20.2 Numerical solution

It is certainly straightforward to numerically solve the two ordinary differential equations of the previous subsection along a characteristic line using say a Runge-Kutta method. You would need to start from some point at which an initial or boundary condition is given.

If you find the solution along each of a densely spaced set of characteristic lines, you have essentially found $u$ everywhere.

Of course, if $a$ is zero somewhere in the region of interest, it may be a better idea to find $u$ and $x$ as functions of $y$ instead of $u$ and $y$ as functions of $x$, by taking suitable ratios from (20.4). Or you could just find all three variables as function of the arc length $s$ along the characteristic lines, by solving

$$\frac{{\rm d}x}{{\rm d}s} = \frac{a}{\sqrt{a^2+b^2}} \qqua... ...\qquad \frac{{\rm d}u}{{\rm d}s} = \frac{c}{\sqrt{a^2+b^2}}$$

This allows either $a$ or $b$ to be zero; it only fails if both are zero at the same point, and that is a true physical problem rather than a mathematical one.