Chapter 6 Partial Differential Equations

40 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS
Thus we need to find a solution ϕ(x, y) of (6.3.19) and then select ψ(x, y) so that
∣
∂(ϕ, ψ) ∣∣∣∣
∂(x, y) = ϕ x
ψ x
ϕ y
ψ y
∣ ∣∣∣∣
= ϕ x ψ y − ψ x ϕ y
= − b a ϕ yψ y − ψ x ϕ y
= −ϕ y
(ψ x + b )
a ψ y ≠0.
We may, for example, take ψ = x and prescribe inital conditions for (6.3.18) so that
ϕ y ≠ 0. With these choices, ã = 0 (since ϕ satisfies (6.3.18)) and from (6.3.10) we find
˜b = aϕx ψ x + b(ϕ x ψ y + ϕ y ψ x )+cϕ y ψ y
(
= a − b ) [(
a ϕ y ψ x + b − b ) ]
a ϕ y ψ y + ϕ y ψ x + cϕ y ψ x
[
]
= ϕ y −bψ x − b2
a ψ y + bψ x + cψ y
[ ]
= −ϕ y ψ y − b2
a + c =0.
Thus with α = ϕ(x, y), β = ψ(x, y), (6.3.4) becomes
˜cu ββ + ˜du α + ẽu β + ˜fu = ˜g.
Furthermore, since b 2 = ac, and with the choice ψ = x we have
(
˜c = aψx 2 +2bψ x ψ y + cψy 2 = a ψ x + b 2
y)
a ψ = a ≠0.
Therefore we can divide by ˜c = a to get the desired form.
If ϕ(x, y) is a solution of (6.3.19) and ϕ(x, y) = constant, then it follows from (6.3.18)
that the characteristics could be obtained from
dy
dx = b a .
Example 6.3.7. Let us transform the equation
u xx − 2xu xy + x 2 u yy − 2u y =0