An initial- and boundary-value problem for Burgers' equation with no solution Prove that there is no solution to the following Cauchy problem:
$$\begin{align}
u_t+uu_x& =0&\quad x&\in(-1,1), t\gt0 \label{1}\tag{1}\\
u(x,0)&=x&\quad x&\in[-1,1] \label{2}\tag{2}\\
u(-1,t)&=-1& \quad t&\geq0 \label{3}\tag{3}\\
u(1,t)&=1&\quad t&\geq0 \label{4}\tag{4}
\end{align}$$
My attempt: Using the method of characteristics, I found the classical solution 
$$
u=\frac{x}{1+t} \quad t\gt{-1}
$$
which satisfies conditions \eqref{1} and \eqref{2}, but not \eqref{3} and \eqref{4}.

But how can I show that there is also no weak solution to this problem? 

Thanks in advance. 
 A: In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:

The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $x\simeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $x\simeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.

*The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=\pm 1$, which have been represented outside the interior domain to keep the figure readable.
**The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.
