I recently posed a system of PDEs to solve on MSE at https://math.stackexchange.com/q/514147/36530. It was quickly solved by a nice pair of subsitutions.

However, in this post, I'd like to show here how I found this system of PDEs and then ask if my approach is known to anyone on MO: $$ \frac{\partial u}{\partial x}=u^2+v^2 \qquad \frac{\partial v}{\partial x} = 2uv $$ where $\frac{\partial u}{\partial x} =\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$.

I solved this problem as follows:

- introduce hyperbolic numbers $\mathcal{H} = \mathbb{R} \oplus j\mathbb{R}$ where $j^2=1$
- recall the derivative with respect to a hyperbolic variable $z$ is realized as partial differentiation with respect to $x$ (taking $x$ as the coordinate of $\mathbb{R}$ and $y$ as the coordinate of $j\mathbb{R}$ hence if $f = u+jv$ then $$ \frac{df}{dz} = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} +j\frac{\partial v}{\partial x}$$
- observe $f^2 = (u+jv)(u+jv) = u^2+v^2+j(2uv)$
- our system of PDEs is captured as a single differential equation with respect to the hyperbolic variable $z$. Setting $w=f$ we have $$ \frac{dw}{dz} = w^2 $$
- solve to find $w = \frac{-1}{z+c}$ where $z = x+jy$ and $c=a+jb$
- express the hyperbolic solution in real notation and find $$ u+jv = \frac{-(x+a)+j(y+b)}{(x+a)^2-(y+b)^2} $$ We can read off the $u$ and $v$ solutions from the hyperbolic equation above. Moreover, you can check that standard methods for solving PDEs reproduce the same: https://math.stackexchange.com/q/514147/36530.

I'm illustrating the method for hyperbolic numbers, but I'm fairly sure I can produce similar results for any associative, semi-simple commutative algebra with unity. I'm curious,**is my approach here at all novel or is this path already well-worn ?**