1
$\begingroup$

Let $\Phi$ be the unique solution of $$\begin{cases} \frac{d}{dt}\Phi(x,t) = f(\Phi(x,t),t) \quad t >0 \\ \Phi(x,0) = x \quad x \in \mathbb{R}^N \end{cases}$$ where we have assumed $f$ smooth.

How do you prove that $\Phi(\cdot, t)$ is smooth and in particular that the following holds?

$$\begin{cases} \frac{d}{dt} \nabla \Phi(x,t) = \nabla_1 f(\Phi(x,t),t)\nabla \Phi(x,t) \quad t>0 \\ \nabla\Phi(x,0) = 1 \quad x \in \mathbb{R}^N \end{cases}$$ and $$\begin{cases} \frac{d}{dt} J \Phi(x,t) = \mathrm{div} f(\Phi(x,t),t)J \Phi(x,t) \quad t>0 \\ \nabla\Phi(x,0) = 1 \quad x \in \mathbb{R}^N \end{cases}$$ where $Jf = det \nabla f$.

$\endgroup$
  • 1
    $\begingroup$ This is more appropriate for math.stackexchange.com. I would suggestion writing everything out in their components $\Phi = (\Phi_1, \dots, \Phi_N)$, $x = (x_1, \dots, x_N)$, etc., and using partial derivatives and the chain rule you learned in calculus. The only hard step is differentiating the determinant, but you should be able to find a reference for that. $\endgroup$ – Deane Yang Apr 10 '19 at 22:27
  • $\begingroup$ @DeaneYang I wasn't able to find any reference for these results. I agree that they should be standard, but I'd like to see a detailed proof of them. $\endgroup$ – user124345 Apr 11 '19 at 11:51
  • $\begingroup$ For example, L. Perko Differential Equations and Dynamical Systems, pp. 80-84, or any other good textbook on ODEs. I wholeheartedly agree that this is a question for MSE. $\endgroup$ – user539887 Apr 11 '19 at 12:15
1
$\begingroup$

The smoothness of $\Phi$ usually is addressed in the textbooks on ODEs (some of them are discussed on MO and on MSE as well).

An application of the chain rule yields the equation for $\nabla \Phi$, as Deane Yang commented. The equation for $\det \nabla \Phi$ is also well-known, but I cannot immediately recall a book where its proof is presented. But the standard proof was already sketched by Deane Yang.

For any differentiable time-dependent matrix $A(t)$ by Jacobi's formula $(\det A(t))' = \mathop{\mathrm{tr}} (\mathop{\mathrm{adj}(A(t)) \cdot A'(t)})$. In particular, if $A'(t) = B(t) A(t)$ for some matrix $B$ then $$\det(A(t))' = \mathop{\mathrm{tr}} (\mathop{\mathrm{adj}(A(t)) \cdot B(t) \cdot A(t)}) = \mathop{\mathrm{tr}} (B(t) \cdot A(t) \cdot \mathop{\mathrm{adj}(A(t))}) = \det(B(t)) \det(A(t)),$$ where a standard property of adjugate matrix was used in the second equality. It then remains to substitute $A_{ij}(t) = \nabla_j \Phi_i(x,t)$ and $B_{ik}(t) = \partial_k f_i(\Phi(x,t),t)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. A side question: $\nabla \Phi$ if $\Phi: \mathbb{R}^N \to \mathbb{R}^N$ really the $N \times N$ Jacobian matrix of $\Phi$, right?(en.wikipedia.org/wiki/Jacobian_matrix_and_determinant) $\endgroup$ – user124345 Apr 11 '19 at 17:07
  • $\begingroup$ @Hiro Actually $\Phi \colon \mathbb R^{N+1} \to \mathbb R^N$ and $\nabla \Phi$ denotes the Jacobian matrix of the function $\Phi(\cdot, t)$ with respect to the spacial variables. Probably in your notation it would be more precise to write $\nabla_1 \Phi$ instead of $\nabla \Phi$. $\endgroup$ – Skeeve Apr 11 '19 at 18:18