Breakdown of solution to inviscid Burgers equation

Question

Let $u = f(x-ut)$ where $f$ is differentiable. Show that $u$ (amost always) satisfies $u_t + uu_x = 0$. What circumstances is it not necessarily satisfied?

This is a question in a tutorial sheet I have been given and I am slightly stuck with the second part. To show that $u$ satisfies the equation I have differentiated it to get:

$u_t = -f'(x-ut)u$

$u_x = f'(x-ut)$

Then I have substituted these results into the original equation. The part I am unsure of is where it is not satisfied. If someone could push me in the right direction it would be much appreciated.

Answer 1

We have \[ u_t = f'(x-ut)(x-ut)_t = -f'(x-ut)(u_t t + u) \iff \bigl(1 + tf'(x-ut)\bigr)u_t = -uf'(x-ut) \] and \[ u_x = f'(x-ut)(x-ut)_x = f'(x-ut)(1 - u_xt) \iff \bigl(1 + tf'(x-ut)\bigr)u_x = f'(x-ut) \] Which gives that \[ \bigl(1 + tf'(x-ut)\bigr)(u_t +uu_x) = 0 \] so at each point either $1 + tf'(x-ut) = 0$ or $u_t + uu_x = 0$.

Answer 2

What do we know about $t$ and $x$?

If $t$ and $x$ are independent variables, then you already have the proof.

Answer 3

Following @martini's comment, and rearranging, you should find $$ (1+tf')u_t=-uf',$$ omitting the argument of $f'$ for convenience. Likewise, $$ (1+tf')u_x=f'.$$ These lead to $$(1+tf')(u_t+uu_x)=0,$$ from which the equation $u_t+uu_x=0$ follows, provided the coefficient $1+tf'(x-ut)$ is non-zero. This is the source of the 'almost always' aspect of the result.