# Derivative and Jacobian determinant of solution of ODE [closed]

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 $$Jf = det \nabla f$$.

• 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. – Deane Yang Apr 10 '19 at 22:27
• @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. – user124345 Apr 11 '19 at 11:51
• 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. – user539887 Apr 11 '19 at 12:15

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 $$B_{ik}(t) = \partial_k f_i(\Phi(x,t),t)$$.
• 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) – user124345 Apr 11 '19 at 17:07
• @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 $\nabla \Phi$. – Skeeve Apr 11 '19 at 18:18