A profile travels to the right without changing shape. At speed 3, the value now observed at position 8 and time 2 came from position 2. Draw the straight line connecting (2,0) to (8,2) in a position–time diagram. The line is a characteristic: a path along which the PDE becomes an ordinary differential equation.
Let u(x,t) be a scalar field and c a constant speed. The equation uₜ + c uₓ = 0 describes ideal translation on the whole line. It does not include diffusion, acceleration or interactions between profile values.
Predict
For initial data f(x) = x², compare f(x − ct) with f(x + ct). Which moves the minimum to the right when c is positive? Locate the minimum first, then differentiate. This prevents a sign error that can survive an otherwise correct chain-rule calculation.
Experiment
With c = 3 and f(x) = x², u(x,t) = (x − 3t)². Follow three initial locations ξ, where ξ labels a characteristic.
ξ
x at t = 1
x at t = 2
Value along the path
−1
2
5
1
0
3
6
0
2
5
8
4
The paths remain parallel and never cross. Add a source F = 2. Along each path the value now increases by 2t, although the path locations stay the same. At (8,2), the new value is 4 + 4 = 8. Recompute another row without using the final formula.
Understand
Choose x(t) with x′(t) = c. The chain rule gives d[u(x(t),t)]/dt = uₜ + c uₓ. For zero right-hand side this derivative vanishes, so u(x,t) = f(x − ct). For a continuous source F and sufficiently smooth data, integration gives
u(x,t) = f(x − ct) + ∫₀ᵗ F(x − c(t − s),s) ds.
Every source contribution is sampled along the backward characteristic, not at the fixed final position x. If the equation additionally contains a u with constant a, the characteristic ODE is v′ + av = F. An integrating factor produces exponential attenuation and a correspondingly weighted source integral.
Data need not be specified on t = 0, but the data curve must be noncharacteristic locally. Parametrize it by (x₀(s),t₀(s)). The map from curve label and travel time to the plane has nonzero Jacobian when x₀′ − c t₀′ is nonzero. If this expression vanishes, the curve follows a characteristic: its data may be incompatible with the ODE or may leave neighboring characteristics unspecified. This is the geometric reason for the condition.
Apply
On a finite interval with c positive, trace characteristics backward to see whether they reach the initial line or the left boundary first. Supply boundary values at that inflow boundary. Explain why arbitrary additional right-boundary values generally conflict with values already transported there. Check continuity where the initial and inflow data meet. Then solve a sourced case by integrating along the actual path.
Continue
Constant-speed characteristics never intersect. In the next lesson the speed depends on the solution itself, so different labels can collide. The chain rule remains valid before that collision, but a smooth single-valued solution may cease to exist afterward.
Apply · explain · revise
Practice and reasoning
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For uₜ+2uₓ=0 with u(x,0)=x², find u(5,2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For uₜ+3uₓ=4 with zero initial data, find u(7,2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Derive u(x,t)=f(x−ct) for uₜ+cuₓ=0.
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Derive the source formula for uₜ+cuₓ=F(x,t), u(x,0)=f(x).
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Why can prescribing arbitrary data on a characteristic curve fail to determine a unique transport solution?
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Solve uₜ+cuₓ=−a u with initial f for constant a, and verify the result.
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Reflect and continue
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