Consider the equation

$$\frac{\partial u}{\partial t} +\frac{\partial}{\partial x}(u^2+uf(x,t)) =0$$

where $f(x,t)$ is a suitably well behaved function. Given the initial condition $u(x,t=0)$, for which family of functions $f(x,t)$ will shocks form?

Note, for $f=0$, we return to the inviscid Burgers' equation, which admits shock solutions.


1 Answer 1


Take one spatial derivative of the equation. You get something like

$$ u_{xt}+u_x^2+uu_{xx}+(u_xf+uf_x)_x=0 $$

Write $v=u_x$. Then,

$$ v_{t}+v^2+(u+f)u_{xx}+2vf_x+uf_{xx}=0. $$

We know that $$ \max_x |u(x,t)|\leq \max_x|u_0(x)|, $$ where $u_0$ is the initial data. Then, we have that if we evaluate the eqution for $v$ at the point where $v$ reaches its minimum, we have $$ \frac{d}{dt}y+y^2+2yf_x+uf_{xx}=0, $$ where $y(t)=\min_x v(x,t)=v(X(t),t)$.

If a shock is formed is because $\liminf_{t\rightarrow T} y(t)=-\infty$, for certain $T$.

So, now you can see for which functions $f$ (and initial datas $y(0)$) the equation for $y$ does NOT blow up. Those are the functions $f$ that prevent the shocks.

Hopefully this helps.

%%%% Expanded answer (see comments below) %%%%

Let me try to give you a family of functions such that there is NO shock. The same idea may be used to show some families of f leading to shocks.

The idea is to get some assumptions on $u_0$ and $f$ such that we prove that $y'>0$. Then, as $y(0)<0$, we obtain a global bound for the minimum of the derivative (so, no shocks are formed).


1) $$ u_0< 0 $$

2) $$ f\leq0 $$

(so, $u(t)$ remains non-positive for all times where the solution exists). To see that notice that the evolution of $$ U(t)=\max_x u(x,t)=u(\tilde{x}(t),t) $$ reads $$ U'(t)=-f_x(\tilde{x}(t),t)U(t). $$

3) Assume also that $\infty>c\geq f_x(x)\geq C>0$ and $f_{xx}(x)\geq 0$ for all $x$.

Then the previous ODE for $y$ implies $$ y'\geq -y(y+2f_x(X(t)). $$ We need $$ y(t)+2f_x(X(t))>y(t)+2C>0. $$ Thus, if $u_0$ is such that $y(0)+2C>0$ (or $0>\min_s (u_0)_x(s)>-2C$, which is a 'smallness' requirement), we have $y'(h)>0$ (for $h$ close to zero), which means $y(t)>y(0)>-2C$ (so, our assumptions on the initial data propagates in time).

  • $\begingroup$ Thank you for your response. This is indeed the way I have been going about this, mostly following Whitham's book (Linear and Nonlinear waves,$ \S$ 13.14). Although this presentation clarifies the problem, I don't see how it can then be exploited to constrain $f$, other then putting in functions and numerically integrating. $\endgroup$
    – Nick P
    Aug 15, 2015 at 17:18
  • $\begingroup$ Ok, let me add another answer because I do not have enough space in this comment. $\endgroup$
    – guacho
    Aug 15, 2015 at 20:52

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