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MA 2930, April 20, 2011
Worksheet 12
1.
Let’s solve a heat equation problem similar to the one on the prelim in a
stepbystep fashion:
u
t
=
u
xx

2
u
x
(a) Separate the time and space variables and obtain two ODEs. How are
the solutions of the two ODEs connected?
(b) Solve the ODEs.
(c) Write down a fundamental solution of the heat equation.
(d) Apply the spatial boundary conditions:
u
x
(0
,t
) = 0 =
u
(5
,t
)
. (To
which of the two ODEs do these conditions apply?) Write down the non
trivial solutions of the ODEs ﬁrst, then the list of fundamental solutions of
the heat equation, and ﬁnally its general solution. In constructing the general
solution from the list of fundamental solutions what is the principle you are
using? What kind of diﬀerential equations does that principle apply to?
(e) Now apply the temporal initial condition:
u
(
x,
0) = 7
e

x
cos(
πx/
10)
.
What is the resulting solution’s behavior as
t
→ ∞
, or in other words, what
would be the ﬁnal temperature distribution along the bar?
(a)
To separate the variables let
u
(
x,t
) =
X
(
x
)
T
(
t
). Then,
u
t
=
u
xx

2
u
x
XT
0
=
X
00
T

2
X
0
T
XT
0
=
T
(
X
00

2
X
0
)
T
0
T
=
X
00

2
X
0
X
Since the two sides are functions of diﬀerent variables and yet they always
manage to remain equal, they must equal a constant, say

λ
. Equating each
side to

λ
gives us two ODEs:
T
0
+
λT
= 0
and
X
00

2
X
0
+
λX
= 0
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View Full Document Note that the two ODEs  and their solutions  are linked through the com
mon parameter
λ
. For each value of
λ
we get a solution of the
x
ODE and of
the
t
ODE. Only these solutions can be multiplied together to get a solution
u
(
x,t
);
x
 and
t
 solutions for diﬀerent values of
λ
do not go together.
(b)
The solutions of the
t
ODE are clearly
T
λ
(
t
) =
ce

λt
where
c
is any constant.
The solutions of the
x
ODE are constructed from its characteristic roots,
which are the roots of
r
2

2
r
+
λ
= 0. So,
r
=

1
±
√
1

λ
. The (real)
solutions are:
X
λ
(
x
) =
c
1
e
(

1+
√
1

λ
)
x
+
c
2
e
(

1

√
1

λ
)
x
,
if
λ <
1
c
1
e

x
+
c
2
xe

x
,
if
λ
= 1
e

x
(
c
1
cos
√
λ

1
x
+
c
2
sin
√
λ

1
x
)
,
if
λ >
1
Note that solutions depend upon the parameter
λ
.
(c)
A fundamental solution of the heat equation is
u
λ
(
x,t
) =
e

λt
[
c
1
e
(

1+
√
1

λ
)
x
+
c
2
e
(

1

√
1

λ
)
x
]
,
if
λ <
1
e

λt
[
c
1
e

x
+
c
2
xe

x
]
,
if
λ
= 1
e

λt
[
e

x
(
c
1
cos
√
λ

1
x
+
c
2
sin
√
λ

1
x
)]
,
if
λ >
1
(d)
We can apply these boundary conditions to
u
or
X
; it’s equivalent
because, e.g.,
u
x
(0
,t
) =
X
0
(0)
T
(
t
) we want
T
λ
(
t
) to be nontrivial. Applying
them to
X
we get
0 =
X
0
λ
(0) =
c
1
(

1 +
√
1

λ
) +
c
2
(

1

√
1

λ
)
,
if
λ <
1

c
1
+
c
2
,
if
λ
= 1
c
1
+
c
2
√
λ

1
,
if
λ >
1
and
0 =
X
(5) =
c
1
e
5(

1+
√
1

λ
)
+
c
2
e
5(

1

√
1

λ
)
,
if
λ <
1
c
1
e

5
+
c
2
5
e

5
,
if
λ
= 1
e

5
(
c
1
cos 5
√
λ

1 +
c
2
sin 5
√
λ

1)
,
if
λ >
1
Solving these two equations for
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This note was uploaded on 06/10/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 TERRELL,R
 Differential Equations, Equations

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