Always start with
X
(
x
)
, do
NOT
touch
T
(
t
)
until right at the end!
Now use the
boundary conditions
in (5.1):
u
(0
, t
) =
X
(0)
T
(
t
) = 0
⇒
X
(0)
T
(
t
) = 0
⇒
X
(0) = 0
u
(1
, t
) =
X
(1)
T
(
t
) = 0
⇒
X
(1)
T
(
t
) = 0
⇒
X
(1) = 0
Hence we get:
X
00
(
x
) =
λX
(
x
)
X
(0) =0
X
(1) =0
(5.6)
Step 3: Eigenvalues/Eigenfunctions
The auxiliary polynomial of (5.6) is
p
(
λ
) =
r
2

λ
Now we need to consider
3
cases:
Case 1:
λ >
0
, then
λ
=
ω
2
, where
ω >
0
Then:
r
2

λ
= 0
⇒
r
2

ω
2
= 0
⇒
r
=
±
ω
Therefore:
X
(
x
) =
Ae
ωx
+
Be

ωx
13
Now use
X
(0) = 0
and
X
(1) = 0
:
X
(0) = 0
⇒
A
+
B
= 0
⇒
B
=

A
⇒
X
(
x
) =
Ae
ωx

Ae

ωx
X
(1) = 0
⇒
Ae
ω

Ae

ω
= 0
⇒
Ae
ω
=
Ae

ω
⇒
e
ω
=
e

ω
⇒
ω
=

ω
⇒
ω
= 0
But this is a
contradiction
, as we want
ω >
0
.
Case 2:
λ
= 0
, then
r
= 0
, and:
X
(
x
) =
Ae
0
x
+
Bxe
0
x
=
A
+
Bx
And:
X
(0) = 0
⇒
A
= 0
⇒
X
(
x
) =
Bx
X
(1) = 0
⇒
B
= 0
⇒
X
(
x
) = 0
Again, a
contradiction
(we want
X
≡
0
, because otherwise
u
(
x, t
)
≡
0
)
Case 3:
λ <
0
, then
λ
=

ω
2
, and:
r
2

λ
= 0
⇒
r
2
+
ω
2
= 0
⇒
r
=
±
ωi
Which gives:
X
(
x
) =
A
cos(
ωx
) +
B
sin(
ωx
)
Again, using
X
(0) = 0
,
X
(1) = 0
, we get:
X
(0) = 0
⇒
A
= 0
⇒
X
(
x
) =
B
sin(
ωx
)
X
(1) = 0
⇒
B
sin(
ω
) = 0
⇒
sin(
ω
) = 0
⇒
ω
=
πm,
(
m
= 1
,
2
,
· · ·
)
This tells us that:
Eigenvalues:
λ
=

ω
2
=

(
πm
)
2
(
m
= 1
,
2
,
· · ·
)
Eigenfunctions:
X
(
x
) = sin(
ωx
) = sin(
πmx
)
(5.7)
14
Step 4:
Deal with (5.5), and remember that
λ
=

(
πm
)
2
:
T
0
(
t
) =
λT
(
t
)
⇒
T
(
t
) =
Ae
λt
=
T
(
t
) =
f
A
m
e

(
πm
)
2
t
m
= 1
,
2
,
· · ·
Note:
Here we use
f
A
m
to emphasize that
f
A
m
depends on
m
.
Step 5:
Take linear combinations:
u
(
x, t
) =
∞
X
m
=1
T
(
t
)
X
(
x
) =
∞
X
m
=1
f
A
m
e

(
πm
)
2
t
sin(
πmx
)
(5.8)
Step 6:
Use the initial condition
u
(
x,
0) =
x
in (5.1):
u
(
x,
0) =
∞
X
m
=1
f
A
m
sin(
πmx
) =
x
on
(0
,
1)
(5.9)
Now we want to express
x
as a linear combination of sines, so we have to use
a
sine series
(that’s why we used
f
A
m
instead of
A
m
):
f
A
m
=
2
1
Z
1
0
x
sin(
πmx
)
dx
= 2

x
cos(
πmx
)
πm
1
0

Z
1
0

cos(
πmx
)
πm
dx
!
= 2

cos(
πm
)
πm
+
Z
1
0
cos(
πmx
)
πm
dx
= 2

(

1)
m
πm
+
sin(
πmx
)
(
πm
)
2
1
0
!
=
2(

1)
m
+1
πm
(
m
= 1
,
2
,
· · ·
)
15
Step 7:
Conclude using (5.10)
u
(
x, t
) =
∞
X
m
=1
2(

1)
m
+1
πm
e

(
πm
)
2
t
sin(
πmx
)
(5.10)
16
6
The Wave equation
Problem:
Solve the following
wave equation
:
∂
2
u
∂t
2
=
∂
2
u
∂x
2
0
< x < π,
t >
0
u
(0
, t
) =
u
(
π, t
) = 0
t >
0
u
(
x,
0) =
sin(4
x
) + 7 sin(5
x
)
0
< x < π
∂u
∂t
(
x,
0) = 2 sin(2
x
) + sin(3
x
)
0
< x < π
(6.1)
Step 1: Separation of variables
Suppose:
u
(
x, t
) =
X
(
x
)
T
(
t
)
(6.2)
Plug (6.2) into the differential equation (6.1), and you get:
(
X
(
x
)
T
(
t
))
tt
= (
X
(
x
)
T
(
t
))
xx
X
(
x
)
T
00
(
t
) =
X
00
(
x
)
T
(
t
)
Rearrange and get:
X
00
(
x
)
X
(
x
)
=
T
00
(
t
)
T
(
t
)
(6.3)
Now
X
00
(
x
)
X
(
x
)
only
depends on
x
, but by (6.3)
only
depends on
t
, hence it is
constant:
X
00
(
x
)
X
(
x
)
=
λ
X
00
(
x
) =
λX
(
x
)
(6.4)
Also, we get:
17
T
00
(
t
)
T
(
t
)
=
λ
T
00
(
t
) =
λT
(
t
)
(6.5)
but we’ll only deal with that later (Step 4)
Step 2:
Consider (6.4):
X
00
(
x
) =
λX
(
x
)
Note:
Always start with
X
(
x
)
, do
NOT
touch
T
(
t
)
until right at the end!
Now use the
boundary conditions
in (6.1):
u
(0
, t
) =
X
(0)
T
(
t
) = 0
⇒
X
(0)
T
(
t
) = 0
⇒
X
(0) = 0
u
(
π, t
) =
X
(
π
)
T
(
t
) = 0
⇒
X
(
π
)
T
(
t
) = 0
⇒
X
(
π
) = 0
Hence we get:
X
00
(
x
) =
λX
(
x
)
X
(0) =0
X
(
π
) =0
(6.6)
Step 3: Eigenvalues/Eigenfunctions
The auxiliary polynomial of (6.6) is
p
(
λ
) =
r
2

λ
Now we need to consider
3
cases:
Case 1:
λ >
0
, then
λ
=
ω
2
, where
ω >
0
Then:
18
r
2

λ
= 0
⇒
r
2

ω
2
= 0
⇒
r
=
±
ω
Therefore:
X
(
x
) =
Ae
ωx
+
Be

ωx
Now use
X
(0) = 0
and
X
(
π
) = 0
:
X
(0) = 0
⇒
A
+
B
= 0
⇒
B
=

A
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 Spring '08
 Chorin
 Differential Equations, Equations, Partial Differential Equations, Sin, BM, Partial differential equation