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Homogeneous linear transport is solved by the inverse characteristic flow

Statement

Assume $a$ is regular enough that for each $(x, t)$ in a region $U$ the unique characteristic $s \mapsto (X (s; t, x), s)$ is defined from $s = t$ to $s = 0$ and its whole segment remains in $U$ , and assume these characteristics have the usual flow consistency. If $u \in C^{1} (U)$ solves

$u_{t} + a \cdot \nabla_{x} u = 0, u (x, 0) = u_{0} (x),$

then

$u (x, t) = u_{0} (X (0; t, x)) .$

Hence there is at most one classical solution on $U$ . Conversely, any $C^{1}$ function on $U$ satisfying this formula is that unique classical solution.

Facts & Assumptions

Given: A classical solution of the homogeneous transport equation on a region $U$ where every characteristic segment from time $t$ to time $0$ remains in $U$ , is unique, and satisfies flow consistency.

[L1]

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

[L2]

Along a characteristic, a transport solution satisfies the corresponding scalar ODE (A transport equation restricts to a linear ODE along each characteristic).

[L3]

The chain rule computes the derivative of a $C^{1}$ function along a $C^{1}$ curve (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Proof

technique · direct

1.1

Fix $(x, t) \in U$ and let $s \mapsto X (s; t, x)$ be its characteristic. Its whole segment to time $0$ lies in $U$ , so [L2] with $c = f = 0$ makes $v (s) = u (X (s; t, x), s)$ satisfy $v^{'} (s) = 0$ . Hence $v$ is constant and $u (x, t) = v (t) = v (0) = u_{0} (X (0; t, x))$ .

L2given

2.1

Step 1.1 proves the representation formula for every classical solution. The assumed characteristic flow is single valued, so two solutions with the same initial datum agree pointwise on $U$ .

givenstep 1.1

3.1

Conversely, suppose a $C^{1}$ function satisfies the displayed formula. Along the characteristic through $(x, t)$ , flow consistency gives $X (0; s, X (s; t, x)) = X (0; t, x) .$ Applying the formula at $(X (s; t, x), s)$ therefore makes $s \mapsto u (X (s; t, x), s)$ constant. By [L3] and the characteristic equation from [L1], its derivative is $u_{t} (X (s; t, x), s) + a (X (s; t, x), s) \cdot \nabla_{x} u (X (s; t, x), s) .$ Evaluating at $s = t$ proves the PDE at $(x, t)$ , while setting $t = 0$ gives the initial condition. Step 2.1 then gives uniqueness.

L1L3givenstep 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources

Victor Ivrii, Partial Differential Equations (standard reference, not scraped)