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Partial differential equations II
Lesson 27 of 75
Easy practice

First-order PDEs

Objective. Solve transport equations by characteristics.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

First-order PDEs link a function's partial derivatives linearly, and the method of characteristics solves them by converting the PDE into ordinary differential equations along special curves. For a transport equation uₜ + c·uₓ = 0, the characteristics are the lines x − ct = constant, along which u stays constant, because du/dt = uₜ + (dx/dt)uₓ vanishes when dx/dt = c. Given initial data u(x,0) = f(x), the solution is simply u(x,t) = f(x − ct), a rigid shift of the initial profile. Example: for uₜ + 2uₓ = 0 with u(x,0) = sin x, the solution is u(x,t) = sin(x − 2t), a wave translating rightward at speed 2.

Worked example

Solve the first-order PDE uₜ + 3uₓ = 0 with initial condition u(x,0) = x² using characteristics.

3 … 0
Illustrated card showing 3 … 0
  1. 1Characteristic curves satisfy dx/dt = 3, so x(t) = x₀ + 3t, meaning x − 3t = x₀ is constant along each curve.
  2. 2Along a characteristic, du/dt = uₜ + 3uₓ = 0, so u is constant on each curve, equal to its value at t = 0.
  3. 3At t = 0, u = x₀² = (x − 3t)², since x₀ = x − 3t.
  4. 4Hence the solution is u(x,t) = (x − 3t)², matching u(x,0) = x² when t = 0.

Answer: u(x,t) = (x − 3t)²

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Common mistake. A common error is solving uₜ + c·uₓ = 0 as u(x,t) = f(x + ct); the sign must match the characteristic direction, since characteristics satisfy x − ct = const, giving a shift by +ct not −ct in the wrong direction.

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