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An entropy solution is a weak solution of a nonlinear scalar conservation law that also satisfies entropy inequalities selecting physically admissible shocks. For convex fluxes, this excludes expansion shocks and restores uniqueness for bounded initial data.

The inviscid Burgers equation is the scalar conservation lawIts discontinuities obey the Rankine-Hugoniot condition $\sigma =(u_{-}+u_{+})/2$. Restricting to entropy solutions restores uniqueness for bounded initial data.

Write the Inviscid Burgers equation in conservation form asA bounded function $u$ is a weak solution with initial datum $u_0$ whenfor every compactly supported $C^1$ test function $\phi$.

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:when $u_{-}\ne u_{+}$. For every $c>0$, defineThe 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 $u_c(t,\cdot )\to 0$ in $L_{\mathrm{loc}}^1$ as $t\downarrow 0$; 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 condition would exclude.

Solved by gpt-5.6-sol high.

SetThe scalar conservation law is $u_t+a(t)f^{\prime }(u)u_x=0$. Its characteristic curve issuing from $\xi$ satisfiesThe Jacobian of the one-dimensional characteristic map isBefore characteristic crossing, differentiation with respect to $\xi$ givesBecause $u_0$ has compact support, $m$ is continuous and vanishes outside a compact set. It therefore attains its minimumby the hypothesis. Since $a(t)>\epsilon$, the function $A$ is strictly increasing and tends to infinity. There is consequently a unique first time $T_{\ast }>0$ satisfyingAt a minimizer of $m$, the numerator $u_0^{\prime }$ is nonzero because $f^{\prime \prime }$ is nonzero, while the denominator tends to zero as $t\uparrow T_{\ast }$. Hence the classical solution has gradient blow-up:

Solved by gpt-5.6-sol high.