CAAM 336
·
DIFFERENTIAL EQUATIONS
Problem Set 11
·
Solutions
Posted Thursday 18 November 2010. Due Tuesday 23 November 2010, 5pm.
This problem set counts for 75 points.
Late problem sets are due by 5pm on Wednesday 24 November 2010.
1. [30 points: 7 points each for (a), (b), (c); 9 points for (d)]
This question concerns the homogeneous wave equation on an unbounded spatial domain:
u
tt
(
x,t
) =
u
xx
(
x,t
)
,
∞
< x <
∞
,
t >
0
.
Find the solution
u
(
x,t
) to this equation with the following initial conditions:
(a)
u
(
x,
0) = 2 sin(
x
)
e

x
2
,
u
t
(
x,
0) = 0;
(b)
u
(
x,
0) = 0,
u
t
(
x,
0) =

2
x
(1 +
x
2
)
2
;
(c)
u
(
x,
0) = 2 sin(
x
)
e

x
2
,
u
t
(
x,
0) =

2
x
(1 +
x
2
)
2
.
(d) Produce a plot (or plots) showing your solution to part (c) over

10
≤
x
≤
10
at times
t
= 0
,
1
,
2
,
3
,
4
,
5.
Solution.
(a) D’Alembert’s solution takes the form
u
(
x,t
) =
1
2
(
ψ
(
x

t
) +
ψ
(
x
+
t
)
)
= sin(
x

t
)
e

(
x

t
)
2
+ sin(
x
+
t
)
e

(
x
+
t
)
2
.
(b) D’Alembert’s solution is now
u
(
x,t
) =
1
2
Z
x
+
t
x

t
γ
(
s
)
ds.
We can compute
Z

2
x
(1 +
x
2
)
2
dx
=
1
1 +
x
2
+ constant
,
and so
u
(
x,t
) =
1
2
(
ψ
(
x

t
) +
ψ
(
x
+
t
)
)
=
1
2
±
1
1 + (
x
+
t
)
2

1
1 + (
x

t
)
2
²
.
(c) When the nonzero boundary conditions from (a) and (b) are combined, we simply sum the solu
tions to the two previous parts:
u
(
x,t
) = sin(
x

t
)
e

(
x

t
)
2
+ sin(
x
+
t
)
e

(
x
+
t
)
2
+
1
2
±
1
1 + (
x
+
t
)
2

1
1 + (
x

t
)
2
²
.
(d) The following plots show the solution in part (c) at times
t
= 0
,
1
,
2
,
3
,
4
,
5, with the code that
produced the plots following.