TheoremStatement: Literature-sourcedProof: AI-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The inhomogeneous linear transport equation has the characteristic integrating-factor formula

Statement

Let $U \subseteq R^{n} \times R$ be a region. Assume $a \in C^{1} (U; R^{n})$ and $c, f \in C (U)$ , and assume that for each $(x, t) \in U$ there is a unique characteristic $X (s; t, x)$ on the closed interval with endpoints $0$ and $t$ , its space-time graph remains in $U$ , and the characteristic family has the usual flow consistency. If $u \in C^{1} (U)$ solves

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

then

$u (x, t) = e^{- \int_{0}^{t} c (X (τ; t, x), τ) d τ} u_{0} (X (0; t, x)) + \int_{0}^{t} e^{- \int_{s}^{t} c (X (τ; t, x), τ) d τ} f (X (s; t, x), s) d s .$

Conversely, any $C^{1}$ function satisfying this formula solves the transport equation on $U$ .

Facts & Assumptions

Given: A $C^{1}$ transport field $a$ , continuous coefficients $c, f$ , a classical transport solution, and a unique flow-consistent characteristic through each $(x, t) \in U$ whose whole segment to time $0$ remains in $U$ .

[L1]

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative makes the stated $C^{1}$ function totally differentiable on the open region; The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ then computes its derivative along a differentiable characteristic curve, whose space-time derivative is $(X^{'} (s), 1)$ .

[L2]

Along a characteristic, the PDE becomes the scalar linear ODE $v^{'} + c v = f$ (A transport equation restricts to a linear ODE along each characteristic).

[L3]

A scalar first-order linear ODE is solved by the integrating-factor formula (A scalar first-order linear ODE has a unique solution given by the integrating-factor formula).

[L4]

A characteristic satisfies $X^{'} (s) = a (X (s), s)$ , and uniqueness makes the characteristic family a single-valued flow (Linear transport equations and their characteristic flow).

Proof

technique · direct

1.1L2given

Fix $(x, t) \in U$ and let $v (s) = u (X (s; t, x), s)$ . The characteristic segment stays in $U$ , so [L2] gives $v^{'} (s) + c (X (s; t, x), s) v (s) = f (X (s; t, x), s)$ and $v (0) = u_{0} (X (0; t, x))$ .

2.1L3givenstep 1.1

If $t = 0$ , the displayed formula is exactly the initial condition. If $t \neq 0$ , apply [L3] to the scalar ODE from step 1.1 on the interval with endpoints $0$ and $t$ ; continuity of $c, f$ , the characteristic, and its in-domain graph makes the two composed coefficients continuous there, while oriented integrals cover either order of the endpoints. This yields the displayed formula for $v (t) = u (x, t)$ .

3.1L1L3L4given∎

Conversely, suppose a $C^{1}$ function satisfies the displayed formula and fix the characteristic $Y (s) = X (s; t, x)$ . Flow consistency gives $X (r; s, Y (s)) = Y (r)$ for every relevant $r, s$ . Substitution in the displayed formula therefore writes $v (s) = u (Y (s), s)$ exactly as the [L3] integrating-factor solution with coefficients $p (s) = c (Y (s), s)$ and $q (s) = f (Y (s), s)$ . Hence $v^{'} + p v = q$ . By [L1] and [L4], $v^{'} = u_{t} + a \cdot \nabla_{x} u$ , proving the PDE at $(x, t)$ . At $t = 0$ , [L4] gives $X (0; 0, x) = x$ and the integral vanishes, so the formula also gives $u (x, 0) = u_{0} (x)$ .

Depends on

Used by

Transport with growth and source along straight characteristics Example

Dependency tree · two levels

24 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

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