It’s true! All the passages in the proof that you already know are reversible.
Edit:
Let $ [u] \stackrel{\text{df}}{=} {u_{+}}(p) - {u_{-}}(p) $ for all $ p \in \gamma_{0} = \{ (\xi(t),t) \mid t \in I \} $.
(R.H.S.): As $ \xi'(t) = \dfrac{[f(u)]}{[u]}(\xi(t),t) $ for all $ t \in I $, we have
\begin{align}
& [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) = 0 \\ \Longrightarrow \quad
& 0 = \int_{I}
\Big( [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) \Big) \cdot
\phi(\xi(t),t) ~
\mathrm{d}{t}.
\end{align}
Let $ v = (v_{1},v_{2}) = (1,- \xi'(t)) $. We can then write the previous formula as follows:
\begin{align}
0
& = \int_{I} ([f(u)] v_{1} + [u] v_{2}) \phi ~ \mathrm{d}{s} \\
& = \int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
\mathrm{d}{s} -
\int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~
\mathrm{d}{s}.
\end{align}
We have taken $ \phi \in {C_{c}^{1}}(\Omega) $, so $ \text{supp}(\phi) = \omega_{-} \cup (\gamma_{0} \cap \text{supp}(\phi)) \cup \omega_{+} $.
This is the crucial point: We are going to use the Divergence Theorem in the ‘non-standard’ way:
\begin{align}
\int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
\mathrm{d}{s}
& = \int_{\partial \omega_{+}}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~\mathrm{d}{s} \\
& = \iint_{\omega_{+}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t} +
\iint_{\omega_{+}}
\left(
\frac{\partial u}{\partial t} + \frac{\partial f(u)}{\partial x}
\right) \phi ~ \mathrm{d}{x} \mathrm{d}{t}.
\end{align}
Note that the last integral is $ 0 $ because our solution is classical outside the shock. We now have the following equations:
$$
\int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
\mathrm{d}{s}
= \iint_{\omega_{+}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t},
$$
and, similarly,
$$
- \int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~ \mathrm{d}{s}
= \iint_{\omega_{-}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t}.
$$
Summing up term by term:
\begin{align}
0
& = \int_{I} \Big( [f(u)] v_{1} + [u] v_{2} \Big) \phi ~ \mathrm{d}{s} \\
& = \iint_{\omega_{+} \cup \omega_{-}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t} \\
& = \int_{\text{supp}(\phi)}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t}.
\end{align}
This is exactly what we were supposed to prove.
Hope it helps!
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
$$
x_s(t) = 2 \sqrt{t}
$$
Best Answer
The classical solution $u=g(x-ut)$ where $g = u(\cdot,0)$ denotes the initial data is correct. One notes that the characteristic curve starting at $x_0=1^-$ intersects the characteristic curve starting at $x_0=1^+$ already at time $t=0$ (see representation of the base characteristics in the $x$-$t$ plane below). Hence, since the classical solution is no longer well-defined in the vicinity of $x=1$, we need to look for weak solutions.
Because the characteristics intersect at $(x,t)=(1,0)$, a shock wave is generated. On the left of the shock located at the position $x=\sigma(t)$, we have $u^-=\frac{x}{1+t}$. On the right of the shock, we have $u^+=0$. The shock trajectory $\sigma(t)$ is given by the Rankine-Hugoniot condition $$ \sigma'(t) = \frac{u^-+u^+}2 =\frac12\left(\frac{\sigma(t)}{1+t} + 0\right) $$ with the initial condition $\sigma(0)=1$. Hence, $\sigma(t)=\sqrt{1+t}$, and $$ u(x,t) =\left\lbrace \begin{aligned} &\tfrac{x}{1+t}, && x< \sqrt{1+t}\\ &0, && x> \sqrt{1+t} \end{aligned}\right. $$