In order to understander the nonlinear elliptic equation with natural boundary condition,
$$\sigma\_2(D^2u)=0 \text{ in } \Omega$$
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 $R^n$, and $\Sigma\subset \Omega$, $\{e\_1,e\_2\}$ 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:
$$\int\_{\gamma}\partial\_{e\_1e\_1} u \, de\_1=\int\_{\partial \gamma} u = \partial\_{e\_1} u(\gamma(1))-\partial\_{e\_1}u(\gamma(0))$$
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,
>  $$E(u,\Sigma)=\int\_\Sigma \det(D^2u|\_{\Sigma}) \, de\_1\wedge de\_2=\int\_{\partial\Sigma}\hat E(u,u\_1,u\_2)?$$
>  $\hat E$ is a functional only relate to $u,\nabla u$. 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 come from the 1 dimensional version is true, and 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 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 
$$\exp(\operatorname{tr}(A))=\det(\exp(A))$$
We try to solve the equation $D^2 u=\exp\operatorname{tr} (A) \tag{$\*$},$ 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,
$$E(u,\Sigma)=\int\_\Sigma e^{\operatorname{tr}\left(\sum\_{k=0}^\infty \frac{(-1)^{k+1}}{k}{D^2(u)}^k\right)} \, de\_1\wedge de\_2$$
It seems much easy to find $\hat E(u)$ 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:


$$L\_i L\_j u(x)-L\_jL\_iu(x)=\sum\_k c\_{ij}^k(x)L\_ku(x)$$
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 only get a part of result(instead of identity, we could only get a inequality). 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:


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


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!



How to understand the integral?
If I understand your question correctly, the answer is affirmative for Ricci-flat (i.e. flat) surfaces (not necessarily embedded in a Euclidean space) according to equation (14) of


*Reilly, Robert C.*, [**Applications of the Hessian operator in a Riemannian manifold**](http://dx.doi.org/10.1512/iumj.1977.26.26036), Indiana Univ. Math. J. 26, 459-472 (1977). [ZBL0391.53019](https://zbmath.org/?q=an:0391.53019).


which is available at <http://www.jstor.org/stable/24891302?seq=1#page_scan_tab_contents> . However for general surfaces there is a Ricci curvature (which is just scalar curvature on 2D) term which is unlikely to be eliminated in general (e.g. for positive curvature or negative curvature surfaces it will have a definite sign).


Added later: actually one can just rewrite the integral using covariant derivatives with the Levi-Civita connection on $\Sigma$ as $\int\_\Sigma \frac{1}{2} (\nabla^\alpha \nabla^\alpha u \nabla\_\beta \nabla\_\beta u - \nabla^\alpha \nabla^\beta u \nabla\_\alpha \nabla\_\beta u)\ dg$, where $dg$ is the Riemannian volume form, and integrate by parts, picking up the curvature term thanks to the lack of commutativity of the covariant derivatives when applied to tensors.


