Classifying PDE

Question

I came across the following PDE while reading a book: $$ u_t + (au)_x + \frac1{2}(b^2u^2)_x + \frac1{2}(b^2u_x)_x=0 , $$ where $b$ and $a$ are smooth functions from $\mathbb{R}$ into itself. It seems that this PDE's solutions exist under some assumptions, but what type of PDE is it?

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What I did:

I did a bit of fiddling and got it down to $$ \frac1{2}(ub^2+1)u_{xx} + (a + 2bb_x)u_x + (a_x+2bb_x)u^2 + u_t=0. $$ It looks like it is parabolic but the presence of $u$ in the $u_xx$'s coefficient makes me feel that that is wrong.

Answer 1

This is a second order quasilinear homogeneous PDE. For a quasilinear equation (one where the terms involving the highest order derivative are linear in the unknown function) you can classify using the standard linear approach, as such so long as $b \neq 0$ the equation is parabolic. If at any point $b=0$ then you will nead to be careful.

Answer 2

Your PDE is nonlinear. The classification you are aiming for is (usually) only applied to linear PDE (often only even with constant coefficients, but one can extend to variable coefficients easily). That's why you can't classify this PDE.

The linear part of this PDE is $u_t + a_x u_x + \frac{1}{2}bb_x u_x + \frac{1}{2} b^2 u_xx = 0$. That is parabolic if $b$ is not zero.