# How to understand the integral?

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!

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.