First Order Partial Differential Equations: Notes and FAQs

A first-order PDE involves the unknown u and its first derivatives, such as a(x,t)·ux + b(x,t)·ut = c(x,t,u). The method of characteristics solves it by finding special curves in the xt-plane — the characteristics — along which the PDE collapses into an ordinary differential equation that can be integrated directly.

The flagship example is the transport equation ut + c·ux = 0, which says a quantity is carried along at speed c without changing: pollution drifting down a river, a traffic density wave moving along a road. Its solution is simply the initial profile shifted by ct — a result the characteristics reveal in one line.

Key Concepts

• Characteristic curves: curves satisfying dx/a = dt/b = du/c; along them, u evolves by an ODE instead of a PDE.
• Characteristic ODEs: the system dx/ds = a, dt/ds = b, du/ds = c parametrised by s along each curve.
• Non-characteristic initial curve: initial data must be prescribed on a curve nowhere tangent to the characteristics, otherwise the data is redundant or contradictory.
• Constant along characteristics: for the homogeneous transport equation, u is literally constant on each characteristic line.
• Shocks: when characteristics cross, the solution becomes multi-valued and a discontinuous shock forms — the mathematics of traffic jams.

How It Works

1. Write the characteristic ODEs: from a·ux + b·ut = c, form dx/a = dt/b = du/c.
2. Solve for the curves: integrate dx/dt = a/b to get the characteristic curves (e.g. x − ct = constant).
3. Propagate u along them: integrate du/ds = c along each curve (for c = 0, u is constant).
4. Trace back to the initial data: for a point (x, t), follow its characteristic back to t = 0 and read off the initial value there.
5. Verify: differentiate the candidate solution and confirm it satisfies both the PDE and the initial condition.

Worked Example

Solve ut + 3ux = 0 with u(x, 0) = sin x. The characteristic ODE is dx/dt = 3, giving x = 3t + s, i.e. each characteristic is the line s = x − 3t = constant. Along it, du/dt = 0, so u is constant: u(x, t) = u(s, 0) = sin(s) = sin(x − 3t). Verification: ut = −3cos(x − 3t) and ux = cos(x − 3t), so ut + 3ux = −3cos(x−3t) + 3cos(x−3t) = 0; and u(x, 0) = sin x. The sine wave drifts rightward at speed 3 with its shape perfectly preserved.

Common Misconceptions

• Characteristics are just a computational trick. They carry physical meaning — the speed at which information and signals actually travel.
• Solutions are always smooth. Crossing characteristics create shocks (sudden jumps), requiring weak-solution theory beyond the classical method.
• Initial data can be given on any curve. Data on a characteristic curve itself is either redundant or inconsistent — the curve must be non-characteristic.

Real-World Applications

• Traffic flow: the Lighthill-Whitham model is a first-order PDE whose shocks are traffic jams.
• Pollutant and sediment transport in rivers follows the advection equation.
• Chromatography: chemical concentrations separate into moving bands described by first-order PDEs.
• Optimal control (Hamilton-Jacobi-Bellman) and geometrical optics (eikonal equation) are nonlinear first-order PDEs solved by generalised characteristics.

First-order PDEs are solved by the method of characteristics: find the curves dx/a = dt/b = du/c along which the PDE becomes an ODE, integrate along them, and trace each point back to the initial data. For ut + c·ux = 0 this gives the travelling profile u(x,t) = u0(x − ct). When characteristics cross, shocks form — the doorway to nonlinear wave theory.