Limiting solution profile of nonuniform transport equation

Question

For an upcoming course I need to know something about solving PDE's. I'm using P. Olver's Introduction ot Partial Differential Equations for this. I'm stuck on this question for a while now(ex. 2.2.18):

Suppose the initial data $u(0,x) = f(x)$ of the nonuniform transport equation $$ u_t+(x^2-1)u_x = 0 $$ is continuous and satisfies $f(x) \to 0$ as $|x|\to\infty$. What is the limiting solution profile as $t\to\infty$ and $t\to-\infty$.

I know that the characteristic curves are the solutions to $$ \frac{dx}{dt} = x^2-1,$$ so $$\frac{1}{2}\log\left|\frac{x-1}{x+1}\right| = t+k. $$

The general solution is $u(t,x) = v\left(\frac{1}{2}\log\left|\frac{x-1}{x+1}\right| - t\right)$ where $v$ is some $C^1$ function. I have no idea where to go from here.

Answer 1

When integration gives $\log|f|=g + k$ it is advisable to apply exponential map right away, and get rid of absolute value: $f = Ae^g$ where $A$ is a new undetermined constant. In this way, $$\frac{1}{2}\log\left|\frac{x-1}{x+1}\right| = t+k$$ turns into $$ \frac{x-1}{x+1} =A e^{2t} $$ leading to a nicer general solution, $$u(t,x) = v\left(e^{-2t}\frac{x-1}{x+1}\right) $$ Here $v((x-1)/(x+1))=f(x)$, hence $v(x) = f((1+x)/(1-x))$. Conclusions about the limiting solution profile:

• As $t\to+\infty$, $u(t,x)\to v(0)=f(1)$.
• As $t\to-\infty$, $u(t,x)\to v(\infty)=f(-1)$.

Both of these make sense if you think of what the transport with velocity $x^2-1$ does. The point $x=1$ is a source and $x=-1$ is a sink: so, as time goes forward, the solution matches the value at the source; and similarly after time reversal.