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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:

  1. introduce hyperbolic numbers $\mathcal{H} = \mathbb{R} \oplus j\mathbb{R}$ where $j^2=1$
  2. 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 $$ \frac{df}{dz} = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} +j\frac{\partial v}{\partial x}$$
  3. observe $f^2 = (u+jv)(u+jv) = u^2+v^2+j(2uv)$
  4. our system of PDEs is captured as a single differential equation with respect to the hyperbolic variable $z$. Setting $w=f$ we have $$ \frac{dw}{dz} = w^2 $$
  5. solve to find $w = \frac{-1}{z+c}$ where $z = x+jy$ and $c=a+jb$
  6. 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 ?

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  • $\begingroup$ The system $u_x = v_y$ and $u_y = v_x$ implies that $u,v$ solve the wave equation. The introduction of the hyperbolic numbers (which is the natural Clifford algebra associated to $\mathbb{R}^{1,1}$) allows you to factor the wave equation into the analogue of the Dirac equation you wrote down. So in the end you end up with $\partial w = w^2$ and $\bar{\partial} w = 0$ where $\partial$ is some notion of a Dirac operator. But all these is after-the-fact (and you of course recognize its similarity to the case of Complex Analysis), so I don't know if it helps you at all. $\endgroup$
    – Willie Wong
    Commented Oct 4, 2013 at 8:27
  • $\begingroup$ @WillieWong that might be a useful line of inquiry. Here's my thinking: if I can identify a certain PDE as allowing hyperbolic-calculus representation then it is likely possible to solve the problem in the hyperbolic notation. So, to make this interesting, I'd like to figure out how to take a given PDE and identify the algebra which might allow a simplification like the one illustrated in my post. Perhaps I should take a long look at Clifford Algebras. I know all the commutative, semi-simple algebras over $\mathbb{R}$ are direct sums of of $\mathbb{R}, \mathbb{C}$ and $\mathcal{H}$.. $\endgroup$
    – James S. Cook
    Commented Oct 4, 2013 at 19:35
  • $\begingroup$ where $\mathcal{H}$ is mostly a convenience since the direct product of $\mathbb{R}$ with itself is isomorphic. In the general case, the assumption of $\mathcal{A}$-differentiability necessitates $n^2-n$ Cauchy Riemann equations, so the systems of PDEs which this idea captures are probably only of a particularly special type. $\endgroup$
    – James S. Cook
    Commented Oct 4, 2013 at 19:38

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