First-Order Linear PDEs and the Method of Characteristics

The method of characteristics is the entry point to the entire theory of first-order PDEs and to hyperbolic PDEs more broadly. In fluid dynamics and gas dynamics, the (nonlinear, quasilinear) analogue of this method — see Quasilinear PDEs and Characteristic Curves — governs shock formation in the Burgers equation and traffic-flow models. In geometric optics, the eikonal equation (a nonlinear first-order PDE for the phase of a wave) is solved by characteristic "light rays," directly generalizing this construction. In control theory, the Hamilton–Jacobi–Bellman equation is solved along characteristic trajectories that are optimal control paths. In numerical analysis, characteristics underlie the CFL condition and semi-Lagrangian schemes for advection-dominated PDEs, where grid points are transported along approximate characteristic paths rather than updated by finite differences alone. In probability, the Kolmogorov forward/backward equations for pure-drift (no-diffusion) Markov processes reduce, via characteristics, to deterministic ODE flows — the bridge from PDEs to stochastic processes. Finally, this article's transversality condition is the seed of the general theory of Well-Posedness of PDE Problems, including the celebrated Cauchy–Kovalevskaya theorem for more general systems.

Ask the tutor: How does this characteristic-curve method change when the coefficients $a, b, c, f$ are allowed to depend on the unknown $u$ itself, as in the quasilinear case?

Picture the plane $D$ filled with a vector field $(a(x,y), b(x,y))$, like arrows showing wind direction at every point. The characteristic curves are the trajectories a leaf would follow if dropped into that wind — they are exactly the flow lines of the vector field, found by solving $dx/dt = a$, $dy/dt = b$ as you would any first-order ODE system. Now watch what the PDE says about $u$ along one of these flow lines. If $(x(t), y(t))$ is a characteristic curve, the chain rule gives

But the left side of the PDE is exactly $a u_x + b u_y$, so along the characteristic,

This is a genuine ODE in the single variable $t$, with $u$ evaluated along the curve playing the role of the unknown. The two-variable PDE has been sliced into a one-parameter family of one-variable ODEs, one for each characteristic curve threading through $D$. Solving the PDE is now a two-step process: find the characteristics (geometry), then solve the ODE along each one (analysis), then use the data on $\Gamma$ to fix the constant of integration for each curve individually. The picture below shows a vector field, one characteristic curve threading through it, and a transversal data curve $\Gamma$ crossing every characteristic exactly once — the geometric condition that makes the Cauchy problem well-posed.

Example 1 (variable coefficients, asymmetric data). Solve

on the region $x>0$. Here $a(x,y) = x$, $b(x,y)=1$, $c=0$, $f(x,y) = x^2$. The data curve is $\Gamma = \{(s,0): s>0\}$ with $x_0(s)=s$, $y_0(s)=0$, $g(s)=s^3$. Transversality check. $J(s) = a\,y_0' - b\,x_0' = (s)(0) - (1)(1) = -1 \neq 0$ for all $s>0$, so the theorem applies and guarantees a unique local $C^1$ solution. Characteristics. Solve $dx/dt = x$, $dy/dt = 1$, with $x(0)=s$, $y(0)=0$. Then $y(t) = t$ and $x(t) = s e^{t} = s e^{y}$. Solving for $s$: $s = x e^{-y}$. ODE along the characteristic. Since $c=0$, along the curve $du/dt = f = x(t)^2 = s^2 e^{2t}$. With $u(0) = g(s) = s^3$,

Back-substitute. Using $t = y$ and $s = xe^{-y}$:

Verification. Compute $u_x = 3x^2 e^{-3y} + x(1-e^{-2y})$ and $u_y = -3x^3 e^{-3y} + x^2 e^{-2y}$. Then

matching $f = x^2$. At $y=0$: $u(x,0) = x^3 \cdot 1 + \frac{x^2}{2}(1-1) = x^3$, matching the data. Both checks confirm the solution.

Example 2 (nonzero $c$, transport-reaction equation). Solve

Here $a=1$, $b=2$, $c=3$, $f(x,y)=e^y$. Data curve: $x_0(s)=0$, $y_0(s)=s$, $g(s)=s^2$. Transversality. $J(s) = a y_0' - b x_0' = (1)(1) - (2)(0) = 1 \neq 0$ everywhere — a globally transversal line, so the theorem gives a solution on all of $\mathbb{R}^2$ here (the characteristics sweep out the whole plane). Characteristics. $dx/dt = 1,\, dy/dt = 2$, $x(0)=0,\, y(0)=s$. So $x(t) = t$, $y(t) = 2t + s$, giving $s = y - 2x$. ODE along the characteristic. Along the curve, $f(x(t),y(t)) = e^{2t+s}$, so

Integrating factor $\mu = e^{3t}$: $\dfrac{d}{dt}(e^{3t}u) = e^{3t}e^{2t+s} = e^{s}e^{5t}$. Integrate:

With $u(0) = g(s) = s^2$:

Back-substitute $t = x$, $s = y-2x$:

Verification. At $x=0$: $u(0,y) = y^2\cdot 1 + \frac{e^y - e^y}{5} = y^2$, matching the data exactly. Differentiating and substituting into $u_x + 2u_y + 3u$ (straightforward but tedious chain-rule bookkeeping) reproduces $e^y$; the reader can confirm the $e^y$ term contributes $e^y + 2e^y+3\cdot\frac{e^y}{5}\cdot(\dots)$ collapsing correctly because the particular solution $\frac{e^y}{5}$ alone already solves $u_x+2u_y+3u=e^y$ when the transient $-\frac{e^{y-5x}}{5}$ term is annihilated by the operator (its characteristic exponent $-5$ matches $3 + a(-2)\cdot(\text{rate})$ exactly by design of the integrating factor), and the homogeneous piece $(y-2x)^2e^{-3x}$ satisfies the homogeneous equation $u_x+2u_y+3u=0$ since it is constant along characteristics up to the factor $e^{-3x}$ decaying at rate matching $c=3$.