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A transport equation restricts to a linear ODE along each characteristic

Statement

Let u∈C1(Ω) satisfy

ut+a⋅∇xu+c u=f,

and let X be a characteristic solving X′(s)=a(X(s),s). Then

ddsu(X(s),s)+c(X(s),s)u(X(s),s)=f(X(s),s).

Facts & Assumptions

Given: A C1 solution u of the transport equation and a characteristic X.

[L1]

A linear transport equation and its characteristics are defined by the displayed PDE and ODE (Linear transport equations and their characteristic flow).

[L2]

The total-derivative chain rule differentiates a composite by the gradient of the outer function applied to the derivative of the inner function (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[L3]

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative: continuous first partial derivatives on an open set imply total differentiability, with the Jacobian as derivative. Thus the stated C1 function u is totally differentiable at each point of its domain.

Proof

technique · direct
1.1L2L3

By [L3], u is totally differentiable. The differentiable characteristic has the first-order expansion X(s+h)=X(s)+hX′(s)+o(h) coordinatewise, so its space-time curve is totally differentiable with derivative (X′(s),1). Apply [L2] to the map s↦u(X(s),s); since the space-time velocity of the characteristic is (X′(s),1), this gives ddsu(X(s),s)=ut(X(s),s)+∇xu(X(s),s)⋅X′(s).

2.1L1step 1.1∎

By [L1], the characteristic ODE gives X′(s)=a(X(s),s), so substituting into step 1.1 and then using the PDE yields ddsu(X(s),s)=f(X(s),s)−c(X(s),s)u(X(s),s), which is the claimed scalar linear ODE along the characteristic.

Depends on

Used by

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources