Walter VanCleave
Homework #2, Prob 2-8
February 3, 2000
3-D Dimensional Homogeneous Equations
Heat Equation:
 2T  2T  2T 1 T



x 2 y 2 z 2  t
B.C.’s
T 0
T 0
T 0
I.C.
T ( x, y, z )  T0
at x = 0
at y = 0
at z = 0
for t = 0
at x > 0,
y > 0,
z>0
t>0
t>0
t>0
t>0
in region
Since this is a multidimensional homogeneous heat conduction problem. The Product
Solution Method can be used to solve. The 3-D problem can be seperated into three 1-D
homogeneous problems, one for each cordinate directions as shown below.
X – Direction
Heat Equation
 2T1 1 T1

x 2  t
B.C.
T1  0
I.C.
T1 ( x )  T0
Y-Direction
Heat Equation
 2T2 1 T2

y 2  t
B.C.
T2  0
I.C.
T2 ( x )  T0
at x > 0
t>0
at x = 0
t>0
for t = 0
at y > 0
at y = 0
for t = 0
in region 0 < x < 
t>0
t>0
in region 0 < y < 
Z-Direction
Heat Equation
 2T3 1 T3

z 2  t
B.C.
T3  0
I.C.
T3 ( x )  T0
at z > 0
t>0
at z = 0
t>0
for t = 0
in region 0 < z < 
The boundary condition for each of the three 1-D problems are the same. This allows the
solution of the three dimensional problem to be written as the product of the three one
dimensional problems above.
T ( x, y, z, t )  T1 ( x, t )  T2 ( y, t )  T3 ( z, t )
Seperation of variables is used to solve each of the 1-D problems. As show below for the
x-direction each direction can be solved.
Assume a seperation
T ( x , t )  X ( x ) ( t )
Substituting into 1-D Heat Equation and setting equal to a constant()
1  2 X ( x)
1 (T )

  2
2
X ( x ) x
(t ) t
The solution for the time dependent function (t)
(t )  e  t
2
The solution for the space dependent function X(x)
X (  , x )  (sin x )
Through superposition of time and space functions
T ( x, t )  

 0
c(  )e  t (sin x )d
2
I.C’s applied

1
c(  ) 
X (  , x ) F ( x )dx 
N (  ) 0
The General solution for the 1-D problem


2
1
T ( x, t )   e  t
X (  , x )  X (  , x ) F ( x )dx d
 0
x  0
N ( )
Table 2-3 shows that B.C.’s are
X (  , x)  sin x,
1
2

N ( ) 
Substituting into the general solution

2
2 
T ( x, t )   F ( x )  e  t (sin x )(sin x )ddx 
 0
 x  0
Integrating in Beta
T ( x, t ) 
1
4t
 ( x  x )2 
   ( x  x)2 



4t 
4t 


e
dx 

F
(
x
)

e
x0





For this constant initial temperature
 ( x  x ) 2 
 ( x  x ) 2 




   4t 
T ( x, t )
1     4t 



e
dx    e
dx 
x0
T0

4t  x0


This can be reduced to
2
T ( x, t )
2 x / 4t 

e d
T0
 0
The solutions to the 1-D problems above take the froms shown here
x
T1 ( x, t )  erf (
)
4t
y
T2 ( y, t )  erf (
)
4t
z
T3 ( z, t )  erf (
)
4t
Therfore using the product solution the answer to the three dimensional homogeneous
problem is:
x
y
z
T1 ( x, y, z, t )  T0 erf (
)erf (
)erf (
)
4t
4t
4t