ODE, extension of the Wronskian

Question

I asked this question exactly about a month ago. Thought of asking again. I've answered part (a) using the definition of the wronskian

$$W(t)=W(t_0)\exp\left(\int^t_{t_0} \mathrm{trace}(A(s))ds\right)$$

Any suggestion to extend this and prove part (b) is appreciated. Does anybody has an approach to solve this problem ?

Answer 1

Let $F(t;y)=\dfrac{\partial x}{\partial y}(t;y)$.

From

\begin{align} x'(t)&=f(x)\\ x(0)&=y \end{align} we have $$ x(t;y) = y+\int_0^t f(x(s;y))ds$$

Taking derivative with respect to $y$ on both sides, we have

$$ \dfrac{\partial x(t;y)}{\partial y} = 1+\int_0^t \nabla f(x(s;y))\dfrac{\partial x}{\partial y}(x;y)ds $$

That is

$$ F(t;y) = I + \int_0^t \nabla f(x(s;y))F(s;y)ds$$

$F(t;y)$ is now the solution to the ODE

\begin{align} F'(t) &= \nabla f(x(t;y)) F(t)\\ F(0) &= I \end{align}

According to the consequence in (1), we have

$$\frac{d}{dt}\ln \det F(t;y) = tr(\nabla f(x(t;y))) = \nabla\cdot f(x(t;y))$$

This is exactly what you want.