Quasilinear PDEs and Characteristic Curves

Quasilinear characteristic theory is the computational backbone of gas dynamics: the Euler equations, when written in one space dimension, admit Riemann invariants that are constant along characteristic curves with state-dependent speed $u\pm c$ (fluid velocity plus/minus sound speed), directly generalizing the scalar theory above. In traffic engineering, the Lighthill–Whitham–Richards model $\rho_t + q(\rho)_x = 0$ is a quasilinear conservation law whose characteristic speed $q'(\rho)$ depends on the density itself, exactly producing the "faster catches slower" crossing seen in Example 1 and explaining real-world traffic jam formation. In chromatography and multiphase flow, quasilinear transport equations with state-dependent speeds model how a chemical concentration profile self-steepens as it moves through a column. Numerically, characteristic tracing underlies Godunov-type and semi-Lagrangian schemes for conservation laws, and the crossing-time computation of Example 1 is precisely how numerical shock-capturing methods know when to switch from smooth interpolation to weak-solution (Rankine–Hugoniot) machinery, foreshadowing Shock Formation and Weak Solutions. In control theory and image processing, quasilinear transport (e.g., the eikonal-type equations in level-set methods) governs front propagation with speed depending on curvature or intensity, again solved by characteristic tracing until crossing.

Ask the tutor: How does the classical characteristic breakdown at $\det DF = 0$ get resolved by weak (discontinuous) solutions and the Rankine–Hugoniot jump condition?

In the linear theory, $a(x,t)$ and $b(x,t)$ were given functions, so the characteristic curves $dx/ds = a$, $dt/ds = b$ could be solved once and for all, independent of $u$. Here the same ODEs appear, but with $a$ and $b$ evaluated at $u(x,t)$ — a quantity we don't yet know. The trick is to treat $u$ along the characteristic as a third unknown function of the parameter $s$, governed by its own ODE $du/ds = c$. This converts one PDE into a closed system of three coupled ODEs in three unknowns $x(s,s_0), t(s,s_0), u(s,s_0)$, launched from each point $s_0$ of the initial curve. The PDE is satisfied automatically once this system is solved, because differentiating $u$ along the characteristic direction reproduces exactly the left side of the PDE. Geometrically, each characteristic is a straight line in space only when $a,b$ are constants; in general it is a curve in $(x,t)$-space whose tangent direction $(a,b)$ is dictated by the value of $u$ carried along it — but that value never changes along the curve unless $c \neq 0$ forces it to. This is the crucial qualitative difference from the constant-coefficient wave equation: because $u$ is transported at a speed depending on $u$, points with larger $u$-values can move faster (or slower) than points with smaller values, so characteristics starting apart can converge and eventually intersect. The picture below shows this for the model equation $u_t + u\,u_x = 0$ (inviscid Burgers' equation), where a characteristic through $(x_0, 0)$ has slope $1/u_0(x_0)$ in the $(x,t)$-plane and taller initial values race ahead only until the crossing time.

:::svg[Characteristics for $u_t + u u_x = 0$ with data $u_0(x) = 2 - x$ on $[0,1]$: lines converge and cross at $t=1$]

Example 1 — genuinely nonlinear flux (traffic-type equation with asymmetric data). Solve

Here $a = u$, $b=1$, $c=0$. The characteristic system with parameter $s = x_0$ (the initial $x$-coordinate) and $\sigma = t$ is

Since $du/dt=0$, $u$ is constant $=u_0(x_0)$ along its characteristic, so $x(t) = x_0 + u_0(x_0)\,t$. On the middle ramp $u_0(x_0) = 3-2x_0$ for $x_0\in(0,1)$, so

Solving for $x_0$ when $t < 1/2$: $x_0 = \dfrac{x - 3t}{1-2t}$, giving

For $x \le 3t$ (behind the ramp, coming from $x_0\le 0$), $u=3$; for $x \ge 1+t$ (ahead, from $x_0\ge 1$), $u=1$. Check continuity at $t=0$: at $x=3t=0$, formula gives $u=3$, matching; at $x=1+t=1$, formula gives $u = (3-2)/1 = 1$, matching. The characteristic from $x_0=0$ has speed $3$ (line $x=3t$); from $x_0=1$ has speed $1$ (line $x=1+t$). Since $3>1$, the fast line behind catches the slow line ahead: they meet when $3t = 1+t$, i.e. $t=1/2$. Indeed the denominator $1-2t$ in our formula vanishes exactly at $t=1/2$, blowing up — the classical solution exists only for $0\le t < 1/2$, exactly matching the crossing predicted geometrically. This is the mechanism of shock formation to be developed subsequently.

Example 2 — quasilinear equation with a genuine source term (asymmetric, nonzero $c$). Solve

Here $a=x$, $b=u$, $c=u^2$; the Cauchy curve is $t_0(s)=1$, $x_0(s)=s$, $u_0(s)=s$ for $s>0$. Check noncharacteristic: $b\,x_0' - a\,t_0' = u_0(s)\cdot 1 - x_0(s)\cdot 0 = s \ne 0$ for $s>0$. Good, the theorem applies on $s>0$. Characteristic ODEs, parameter $\sigma$, with $x(0)=s$, $t(0)=1$, $u(0)=s$:

Now invert. From $x=se^\sigma$, we get $s = x e^{-\sigma}$. Substitute into the $t$-equation:

This transcendental relation implicitly defines $\sigma = \sigma(x,t)$; but we can instead express $u$ directly. From $u = s/(1-s\sigma)$ and $x = se^\sigma$, note $1-s\sigma = e^{1-t}$ (from the $t$-formula, $t-1 = -\ln(1-s\sigma)$, so $1-s\sigma = e^{1-t}$). Hence

Also $s = x e^{-\sigma}$, and we still need $\sigma$ in terms of $x,t$. But actually we can eliminate $\sigma$ differently: from $u = s e^{t-1}$, so $s = u e^{1-t}$; and from $1-s\sigma = e^{1-t}$, $\sigma = \dfrac{1-e^{1-t}}{s} = \dfrac{1-e^{1-t}}{u\,e^{1-t}}$. Substitute into $x=se^\sigma$:

This is an implicit equation $F(x,t,u)=0$ determining $u$ near $t=1$; it is exactly of the type guaranteed solvable by the implicit function theorem (part 2–3 of the theorem), since at $t=1,\ \sigma=0$: $x = u\cdot 1\cdot \exp(0) = u$, recovering the data $u=x$ correctly, and $\partial u/\partial x$ there is finite (checked directly since the exponent's coefficient $\to$ a removable form as $t\to1$, $\sigma\to 0$). For instance, at $s=2$ (so $u_0=2$), the characteristic is $x(\sigma)=2e^\sigma$, $u(\sigma) = 2/(1-2\sigma)$, $t(\sigma) = 1-\ln(1-2\sigma)$; at $\sigma = 0.2$: $x = 2e^{0.2}\approx 2.443$, $u = 2/(1-0.4) = 3.333$, $t = 1-\ln(0.6)\approx 1.511$. One checks directly $x\,u_x+u\,u_t=u^2$ holds along this family by the general proof (Step 4), without needing a closed form for $u(x,t)$.