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Proof 6.6.1 Method of Characteristic Coordinates. We intro-

duce the usual characteristic coordinates ξ = x + ct, η = x − ct, (see
Figure 6.3). As above, we have

ξ + η ξ − η
2
2
Lu = u − c u xx = −4c u ξη = f , .
tt
2 2c
We integrate this equation with respect to η, leaving ξ as a constant.
R η
2
Thus u = −(1/4c ) fdη. Then we integrate with respect to ξ to get
ξ
ξ
Z Z η
1
u = − fdηdξ. (6.13)
4c 2
The lower limits of integration here are arbitrary: They correspond to
constants of integration. The calculation is much easier to understand

if we fix a point P with coordinates x , t and
0
0
0
ξ = x + ct , η = x − ct .
0
0
0
0
0
0
We evaluate (6.13) at P and make a particular choice of the lower
0
limits. Thus,
Z Z
1 xi 0 η 0 ξ + η ξ − η
u(P ) = − f , dξdη = (6.14)
0
4c 2 2 2c
η 0 ξ
Z Z ξ
1 xi 0 ξ + η ξ − η
= + f , dξdη (6.15)
4c 2 2 2c
η 0 η 0
is a particular solution. η now represents a variable going along a line
segment from the base η = ξ of the triangle ∆ to the left-hand edge
η = η , while ξ runs from the left-hand comer to the right-hand edge.
0
Thus we have integrated over the whole triangle ∆.
The iterated integral, is however, not exactly the double integral
over ∆ because the coordinate axes are not orthogonal. The original
axes (xand t) are orthogonal, so we make a change of variables back to
x and t.
ξ+η ξ−η
This amounts to substituting back x = , t = .
2 2c
The change in its area is measured by the Jacobian determinant J.
Since our change of variable is a linear transformation, the Jacobian is

just the determinant of its coefficient matrix:

∂ξ ∂ξ
1 c
J = det ∂x ∂t = det = 2c.

∂η ∂η 1 −c
∂x ∂t
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