2. INTRODUCTION
The basic partial differential equations of heat conduction, wave
propagation, and potential theory that we discuss in this chapter are
associated with three distinct types of physical phenomena: diffusive
processes, oscillatory processes, and time-independent or steady processes.
Consequently, they are of fundamental importance in many branches of
physics, and are significant mathematically.
The partial differential equations whose theory is best developed and
whose applications are most significant and varied are the linear
equations of second order.
All such equations can be classified as one of three types:
The heat equation, the wave equation, and the potential equation, are
prototypes of these categories.
3. HEAT CONDUCTION IN A ROD: ASSUMPTIONS
Consider a heat equation conduction problem for a straight bar of
uniform cross section and homogeneous material.
Let the x-axis be chosen to lie along the axis of the bar, and let x
= 0 and x = L denote the ends of the bar. See figure below.
Suppose that the sides of the bar are perfectly insulated so that
no heat passes through them.
Assume the cross-sectional dimensions are so small that the
temperature u can be considered constant on the cross sections.
Then u is a function only of the axial coordinate x and time t.
4. HEAT CONDUCTION:
INITIAL AND BOUNDARY CONDITIONS
In addition, we assume that the initial temperature
distribution in the bar is given, and hence
where f is a given function.
Finally, we assume that the ends of the bar are held at
fixed temperatures: the temperature T1 at x = 0 and T2 at
x = L.
However, as will be shown in Section 10.6, we need only
consider T1 = T2 = 0.
Thus we have the boundary conditions
L
x
x
f
x
u
0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
t
t
L
u
t
u
5. HEAT CONDUCTION PROBLEM
Thus the fundamental problem of heat conduction is to find u(x,t)
satisfying
With respect to the time variable t, this is an initial value problem;
an initial condition is given and the differential equation governs
what happens later.
With respect to the spatial variable x, it is a boundary value
problem; boundary conditions are imposed at each end of the bar
and the differential equation describes the evolution of the
temperature in the interval between them.
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx
0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2
6. HEAT CONDUCTION: BOUNDARY
PROBLEM
Alternatively, we can consider the problem as a boundary
value problem in the xt-plane, see figure below.
The solution u(x,t) satisfying the heat conduction problem
is sought in the semi-definite strip 0 < x < L, t > 0, subject
to the requirement that u(x,t) must assume a prescribed
value at each point on the boundary of this strip.
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx
0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2
7. HEAT CONDUCTION:
LINEAR HOMOGENEOUS EQUATION
The heat conduction problem
is linear since u appears only to the first power
throughout.
The differential equation and boundary conditions are
also homogeneous.
This suggests that we might approach the problem by
seeking solutions of the differential equation and
boundary conditions, and then superposing them to
satisfy the initial condition.
We next describe how this plan can be implemented.
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx
0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2
8. SEPARATION OF VARIABLES
METHOD
Our goal is to find nontrivial solutions to the differential
equation and boundary conditions.
We begin by assuming that the solution u(x,t) has the
form
Substituting this into our differential equation
we obtain
or
)
(
)
(
)
,
( t
T
x
X
t
x
u
t
xx u
u
2
T
X
T
X
2
T
T
X
X
2
1
9. ORDINARY DIFFERENTIAL EQUATIONS
We have
Note left side depends only on x and right side only on t.
Thus in order for this equation to be valid for 0 < x < L, t >
0, it is necessary for both sides of this equation to equal
the same constant, call it -. Then
Thus the partial differential equation is replaced by two
ordinary differential equations.
T
T
X
X
2
1
0
0
1
2
2
T
T
X
X
T
T
X
X
10. BOUNDARY CONDITIONS
Recall our original problem
Substituting u(x,t) = X(x)T(t) into boundary condition at x
= 0,
Since we are interested in nontrivial solutions, we
require X(0) = 0 instead of T(t) = 0 for t > 0. Similarly,
X(L) = 0.
We therefore have the following boundary value problem
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx
0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2
0
)
(
)
0
(
)
,
0
(
t
T
X
t
u
0
)
(
)
0
(
,
0
L
X
X
X
X
11. EIGENVALUES AND EIGEN FUNCTIONS
Thus
From Section 10.1, the only nontrivial solutions to this
boundary value problem are the eigen functions
associated with the eigen values
With these values for , the solution to the first order
equation
is
0
)
(
)
0
(
,
0
L
X
X
X
X
,
3
,
2
,
1
,
/
sin
)
(
n
L
x
n
x
Xn
,
3
,
2
,
1
,
/ 2
2
2
n
L
n
n
0
2
T
T
constant.
,
2
)
/
(
n
t
L
n
n
n k
e
k
T
12. FUNDAMENTAL SOLUTIONS
Thus our fundamental solutions have the form
where we neglect arbitrary constants of proportionality.
The functions un are sometimes called fundamental solutions
of the heat conduction problem.
It remains only to satisfy the initial condition
Recall that we have often solved initial value problems by
forming linear combinations of fundamental solutions and then
choosing the coefficients to satisfy the initial conditions.
Here, we have infinitely many fundamental solutions.
,
,
3
,
2
,
1
,
/
sin
)
,
(
2
)
/
(
n
L
x
n
e
t
x
u t
L
n
n
L
x
x
f
x
u
0
),
(
)
0
,
(
13. FOURIER COEFFICIENTS
Our fundamental solutions are
Recall the initial condition
We therefore assume that
where the cn are chosen so that the initial condition is
satisfied:
Thus we choose the coefficients cn for a Fourier sine
series.
,
,
3
,
2
,
1
,
/
sin
)
,
(
2
)
/
(
n
L
x
n
e
t
x
u t
L
n
n
L
x
x
f
x
un
0
),
(
)
0
,
(
1
)
/
(
1
/
sin
)
,
(
)
,
(
2
n
t
L
n
n
n
n
n L
x
n
e
c
t
x
u
c
t
x
u
1
/
sin
)
(
)
0
,
(
n
n L
x
n
c
x
f
x
u
14. SOLUTION
Therefore the solution to the heat conduction problem
is given by
where
1
)
/
(
/
sin
)
,
(
2
n
t
L
n
n L
x
n
e
c
t
x
u
L
n dx
L
x
n
x
f
L
c
0
/
sin
)
(
2
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx
0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2
15. EXAMPLE 1: HEAT
CONDUCTION PROBLEM
Find the temperature u(x,t) at any time in a metal rod 50
cm long, insulated on the sides, which initially has a
uniform temperature of 20° C throughout and whose ends
are maintained at 0° C for all t > 0.
This heat conduction problem has the form
50
0
,
20
)
0
,
(
0
,
0
)
,
50
(
,
0
)
,
0
(
0
,
50
0
,
2
x
x
u
t
t
u
t
u
t
x
u
u t
xx
16. EXAMPLE 1: SOLUTION
The solution to our heat conduction problem is
where
Thus
1
)
50
/
(
50
/
sin
)
,
(
2
n
t
n
n x
n
e
c
t
x
u
even
,
0
odd
,
/
80
cos
1
40
50
/
sin
5
4
50
/
sin
20
50
2
/
sin
)
(
2
50
0
50
0
0
n
n
n
n
n
dx
x
n
dx
x
n
dx
L
x
n
x
f
L
c
L
n
17. EXAMPLE 1: RAPID
CONVERGENCE
Thus the temperature along the rod is given by
The negative exponential factor in each term cause the
series to converge rapidly, except for small values of t or
2
.
Therefore accurate results can usually be obtained by
using only a few terms of the series.
In order to display quantitative results, let t be measured
in seconds; then 2
has the units cm2
/sec.
If we choose 2
= 1 for convenience, then the rod is of a
material whose properties are somewhere between copper
and aluminum.