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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.

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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 $g(r)$?

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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.

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