= Solution
Set
$$
A(t)=\int_0^t a(s)\,ds.
$$
The <scalar conservation law> is $u_t+a(t)f'(u)u_x=0$. Its <characteristic curve> issuing from $\xi$ satisfies
$$
X(t;\xi)=\xi+A(t)f'(u_0(\xi)),
\qquad
u(t,X(t;\xi))=u_0(\xi).
$$
The <Jacobian matrix>[Jacobian] of the one-dimensional characteristic map is
$$
X_\xi(t;\xi)=1+A(t)m(\xi),
\qquad
m(\xi)=f''(u_0(\xi))u_0'(\xi).
$$
Before <characteristic crossing>, differentiation with respect to $\xi$ gives
$$
u_x(t,X(t;\xi))
=\frac{u_0'(\xi)}{1+A(t)m(\xi)}.
$$
Because $u_0$ has <compact support>, $m$ is continuous and vanishes outside a compact set. It therefore attains its minimum
$$
m_*=\min_{\xi\in\mathbb R}m(\xi)<0
$$
by the hypothesis. Since $a(t)>\varepsilon$, the function $A$ is strictly increasing and tends to infinity. There is consequently a unique first time $T_*>0$ satisfying
$$
\int_0^{T_*}a(s)\,ds=A(T_*)=-\frac1{m_*}.
$$
At a minimizer of $m$, the numerator $u_0'$ is nonzero because $f''$ is nonzero, while the denominator tends to zero as $t\uparrow T_*$. Hence the classical solution has <gradient blow-up>:
$$
\lVert u_x(t,\cdot)\rVert_{L^\infty(\mathbb R)}\longrightarrow\infty
\qquad(t\uparrow T_*).
$$
Solved by gpt-5.6-sol high.
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