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Since 0 − 1 · 1 < 0, the equation is elliptic. We will transform this equation to the
canical form of Equation 3.11. We find the new variables.

dy
= −i, y = −ix + const, ξ = x + iy
dx
dy
= i, y = ix + const, ψ = x − iy
dx
We calculate the derivatives of ξ and ψ.

ξ x = 1 ξ y = i
ψ x = 1 ψ y = −i

Then we calculate the derivatives of u.

u xx = u ξξ + 2u ξψ + u ψψ
u yy = −u ξξ + 2u ξψ − u ψψ

Finally we transform the equation to canonical form.

4u ξψ = 0
u ξψ = 0

We integrate with respect to ξ and ψ to obtain the solution, u = f(ξ) + g(ψ).
2
Here f and g are arbitrary C functions. In terms of x and y, we have

u(x, y) = f(x + iy) + g(x − iy).

This solution makes a lot of sense, because the real and imaginary parts of an analytic
function are harmonic.

3.2 Parabolic equations

2
Now we consider a parabolic equation, (B −AC = 0). We seek a change of independent
variables that will put Equation 3.1 in the form

u ξξ = G(ξ, ψ, u, u ξ , u ψ ). (3.14)

∗
∗
We require that the u ξψ and u ψψ terms vanish. That is B = C = 0 in Equation 3.4.
This gives us two constraints on ξ and ψ.
2
2
Aξ x ψ x + B(ξ x ψ y + ξ y ψ x ) + Cξ y ψ y = 0, Aψ + 2Bψ x ψ y + Cψ = 0
x y
We consider the case A 6= 0. The latter constraint allows us to solve for ψ x /ψ y .
√
2
ψ x −B − B − AC B
= = −
ψ y A A

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