D.5 2D elliptical transformation

To bring two-dimensional elliptical equations in the two-dimensional canonical form, you need to solve, say, the ordinary differential equation

$$\frac{{\rm d}y}{{\rm d}x} = \frac{b + i \sqrt{ac-b^2}}{a}$$

Note now that even if you take ${\rm d}{x}$ to be real, ${\rm d}{y}$ will be complex. And that means that you need to know what happens to the coefficients $a$, $b$, and $c$ when you depart the real $x,y$-plane into the complex domain. That may be fine if you know the coefficients analytically, but otherwise it is a problem.

Assuming that you can solve the system, call the integration constant $\tilde\xi$. Assuming that it is a differentiable function of $x$ and $y$, it will satisfy

$$a (\tilde\xi_x)^2 + 2 b \tilde\xi_x\tilde\xi_y + c(\tilde\xi_y)^2=0$$

Now set

$$\tilde\xi = \xi + {\rm i}\eta$$

If you plug that in the equation above and multiply out, you get

$$\begin{eqnarray*} && [a (\xi_x)^2 + 2 b \xi_x\xi_y + c(\xi_y)^2] - [a (\eta_... ... \xi_x\eta_x + b \xi_x\eta_y + b \xi_y\eta_x + c\xi_y\eta_y] =0 \end{eqnarray*}$$

Now within the square brackets above, you find the generic expressions for the coefficients $a'$, $c'$ and $b'$, respectively, of the transformed partial diffential equation. For a complex number to be zero, both its real and its imaginary part must be zero. It follows that $a'=c'$ and that $b'=0$.