are arbitrary constants. Since we are only looking for a
particular solution we choose cn ≡ 0.
Therefore
w(x, t) =
∞
n=1
⎞
⎛ t
nπ 2
⎝ e−α2 ( L ) (t−τ ) ŝn (τ ) dτ ⎠ sin nπx .
L
(17.28)
0
2. Now that we have a particular solution we exploit the fact that the
Problem (17.17)-(17.18) is linear and use superposition. Let
u(x, t) = w(x, t) + v(x, t)
(17.29)
2
t +vt = α (w
xx +vxx ) +
s (x, t)
ut = w
2
⇒ vt = α vxx .
(17.30)
(17.31)
A = u(0, t) = w(0, t) + v(0, t) = v(0, t) since
B = u(0, t) = w(L, t) + v(L, t) = v(L, t) since
w(0, t) = 0
w(L, t) = 0.
f (x) = u(x, 0) = w(x, 0) + v(x, 0) ⇒ v(x, 0) = f (x) − w(x, 0) (17.32)
thus v(x, t) satisfies:
⎫
vt = α2 vxx
⎬
v(0, t) = A v(L, t) = B
.
⎭
v(x, 0) = f (x) − w(x, 0)
BC:
IC:
(17.33)
Now the boundary value Problem (2) was solved on pg. 76 of the notes.
Therefore
u(x, t) =
108
B−A
L
x+A+
∞
n=1
⎫
⎧
t
⎬
⎨
2
α2 ( nπ
τ
)
)
L
ŝn (τ ) dx
b + e
⎭
⎩ n
0
nπx sin
(17.34)
L
2
−α2 nπ
t
L
e
(
where
2
bn =
L
L nπx x
f (x) − w(x, 0) − (B − A) + A sin
dx.
(17.35)
L
L
0
109