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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!

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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, Indiana Univ. Math. J. 26, 459-472 (1977). ZBL0391.53019.

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.

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  • $\begingroup$ thanks a lot for the reference and argument! The form of this theorem in the refference is exactly what I expect and help me a lot. In fact It does not matter for me to add the condition of Ricci flat, the residue difficulties for me is to get fully understanding with the term integral on boundary ... $\endgroup$ – Hu xiyu Dec 13 '17 at 15:37

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