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Solve the adjoint transport equationbackward with terminal value zero. Along the characteristic flow map $Z_{t,s}(x)=e^{s-t}x$, the required solution isDifferentiation under the integral verifies the equation. If $\psi$ has compact support in $[0,T]\times [-R,R]$, then $\phi$ vanishes for $t>T$ and for $|x|>R$, so $\phi \in C_c^1$ as required.

When the initial datum is zero, inserting this $\phi$ into the weak formulation givesfor every $\psi \in C_c^1$. Thus $u=0$ almost everywhere. The difference of two bounded weak solutions has zero initial datum, so this proves uniqueness.

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.

A test function is a smooth function with compact support. Weak and distributional equations are identities obtained by integrating against every test function.