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Solved Problems

and C characteristics through (x , t ) and let them intersect the x-axis at the
0
−
0
points A,B. Use the properties of these curves derived in part (a) to determine
u (x , t ) in terms of initial data at points A and B. Using a similar technique
t
0
0
to obtain u (x , τ) with 0 < τ < t, determine u(x , t ) by integration with
t
0
0
0
respect to τ, and compare this with the solution derived in class:
Z x+ct
1 1
u(x, t) = (u(x + ct, 0) + u(x − ct, 0)) + u (τ, 0)dτ.
t
2 2c x−ct
Observe that this “method of characteristics” again shows that u(x , t ) de-
0
0
pends only on that part of the initial data between points A and B.
Solution:
(a) We write the value of u along the line x−ct = k as a function of t: u(k+ct, t).
We differentiate u − cu with respect to t to see how the quantity varies.
t
x
d 2
(u (k + ct, t) − cu (k + ct, t)) = cu + u − c u xx − cu xt
xt
x
tt
t
dt
2
= u − c u xx
tt
= 0
Thus u −cu is constant along the line x−ct = k. Now we examine u +cu x
x
t
t
along the line x + ct = k.
d
2
(u (k − ct, t) + cu (k − ct, t)) = −cu + u − c u xx + cu xt
tt
t
xt
x
dt
2
= u − c u xx
tt
= 0
u + cu is constant along the line x + ct = k.
t
x
(b) From part (a) we know
u (x , t ) − cu (x , t ) = u (x − ct , 0) − cu (x − ct , 0)
x
0
0
0
0
t
0
0
x
0
t
0
u (x , t ) + cu (x , t ) = u (x + ct , 0) + cu (x + ct , 0).
x
0
t
0
0
t
0
0
x
0
0
0
We add these equations to find u (x , t ).
0
t
0
1
u (x , t )= (u (x − ct , 0)−cu (x −ct , 0)u (x + ct , 0)+cu (x +ct , 0))
t
0
0
0
t
x
0
0
0
t
0
0
0
0
x
2
Since t was arbitrary, we have
0
1
u (x , τ)= (u (x − cτ, 0)−cu (x − cτ, 0)u (x + cτ, 0)+cu (x + cτ, 0))
x
t
0
x
0
t
t
0
0
0
2
for 0 < τ < t . We integrate with respect to τ to determine u(x , t ).
0
0
0
Z
t 0 1
u(x , t ) = u(x , 0) + (u (x − cτ, 0) − cu (x − cτ, 0)u (x + cτ, 0)+
0
0
x
0
0
t
t
0
0
0 2
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