# Does a first integral of $\dot x = f(x)$ satisfy $\nabla H \cdot f = 0$?

Let $$\dot x = f(x)$$ be an ODE, where $$f: U\to \mathbb{R}^n$$, and $$U\subset \mathbb{R}^n$$ is open. Then I know that a first integral is a function $$H: U\to \mathbb{R}^n$$ so that $$H(\varphi(t,x_0))=const$$ for every solution $$\varphi(t,x_0)$$ of the ODE.

My notes prove that $$\nabla H \cdot f = 0 \Rightarrow H$$ is a first integral.

My question is whether the reciprocal is true, that is, if $$H$$ is a first integral does it satisfy $$\nabla H \cdot f = 0$$? I cannot find any clues online.

My notes say that this is true for a planar complex polynomial system, that is, $$n=2$$ and $$f(x,y)=\big(P(x,y),Q(x,y)\big)$$, where $$P,Q$$ are polynomials. Is this true for an $$n-$$dimensional polynomial system? How about for an arbitrary function $$f$$?

Thank you very much!

Indeed,$$0=\frac{d}{dt}H(\varphi(t,x_0))\Big|_{t=0}=\nabla H(\varphi(t,x_0))\Big|_{t=0}\cdot \frac{d}{dt}\varphi(t,x_0)\Big|_{t=0}=\nabla H(x_0)\cdot f(x_0),$$ since $$\frac{d}{dt}\varphi(t,x_0)=f(\varphi(t,x_0)).$$
• Awesome! Then the result holds because $x_0$ can take any value in $U$, right? Thank you so much :)