Normalizing a geometric flow liking mean curvature flow

Question

I try to normalize a flow, which is slight modification of mean curvature flow, $$ \partial_t F = -(H-1)\nu $$ where $F$ is position vector, $H$ is mean curvature, $\nu$ is normal vector. According to the Huisken's way, let $\tilde F = \psi F$ such that $$ \int d\tilde\mu_t = \int d\mu_0 $$ i.e. keep the area constant. So, we have $$ \tilde g_{ij} = \psi^2 g_{ij} ~~~~\sqrt {\tilde g} = \psi ^n \sqrt g $$ and since $$ \partial \sqrt g =(H-H^2)\sqrt g $$ we have $$ 0=\partial_t \int \psi^n \sqrt g \\ =\int \psi^{n-1}\psi' \sqrt g +\int \psi^n(H-H^2)\sqrt g $$ i.e. $$ \frac{\psi'}{\psi} = \frac{1}{n}\frac{\int (H^2 -H)\sqrt g}{\int \sqrt g} $$ let $$ h_1= \frac{\int H \sqrt g}{\int \sqrt g} ~~~~~~~~ h_2 = \frac{\int H^2 \sqrt g}{\int \sqrt g} $$ so $$ \tilde h_2 = \psi^{-2} h_2 ~~~~~~~~ \tilde h_1 = \psi^{-1}h_1 $$ Generally, for offset $\psi$ we must let $\tilde t =\tilde t(t)$, the way of Huisken is $$ \tilde t(t) = \int_0^t \psi^2(\tau) d\tau ~~~~~ \frac{dt}{d\tilde t}=\psi^{-2} $$ Then, we have $$ \partial _t \tilde F =\frac{dt}{d\tilde t}[\psi' F + \psi \partial_t F] =\frac{dt}{d\tilde t} [ \frac{\psi}{n} h_2 F -\frac{\psi}{n} h_1 F -\psi H \nu +\psi \nu ] $$ But the $\psi$ can't be completely offset. How to do the normalize of this flow ?