Page 30 - 6099

We can transform Equation 3.11 to this form with the change of variables

σ = ξ + ψ, τ = ξ − ψ.

Equation 3.11 becomes

σ + τ σ − τ
u σσ − u ττ = G , , u, u σ + u τ , u σ − u τ .
2 2
Example 3.1 Consider the wave equation with a source.

2
u tt − c u xx = s(x, t)

2
Since 0 − (1)(−c ) > 0, the equation is hyperbolic. We find the new variables.
dx
= −c, x = −ct + const, ξ = x + ct
dt
dx
= c, x = ct + const, ψ = x − ct
dt
Then we determine t and x in terms of ξ and ψ.

ξ − ψ ξ + ψ
t = , x =
2c 2

We calculate the derivatives of ξ and ψ.

ξ t = c ξ x = 1
ψ t = −c ψ x = 1

Then we calculate the derivatives of u.

2 2 2
u tt = c u ξξ − 2c u ξψ + c u ψψ
u xx = u ξξ + u ψψ
Finally we transform the equation to canonical form.

ξ + ψ ξ − ψ
2
−2c u ξψ = s ,
2 2c

1 ξ + ψ ξ − ψ
u ξψ = − s ,
2c 2 2 2c
If s(x, t) = 0, then the equation is u ξψ = 0 we can integrate with respect to ξ and
2
ψ to obtain the solution, u = f(ξ) + g(ψ). Here f and g are arbitrary C functions.
In terms of t and x, we have

u(x, t) = f(x + ct) + g(x − ct).

23