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
$$
x_s(t) = 2 \sqrt{t}
$$
$u_t+u_x^2=t$
$u_{tx}+2u_xu_{xx}=0$
Let $v=u_x$ ,
Then $v_t+2vv_x=0$ with $v(x,0)=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{ds}=2v=2v_0$ , letting $x(0)=f(v_0)$ , we have $x=2v_0s+f(v_0)=2vt+f(v)$ , i.e. $v=F(x-2vt)$
$v(x,0)=0$ :
$F(x)=0$
$\therefore v=0$
$u_x=0$
$u(x,t)=g(t)$
$u_t=g_t(t)$
$\therefore g_t(t)=t$
$g(t)=\dfrac{t^2}{2}+C$
$\therefore u(x,t)=\dfrac{t^2}{2}+C$
$u(x,0)=0$ :
$C=0$
$\therefore u(x,t)=\dfrac{t^2}{2}$
Best Answer
Characteristic lines are paths that carry information. If two characteristics collide, there will be two contradictory accounts of the solution there, and the continuum description will break down. At that point, usually $\partial_xu$ becomes infinite.
From the solution given, we know that along a characteristic $$\frac{dx}{dt}=u,\qquad\frac{du}{dt}=-\frac{u}{2}$$ so that the location of the characteristic originating at $x_0$ is $$x=x_0+2u_0(1-\exp{(-t/2)})$$ and$$\frac{dx}{dx_0}=1+2\frac{du_0}{dx_0}(1-\exp{(-t/2)})$$ We want to know the time and place where this first vanishes. This will be where $\frac{du_0}{dx_0}$ takes its most negative value, of $-1$ at $x_0=\pi$, and when $$1-2(1-\exp{(-t/2)})=0$$ which gives you $t_c$.
It might be instructive if you used Mathematica or something to plot the characteristic paths. You would see how they intersect and how they then produce a double covering of the plane.