Method of Characteristics, Quasilinear pde I am attempting to solve the quasilinear pde:
$u_t+uu_x=-u$;  $x\in \mathbb{R}$, $t>0$
$u(x,0)=-x/2$ ; $x\in \mathbb{R}$
I understand everything except from the last step.

If I make $\xi$ the subject of (s2) I get:
$\xi=\frac{2x}{1+e^{-t}}$, then subbing into (s1)
$u(\frac{2x}{1+e^{-t}},t)=-\frac{xe^{-t}}{1+e^{-t}}$
How do I get to what is written in the green box?
 A: I believe the confusion here is one of notation, in connection with the method of characteristics itself.
In doing the method of characteristics, one begins, in effect, by letting $x=x(\xi,t)$, so that
$$
u(x,t) = u(x(\xi,\hat t),\hat t) = \hat u(\xi,\hat t) \tag{1}
$$
Note that I've put a hat on the modified variables - that is, $t=\hat t$ and $u=\hat u$, but the hatted variables are for the $(\xi,\hat t)$ coordinates, while the unhatted variables are for the $(x,t)$ coordinates. This matters when dealing with $u_t$, as it is not the same as $\hat u_\hat t$. It is simply helpful when dealing with the mixed notation.
Now, we have
$$
\frac{\partial \hat u}{\partial \hat t} = \frac{\partial u}{\partial t}\frac{\partial t}{\partial \hat t} + \frac{\partial u}{\partial x}\frac{\partial x}{\partial \hat t} = \frac{\partial u}{\partial t} + \frac{\partial x}{\partial \hat t}\frac{\partial u}{\partial x}
$$
and the right-most expression matches the left side of our PDE if we require that
$$
\frac{\partial x}{\partial \hat t} = \hat u(\xi,\hat t)
$$
and this definition is the definition of the "characteristic". And using this definition, we get
$$
\frac{\partial \hat u}{\partial \hat t} = - \hat u
$$
Now, the work from here is provided in the image provided in the question, so we turn our attention to the last step. Here, we have two expressions which, when written in this clarified form, say
$$
\hat u(\xi,\hat t) = -\frac\xi2e^{-\hat t}\\
x(\xi,\hat t) = \frac\xi2(1+e^{-\hat t})
$$
However, we want $u(x,t)$, and so, noting (1), we have
$$
u(x,t) = u(x(\xi,\hat t),\hat t) = -\frac\xi2e^{-\hat t} = -\frac{x(\xi,\hat t)e^{-\hat t}}{1+e^{-\hat t}} = -\frac{xe^{-t}}{1+e^{-t}}
$$
To explain why the hatted notation isn't generally used - it's because, when performing the method of characteristics, the $\xi$ variable is simply used as a way to reference the characteristic being examined; that is, there are no derivatives with respect to it, and thus we can solve the system along a characteristic while keeping $\xi$ constant. Thus we could consider $\hat u(\xi,\hat t) = U(t)$ for any given $\xi$, and solve for $U(t)$. And so, it is not unusual, while doing a method of characteristics, to treat derivatives as being full derivatives in $t$, rather than "partial derivative in $t$ keeping $\xi$ constant".
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=-u$ , letting $u(0)=u_0$ , we have $u=u_0e^{-s}=u_0e^{-t}$
$\dfrac{dx}{ds}=u=u_0e^{-s}$ , letting $x(0)=f(u_0)$ , we have $x=f(u_0)+u_0(1-e^{-s})=f(ue^t)+u(e^t-1)$ , i.e. $u=e^{-t}F(x+u(1-e^t))$
$u(x,0)=-\dfrac{x}{2}$ :
$F(x)=-\dfrac{x}{2}$
$\therefore u=-\dfrac{e^{-t}(x+u(1-e^t))}{2}$
$-2u=xe^{-t}+u(e^{-t}-1)$
$u(1+e^{-t})=-xe^{-t}$
$u(x,t)=-\dfrac{xe^{-t}}{1+e^{-t}}=-\dfrac{x}{e^t+1}$
