Heat Conduction in a Rod with a Non-uniform Thermal Di¤usivity
J. L. Taylor1
Abstract
Let L be the length of a rod and u (x; t) be its temperature for (x; t) 2
[0; L] [0; 1). We assume that the initial and boundary temperatures of the
rod are f (x) and 0 respectively. This heat conduction problem is formulated as
the following …rst initial-boundary value problem:
ut =
1
u(x; 0) = f (x) for 0
uxx for 0 < x < L; t > 0;
x
L; u(0; t) = 0 = u(L; t) for t > 0:
where is a positive function and f is a nonnegative function. In this paper,
we study an approximated solution to the above problem.
1. Introduction
The heat equation can be used to describe the transfer of heat in a given
material. The transfer of heat from one material to another creates problems
in the machines’functions. This is the motivation to study heat conduction in
materials.
Suppose that a rod is placed along the x-axis with x = 0 at the left end of
the rod and x = L at the right end. The heat conduction problem of the rod is
described by (cf. Nagle et al. [3, p. 578]):
c
@u (x; t)
@
=
@t
@x
K0
@u (x; t)
@x
+ Q (x; t)
where c, , K0 , Q, and u are corresponding to speci…c heat, mass density, thermal conductivity, heat source, and temperature respectively. In the following
discussion, we assume that there is no heat source, and K0 is a positive constant.
Then, the above expression becomes
ut =
1
uxx
where = c =K0 and 1= is called the thermal di¤usivity.
We assume that is a twice-di¤erentiable positive function of x. Furthermore, the rod satis…es the homogeneous boundary condition. That is, the temperature of the rod at each end is 0. Also, the rod has an initial temperature
1 This paper is the research work of the independent study in Spring 2006 semester under
the supervision of Prof. W. Y. Chan in the Department of Mathematics at Southeast Missouri
State University.
Email Address: taylorseries1580@yahoo.com
1
f (x) which is a di¤erentiable function. Then, the following …rst initial-boundary
value problem is formulated:
ut =
1
u(x; 0) = f (x) for 0
uxx for 0 < x < L; t > 0;
x
L; u(0; t) = 0 = u(L; t) for t > 0:
(1)
(2)
To solve the problem (1)-(2), the method of separation of variables is used. We
assume that u(x; t) = (x)h(t). Substitute this expression into (1); it leads to
the following two ordinary di¤erential equations:
dh
+ h = 0;
dt
(3)
d2
+
= 0; (0) = 0 = (L);
(4)
dx2
where is a constant. Equation (4) is called Sturm-Liouville eigenvalue problem. Let n and n be the corresponding eigenvalues and eigenfunctions of (4)
respectively. f n g forms an orthogonal set with the weight function (x). That
is,
Z L
(x) n (x) m (x) dx = 0 for n 6= m:
0
To each n , the solution of (3) is hn (t) = e n t . From the result of Haberman
[2, p. 142], the solution of the problem (1)-(2) is given by the following in…nite
series:
1
X
an n (x) e n t
(5)
u (x; t) =
n=1
where
an =
RL
0
(x) f (x) n (x) dx
:
RL
(x) 2n (x) dx
0
(6)
However, the analytic solution to n (x) in (4) may not be obtained for any given
(x). In Section 2, we study the approximated solution to n (x). In Section 3,
we give an example to show the approximated solution of the problem (1)-(2)
with given functions (x) and f (x).
2. Sturm-Liouville eigenvalue problem
The general form of Sturm-Liouville eigenvalue problem (cf. Gustafson [1,
p.175]) is given by
d
d
p
+q + r =0
dx
dx
where p, q, and r are functions of x, and
q = 0, and r = , then it gives
d2
+
dx2
is the eigenvalue. If we let p = K0 ,
= 0:
2
(7)
The Liouville-Green approximation is used to determine the approximated value
of (x) when is large. The following theorem is from Olver [4, pp. 190-191].
The book only provide the outline of the proof. We give the detail of the proof.
Theorem 1. If
1, the approximation for (x) is given by
(x)
A~
1
4
Rx
e
0
(
)1=2 ds
1
4
~
+B
e
Rx
0
(
)1=2 ds
~ are arbitrary constants.
where A~ and B
Proof. Let g (x) =
(x), then (7) becomes
d2
= g (x) :
dx2
Let (x) =
and
Rx
0
g 1=2 (s) ds and
(x) = ( 0 (x))
1=2
(8)
W ( ). Then,
d
= g 1=2
dx
dW 0
d
=
( )
dx
dx
W 0
( )
2
1
2
3
2
00
:
Also,
W d h 0 23 00 i
( )
2 dx
3 0 25 00 2
W
( ) ( ) + ( 0)
2
2
dW d h 0 23 00 i
( )
d dx
d2 W 0 21
dW 0 23 00
d2
=
( )
( )
dx2
dx2
dx
dW 0 23 00
d2 W 0 21
( )
( )
=
2
dx
dx
1
d dW d
d
=
( 0) 2
d
d dx
dx
3W 0 25 00 2 W 0 23
( ) ( )
( )
+
4
2
000
3
:
3
2
000
By
1
d
d2
= g
= g 1=2 ,
dx
dx2
2
1=2 0
g , and
1
d3
=
g
dx3
4
3=2
1
2
(g 0 ) + g
2
1
1
d dW 1
dW
3W
d2
1 1 0
=
g2 g4
g 4
g 2 g +
g
2
dx
d
d
d
2
4
1 3 0 2 1 1 00
W 3
g 4
g 2 (g ) + g 2 g
2
4
2
1 1 dg
d2 W 3
dW
1 3 dg dW
=
g4 +
g 4
g 4
2
d
d
2d
2
dx d
W
3 9 0 2 1 9 0 2 1 5 00
g 4 (g )
g 4 (g ) + g 4 g
2
8
4
2
d2 W 3
1 1 dW dg 1 3 dg dW d
g4 + g 4
g 4
=
d 2
2
d d
2
d d dx
5 9 0 2 1 5 00
W
g 4 (g ) + g 4 g
2
8
2
2
d W 3
W
5 9 0 2 1 5 00
=
g4
g 4 (g ) + g 4 g :
2
d
2
8
2
5
4
1=2 00
1
g
4
g , we have
1
2
(g 0 )
Then (8) becomes,
d2 W 3
g4
d 2
W
2
5
g
8
9
4
1
2
(g 0 ) + g
2
5
4
3
g 00 = g 4 W:
It is equivalent to
d2 W
=W
d 2
Let ' =
g
3=4
d2
g
dx2
1=4
5
g
16
3
. By g =
1
2
(g 0 ) + g
4
2 00
g +1 :
(9)
, we have
5 3 0 2 1
g (g ) + g
16
4
2
00
5 ( 0)
=
:
16 3
4 2
'=
2 00
g
Therefore, (9) becomes
d2 W
= (1 + ') W:
d 2
Now, if
1, then ' is negligible. The approximated equation of (10) is
d2 W
d 2
W:
Therefore, the approximated solution to W is
W
Ae + Be
4
(10)
where A and B are arbitrary constants. By
1=2
( 0 (x))
W , we obtain
(x)
By g =
1
4
Ag
Rx
e
0
g 1=2 ds
(x) =
1
4
+ Bg
Rx
0
e
Rx
e
Rx
0
g 1=2 (s) ds and
g 1=2 ds
(x) =
:
, we get
(x)
where A~ = A (
)
A~
1=4
1
4
Rx
e
0
)1=2 ds
(
~ =B(
and B
1
4
~
+B
1=4
)
0
(
)1=2 ds
;
. The proof is completed.
Adjustment can be made to the Liouville-Green approximation. This adjustment is from Haberman [2, p. 184], and it deals with the envelope function,
~ 1=4 . The idea is to set the envelope function as an in…nite
namely A~ 1=4 or B
series.
Theorem 2. If
1, there exists an in…nite series
E(x) =
1
X
n
2
En (x)
(11)
n=0
where En satis…es
En (x) =
i
2
1
4
Z
1=2
1
4
Rx
1=2
1=4
0
with E0 =
such that E(x)e i
(7).
Rx
Proof. Let ! = i 1=2 0 1=2 (s)ds, then
d!
=i
dx
and
If
d2 !
=i
dx2
(x)
En00
1
2
ds
1
2
1
2
2
(x) dx
is an approximation solution to
;
0
1
2
1
(12)
:
(13)
E (x) e! , then
d
dx
E 0 e! + Ee!
d!
;
dx
d!
+ Ee!
dx
d!
dx
and
d2
dx2
E 00 e! + 2E 0 e!
2
+ Ee!
d2 !
:
dx2
Thus, (7) is approximated by
"
d!
E 00 e! + 2E 0 e!
+ E e!
dx
d!
dx
5
2
#
2
d
!
+ e! 2 +
dx
Ee!
0:
By (12), we have
E 00 e! + 2i
1
2
1
2
E 0 e! + Ee!
+
d2 !
+
dx2
0:
By (13), the above expression becomes
1
2
00
E + 2i
Now, use E(x) =
1
P
n
2
1
2
1
2
i
0
E +
0
1
2
2
E
0:
En (x), we get
n=0
(E000 +
1
+2i 2
1
+i 2
0:
2
1
2
E100 +
1
2 (E 0 +
0
0
(E
1
0+
2
n
1
2 E 00
E200 + ::: +
n
1
0
1
0
2E +
E2 + ::: +
1
1
1
2E +
E2 + ::: +
1
1
n
2
En00 + :::)
n 1
n
0
2 E
2 E 0 + :::)
n
n 1+
n 1
n
2 E
2 E
+
n 1
n + :::)
+
1
9
>
>
>
=
>
>
>
;
(14)
Again, using the fact that
1; we can say that higher powers of will give
the most important terms in our approximation. Equate the coe¢ cient of 1=2
to 0, we have
2
1
2
0
E00 +
1 E0 = 0:
2 2
Integrate the above expression with respect to x, then
1=4
E0 =
In (14), equate the coe¢ cient of
E000 + 2i
0
1
2
to 0, it gives
E10 +
n
Similarly, equate the coe¢ cient of
En00
1 + 2i
:
1
2
1
2
i
0
2
1
2
E1 = 0:
to 0, we obtain
En0 +
i
0
2
1
2
En = 0:
The above expression is equivalent to
2i
1
2
En0 + i
0
1
2
2
En0 +
0
4
En00
En =
En =
Multiplying both side by the integrating factor
d
En
dx
1=4
=
6
i
2
1
iEn00 1
1=4
1=2
1
2
2
1
4
En00
:
, we have
1:
Integrating both side with respect to x,
Z
1
i
4
En =
2
1
4
Similarly, the above proof is true for ! =
completed.
En00
1
2
i
1 dx:
Rx
1
2
0
(t)dt. Thus, the proof is
Note that E (x) is precisely the Liouville-Green approximation if the …rst
term in the in…nite series in (11) is used. We remark that a better approximation
of (x) can be obtained if more terms in the in…nite series of (11) is used.
From the result of Theorem 1,
(x)
A~
1=4 i
e
1=2
Rx
1=2
0
ds
~
+B
1=4
e
i
1=2
Rx
0
1=2
ds
:
We use sine and cosine functions instead of exponential function, it leads to
Z x
Z x
1
1
1
1
1
1
4
2
2
4
2
2 ds)
cos(
ds) + C2
sin(
(x) C1
0
0
where C1 and C2 are arbitrary constants. Substitute x = 0 in the above expression and by the boundary condition in (4), (0) = 0, it gives
1
4
C1
=0
which implies C1 = 0. Therefore,
(x)
C2
1
4
1
2
sin(
Z
x
1
2
ds):
0
Substitute x = L in the above expression and by another boundary condition
in (4), (L) = 0, we have
1
4
C2
sin(
Z
1
2
L
1
2
ds) = 0:
0
As C2 6= 0, otherwise
(x)
1
2
Z
0, we obtain
L
1
2
ds = n for n = 1; 2; 3::::
(15)
0
This expression is equivalent to
n
=
RL
n
0
Therefore,
n (x)
C2n
1
4
1
2
sin(
ds
1
2
n
!2
Z
0
7
:
x
(16)
1
2
ds):
We choose C2n such that f
Z
ng
forms an orthonormal set, that is,
L
0 if n 6= m
:
1 if n = m
dx =
n m
0
Thus, if n = m, then
2
(C2n )
Z
L
1
2
Z
1
2
sin2
n
0
Let u =
1=2 R x
n
0
1=2
x
1
2
ds dx
1:
0
ds, then du =
Z n
2
(C2n )
1=2 1=2
dx.
n
2
n
2
n
1
2
By (15), we have
sin2 udu
1:
0
It implies
1
2
(C2n )
Rn
sin udu
0
=
n
2
1
2
:
n
From (16), we obtain
2
(C2n )
RL
0
2
1
2
:
ds
This gives our approximation to the solution of (7)
n (x)
RL
2
1
2
0
ds
! 12
1
4
sin n
Rx
0
RL
0
1
2
ds
1
2
ds
!
:
(17)
3. An Example
In this section, we let L = 1. We want to look at the behavior of the
approximated solution to the problem (1)-(2) for some (x) and f (x).
Let (x) = 1 + 50x and f (x) = x (1 x). From (4),
d2
+ (1 + 50x)
dx2
= 0;
(0) = 0 = (1):
Using (16), we have
n
We note that f
1
n gn=1
75n
513=2 1
=
2
:
is an increasing sequence. That is,
1
<
2
< ::: <
8
n
< ::::
(18)
From (17),
1=2
150
n (x)
513=2
(1 + 50x)
1
Let n = 10, then
10
=
513=2
1
sin
75n
513=2 1
Z
x
1
(1 + 50s) 2 ds : (19)
0
2
and
1=2
150
10 (x)
750
513=2 1
1
4
(1 + 50x)
1
4
750
513=2
sin
1
Z
x
1
(1 + 50s) 2 ds :
0
The graph of this approximated eigenfunction is:
y
0.5
0.25
0
0
0.25
0.5
0.75
1
x
-0.25
-0.5
In the above graph, there are exactly 9 roots in the interval (0; 1). This matches
to the result of the following theorem (cf. Haberman [2, p. 135]).
Theorem 3. The eigenfunction n (x) has exactly n 1 roots in the interval
0 < x < L.
Substitute (18) and (19) into (5), we get the approximated solution to the
problem (1)-(2)
u (x; t)
1
X
an
150
513=2 1
1=2
(1 + 50x)
1
4
75n
513=2 1
sin
n=1
Z
x
1
(1 + 50s) 2 ds e
75n
513=2
2
1
0
Rewrite the above expression, it yields
u (x; t)
1
X
n=1
a
~n (1 + 50x)
1
4
sin
75n
513=2 1
9
Z
0
x
1
(1 + 50s) 2 ds e
75n
513=2
2
1
t
t
:
where a
~n = an
When t = 0,
1
X
1=2
150
513=2 1
. To solve for a
~n , we use the initial condition of (2).
1
4
a
~n (1 + 50x)
n=1
sin
75n
513=2 1
Z
x
1
(1 + 50s) 2 ds
u (x; 0)
0
= f (x)
= x (1 x) :
From (6), a
~n is given by
R1
1
(1 + 50x) x (1 x) (1 + 50x) 4 sin
0
a
~n
h
R1
1
(1 + 50x) (1 + 50x) 4 sin 5175n
3=2
0
75n
513=2 1
Simplify the above expression, we have
R1
0
a
~n
3
(1 + 50x) 4 x (1
R1
0
1
2
x) sin
2
(1 + 50x) sin
Rx
1
(1 + 50s) 2 ds dx
:
i2
Rx
1
2
dx
(1 + 50s) ds
1 0
0
[(1+50x)3=2
1]n
513=2 1
[(1+50x)3=2
1]n
513=2 1
dx
:
dx
Since the analytic solution of the right-hand side of the above expression cannot
be determined, we use numerical method to compute the approximated solution.
Adaptive Gaussian Quadrature is used to compute the value of the numerator
and denominator, this numerical integration method is built in Maple R 2 version
9.03. We use the …nite sum
Z x
100
2
X
75n
1
1
75n
t
u
~ (x; t) =
a
~n (1 + 50x) 4 sin
(1 + 50s) 2 ds e 513=2 1
3=2
51
1
0
n=1
to approximate u (x; t). The table below shows that the values of f (x) =
x (1 x) and u
~ (x; 0) are close to each other at x = 0; 0:1; 0:2; 0:3; :::; 1:
x
0
0:1
0:2
0:3
0:4
0:5
0:6
0:7
0:8
0:9
1
2 Maple R
f (x)
0
0:09
0:16
0:21
0:24
0:25
0:24
0:21
0:16
0:09
0
u
~(x; 0)
0
0:08997841586
0:1599906913
0:2100016569
0:2399981029
0:2499991097
0:2400001991
0:2100001255
0:1599993234
0:08999867687
9:377258464 10 12
is a registered trademark of Waterloo Maple Inc., Waterloo, Ontario, Canada
10
The graphs of u
~ (x; t) when t = 0, 0:5, and 2 are:
t=0
t = 0:5
t=2
The graphs above show that u
~ (x; t) skews to the right-hand side when t is
increasing. For each x 2 [0; 1], u
~ (x; t) is a non-increasing function of t.
If (x) = 1 and f (x) = x (1 x), the exact solution to problem (1)-(2) is
1 Z 1
X
2
u (x; t) = 2
x (1 x) sin (n x) dx sin (n x) e (n ) t
(20)
n=1
0
(cf. Haberman [2, pp. 42-43]). We use the …nite sum
100 Z 1
X
u
^ (x; t) = 2
x (1 x) sin (n x) dx sin (n x) e
n=1
(n )2 t
0
to approximate the solution in (20). Use Maple R to graph u
^ (x; t) for t = 0;
0:5; and 2. The graphs are:
t=0
t = 0:5
11
t=2
The graphs above show that u
^ (x; t) is symmetric with respect to the line
x = 0:5 for all t. For each x 2 [0; 1], u
^ (x; t) is a non-increasing function of
t. Compare the graphs of u
~ (x; t) and u
^ (x; t), the rate of decrease of u
^ (x; t) is
faster than u
~ (x; t).
References
[1] K. Gustafson, Introduction to Partial Di¤ erential Equations and Hilbert
Space Methods, Mineola, NY, Dover, 1999, p. 175.
[2] R. Haberman, Elementary Applied Partial Di¤ erential Equations, Englewood Cli¤s, NJ, Prentice-Hall, 1987, pp. 42-43, 135, 142, and 184.
[3] R. Nagle, E. Sa¤, and A. Snider, Fundamentals of Di¤ erential Equations
and Boundary Value Problems, New York, NY, Addison Wesley, 2004, p.
578.
[4] F. Olver, Asymptotics and Special Functions, New York, NY, Academic
Press, 1974, pp. 190-191.
12