I am trying to minimize the following equation
$$ C(\rho) = \| I-\sqrt\rho \nabla. \frac{1}{\rho} \nabla \sqrt\rho\|$$
where $I(x,y)$ and $\rho(x,y)$ are functions of x and y.
I found solution for this equation in a paper where new variables were introduced
$$ \alpha^4 = \rho$$ $$ \vec{F} = \frac{1}{\beta}\nabla \beta$$
then the equation is transformed to $ C(\rho, \alpha, \vec{F})= min$
and the solution is
$$ C = \|I+|\vec{F}|^2 - \nabla . \vec{F}\|^2 + \|\vec{F}- \frac{\nabla \alpha^2}{\alpha^2}\|^2 + \epsilon \|\vec{F}\|^2$$
where $\epsilon$ is positive real penalization parameters. The details of this derivation is not included in the paper.
I want to understand how this solution is derived.
it seems to me if I solve
$$ I-\sqrt\rho \nabla. \frac{1}{\rho} \nabla \sqrt\rho = 0 $$
It is easier because this is Poisson equation $$\sqrt\rho \nabla. \frac{1}{\rho} \nabla \sqrt\rho = I$$
$$\sqrt\rho \nabla^2 \frac{\sqrt\rho}{\rho} = I$$
Are the two solutions are similar or Am I am missing something