LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedaudited 2026-09-06

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The Burgers slope obeys a Riccati law along characteristics

Statement

Let $u \in C^{2} (Ω)$ solve $u_{t} + u u_{x} = 0$ on an open $Ω \subseteq R^{2}$ . Let $I$ be an open interval containing zero and $X \in C^{1} (I)$ satisfy $(t, X (t)) \in Ω$ and $\dot{X} = u (t, X)$ . Then $q (t) : = u_{x} (t, X (t))$ satisfies

$\dot{q} = - q^{2}, q (t) = \frac{q (0)}{1 + t q (0)}$

for every $t \in I$ ; in particular $1 + t q (0)$ cannot vanish in $I$ .

Facts & Assumptions

Given: A $C^{2}$ classical Burgers solution and one of its projected characteristics.

Proof

technique · direct

1.1

Put $r (t, x) = u_{x} (t, x)$ . Differentiate $u_{t} + u u_{x} = 0$ in $x$ to obtain $r_{t} + u r_{x} + r^{2} = 0$ ; equality of mixed partials holds since $u$ is $C^{2}$ .

givenalgebra

2.1

Along $\dot{X} = u$ , the total-derivative chain rule gives $\dot{q} = r_{t} (t, X) + u (t, X) r_{x} (t, X)$ , so $\dot{q} = - q^{2}$ .

step 1.1givenalgebra

3.1

Set $c = q (0)$ and $h (t) = (1 + t c) q (t) - c$ . Then $h (0) = 0$ and $h^{'} = - q h$ , so differentiation shows $h (t) \exp (\int_{0}^{t} q (s) d s)$ is constant and hence zero. Thus $(1 + t c) q (t) = c$ . If $c = 0$ , this gives $q = 0$ ; if $c \neq 0$ , the identity rules out a zero denominator in $I$ and yields the displayed formula.

step 2.1algebra∎

Depends on

Used by

Dependency tree · two levels

7 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

Part III: Analysis of Partial Differential Equations (standard reference, not scraped)