C. Hinsley

14 March 2023

I wrote this post to address an unclear point in the derivation and use of the Lagrange-Charpit equations in the method of characteristics commonly seen in books like Hydon [2], Olver [3], and physics texts. Namely, it was not clear to me how one goes from the system of ODEs in this post to the Lagrange-Charpit equations, nor was it clear exactly how this helps to produce the solution to the original PDE by solving the Lagrange-Charpit equations. More detailed analyses are available elsewhere, but this was the result of a few minutes’ discussion with a friend on the topic, sufficient to satisfy my curiosity.

Partial differential equations (PDEs) are equations that involve derivatives of one or more unknown functions (also referred to as dependent variables) with respect to more than one independent variable. They are widely used to model various phenomena in physics, engineering, biology, and other fields. When we speak of “solving” PDEs, we are usually referring to finding analytical solutions of the unknown functions.

One method that can be used to solve some classes of PDEs is the method of characteristics [1]. This method transforms a PDE into a system of ordinary differential equations (ODEs) along certain curves in the domain of the unknown function. These curves are called characteristics and they have the property that along them, the PDE reduces to an ODE.

The usual application of the method of characteristics is in solving linear first-order PDEs. These are PDEs resembling the equation

$$a(x,y,z)z_x + b(x,y,z)z_y = c(x,y,z),$$

where $z = f(x, y)$ is the dependent variable, and $z_x$ and $z_y$ denote its partial derivatives with respect to $x$ and $y$ respectively: $z_x \equiv \frac{\partial f}{\partial x}, z_y \equiv \frac{\partial f}{\partial y}$. The coefficients $a$, $b$, and $c$ are given functions that may depend explicitly on $x$, $y$, and $z$.

It can be helpful to define a function $u(x, y, w) = f(x, y) - w$ with $z$ being thought of as independent from $x$ and $y$; this function has partial derivatives

$$u_x \equiv \frac{\partial u}{\partial x} = \frac{\partial f}{\partial x} \equiv z_x,$$

$$u_y \equiv \frac{\partial u}{\partial y} = \frac{\partial f}{\partial y} \equiv z_y,$$

$$u_w \equiv \frac{\partial u}{\partial w} = -1.$$

It then follows by substituting into the original PDE that

$$a(x, y, z)u_x + b(x, y, z)u_y + c(x, y, z)u_w = 0.$$

We may now define a $C^1$ curve $(x(t), y(t), f(x(t), y(t))) \subset \R^3$ by choosing some functions $x(t), y(t)$ according to the system of ODEs

$$x'(t) = a(x(t), y(t), f(x(t), y(t))),$$

$$y'(t) = b(x(t), y(t), f(x(t), y(t))),$$

$$z'(t) = c(x(t), y(t), f(x(t), y(t))).$$

Along this curve our PDE takes the form