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Before characteristic crossing, the method of characteristics for the Inviscid Burgers equation transports each initial value without changing it:Here $\xi$ labels the starting position. Suppose $u_0$ is a bounded continuously differentiable function with bounded derivative. As long as $1+t{u}_0^{\prime }(\xi )>0$ for every label, the characteristic flow map is invertible and differentiating it gives the displayed formula for $u_x$.^[ref]

If the negative minimum $m={\min }_{\xi }{u}_0^{\prime }(\xi )$ is attained, the first breaking time is $t_{\ast }=-1/m$. Two labels ${\xi }_1<{\xi }_2$ with $u_0({\xi }_1)>u_0({\xi }_2)$ have formal intersection timeTheir neighboring-label limit explains why the most negative initial slope breaks first. As $t\uparrow t_{\ast }$, the amplitude remains bounded but $u_x$ diverges: this is gradient blow-up. Continuing intersecting lines would assign incompatible values at one position; the post-breaking entropy solution instead uses shock waves. Nondecreasing data have no forward characteristic crossing by this mechanism.

For a concrete calculation in dimensionless quantities on the infinite line, choose $u_0(\xi )=1-\tanh \xi$, with limiting states $2$ and $0$. Then $u_0^{\prime }(\xi )=-(1-{\tanh }^2\xi )$ has minimum $-1$ at $\xi =0$, so $t_{\ast }=1$ and the breaking point is $(x,t)=(1,1)$. Along the central characteristic curve $x=t$, $u=1$ and $u_x=-1/(1-t)$ before breaking. The labels $\xi =\pm 1$ give intersecting formal lines at $t=1/\tanh 1\approx 1.313$, after the classical solution has already broken.

For $u_t+(c_0+bu)u_x=-\alpha u$, $b,\alpha >0$, the characteristic curves are $u=e^{-\alpha t}{u}_0(\xi )$ and $x=c_{0}t+\xi +(b/\alpha )(1-e^{-\alpha t})u_0(\xi )$. If $m=\max [-u_0^{\prime }]>0$, finite-time gradient blow-up occurs precisely for $\alpha <bm$, at $t_s=-{\alpha }^{-1}\log (1-\alpha /(bm))$. Equality $\alpha =bm$ still gives a positive Jacobian at each finite time; only the limiting map degenerates.

For isentropic gas with $P=K{\rho }^{\gamma }$, $\gamma >1$, the sound speed obeys $c^2=P^{\prime }(\rho )$. The one-dimensional mass and momentum equations areSince $c\propto {\rho }^{(\gamma -1)/2}$, these implyAdding and subtracting yields the Riemann invariantsA simple wave has one invariant constant throughout. For a right-moving wave into a resting state with sound speed $c_0$, $R_{-}=0$, so $c=c_0+(\gamma -1)u/2$ and the right characteristic speed is $c_0+bu$, $\boxed{b=(\gamma +1)/2}$.

For the damped scalar equation, characteristics from $\xi$ satisfyAssume smooth bounded initial data with bounded derivative, and the stated maximal compressive slope exists. ThenFor the damping threshold for a simple wave, if $m={\max }_{\xi }[-u_0^{\prime }(\xi )]>0$, put $\boxed{{\alpha }_c=bm}$. For $0<\alpha <{\alpha }_c$, a gradient blow-up first occurs atFor $\alpha >{\alpha }_c$, $J$ stays uniformly positive. At $\alpha ={\alpha }_c$ it still satisfies $J\ge e^{-\alpha t}>0$ for every finite time. At a point attaining the minimum slope, $u_x=u_0^{\prime }$ for all time, rather than blowing up. Hence the precise conclusion is no finite-time shock occurs when $\alpha \ge {\alpha }_c$, not the printed strict “if and only if $\alpha >{\alpha }_c$”. Equality only makes the limiting characteristic map degenerate as $t\to \infty$. The profile $u_0(\xi )=-\sin \xi$ directly verifies this endpoint distinction. If there is no compressive slope, take ${\alpha }_c=0$; no finite-time shock occurs for positive damping. A nonattained maximal slope is interpreted with a supremum and the corresponding infimum of breakdown times. The gas interpretation also requires $c_0+(\gamma -1)u/2>0$.

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: