关于PDE论文中行列式Q对t的微分表达式推导的技术问询

Hey there, let's walk through this step by step—this is a standard result about ODE flow maps, so it's totally okay to get stuck on the determinant derivative at first! Here's how to unpack it:

1. Start with the core determinant derivative formula #

First, remember this key identity for any smooth, invertible matrix-valued function (A(t)):
$$
\frac{d}{dt} \det(A(t)) = \det(A(t)) \cdot \text{tr}\left( A(t)^{-1} \cdot \frac{d}{dt}A(t) \right)
$$
This comes from the cofactor expansion of the determinant, but the trace-based form is the most useful here. For our problem, (A(t) = \nabla_\alpha X^t(\alpha))—the Jacobian matrix of (X^t) with respect to (\alpha).

2. Swap derivatives to simplify the Jacobian's time derivative #

Since (X^t(\alpha)) is smooth (a standard assumption in PDE/ODE contexts), we can interchange the time derivative and spatial gradient:
$$
\frac{d}{dt} \nabla_\alpha X^t(\alpha) = \nabla_\alpha \frac{d}{dt} X^t(\alpha)
$$
From the original ODE, we know (\frac{d}{dt}X^t(\alpha) = v(X^t(\alpha), t)). Apply the chain rule to the right-hand side:
$$
\nabla_\alpha v(X^t(\alpha), t) = Dv(X^t(\alpha), t) \cdot \nabla_\alpha X^t(\alpha)
$$
Here, (Dv) is the Jacobian matrix of the vector field (v) (i.e., (Dv_{ij} = \partial v_i / \partial x_j)).

3. Plug into the determinant formula and simplify the trace #

Substitute our expressions into the determinant identity:
$$
\frac{d}{dt} Q(\alpha,t) = \det(\nabla_\alpha X^t(\alpha)) \cdot \text{tr}\left( \left(\nabla_\alpha X^{t(\alpha)\right)}{-1} \cdot Dv(X^t(\alpha),t) \cdot \nabla_\alpha X^t(\alpha) \right)
$$
Now use a critical trace property: trace is invariant under similarity transformations. That means (\text{tr}(A^{-1}BA) = \text{tr}(B)) for any invertible (A) and square (B). Applying this simplifies the trace term to (\text{tr}(Dv(X^t(\alpha),t))).

4. Link the trace to divergence #

The trace of (Dv) is exactly the divergence of (v)! By definition:
$$
\text{div } v(x,t) = \sum_{i=1}^d \frac{\partial v_i}{\partial x_i}(x,t) = \text{tr}(Dv(x,t))
$$

Putting it all together, we arrive at the desired equation:
$$
\frac{d}{dt}Q(\alpha,t) = (\text{div } v)(X^t(\alpha),t) Q(\alpha,t)
$$

内容的提问来源于stack exchange，提问作者Eugene