Limit of the solution of $u_t + (x^2 - 1)u_x = 0$ I was trying to find the limit when $t \to +/-\infty$ of the solution of the pde $u_t + (x^2 - 1)u_x = 0$ with initial condition $u(0,x) = f(x)$ where $f$ is continuous and $f \to 0$ when $|x| \to \infty$.
So I found the characteristic $\xi = \frac{1}{2} \ln \left|\frac{x-1}{x+1} \right| - t$, which gave me the solution
$u(t,x) = f \left( \frac{x+1+(x-1)e^{-2t}}{x+1-(x-1)e^{-2t}} \right)$
So I concluded that $u(t,x) \to f(1)$ as $t \to \infty$ and $x \neq -1$, and $u(t,x) \to f(-1)$ as $t \to \infty$ and $x = -1$. But the solution book states:
$u(t,x) \to f(1)$ as $t \to \infty$ and $x > -1$
$u(t,x) \to f(-1)$ as $t \to \infty$ and $x = -1$
$u(t,x) \to 0$ as $t \to \infty$ and $x < -1$
Where is my mistake? What do I do for $t \to -\infty$?
Thank you very much.
 A: You wrote : I found the characteristic $\xi = \frac{1}{2} \ln \left|\frac{x-1}{x+1} \right| - t \tag 1$, which gave me the solution $u(t,x) = f \left( \frac{x+1+(x-1)e^{-2t}}{x+1-(x-1)e^{-2t}} \right) \tag 2$
Without the intermediate steps between $(1)$ and $(2)$ it isn't possible to point out where exactly the mistake is. Probably, the treatment of the absolute value isn't correct.
I agree with the equation of the characteristic curves :
$$\frac{1}{2} \ln \left|\frac{x-1}{x+1} \right| - t=c$$
So, a form of general solution is :
$$u(x,t)=\Phi\left( \frac{1}{2} \ln \left|\frac{x-1}{x+1} \right| - t\right)$$
where $\Phi(X)$ is any differentiable function.
An equivalent form, with the related function $F$ :
$$\Phi(X)=F\Big(\exp(2X)\Big)$$
$X=\frac{1}{2} \ln \left|\frac{x-1}{x+1} \right| - t \quad\to\quad \exp(2X)=\left|\frac{x-1}{x+1} \right|e^{-2t}$
$$u(x,t)=F\left(\left|\frac{x-1}{x+1} \right|e^{-2t} \right) \tag 3$$
where $F$ is any differentiable function.
Condition : $u(0,x)=f(x)=F\left(\left|\frac{x-1}{x+1} \right| \right)$
Let $\quad \left|\frac{x-1}{x+1} \right|=\chi \quad 
\begin{cases}
\text{if } |x|>1 \quad\to\quad \chi=\frac{x-1}{x+1} \quad\to\quad x=\frac{1+\chi}{1-\chi} \quad\to\quad F(\chi)=f\left(\frac{1+\chi}{1-\chi}\right)\\
\text{if } |x|<1 \quad\to\quad \chi=-\frac{x-1}{x+1} \quad\to\quad x=\frac{1-\chi}{1+\chi} \quad\to\quad F(\chi)=f\left(\frac{1-\chi}{1+\chi}\right)\\
\end{cases}$
Now, the function $F(\chi)$ is determined. You just have to put it into $(3)$ 
with $\chi=\left|\frac{x-1}{x+1}\right| e^{-2t} $ and simplify in both cases. 
IN ADDITION, after further calculus I also got to :
$$u(t,x) = f \left( \frac{x+1+(x-1)e^{-2t}}{x+1-(x-1)e^{-2t}} \right)\quad\text{if}\quad x+1-(x-1)e^{-2t} \neq 0 \tag 4$$


*

*Case $x>-1\quad;\quad$  $t$ starting from $0$ and $t\to\infty$


$x+1>(x-1)e^{-2t} \quad\implies\quad $ The condition $(4)$ is always fulfilled any $t\geq 0$.
Thus, the above solution $(4)$ is valid and $e^{-2t}\to 0 \quad\implies\quad u(t,x)\to f(1)$


*

*Case $x=-1$


Again, the condition $4$ is fulfilled. The above solution simplifies to $\quad u(t,-1)= f(-1)\quad $ any $t$.


*

*Case $x<-1\quad;\quad$  $t$ starting from $0$ and $t\to\infty$


The condition $(4)$ is not fulfilled any $t$. 
For a given $x<-1$, the function $\quad y(t)=x+1-(x-1)e^{-2t}\quad$ starts from $y(0)=2$ and decreases to $(x+1)<0$. Hence, $y=0$ for a value $t_m$.
When $t\to t_m$ the argument of the function $f$ in equation $(4)$ tends to infinity. In the wording of the problem, it is specified that $f(\infty)=0$.
Hence $\quad u(t_m,x)=0$.
Beyond this point $t>t_m\;,\quad$ the PDE $\quad u_t+(x^2-1)u_x=0\quad$ is satisfied by the trivial solution $\quad u(t,x)=0$, with initial condition $u(t_m,x)=0$.
Thus, in this case, $u(t,x)=0$ for $t\to\infty$.
