Chapter 6 Partial Differential Equations

6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 35
where v could represent either ϕ or ψ. We note that (6.3.11) can be factored into
[ ( √ ) ][ ( √ ) ]
−b + b2 − ac
−b − b2 − ac
a v x −
v y v x −
v y . (6.3.12)
a
a
Thus from (6.3.12) we seek ϕ and ψ so that, for example,
[ ( √ ) ]
−b + b2 − ac
ϕ x −
ϕ y =0, (6.3.13)
a
and
[
ψ x −
( √ ) ]
−b − b2 − ac
ψ y =0. (6.3.14)
a
With these choices (6.3.11) is satisfied with v given by ϕ and ψ. In addition we must
impose a noncharacteristic solvability condition
∣
∂(ϕ, ψ) ∣∣∣∣
∂(x, y) = ϕ x
ψ x
ϕ y
ψ y
∣ ∣∣∣∣
( √ )
−b + b2 − ac
=
ϕ y ψ y −
a
( −b −
√
b2 − ac
a
)
ϕ y ψ y
= 2 a
√
b2 − ac ϕ y ψ y ≠0. (6.3.15)
A solution of (6.3.13) is found by solving
dx
dt =1,
( √ )
dy −b +
dt = − b2 − ac
,
a
dϕ
dt =0.
For this problem we take the initial conditions
x 0 (s) =x 0 ,y 0 (s) =y 0 + s,
ϕ 0 (s) =s.
With this choice we find (in our earlier notation we used a and b which are not the
same as those occuring in the present problem)
∣ ∣∣∣∣∣∣ a x ′ 1 0
0(s)
∣b
y 0(s)
′ ∣ = ( √ )
b − b2 − ac
≠0. (6.3.16)
1
a ∣