The Method of Characteristics

Why did u(x, t) = f(x - ct) solve the transport equation? Because u stays constant along the moving lines x - ct = \text{const}. The method of characteristics turns this observation into a technique: find the curves along which a first-order PDE becomes an ordinary one.

Along a path x(t), the chain rule gives the rate of change of u as you ride it:

\frac{d}{dt}\,u\big(x(t), t\big) = u_t + \frac{dx}{dt}\,u_x.

Compare with u_t + c\,u_x = 0: if we choose the path so that \frac{dx}{dt} = c, the right-hand side is the PDE, so \frac{du}{dt} = 0. The unknown is constant along that path.

The recipe

A first-order PDE is replaced by a pair of ODEs — one for the curve, one for the solution along it:

\frac{dx}{dt} = c \;\Rightarrow\; x = x_0 + ct \qquad\text{(the characteristics)},

\frac{du}{dt} = 0 \;\Rightarrow\; u = u(x_0, 0) = f(x_0) \quad\text{(constant along each one)}.

Eliminating the label x_0 = x - ct recovers u(x, t) = f(x - ct). The information simply rides the characteristic lines out of the initial data.

The characteristics are the curves dx/dt = c in the (x, t) plane.
Along each characteristic the PDE collapses to an ODE — for the transport equation, simply du/dt = 0, so u is carried unchanged.
The solution at (x, t) equals the initial value at the foot of the characteristic through that point.

The lines that carry the solution

Each line below is a characteristic x = x_0 + ct rising through the (x, t) plane; u never changes as you travel up one. Change the speed c and the whole family tilts — a larger speed leans the lines further over, carrying the data faster. The highlighted line is the one starting at x_0 = 0.