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Let $U : \mathbb R^d \to \mathbb R^d$ be a smooth vector field, and let $F_t : \mathbb R \times \mathbb R^d \to \mathbb R^d$ be the corresponding smooth flow, defined by the differential equation $$\tfrac{d}{dt} F_t(x) = U(F_t(x)).$$ Let $f : \mathbb R^d \to \mathbb R$ be an integrable function, and consider the integral $$\int_{\mathbb R^d} f( F_t(x) ) ~dx.$$ To calculate this, I would make the change of variables $y = F_t(x)$, making the ansatz that there exists some function $\rho(y)$ so that the integral equals $$\int_{\mathbb R^d} f(y) \rho(y) ~dy.$$ From what I can gather from the paper I'm reading, it's common knowledge that this function $\rho$ has the form $$\rho(y) = \tfrac{1}{J_t(F_t^{-1}(y))} = \exp \left(-\int_0^t (\operatorname{div} U)(F_{s-t}(y) ) ~ds \right),$$ where of course $J_t$ stands for the Jacobian of the flow.

Could you please point me to a reference for this formula?

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up vote 4 down vote accepted

This is the Liouville formula. It is explained nicely in Ordinary Differential Equations by Arnold.

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Bravo, I knew it would be simple. Thanks, Andrey. –  Tom LaGatta Jun 8 '10 at 4:19
    
You're welcome. –  Andrey Rekalo Jun 8 '10 at 4:26
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For future finders of this question: the precise location Section 27.6 (Liouville's Theorem) on page 195 of Arnold's Ordinary Differential Equations (the green book, translated by Richard Silverman, not the yellow one translated by Cooke). –  Tom LaGatta Jun 8 '10 at 4:33
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