MATH20411
Problem sheet 6: Separation of Variables
1.
a)
u
tt

u
xx
= 0 with
u
(0
,t
) =
u
(ˆL
,t
) = 0,
u
(
x,
0) =
f
(
x
) and
u
t
(
x,
0) =
g
(
x
):
•
Assuming
u
=
X
(
x
)
T
(
t
) we can use the PDE with
u
xx
=
X
00
(
x
)
T
(
t
) and
u
tt
=
X
(
x
)
T
00
(
t
)
to ﬁnd
u
tt

u
xx
=
XT
00

TX
00
= 0. Dividing by
XT
(assumed not equal to zero) gives
T
00
T

X
00
X
=
λ
. Thus we have
X
00

λX
= 0 and
T
00

λT
= 0 with
λ
constant because
X
00
X
is independent of
t
and
T
00
T
is independent of
x
. The homogeneous boundary conditions
give
u
(0
,t
) =
X
(0)
T
(
t
) = 0,
u
(
L,t
) =
X
(
L
)
T
(
t
) = 0
.
Since we seek
XT
6
= 0 this gives
X
(0) =
X
(
L
) = 0
.
•
Solving
X
00

λX
= 0 with
X
(0) =
X
(
L
) = 0:
λ
= 0
:
we have
X
00
= 0 giving
X
=
A
+
Bx
.
X
(0) =
X
(
L
) = 0 give
A
= 0 and
BL
= 0 so that
A
=
B
= 0.
Only the trivial solution arises for
λ
= 0.
λ
=
ω
2
>
0
:
we have
X
00

ω
2
X
= 0 giving
X
=
Ae
ωx
+
Be

ωx
.
X
(0) =
X
(
L
) = 0 gives
A
+
B
= 0 and
Ae
ωL
+
Be

ωL
= 0,
i.e. (substituting)
A
(
e
ωL

e

ωL
) = 0 so that
A
=
B
= 0 since
e
ωL

e

ωL
6
= 0
.
Only trivial solutions arise for
λ >
0.
λ
=

ω
2
<
0
:
we have
X
00
+
ω
2
X
= 0 giving
X
=
A
cos(
ωx
) +
B
sin(
ωx
).
X
(0) =
X
(
L
) = 0 give
A
= 0 and
B
sin(
ωL
) = 0.
Thus we can have
B
6
= 0 if and only if sin(
ωL
) = 0
or
ωL
=
nπ
for any
n
= 1, 2, 3, .
...
Hence the only eigenvalues are
λ
=
λ
n
=

(
nπ
L
)
2
for
n
∈
N
with corresponding eigenfunc
tions
X
=
X
n
= sin(
nπx
L
).
•
Solving
T
00

c
2
λT
= 0 :
For any
λ
=

(
nπ
L
)
2
we have
T
00
+
(
nπ
L
)
2
T
= 0. Thus
T
=
A
cos
(
nπt
L
)
+
B
sin
(
nπt
L
)
.
•
Since the solutions
u
=
X
n
(
x
)
T
n
(
t
) all satisfy a homogeneous PDE with homogeneous
boundary conditions, the principle of superposition means that any linear combination of
such solutions is also a solution. Thus a convergent sum:
u
=
∞
X
n
=1
A
n
X
n
T
n
=
∞
X
n
=1
±
A
n
cos
±
nπt
L
¶
+
B
n
sin
±
nπt
L
¶¶
sin
‡
nπx
L
·
is also a solution, for constants
A
n
and
B
n
.
•
At
t
= 0 we have
u
(
x,
0) =
f
(
x
) and
u
t
(
x,
0) =
g
(
x
) so that:
f
(
x
) =
∞
X
n
=1
A
n
X
n
T
n
(0) =
∞
X
n
=1
A
n
sin
‡
nπx
L
·
g
(
x
) =
∞
X
n
=1
±

A
n
‡
nπ
L
·
sin
±
nπ
0
L
¶
+
B
n
‡
nπ
L
·
cos
±
nπ
0
L
¶¶
sin
‡
nπx
L
·
=
∞
X
n
=1
B
n
‡
nπ
L
·
sin
‡
nπx
L
·
Since the eigenfunctions
X
n
= sin
(
nπx
L
)
are orthogonal under the inner product
(
f,g
) =
R
L
0
f
(
x
)
g
(
x
)
dx
we have:
A
n
R
L
0
sin
2
(
πnx
L
)
dx
=
R
L
0
f
(
x
) sin
(
πnx
L
)
dx
.
Integrating:
R
L
0
sin
2
(
πnx
L
)
dx
=
R
L
0
1
2
(1

cos
(
2
πnx
L
)
)
dx
=
L
2

[
L
2
πn
sin
(
2
πnx
L
)
]
L
0
=
L
2
Thus
A
n
=
2
L
R
L
0
f
(
x
)sin
(
πnx
L
)
, n
= 1
,
2
,....
Similarly, for
B
n
,
we obtain the expression,
B
n
=
2
L
(
L
nπ
)R
L
0
g
(
x
)sin
(
πnx
L
)
=
2
nπ
R
L
0
g
(
x
)sin
(
πnx
L
)
n
= 1
,
2
,....
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 Spring '09
 MarniMishna
 Sin, Tn, Boundary value problem, λ, Boundary conditions

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