Solution of the Heat Equation by Separation of Variables
The Problem
Let
u
(
x, t
) denote the temperature at position
x
and time
t
in a long, thin rod of length
ℓ
that runs
from
x
= 0 to
x
=
ℓ
. Assume that the sides of the rod are insulated so that heat energy neither enters nor
leaves the rod through its sides. Also assume that heat energy is neither created nor destroyed (for example
by chemical reactions) in the interior of the rod. Then
u
(
x, t
) obeys the heat equation
∂u
∂ t
(
x, t
) =
α
2
∂
2
u
∂x
2
(
x, t
)
for all 0
< x < ℓ
and
t >
0
(1)
This equation was derived in the notes “The Heat Equation (One Space Dimension)”.
Suppose further that the temperature at the ends of the rod is held fixed at 0.
This information is
encoded in the “boundary conditions”
u
(0
, t
) = 0
for all
t >
0
(2)
u
(
ℓ, t
) = 0
for all
t >
0
(3)
Finally, also assume that we know the temperature throughout the rod time 0.
So there is some given
function
f
(
x
) such that the “initial condition”
u
(
x,
0) =
f
(
x
)
for all 0
< x < ℓ
(4)
is satisfied. The problem is to determine
u
(
x, t
) for all
x
and
t
.
Outline of the Method of Separation of Variables
We are going to solve this problem using the same three steps that we used in solving the wave equation.
Step 1 In the first step, we find all solutions of (1) that are of the special form
u
(
x, t
) =
X
(
x
)
T
(
t
) for
some function
X
(
x
) that depends on
x
but not
t
and some function
T
(
t
) that depends on
t
but
not
x
. Once again, if we find a bunch of solutions
X
i
(
x
)
T
i
(
t
) of this form, then since (1) is a
linear equation,
∑
i
a
i
X
i
(
x
)
T
i
(
t
) is also a solution for any choice of the constants
a
i
.
Step 2 We impose the boundary conditions (2) and (3).
Step 3 We impose the initial condition (4).
The First Step – Finding Factorized Solutions
The factorized function
u
(
x, t
) =
X
(
x
)
T
(
t
) is a solution to the heat equation (1) if and only if
X
(
x
)
T
′
(
t
) =
α
2
X
′′
(
x
)
T
(
t
)
⇐⇒
X
′′
(
x
)
X
(
x
)
=
1
α
2
T
′
(
t
)
T
(
t
)
The left hand side is independent of
t
. The right hand side is independent of
x
. The two sides are equal. So
both sides must be independent of both
x
and
t
and hence equal to some constant, say
σ
. So we have
X
′′
(
x
)
X
(
x
)
=
σ
1
α
2
T
′
(
t
)
T
(
t
)
=
σ
⇐⇒
X
′′
(
x
)
−
σX
(
x
) = 0
T
′
(
t
)
−
α
2
σT
(
t
) = 0
(5)
If
σ
negationslash
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 Spring '10
 YILMAZ
 Thermodynamics, Boundary value problem, Partial differential equation, Boundary conditions

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