Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The method of characteristics is a technique to solve quasi-linear 1st order PDEs. In An Introduction To Partial Differential Equations by Ruebenstein it is stated that difficulties can arise when a characteristic solution intersects an initial curve more than once. I am having trouble seeing why this is the case.

share|cite|improve this question
up vote 4 down vote accepted

Suppose for example the method of characteristics says the solution $f(x,y)$ is constant on the curves $x^2 + y^2 = r^2$ (e.g. for the equation $y \frac{\partial f}{\partial x} - x \frac{\partial f}{\partial y} = 0$), and your initial condition says $f(x,0) = g(x)$. The initial curve $y=0$ intersects the characteristic curve $x^2 + y^2 = r^2$ at two points $(-r,0)$ and $(r,0)$. But if $g(-r) \ne g(r)$ we're in trouble: if the solution is constant on the characteristic curve it must have the same value at $(-r,0)$ as it does at $(r,0)$. So which is it, $g(-r)$ or $g(r)$?

share|cite|improve this answer
Suppose $g(r) =g(-r)$. Do the difficulties necessarily go away? – Digital Gal Mar 31 '12 at 2:32
Yes, of course. – Robert Israel Apr 1 '12 at 6:34

I think I should make one clarification here. The value of f is not always constant on each characteristic, so in general we need g(-r) to be consistent with g(r)... but that doesn't necessarily mean that it's equal.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.