Problems from Strauss, 2.1:
# 3: The disturbance will reach the flea after time
t
=
ρ
T
(
l
4

a
).
# 5: As in the hint we have:
u
(
x, t
) =
1
2
c
{
length of (
x

ct, x
+
ct
)
∩
(

a, a
)
}
When
t
=
a
2
c
:
u
(
x, t
) =
1
2
c
{
length of (
x

a
2
, x
+
a
2
)
∩
(

a, a
)
}
=
1
2
c
×
a

x
 ≤
a
2
3
a
2
 
x

a
2
≤ 
x
 ≤
3
a
2
0

x
 ≥
3
a
2
When
t
=
a
c
:
u
(
x, t
) =
1
2
c
{
length of (
x

a, x
+
a
)
∩
(

a, a
)
}
=
1
2
c
×
2
a
 
x


x
 ≤
2
a
0

x
 ≥
2
a
When
t
=
3
a
2
c
:
u
(
x, t
) =
1
2
c
{
length of (
x

3
a
2
, x
+
3
a
2
)
∩
(

a, a
)
}
=
1
2
c
×
2
a

x
 ≤
a
2
5
a
2
 
x

a
2
≤ 
x
 ≤
5
a
2
0

x
 ≥
5
a
2
When
t
=
2
a
c
:
u
(
x, t
) =
1
2
c
{
length of (
x

2
a, x
+ 2
a
)
∩
(

a, a
)
}
=
1
2
c
×
2
a

x
 ≤
a
3
a
 
x

a
≤ 
x
 ≤
3
a
0

x
 ≥
3
a
Similar formulas can be developed for
t
=
5
a
2
c
.
# 9 Following the hint in the book should lead to search for solutions of the form:
u
(
x, t
) =
f
(
x
+ 4
t
) +
g
(
x

t
)
Match this up with the initial data to get a formula that solves the PDE.
Problems for Haberman, 4.4:
# 3b: Separating variables
u
(
x, t
) =
h
(
t
)
φ
(
x
) leads you to solve
φ
(
x
) =

λφ
(
x
)
φ
(0) =
φ
(
L
) = 0
h
(
t
) +
β
ρ
0
h
(
t
) +
λ
T
0
ρ
0
h
(
t
) = 0
The solutions to
φ
(
x
) take the form
λ
= (
nπ
L
)
2
,
φ
(
x
) =
C
n
sin(
nπx
L
). To solve the
ODE in time, plug in
λ
= (
nπ
L
)
2
and consider the characteristic equation
r
2
+
β
ρ
0
r
+
nπ
L
2
T
0
ρ
0
= 0
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 Spring '10
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 bn sin, square root factor, general initial conditions

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