Page 615 - Advanced_Engineering_Mathematics o'neil

16.6 Characteristics and d’Alembert’s Solution 595

The transformation is invertible, since we can solve for x and t to get
1 1
x = (ξ + η),t = (−ξ + η).
2 2c
Define

U(ξ,η) = u((ξ + η)/2,(−ξ + η)/2c).
We must compute chain rule derivatives:
u x = U ξ ξ x + U η η x = U ξ + U η ,
and
u xx = U ξξ ξ x + U ξη η x + U ηξ ξ x + U ηη η x
= U ξξ + 2U ξη + U ηη .
Similarly,
2
2
2
u tt = c U ξξ − 2c U ξη + c U ηη .
The wave equation transforms to
2
2
2
2
2
u tt − c u xx = 0 =[c U ξξ − 2c U ξη + c U ηη ]− c [U ξξ + 2U ξη + U ηη ],
or
U ξη = 0.
Now we see the rationale for this change of variables. The transformed equation U ξη = 0 is easy
to solve. First, (U η ) ξ = 0 means that U η is independent of ξ, so for some function h,
U η = h(η).
Then

U(ξ,η) = h(η)dη + F(ξ)

in which this integration with respect to η may have ξ in its “constant” of integration. Now

h(η)dη is just another function of η, so we conclude that U(ξ,η) must be a sum of a function
just of ξ and a function just of η:
U(ξ,η) = F(ξ) + G(η).
This function satisfies the transformed wave equation for any twice differentiable functions F
and G of one variable and, conversely, every solution of the transformed wave equation has this
form. In terms of x and t, this means that every solution of the one-dimensional (unforced) wave
equation has the form

u(x,t) = F(x − ct) + G(x + ct). (16.12)
Thus far, we have dealt with just the partial differential equation. The idea now is to choose
F and G to obtain a solution satisfying the initial conditions y(x,0) = f (x) and y t (x,0) = g(x).
First we need
u(x,0) = F(x) + G(x) = f (x) (16.13)
and
u t (x,0) =−cF (x) + cG (x) = g(x). (16.14)

Integrate equation (16.14) and rearrange terms to get
1 x
−F(x) + G(x) = g(w)dw − F(0) + G(0).
c 0

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October 14, 2010 15:23 THM/NEIL Page-595 27410_16_ch16_p563-610