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I think I have found a solution for a PDE of the form

  1. $u_t + g(u)u_x = 0$

  2. where $u(x, 0) = g^{-1}(x)$

The solution is $u(x,y) = g^{-1}\left(\frac x{t+1}\right)$

This solution satisfies 1 and 2 under the assumption that $\forall z, g\left(g(z)^{-1}\right) = z$

However I am worried about the effects of discontinuities in $g$ or its inverse, and issues where the function is not 1-1.

What sort of problems should I watch out for and how can I get around them?

Yiorgos S. Smyrlis
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kleineg
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    The tag [tag:differential-equations] is intended for questions about ordinary differential equations, there is a separate tag for [tag:pde]s; see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) – Alex R. Jan 13 '14 at 16:51
  • The original place I found it had $g(x) = e^x$ which gave an answer $u(x,y) = ln(\frac x{t+1})$ But as far as I know the solution would hold for any smooth function with an inverse, that is what I am trying to determine. – kleineg Jan 15 '14 at 19:39

2 Answers2

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The classical solution $$ u(x,t) = g^{-1}\left(\frac{x}{t+1}\right) $$ follows from the method of characteristics, which requires that the solutions remains smooth along the characteristic curve as time increases. Smoothness is lost at the breaking time $$ t_b = \inf \frac{-1}{(g\circ g^{-1})'(x)} = -1 \, . $$ This can be seen in the solution's denominator which vanishes at $t=-1$. Therefore, if only positive times are considered, no shock wave will form! Now, if there are some discontinuities in $g$ or $g^{-1}$, the problem statement makes little sense as $g$ is no longer invertible over its whole domain of definition.

EditPiAf
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This equation you are dealing with, although it looks simple, it is a very problematic one: it is a nonlinear conservation law. These equation DO NOT have smooth solutions for all $t>0$. Their solution break down after a short time and the develop "shock waves" - discontinuous weak solutions.

The simpler such equation is Burger's Equation ($g(u)=u$) $$ u_t+uu_x=0, $$ and there is a written about it as it is the simplest model for shock formation.

Yiorgos S. Smyrlis
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    In the present case, the classical solution does not break down at positive times (cf. my answer). – EditPiAf Dec 01 '20 at 21:37