My question is a follow-up on
Must a function be convex to be polyconvex?
Given a scalar function $\tilde H\colon [0,\infty) \times (0,\infty) \to \Bbb R$ defined as
$\tilde H(x,y) = \sum\limits_{a=1}^n N_a(x,y)E_a$
where $E_a$ are positive real numbers but $N_a$ are not convex. The $N_a$ depend on the choice of $n$. Think of them as piecewise polynomials and the $E_a$ are the interpolation values. For the remainder, the $N_a$'s are assumed to be twice continuously differentiable. In my framework, the $E_a$'s are determined via non-linear optimization and are the parameters of my optimization problem. For this to work, the function $\tilde H$ must be convex on its domain. Since the $N_a$ are not convex and the $E_a$ can take arbitrary values during iterations of the optimization problem, I have to implement constraints to make $\tilde H$ convex, which I want to ensure by considering the determinant of the Hessian. Example: For $n=1$ the determinant of the Hessian is given by
$$ E_1^2 \left[\frac{\partial^2 N_1}{\partial x \partial x} \frac{\partial^2 N_1}{\partial y \partial y} - \left[\frac{\partial^2 N_1}{\partial x \partial y}\right]^2 \right] = E_1^2 C(x,y)$$
and I would implement the non-linear constraint $E_1^2 C(x,y) > 0$. Note that I require the positive determinant of the Hessian only at the $n$ support points in the domain of $\tilde H$ such that $C(x,y)$ is a constant when writing down the inequality.
As I said, $n$ is a model parameter and I am having troubles to derive the formula for the determinant of the Hessian for $\tilde H = \sum\limits_{a=1}^n N_a(x,y)E_a$.
I would appreciate to get some help to find the analytical expression of the determinant of the Hessian -- in case this is tractable.