Burgers characteristic breaking ($x=\xi +t{u}_0(\xi ),u_x=\frac{u_0^{\prime }(\xi )}{1+t{u}_0^{\prime }(\xi )}$)

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.