= Solution
Write the <Inviscid Burgers equation> in <scalar conservation law>[conservation form] as
$$
u_t+\left(\frac{u^2}{2}\right)_x=0.
$$
A bounded function $u$ is a weak solution with initial datum $u_0$ when
$$
\int_0^\infty\!\int_{\mathbb R}
\left(u\varphi_t+\frac{u^2}{2}\varphi_x\right)dx\,dt
+\int_{\mathbb R}u_0(x)\varphi(0,x)\,dx=0
$$
for every compactly supported $C^1$ <test function> $\varphi$.
Across a straight discontinuity $x=\sigma t$, <integration by parts> on its two sides shows that the boundary terms cancel exactly when the <Rankine-Hugoniot condition> holds:
$$
\sigma(u_+-u_-)=\frac{u_+^2-u_-^2}{2},
\qquad
\sigma=\frac{u_-+u_+}{2}
$$
when $u_-\ne u_+$. For every $c>0$, define
$$
u_c(t,x)=
\begin{cases}
-c,&-ct/2<x<0,\\
c,&0<x<ct/2,\\
0,&\text{otherwise}.
\end{cases}
$$
The three jumps have left and right states $(0,-c)$, $(-c,c)$, and $(c,0)$, so their Rankine-Hugoniot speeds are respectively $-c/2$, $0$, and $c/2$, exactly the speeds of the displayed lines. Hence each $u_c$ satisfies the weak equation away from the origin and across every jump. Moreover, its nonzero support at time $t$ has length $ct$, so $u_c(t,\cdot)\to0$ in $L^1_{\mathrm{loc}}$ as $t\downarrow0$; its initial datum is therefore zero in the weak identity.
The zero function and all the distinct functions $u_c$ are bounded weak solutions with the same zero initial datum. Thus weak solutions are not unique. The central jump from $-c$ to $c$ is an expansion shock, which an <entropy solution>[entropy condition] would exclude.
Solved by gpt-5.6-sol high.
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