1
c Anthony Peirce.
Introductory lecture notes on Partial Differential Equations - °
Not to be copied, used, or revised without explicit written permission from the copyright owner.
Lecture 19: Heat conduction with distributed sources/sinks
(Compiled 3 March 2014)
In this lecture we consider heat conduction problems in which there is a distributed source or sink function s(x, t) that
applies throughout the domain. We first consider the case in which the source is time-independent, i.e., s(x, t) = s(x). In
this case the effect of the source can be dealt with entirely by determining an appropriate steady-state solution. Using
this particular solution, we can reduce the problem to one of the standard homogeneous boundary value problems we
encountered when we introduced separation of variables. Secondly, we consider a fully time-dependent source. In this
case we have to resort to the method of eigenfunction expansions.
Key Concepts: Distributed sources or sinks, Particular Solutions, Steady state Solutions; Separation
of variables, Eigenvalues and Eigenfunctions, Method of Eigenfunction Expansions.
19 Heat Conduction Problems with distributed sources
19.1 Heat Conduction Problems with distributed time-independent sources
Example 19.1 A bar with an external heat source s(x) = x.
ut = α2 uxx + x
BC: u(0, t) = 0
u(L, t) = B
IC: u(x, 0) = g(x).
0<x<L
(19.1)
(19.2)
(19.3)
Figure 1. Bar subject to a time-independent heat source distributed along its length with inhomogeneous Dirichlet BC
2
Steady state problem ut = 0:
0 = α2 u00∞ + x
u∞ (0) = 0 u∞ (L) = B
(19.4)
x
x2
x3
u0∞ = − 2 + a u∞ = − 2 + ax + b
2
α
2α
6α
L3
L2
B
u∞ (0) = b = 0 u∞ (L) = − 2 + aL = B ⇒ a =
+ 2
6α
L
6α
(19.5)
u00∞ = −
Therefore
u∞ (x) = −
x3
+
6α2
µ
L2
B
+ 2
L
6α
¶
½
x=x
¾
1
B
+ 2 (L2 − x2 ) .
L
6α
(19.6)
Let u(x, t) = u∞ (x) + v(x, t).
ut = α2 uxx + x
u(0, t) = 0
u(L, t) = B
u(x, 0) = g(x)
⇒ (u∞% +v)t = α2 (u↓∞ +v)xx + x
↓
⇒ u∞% (0) + v(0, t) = 0
⇒ u∞ (L) + v(L, t) = B
⇒ u∞ (x) + v(x, 0) = g(x)
⇒ vt = α2 vxx
⇒ v(0, t) = 0
⇒ v(L, t) = 0
⇒ v(x, 0) = g(x) − u∞ (x).
(19.7)
Separation of variables yields:
v(x, t) =
∞
X
bn e−(
nπ
L
2
)
α2 t
sin
³ nπx ´
(19.8)
L
n=1
where
2
bn =
L
Therefore
½
u(x, t) = x
ZL
{g(x) − u∞ (x)} sin
0
³ nπx ´
L
dx.
(19.9)
¾ X
∞
³ nπx ´
2 nπ 2
B
1
+ 2 (L2 − x2 ) +
bn e−α ( L ) t sin
L
6α
L
n=1
↑
↑
steady
Note:
½
lim u(x, t) = x
x→∞
(19.10)
transient
¾
B
1
2
2
+ 2 (L − x ) .
L
6α
(19.11)
19.2 Distributed time-dependent heat sources/sinks - eigenfunction expansions
Example 19.2 A Bar with a Time-Varying External Heat Source:
µ
¶
2πx
2
−t
ut = α uxx + e sin
0 < x < L,
L
BC: u(0, t) = 0; u(L, t) = L
t>0
IC: u(x, 0) = x.
Consider the function w(x) = x which satisfies the BC as well as the homogeneous version of the PDE.
(19.12)
(19.13)
(19.14)
Further Heat Conduction Problems
Now let u(x, t) = w(x) + v(x, t).
µ
¶
2πx
ut = (w
% +v)t = α2 (w
% +v)xx + e−t sin
L
¶
µ
2πx
⇒ vt = α2 vxx + e−t sin
L
u(0, t) = w(0)
% +v(0, t) = 0 ⇒ v(0, t) = 0
u(L, t) = w(L)
% +v(L, t) = %⇒
L v(L, t) = 0
x = u(x, 0) = w(x) + v(x, 0) = x + v(x, 0) ⇒ v(x, 0) = 0.
Now assume that v(x, t) =
∞
X
v̂n (t) sin
³ nπx ´
L
n=1
vt − α2 vxx =
∞ ½
X
dv̂n
n=1
(19.15)
(19.16)
(19.17)
(19.18)
(19.19)
.
½ ³ ´
∞
∞
³ nπx ´ ∂ 2 v
³ nπx ´¾
X
X
∂v
dv̂n
nπ 2
=
(t) sin
sin
.
=
v̂n (t) −
∂t
dt
L
∂x2
L
L
n=1
n=1
Therefore
3
dt
+ α2
³ nπ ´2
L
¾
³ nπx ´
v̂n − e−t δ2n sin
= 0.
L
(19.20)
(19.21)
Therefore
³ nπ ´2
dv̂n
+ α2
v̂n = e−t δ2n
dt
L
i
h
i
2
2
d h α2 ( nπ
α2 ( nπ
t
)
L ) −1 t
L
e
δ2n .
v̂n = e
dt
Therefore
2
cn arbitrary
2 nπ 2
e−t δ2n
+ e−α ( L ) t cn
¡ nπ ¢2
α2 L − 1
δ2n
v(x, 0) = 0 ⇒ v̂n (0) = 0 = ¡ ¢2
+ cn ⇒
nπ
2
α L −1
(
0
n 6= 2
cn =
1
− 2 2π 2
n=2
α ( L ) −1
¶
µ
n
o
2 2π 2
1
2πx
e−t − e−α ( L ) t sin
v(x, t) = ¡ ¢2
2π
2
L
α L −1
!
Ã
2
2 2π
µ
¶
2πx
e−t − e−α ( L ) t
sin
u(x, t) = x + v(x, t) = x +
.
¡ ¢2
L
α2 2π
−1
L
v̂n (t) =
(19.23)
h
i
2
α2 ( nπ
L ) −1 t
nπ
e
e−α ( L ) t v̂n =
δ2n + cn
¡ ¢2
α2 nπ
−1
L
2
(19.22)
(19.24)
(19.25)
(19.26)
(19.27)
4
Example 19.3 A bar with a general external heat source s(x, t)
ut = α2 uxx + s(x, t)
BC: u(0, t) = A
(19.28)
u(L, t) = B
(19.29)
IC: u(x, 0) = f (x).
(19.30)
Figure 2. Bar subject to a time dependent heat source distributed along its length with inhomogeneous Dirichlet BC
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 (19.29). Thus if we add w(x, t) to a solution
of (19.28)-(19.29) without the source (i.e. with s(x, t) = 0) we will not affect the BC.
(1) Eigenfunction Expansion:
Let
s(x, t) =
∞
X
ŝn (t) sin
³ nπx ´
(19.31)
L
n=1
where
2
ŝn (t) =
L
ZL
s(x, t) sin
³ nπx ´
L
0
dx.
(19.32)
If we assume
w(x, t) =
∞
X
ŵn (t) sin
³ nπx ´
(19.33)
L
n=1
then
wt =
∞
X
ŵn0 (t) sin
n=1
∞
X
wxx = −
n=1
ŵn
³ nπx ´
³ nπ ´2
L
(19.34)
L
sin
³ nπx ´
L
.
(19.35)
Further Heat Conduction Problems
5
Therefore substituting these expansions into wt = α2 wxx + s(x, t) we obtain:
¾
∞ ½
³ nπx ´
³ ´2
X
0
2 nπ
ŵn − ŝn (t) sin
= 0.
ŵn + α
L
L
n=1
Therefore
ŵn0 (t) = −α2
³ nπ ´2
L
α2
(19.36)
ŵn (t) + ŝn (t).
µ nπ ¶2
This is a linear 1st order ODE with integrating factor e
L
t
(19.37)
.
Therefore
Zt
wn (t) =
2
2 nπ
2 nπ
e−α ( L ) (t − τ )ŝn (τ ) dτ + cn e−α ( L )
2
t
(19.38)
0
where the cn are arbitrary constants. Since we are only looking for a particular solution we choose cn ≡ 0.
Therefore

 t
Z
³
´
2 nπ 2
 e−α ( L ) (t−τ ) ŝn (τ ) dτ  sin nπx .
w(x, t) =
L
n=1
∞
X
(19.39)
0
(2) Now that we have a particular solution we exploit the fact that the Problem (19.28)-(19.29) is linear and use
superposition. Let
u(x, t) = w(x, t) + v(x, t)
(19.40)
ut = w
%t +vt = α2 (w
%xx +vxx ) +%
s (x, t)
(19.41)
2
⇒ vt = α vxx .
A
B
= u(0, t)
= u(0, t)
(19.42)
= w(0, t) + v(0, t) = v(0, t) since
= 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)
(19.43)
thus v(x, t) satisfies:
BC:
IC:

vt = α2 vxx

v(0, t) = A v(L, t) = B

v(x, 0) = f (x) − w(x, 0)
(19.44)
Now the boundary value problem(19.44) was solved in lecture 19 of the notes.
Therefore
µ
u(x, t) =
B−A
L
¶
x+A+
∞
X
n=1
e−α (
2
nπ
L


Zt


³
´
2 nπ 2
) t b + eα ( L ) τ ŝ (τ ) dx sin nπx
n
 n

L
2
(19.45)
0
where
2
bn =
L
ZL n
h
io
³ nπx ´
x
f (x) − w(x, 0) − (B − A) + A sin
dx.
L
L
0
(19.46)
6
Example 19.4 A bar with an external heat source dependent on x and t: S(x, t) = xt
ut = α2 uxx + xt
u(0, t) = A
u(L, t) = B
u(x, t) = f (x)
Let q(x) =
¡ B−A ¢
L
x + A and u(x, t) = q(x) + v(x, t) then
vt = α2 vxx + xt
v(0, t) = 0 = v(L, t)
v(x, 0) = f (x) − q(x).
Expanding the source S(x, t) in terms of the eigenfunctions of the problem with homogeneous boundary conditions
S(x, t) = xt =
∞
X
Ŝn (t) sin(λn x)
n=1
2
Ŝn (t) =
L
ZL
xt sin
³ nπx ´
L
0
dx


¡
¢ ¯¯L
µ ¶ ZL
³ nπx ´ 
2t  cos nπx
L
¯
−x ¡ nπL¢ ¯ +
=
cos
dx

¯
L
nπ
L
L
%
0
0
½
µ 2 ¶¾ µ ¶
2t
L
2L
Ŝn (t) =
(−1)n+1
=
(−1)n+1 t
L
nπ
nπ
0 = vt − α2 vxx − xt =
∞ n
o
X
v̂˙ n + α2 λ2n v̂n − Ŝn (t) sin xλn x?
n=1
µ
v̂˙ n + α2 λ2n v̂n =
³
α2 λ2n t
e
´
µ
v̂n =
2L
nπ
2L
nπ
¶
(−1)n+1 t
"
¶
n+1
(−1)
¶
"
ZL ·
½µ
2
2
2
2
teα λn t
eα λn t
−
α2 λ2n
(α2 λ2n )2
#t
+ cn
0
#
α2 λ2n t
α2 λ2n t
te
(e
−
1)
(−1)n+1
=
−
+ cn
α2 λ2n
α4 λ4n
"
#
µ ¶
2 2
−α2 λ2n t
2 2
−1
2L
n t(α λn ) + e
(−1)
+ cn e−α λn t
v̂n (t) =
2
2
2
nπ
(α λn )
(µ ¶
"
#
)
∞
2 2
−α2 λ2n t
X
2 2
2L
t(α
λ
)
−
1
+
e
n
v(x, t) =
(−1)n
+ cn e−α λn t sin(λn x)
nπ
(α4 λ4n )
n=1
¶
∞ µ
X
2L
(−1)n 0 + cn sin(λn x)
f (x) − q(x) = v(x, 0) =
nπ
n=1
µ
2
cn =
L
2L
nπ
f (x) −
0
B−A
L
¶
¾¸
x+A
sin(λn x) dx