I am attempting to determine the behaviour of characteristic lines that arise from my solution to the PDE:
$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0$
Assuming that $u(x, 0)$ has a typical size U and varies over a typical length scale L (where both are positive constants)
Where I then nondimensionalise the PDE using the variables
$\tau=\frac{Ut}{L}, \zeta=\frac{x}{L}, \phi=\frac{u}{U}$, and then I get the nondimensional PDE:
$\frac{\partial\phi}{\partial\tau} + \phi\frac{\partial\phi}{\partial\zeta} = 0$
The general solution I get is:
$\phi(\zeta,\tau)=F(\zeta-\phi\tau)$ for arbitrary F.
For $\phi(\zeta,\tau)$ I have got the solution, given intial conditions $\phi(\zeta,\tau)=\phi_0(\zeta)$ at $\tau=0$:
$\phi=\frac{\zeta}{1+\tau}$
Given the slope $\zeta'(\tau)$ extends from $-2, -1, 0, 1, 2$ on the $\zeta$ axis (think of $\zeta$ on the x-axis, and $\tau$ on the y-axis). I am now asked to determine what happens when
$\phi_0(\zeta)=sgn(\zeta)$, and it says to consider the function which is equal to
$f(x)=-1$ when $x<-1$, $f(x)=x$ when $-1<x<1$ and $f(x)=1$ for $x>1$
and to look back at the original variables and consider what happens with the length scale variable "L".
I thought that it would go something like this
$sgn(\zeta-\phi\tau)=sgn(\frac{x-ut}{L})$ and then we try and determine what happens to the characteristic lines as L $\rightarrow$ $0$.