I wish to understand the following integral,
$$E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2$$
Where $u\in C^2(\Omega)$ is a solution of some elliptic equation, so we have $D^2u$ is semi-positive, $\Omega\subset R^n$ is a open set, $\Sigma$ is a arbitrary smooth surface equipped with the induce metric from $R^n$, $\{e_1,e_2\}$ is a pair of orthogonal biases of $\Sigma$.

I wish to understand if there is a similar result like Newton-Leibniz formula in the one dimensional case, which is the following:
$$\int_{\gamma}\partial_{e_1e_1}ude_1=\int_{\partial \gamma}u=\partial_{e_1}u(\gamma(1))-\partial_{e_1}u(\gamma(0))$$
Where $\gamma: [0,1]\to \Omega$ is a $C^2$ curve and $e_1$ is the gradient direction along the curve.

So my problem is following:

>**problem**
Can we get some integral expression for $E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2$ use just $u_1=\partial_{e_1}u,u_{2}=\partial_{e_2}u $, and the integral domain of the expression is $\partial \Sigma$? more precisely I wish we could find a functional $\hat E(u,u_1,u_2)$ such that,
$$E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)?$$
Because of $\hat E$ only related with $u,\nabla u$, I would like to say it is the lift of $E(u,\Sigma)$. And may be this is only true for some special domain $\Sigma$, for example $\Sigma$ is a Ball, this is exactly what I excepted.


**Motivation**

Now I need to explain my motivation why I except this is true and why I need this or it variation is true.
The motivation is this is the 1 dimensional version is true, which is crucial to establish **the mean value principle** for Laplace equation,

$$\frac{1}{\mu(B)}\int_{\partial B(r,x_0)}u(x)dx=u(x_0)$$

I just wish to generated the mean value principle to some special nonlinear elliptic equation by this way, although this mean value principle may be not exist, I still wish to explain why the mean value property failed by this way.

**Attempt**

I have four ways to attempt this problem, but there always emerge difficulties I could not settle.


1. Use the identity 
$$exp(tr(A))=det(exp(A))$$
We try to solve the equation $D^2 u=exp^{tr (A)} ...(*)$, we pretend it could be solved then we have: $A=log(D^2(u))=\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{k}{D^2(u)}^k$, so we have,
$$E(u,\Sigma)=\int_{\Sigma}e^{tr(\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{k}{D^2(u)}^k)}de_1\wedge de_2$$
It seems much easy to fine $\hat E(u)$ use stokes theorem with $E(u)$ under this form.
But there exists two problem, one is that the solvable of $(*)$ and there exists infinity many different solution of $(*)$ if my insight is right and I do not know how to proof the identity we could proved in this way is independent with the choice.


2.

We discretization the problem and consider it in $\mathbb Z^2$ which is a two dimensional affine subspace of $\mathbb Z^n$.

The advantage of discretization is that we can explicate calculate $u_{11},u_{12},u_{21},u_{22}$ now, in fact,

$$u_{11}(x,y) = h^2(u(x+2h,y)+u(x,y)-u(x+h,y)-u(x+h,y))        $$
$$u_{12}(x,y) = h^2(u(x+h,y+h)+u(x,y)-u(x+h,y)-u(x,y+h))       $$
$$u_{21}(x,y) = h^2(u(x+h,y+h)+u(x,y)-u(x+h,y)-u(x,y+h))       $$
$$u_{22}(x,y) = h^2(u(x,y+2h)+u(x,y)-u(x,y+h)-u(x,y+h))        $$

and we could use this to calculate $det(D^2u)$, but after calculate I do not find general principle and what shape should $\Sigma$ be to make the identity,
$$\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)$$
make sense in this way.

3.
Investigate the Frobenius integrable condition, which is just mean:

$$L_iL_ju(x)-L_jL_iu(x)=\sum_k c_{ij}^k(x)L_ku(x)$$
should be true, I tried to split $E(u,\Sigma)$ into several parts, and every part of it satisfied the Frobenius integrable condition, i.e.

$$E(u,\Sigma)=\sum_{i=1}^k E_i(u,\Sigma)$$

and $E_i$ satisfied the Frobenius integrable condition. And we investigate each $E_i$ first and combine the result we got together to establish a result for $E(u,\Sigma)$.

But the difficulties comes from that I do not know how to decompose $E(u)$ at all!

4.
The last strategy could not get a part of result. thanks to the elliptic condition we know $D^2u$ is semi-positive and the Principal minors of $D^2u$ is also semi-positive, so $D^2u|_{\Sigma}$ is semi-positive. we could consider the function $f(x)=Det(D^2(u)|_{\Sigma})^{1/2}$, which is a concave function so use Jensen inequality we could get following result:

$$\frac1{{\rm vol}\Sigma}\int_{\Sigma}(\det D^2u|_{\Sigma})^{1/2}\le\det\left(\frac1{{\rm vol}\Sigma}\int_{\Sigma} D^2u|_{\Sigma}(x)\right)^{1/2}.$$

May be according this could gain a mean-value inequality but I am not very sure.



May be all of these approaches are useless. In any case, I wish some result could be establish, whatever positive answer or negative answer. I will appreciate to any valuable advice or new idea, thank you very much!