ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Transport with growth and source along straight characteristics

Example

Let $b \in R^{n}$ be constant and let $q : R^{n} \to R$ be $C^{1}$ , and let $u_{0} \in C^{1} (R^{n})$ . The equation

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

has the explicit solution

$u (x, t) = e^{- t} u_{0} (x - b t) + (1 - e^{- t}) q (x - b t) .$

Facts & Assumptions

Given: The transport equation with constant velocity $b$ , unit zeroth-order coefficient, source $q (x - b t)$ , and datum $u_{0} \in C^{1} (R^{n})$ .

[L1]

The inhomogeneous transport equation is solved by the characteristic integrating-factor formula (The inhomogeneous linear transport equation has the characteristic integrating-factor formula).

Verification

technique · direct

1.1given

The characteristics are the straight lines $X (s; t, x) = x + (s - t) b$ , so $X (0; t, x) = x - b t$ , and along such a curve the source becomes $q (X (s; t, x) - b s) = q (x - b t)$ , which is independent of $s$ .

2.1L1step 1.1∎

Applying [L1] with $c \equiv 1$ gives $u (x, t) = e^{- t} u_{0} (x - b t) + \int_{0}^{t} e^{- (t - s)} q (x - b t) d s$ , and the integral equals $(1 - e^{- t}) q (x - b t)$ , so the displayed closed form follows.

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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)