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Solved Problems

(b) Regardless of the initial condition, the solution has the following form.

(
f (x − c t) + g (x + c t), x < 0
1
1
1
1
u(x, t) =
f (x − c t) + g (x + c t), x > 0
2
2
1
2
For x < 0, the right-moving wave is F(x−c t) and the left-moving wave is zero
1
for x < −c t. For x > 0, there is no left-moving wave and the right-moving
1
wave is zero for x > c t. We apply these restrictions to the solution.
2
(
F(x − c t) + g(x + c t), x < 0
1
1
u(x, t) =
f(x − c t), x > 0
2
We use the continuity of u and u at x = 0 to solve for f and g.
x
F(−c t) + g(c t) = f(−c t)
1
2
1
0
0
0
F (−c t) + g (c t) = f (−c t)
2
1
1
We integrate the second equation.
F(−t) + g(t) = f(−c t/c )
1
2
c 1
−F(−t) + g(t) = − f(−c t/c ) + a
2
1
c 2
We solve for f for x < c t and for g for x > −c t.
1
2
2c 2 c − c 1
2
f(−c t/c ) = F(−t) + b, g(t) = F(−t) + b
2
1
c + c 2 c + c 2
1
1
By considering the case that the solution is continuous, F(0) = 0, we conclude
that b = 0 since f(0) = g(0) = 0.
2c 2 c − c 1
2
f(t) = F(c t/c ), g(t) = F(−t)
2
1
c + c 2 c + c 2
1
1
Now we can write the solution for u(x, t) for t > 0.
(
F(x − c t) + c 2 −c 1 F(−x − c t)H(x + c t), x < 0
1
1
1
u(x, t) = c 1 +c 2
2c 2 c 1
F (x − c t) H(c t − x), x > 0
c 1 +c 2 c 2 2 2
(c) The incident, reflected and transmitted waves are, respectively,

c − c 1 2c 2 c 1
2
F(x−c t), F(−x−c t)H(x+c t), F (x− c t) H(c t−x).
2
2
1
1
1
c + c 2 c + c 2 c 2
1
1
The reflection and transmission coefficients are, respectively,
c − c 2 2c 2
1
, .
c + c 2 c + c 2
1
1
In the limit as c → c , the reflection coefficient vanishes and the transmission
1
2
coefficient tends to unity.
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