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Solved Problems

(b) For x < 0, the equation is hyperbolic. We find the new independent variables.

dy √ 2 √ 2 √
= −x, y = x −x + c, ξ = x −x − y
dx 3 3
dy √ 2 √ 2 √
= − −x, y = − x −x + c, ψ = x −x + y
dx 3 3
Next we determine x and y in terms of ξ and ψ.

1/3
3 ψ − ξ
x = − (ξ + ψ) , y =
4 2

We calculate the derivatives of ξ and ψ.

√ 3 1/6
ξ = −x = (ξ + ψ) , ξ = −1
x
y
4
1/6
3
ψ = (ξ + ψ) , ψ = 1
y
x
4
Then we calculate the derivatives of φ.

1/6
3
φ = (ξ + ψ) (φ + φ )
ψ
ξ
x
4
φ = −φ + φ ψ
y
ξ
1/3
3
φ xx = (ξ + ψ) (φ + φ ) + (6(ξ + ψ)) 1/3 φ ξψ + (6(ξ + ψ)) −2/3 (φ + φ )
ξξ
ξ
ψ
ψψ
4
φ yy = φ − 2φ ξψ + φ ψψ
ξξ
Finally we transform the equation to canonical form.
φ xx + xφ yy = 0
(6(ξ + ψ)) 1/3 φ ξψ + (6(ξ + ψ)) 1/3 φ ξψ + (6(ξ + ψ)) −2/3 (φ + φ ) = 0
ψ
ξ
φ + φ ψ
ξ
φ ξψ = −
12(ξ + ψ)

For x > 0, the equation is elliptic. The variables we defined before are complex-

valued.
2 2
ξ = i x 3/2 − y, ψ = i x 3/2 + y
3 3
We choose the new real-valued variables.

α = ξ − ψ, β = −i(ξ + ψ)

We write the derivatives in terms of α and β.

φ = φ − iφ β
ξ
α
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