Since the question is old, I'll give the complete solution, not just the $x_s$ part.
Before $T_c$
Draw the $xt$ coordinates, with $t$-axis pointing up. From the given $u(x,0)$, we find three families of characteristic lines:
- vertical lines $x=x_0$ with $x_0\le 0$
- slope $1/2$ lines $t=\frac12(x-x_0)$ with $0<x_0<1$
- vertical lines $x=x_0$ with $x_0\ge 1$
There is a gap between the families 1 and 2: it's a triangular region bounded by $x=0$ and $t=\frac12 x$.
This is the rarefaction wave: within this region, all characteristic lines go through $(0,0)$, the origin of the wave.
Therefore, the line through $(x,t)$ has slope $t/x$, which yields
$$
u(x,t)=x/t ,\qquad 0<x< 2t
$$
This is family 4 of characteristics.
The families 2 and 3 appear to overlap. This means they are separated by shock wave, which originates at $x=1$ at time $t=0$ and
moves to the right with velocity $\frac12(2+0)=1$ (the mean of velocities in front and behind the shock, as the jump condition says). Its trajectory is $x=1+t$.
The front edge of rarefaction wave $x=2t$ meets the shock wave $x=1+t$ when $t=1$. Thus, $T_c=1$. The meeting point is $(2,1)$.
After $T_c$
The characteristics of family 2 are no more; the shock is between families 3 and 4. The speed of shock wave as it passes through $(x,t)$ is the mean
of velocities in front and behind the shock:
$$
\frac{1}{2}\left( \frac{x}{t} +0 \right) = \frac{x}{2t}
$$
Therefore, the trajectory of shock wave is described by the ODE
$$
\frac{dx}{dt}=\frac{x}{2t}
$$
Solve this separable ODE with the initial condition $x(1)=2$ to get
$$
x_s(t) = 2 \sqrt{t}
$$
This is the trajectory of shock for $t>1$.
Second method, conservation law
Introduce the quantity $P(t)=\int_{-\infty}^\infty u(x,t)\,dx$. It is actually independent of $t$ because
$$\frac{dP}{dt} = \int_{-\infty}^\infty u_t \,dx = - \int_{-\infty}^\infty (u^2/2) _x \,dx = (u^2/2) \bigg|_{-\infty}^\infty =0$$
Since $P =2$ at $t=0$, it stays at $2$ for all times. At time $t>T_c$ the function $u$ is equal to $x/t$ for $0< x <x_s$, and is zero otherwise. Thus, its integral is
$$
\frac{1}{t} \frac{x_s^2}{2}
$$
Equating the above to $2$, we once again get
$$
x_s(t) = 2 \sqrt{t}
$$
Here is a sketch of the characteristic curves in the $x$-$t$ plane, which equation is given by
$$
\begin{aligned}
x'(t) &= u(x(t),t) \\
&= u(x(0),0)
\end{aligned}
$$
Those curves intersect already at time zero:
Thus, a shock wave arises.
The Burgers' equation is rewritten in the conservative form $u_t + f(u)_x = 0$, where $f(u) = \frac{1}{2}u^2$. The shock wave with speed $s$, left state $u_L = 1$ and right state $u_R = 0$ writes
$$
u(x,t) =
\left\lbrace
\begin{aligned}
&1 &&\text{if}\quad x<st \\
&0 &&\text{if}\quad st<x \, .
\end{aligned}
\right.
$$
The speed of shock must satisfy the Rankine-Hugoniot condition $s = \frac{f(u_R) - f(u_L)}{u_R - u_L}$, i.e.
$s = \frac{1}{2}$.
Here is a modified sketch of the $x$-$t$ plane, which accounts for the shock:
Best Answer
$$u_t+C(u)u_x=0\quad\text{where }C(u)\text{ is a given function}$$ GENERAL SOLUTION :
The system of characteristic differential equations is : $$\frac{dt}{1}=\frac{dx}{C(u)}=\frac{du}{0}$$ A first equation of characteristic cuves comes from $du=0\quad\to\quad u=c_1$ .
A second equation of characteristic cuves comes from $\frac{dt}{1}=\frac{dx}{C(c_1)}\quad\to\quad x-C(c_1)t=c_2$
The general solution of the PDE is expressed on the form of implicit equation $\Phi\left(c_1,c_2\right)=0$ where $\Phi$ is any differentiable function of two variables. $$\Phi\left(u\:,\:x-C(u)t\right)=0$$ This is a manner to express any relationship between the two variables. This is equivalent to express the relationship by any function $F$ : $$x-C(u)t=F(u)$$ where $F$ is any derivable function.
In general, this implicit equation cannot be solved for $u$ on closed form.
DETERMINATION OF THE FUNCTION $F$ according to the condition $u(0,t)=g(t)$ :
$0-C\left(g(t)\right)t=F\left(g(t)\right)$
Let $g(t)=X \quad\to\quad t=g^{-1}(X)\quad$ where $g^{-1}$ is the inverse function of $g$.
$$-C(X)g^{-1}(X)=F(X)$$ Thus the function $F$ is now determined, given the functions $C$ and $g$.
PARTICULAR SOLUTION FITTED WITH THE GIVEN CONDITION :
With the particular function $F$ found above :
$x-C(u)t=F(u)$ with $F(u)=-C(u)g^{-1}(u)$
$$x-C(u)t=-C(u)g^{-1}(u)$$ $$x+C(u)\left(g^{-1}(u)-t \right)=0$$ $g^{-1}(u)=t-\frac{x}{C(u)}$ $$u=g\left(t-\frac{x}{C(u)}\right)$$ The result is on the form of implicit equation. Solving for $u$ in order to obtain an explicit form $u(x,t)$ is generally not possible, except in case of particular functions $C$ and $g$.