In order to understander the nonlinear elliptic equation with natural boundary condition, $$\sigma_2(D^2u)=0 \ in \ \Omega$$$$\sigma_2(D^2u)=0 \text{ in } \Omega$$ I wish to understand the following integral, $$E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2$$$$E(u,\Sigma)=\int_\Sigma \det(D^2u|_\Sigma) \, de_1\wedge de_2$$ 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: $$\int_{\gamma}\partial_{e_1e_1}ude_1=\int_{\partial \gamma}u=\partial_{e_1}u(\gamma(1))-\partial_{e_1}u(\gamma(0))$$$$\int_{\gamma}\partial_{e_1e_1} u \, de_1=\int_{\partial \gamma} u = \partial_{e_1} u(\gamma(1))-\partial_{e_1}u(\gamma(0))$$ Where $\gamma: [0,1]\to \Omega$ is a $C^2$ curve and $e_1$ is the gradient direction along the curve.
Problem Can we get some integral expression for $E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2$$E(u,\Sigma)=\int_\Sigma \det(D^2u|_\Sigma) \, de_1\wedge de_2$ use just $u_1=\partial_{e_1}u,u_{2}=\partial_{e_2}u $$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, $$E(u,\Sigma)=\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)?$$$$E(u,\Sigma)=\int_\Sigma \det(D^2u|_{\Sigma}) \, de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)?$$ $\hat E$ 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.
$$\frac{1}{\mu(B)}\int_{\partial B(r,x_0)}u(x)dx=u(x_0)$$$$\frac{1}{\mu(B)}\int_{\partial B(r,x_0)} u(x) \, dx=u(x_0)$$
Use the identity $$exp(tr(A))=det(exp(A))$$$$\exp(\operatorname{tr}(A))=\det(\exp(A))$$ We try to solve the equation $D^2 u=exp^{tr (A)} ...(*)$,$D^2 u=\exp\operatorname{tr} (A) \tag{$*$},$ 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$$A=\log(D^2(u))=\sum_{k=0}^\infty \frac{(-1)^{k+1}}{k} {D^2(u)}^k$, so we have, $$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$$$$E(u,\Sigma)=\int_\Sigma e^{\operatorname{tr}\left(\sum_{k=0}^\infty \frac{(-1)^{k+1}}{k}{D^2(u)}^k\right)} \, de_1\wedge de_2$$ It seems much easy to finefind $\hat E(u)$ 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.
and we could use this to calculate $det(D^2u)$$\det(D^2u)$, but after calculate I do not find general principle and what shape should $\Sigma$ be to make the identity, $$\int_{\Sigma}det(D^2u|_{\Sigma})de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)$$$$\int_\Sigma \det(D^2u|_\Sigma) \, de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)$$ make sense in this way.
$$L_iL_ju(x)-L_jL_iu(x)=\sum_k c_{ij}^k(x)L_ku(x)$$$$L_i L_j u(x)-L_jL_iu(x)=\sum_k c_{ij}^k(x)L_ku(x)$$ 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.
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}$$f(x)=\det(D^2(u)|_\Sigma)^{1/2}$, which is a concave function so use Jensen inequality we could get following result:
$$\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}.$$$$\frac1{\operatorname{vol}\Sigma}\int_{\Sigma}(\det D^2u|_\Sigma)^{1/2} \le \det \left(\frac1{\operatorname{vol}\Sigma} \int_\Sigma D^2u|_\Sigma(x) \right)^{1/2}.$$
