= Solution
With <diffusion> absent, the <characteristic equations for a first-order quasilinear equation> are $du/dt=0$, $dx/dt=q'(u)$. A <characteristic curve> labelled by its starting point $s$ is
$$
\boxed{u=u_0(s),\quad x=s+tq'[u_0(s)],\qquad
u(x,t)=u_0\!\left(x-tq'[u(x,t)]\right).}
$$
This determines a smooth single-valued solution while $s\mapsto x$ is one-to-one. Its <derivative> is
$$
u_x=\frac{u_0'(s)}{1+tq''[u_0(s)]u_0'(s)}.
$$
All <characteristic curves> move right because $q'>0$. Because $q''<0$, increasing portions of the initial profile have slower <characteristic curves> ahead than behind, so they steepen. The first positive <gradient catastrophe> occurs at
$$
\boxed{t_* =\inf_{s:u_0'(s)>0}\frac{-1}{u_0'(s)q''[u_0(s)]}.}
$$
This is a minimum when attained. The numerator at such a point is positive and the denominator decreases to zero from above, so $u_x\to+\infty$.
Resolve the thin transition with $X=(x-x_s(t))/\epsilon$ and $u\simeq U(X)$. The leading balance is $(-c+q'(U))U'=(D(U)U')'$, where $c=\dot x_s$. Integrating once and matching to the state behind gives
$$
\boxed{D(U)U'=F(U)=q(U)-q(u_-)-c(U-u_-).}
$$
Matching the state ahead requires $F(u_+)=0$ and gives the Rankine-Hugoniot speed
$$
\boxed{c=\frac{q(u_+)-q(u_-)}{u_+-u_-}.}
$$
The strictly <concave> graph of $F$ has zeros at the two states and lies above its chord, hence is strictly positive between them. Since $D>0$, $U'>0$ through the layer. A heteroclinic profile increasing with $X$ can therefore connect the rear state at $X=-\infty$ to the front state at $X=+\infty$ only when $\boxed{u_-<u_+}$. The <concavity> also gives $q'(u_-)>c>q'(u_+)$: <characteristic curves> enter the <viscous shock> from both sides. Reversing the states would demand a decreasing profile, contradicting the sign of $F/D$.
Back to article page
