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Substituting this into equation (3.9), we arrive at the following equation for the slope of
the characteristic curve
2
dy dy
A − 2B + C = 0
dx dx
Since the above is a quadratic equation, it has two, one, or none real solutions, depending
2
on the sign of the discriminant, B − AC, and the solutions are given by the quadratic
formulas √
2
dy B ± B − AC
= (3.10)
dx A

3.1 Hyperbolic Equations

2
We start with a hyperbolic equation, (B − AC > 0). We seek a change of independent
variables that will put Equation 3.1 in the form

u ξψ = G(ξ, ψ, u, u ξ , u ψ ) (3.11)

∗
∗
We require that the u ξξ and u ψψ terms vanish. That is A = B = 0 in Equations (3.5)
and (3.7). This gives us two constraints on ξ and ψ.
2
2
2
2
Aξ + 2Bξ x ξ y + Cξ = 0, Aψ + 2Bψ x ψ y + Cψ = 0 (3.12)
y
x
y
x
√ √
2
2
ξ x −B + B − AC ψ x −B − B − AC
= , =
ξ y A ψ y A
√ √
2
2
B − B − AC B + B − AC
ξ x + ξ y = 0, ψ x + ψ y = 0
A A
Here we chose the signs in the quadratic formulas to get different solutions for ξ and ψ.
Now we have first order quasi-linear partial differential equations for the coordinates ξ
and ψ. We solve these equations with the method of characteristics. The characteristic
equations for ξ are
√
2
dy B − B − AC d
= , ξ(x, y(x)) = 0
dx A dx
Solving the differential equation for y(x) determines ξ(x, y). We just write the solution
for y(x) in the form F(x, y(x)) = const. Since the solution of the differential equation
for ξ is ξ(x, y(x)) = const, we then have ξ = F(x, y). Upon solving for ξ and ψ we divide
∗
Equation 3.4 by B to obtain the canonical form.
Note that we could have solved for ξ y /ξ x in Equation 3.12.
√
2
dx ξ y B − B − AC
= − =
dy ξ x c
This form is useful if A vanishes.
Another canonical form for hyperbolic equations is

u σσ − u ττ = K(σ, τ, u, u σ , u τ ). (3.13)

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