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The characteristic equations for a first-order quasilinear equation are $\dot{x}=x$, $\dot{y}=y$, $\dot{u}=2u$, with characteristic parameter $s$. Through the prescribed point they giveFor a nonzero projected initial point, the characteristic projections are rays from the origin, not trajectories crossing through the origin to the opposite ray. If $x_0=y_0=0$, a classical solution graph requires $u_0=0$, because the original equation reduces there to $0=2u$.

Parameterize the initial line by $(x,y,u)=(a,1-a,a^3)$. Along its characteristic, $x+y=e^s>0$, $a=x/(x+y)$ and $u=a^3{e}^{2s}$. ThereforeDifferentiation verifies the equation, and substitution of $x+y=1$ verifies the initial data. Characteristic reachability for radial transport shows why the domain is this positive half-plane: each point there lies on precisely one characteristic from the data line. Rays in $x+y<0$ never meet that line, so their values are not determined by these data. Allowing a negative scaling factor would incorrectly join two different characteristic rays through a degenerate origin.

With diffusion absent, the characteristic equations for a first-order quasilinear equation are $du/dt=0$, $dx/dt=q^{\prime }(u)$. A characteristic curve labelled by its starting point $s$ isThis determines a smooth single-valued solution while $s\mapsto x$ is one-to-one. Its derivative isAll characteristic curves move right because $q^{\prime }>0$. Because $q^{\prime \prime }<0$, increasing portions of the initial profile have slower characteristic curves ahead than behind, so they steepen. The first positive gradient catastrophe occurs atThis 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 $X=(x-x_s(t))/ϵ$ and $u\simeq U(X)$. The leading balance is $(-c+q^{\prime }(U))U^{\prime }=(D(U)U^{\prime }{)}^{\prime }$, where $c={\dot{x}}_s$. Integrating once and matching to the state behind givesMatching the state ahead requires $F(u_{+})=0$ and gives the Rankine-Hugoniot speedThe strictly concave graph of $F$ has zeros at the two states and lies above its chord, hence is strictly positive between them. Since $D>0$, $U^{\prime }>0$ 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 $q^{\prime }(u_{-})>c>q^{\prime }(u_{+})$: characteristic curves enter the viscous shock from both sides. Reversing the states would demand a decreasing profile, contradicting the sign of $F/D$.