Mth3006 Lecture 8

• Method of characteristics

Missed lecture: auto-generated.

How Do We Derive the Characteristic Equations $\frac{du}{F}=\frac{dx}{A}=\frac{dy}{B}$?

Consider the first-order partial differential equation:

$$A(x,y)\frac{\partial u}{\partial x}+B(x,y)\frac{\partial u}{\partial y}=F(x,y)$$

We seek curves in the $(x,y)$ plane along which the partial differential equation reduces to an ordinary differential equation. Parameterize such a curve by a parameter $s$ (which could be arc length or any other convenient parameter), so that $x=x(s)$, $y=y(s)$, and $u=u(x(s),y(s))$.

By the chain rule, the total derivative of $u$ along the curve is:

$$\frac{du}{ds}=\frac{\partial u}{\partial x}\frac{dx}{ds}+\frac{\partial u}{\partial y}\frac{dy}{ds}$$

Now, choose the curve so that:

$$\frac{dx}{ds}=A(x,y),\frac{dy}{ds}=B(x,y)$$

Then, substituting into the chain rule:

$$\frac{du}{ds}=A(x,y)\frac{\partial u}{\partial x}+B(x,y)\frac{\partial u}{\partial y}$$

But from the original partial differential equation, the right-hand side is exactly $F(x,y)$, so:

$$\frac{du}{ds}=F(x,y)$$

We now have a system of three ordinary differential equations:

$$\frac{dx}{ds}=A(x,y),\frac{dy}{ds}=B(x,y),\frac{du}{ds}=F(x,y)$$

These can be written in the symmetric form:

$$\frac{dx}{A(x,y)}=\frac{dy}{B(x,y)}=\frac{du}{F(x,y)}=ds$$

or more commonly:

$$\frac{dx}{A(x,y)}=\frac{dy}{B(x,y)}=\frac{du}{F(x,y)}$$

These are the characteristic equations of the partial differential equation. The curves in the $(x,y)$ plane defined by solving $\frac{dx}{A}=\frac{dy}{B}$ are called characteristic curves or simply characteristics.

Alternative Parameterisation Using Arc Length

The element of arc length along a curve is defined by:

$$d{s}^2=d{x}^2+d{y}^2$$

Since $\frac{dx}{ds}=A$ and $\frac{dy}{ds}=B$, we have:

$$d{s}^2=A^{2}d{s}^2+B^{2}d{s}^2$$

This is inconsistent unless we interpret $s$ differently. More precisely, if we parameterise by a general parameter $t$, then:

$$d{s}^2=d{x}^2+d{y}^2={\left(\frac{dx}{dt}\right)}^{2}d{t}^2+{\left(\frac{dy}{dt}\right)}^{2}d{t}^2=(A^2+B^2)d{t}^2$$

So:

$$ds=\sqrt{A^2+B^2}dt$$

And:

$$\frac{du}{dt}=F(x,y)$$

becomes:

$$\frac{du}{ds}=\frac{du/dt}{ds/dt}=\frac{F}{\sqrt{A^2+B^2}}$$

This is the form sometimes presented when using arc length as the natural parameter along the characteristic curve.

In practical calculations, we usually work with the simpler form $\frac{dx}{A}=\frac{dy}{B}$ to find the characteristic curves, and then solve $\frac{du}{ds}=F/\sqrt{A^2+B^2}$ if needed.