Chem 3322 homework #1 solutions, out of 18 marks
Problem 1
– 8 marks
Prove that, if
u
1
(
x, t
) and
u
2
(
x, t
) are, separately, solutions to the classical wave equation
with propagation speeds of
v
1
and
v
2
respectively, then the function
u
(
x, t
) =
au
1
(
x, t
) +
bu
2
(
x, t
) is also a solution, where
a
and
b
are real numbers, only if
v
1
=
v
2
. Make it clear
what happens if
v
1
6
=
v
2
.
Solution:
We are told that
∂
2
u
1
∂x
2
=
1
v
2
1
∂
2
u
1
∂t
2
(1)
and
∂
2
u
2
∂x
2
=
1
v
2
2
∂
2
u
2
∂t
2
(2)
Now, let us consider the function
u
(
x, t
) =
au
1
(
x, t
) +
bu
2
(
x, t
):
∂
2
u
∂x
2
=
a
∂
2
u
1
∂x
2
+
b
∂
2
u
2
∂x
2
(3)
(here we used the linearity property of the derivative operator)
=
a
1
v
2
1
∂
2
u
1
∂t
2
+
b
1
v
2
2
∂
2
u
2
∂t
2
(4)
using Eqns. (1) and (2). Now, for
u
to solve the wave equation, we must have
∂
2
u
∂x
2
=
1
v
2
∂
2
u
∂t
2
(5)
for some propagation speed
v
. Clearly this holds if
v
1
=
v
2
, since we can factor a common
velocity out of the right hand side of Eqn. (4). On the other hand, if
v
1
6
=
v
2
, we cannot
factor a common velocity out of Eqn. (4) and hence we cannot rearrange this equation into
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Physical chemistry, Derivative, pH, Partial differential equation, 2m, Eqns., Schoedinger

Click to edit the document details