I am trying to study some methods of resolution of PDEs, for my exam of mathematical methods for physics. Currently I am reading “A guide to mathematical methods for physicists” (volume 2) by Petrini, Pradisi and Zaffaroni. I find that the book is too concise in explaining the most important methods for solving PDEs. For example, in an example (4.7, one-dimensional wave equation), it solves the wave equation using a pretty standard procedure: given \begin{align} & \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} = 0 \\[7pt] & u(x, 0) = h(x) \\[7pt] & \partial_t u(x, 0) = v(x) \end{align} a particular change of variables is performed: \begin{align} \xi_+ = x + t && \xi_- = x - t \end{align} so that the PDE becomes, using the new variables: \begin{align} \frac{\partial^2 u}{\partial \xi_+ \partial \xi_-} = 0 \end{align} This change of coordinates just seems to me completely arbitrary. I do not understand why someone would want to do that without already knowing how the solution physically behaves. In fact, I understand that this change of variables could be hinted at by the physics of the phenomena: there are two identical waves, one that propagates forwards and the other that propagates backwards. However, purely mathematically, why would one want to change variables? Up until now my guesses are two: it is simply an ansatz, therefore no more motivation is needed; or it has something to do with the symmetries of the equation. Could you explain me the general logic behind change of variables in PDEs, and in the case where it has something to do with the symmetries of the equation, how, usually, one searches for such advantageous changes of variables?
- $\begingroup$ The physical explanation is often a good source of inspiration, it was probably the cas for this one since pde mostly arise from physics. However there is one explanation in terms of diagonalisation of an operator $\endgroup$julio_es_sui_glace– julio_es_sui_glace2025-08-27 17:19:43 +00:00Commented Aug 27 at 17:19
1 Answer
It's a difference-of-squares factorization. Rewrite the equation in terms of differential operators as
$$\left( \frac{\partial}{\partial t}^2 - \frac{\partial}{\partial x}^2 \right) u = 0.$$
Differential operators form a ring, and in this ring the operators $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial x}$ commute, so this difference of squares can be factored as
$$\left( \frac{\partial}{\partial t} - \frac{\partial}{\partial x} \right) \left( \frac{\partial}{\partial t} + \frac{\partial}{\partial x} \right) u = 0.$$
The same argument applied to Laplace's equation
$$\left( \frac{\partial}{\partial x}^2 + \frac{\partial}{\partial y}^2 \right) u = 0$$
shows that if we work over $\mathbb{C}$ then we can factor the Laplacian $\Delta$ as
$$\left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) u = 0.$$
This motivates the introduction of the Wirtinger derivatives and leads to the rich connection between harmonic functions in two variables and holomorphic functions. A similar argument in more variables famously led Dirac to Dirac operators.
This technique of treating differential operators as themselves algebraic objects that can be manipulated using the rules of ordinary algebra can already be profitably applied in the ODE case. For example we can solve the differential equation $\frac{d^2 f}{dx^2} = a f$ by factoring it as
$$\left( \frac{d}{dx} - \sqrt{a} \right) \left( \frac{d}{dx} + \sqrt{a} \right) f = 0$$
which immediately reveals that $f(x) = e^{\pm \sqrt{a} x}$ are solutions. When $a = -1$ we immediately see a connection between the trigonometric functions $\sin x, \cos x$ and the complex exponentials $e^{\pm ix}$.
- $\begingroup$ Thank you very much for your help! Symmetries don't take place in this kind of reasoning then, do they? $\endgroup$Luke__– Luke__2025-08-28 08:24:54 +00:00Commented Aug 28 at 8:24
- $\begingroup$ @Luke__ what symmetry do you have in mind in this specific case? $\endgroup$infinitezero– infinitezero2025-08-28 09:21:42 +00:00Commented Aug 28 at 9:21
- $\begingroup$ @Luke__: this argument as presented does not require thinking in terms of symmetries, yes. $\endgroup$Qiaochu Yuan– Qiaochu Yuan2025-08-28 16:08:37 +00:00Commented Aug 28 at 16:08
- $\begingroup$ @infinitezero Sorry for the late reply. I was thinking of something like invariance under parity, $x \to -x$ and $t \to -t$, because of the second order in the derivatives. It would be $u = -x$, so that $\partial_x = \partial_u (\partial_x u) = - \partial_u \Longrightarrow \partial^2_u = (-\partial_u)(-\partial_u)= \partial^2_u$ $\endgroup$Luke__– Luke__2025-08-28 22:24:08 +00:00Commented Aug 28 at 22:24