PDE / Calculus of Variations problem from Evans I'm trying to learn Calculus of Variations and I can't solve the following problem from Chapter 8 of Larry Evans' PDE book. Could someone please help with a solution?
Let $\Sigma \subset \mathbb{R}^3$ denote the graph of the smooth function $u: U \to \mathbb{R},$ $U \subset \mathbb{R}^2.$ Then $$\int_U (1+|Du|^2)^{-2} \; \text{det}(D^2u) \; dx$$ represents the integral of the Gauss curvature over $\Sigma.$ Prove this expression depends only on $Du$ restricted to $\partial U.$
 A: That can be done, though it is a bit messy. In the following we let $f_{ij}$ denotes the partial derivative $\frac{\partial^2 f}{\partial x^i \partial x^j}$. Basically you want to show that for any $u$, 
$$ \frac{d}{dt}\bigg|_{t=0} \mathcal F (u_t) = 0, $$
where $u_t = u+ t\varphi$ for any $\varphi \in C^\infty_0(\Omega)$ and $\mathcal F (f) = \int_{\Omega} (1+ |Df|^2)^{-a} \det (D^2 f) \ \mathrm dx$. 
Direct calculations give 
\begin{align}
&\frac{d}{dt}\bigg|_{t=0} \mathcal F (u_t) \\
&=\int_\Omega \left(\frac{-2a \det (D^2 u)}{(1+ |du|^2)^{a+1}} \langle D u , D \varphi\rangle+ \frac{1}{(1+ |Du|^2)^a } ( \varphi_{11} u_{22} + \varphi_{22} u_{11} - 2\varphi_{12}u_{12})\right) \mathrm dx.\end{align}
For the second term, one use integration by part twice to get 
$$\int_\Omega \left( \frac{\varphi_{11} u_{22} - \varphi_{12} u_{12} }{(1+|Du|^2)^a}\right) \mathrm  dx = \int_\Omega \frac{2a\det D^2 u}{(1+ |Du|^2)^{a+1}} \varphi_1 u_1\mathrm d x.$$
and similarly
$$\int_\Omega \left( \frac{\varphi_{22} u_{11} - \varphi_{12} u_{12} }{(1+|Du|^2)^a}\right) \mathrm  dx = \int_\Omega \frac{2a\det D^2 u}{(1+ |Du|^2)^{a+1}} \varphi_2 u_2\mathrm d x.$$
thus these two terms cancel the first term and thus the derivative is zero. This implies $\mathcal F(u)$ is invariant under any interior perturbation and thus depends only on its value on the boundary. Clearly it does not depend on $u$. But it is not clear to me why it depends only on $Du$ (even when $a=2$ or $3/2$). 
A: The question includes a hint, which alludes to the Gauss-Bonnet theorem,
$$ \int_M K \,\mathrm{d}A + \int_{\partial M} k_g \,\mathrm{d}s = 2\pi \chi(M). $$
Since the integral corresponds to the first term with $M = \text{graph of } u$ which is homotopy equivalent to $U$ (so $\chi(M)=\chi(U)$), the result follows. 
So it remains to show that,
$$ \int_M K \,\mathrm{d}A = \int_U (1+|Du|^2)^{-3/2} \mathrm{det}(D^2u) \,\mathrm{d}x.$$
This follows from the fact that
\begin{align}
  K &= \frac{\mathrm{det}(D^2u)}{1+|Du|^2},\\
  \mathrm{d}A &= \sqrt{1+|Du|^2}\,\mathrm{d}x,
\end{align}
 with respect to the local paramerization $\varphi(x) = (x,u(x)).$ In theory this is not difficult to show, as you have an explicit formula for the Gaussian curvature in terms of $\varphi$ and its derivatives. The computation however is quite painful and I don't know if there are ways to simplify it.

Comment: One may wonder if there's an easier way of showing this. Since the quantity only depends on the values of $u$ on $\partial U,$ we may try to write it as the divergence of something and apply the divergence theorem.
Unfortunately, the Gauss-Bonnet theorem suggests this isn't possible. The RHS includes $\chi(M) = \chi(U),$ which is a topological invariant which doesn't straightforwardly depend on only it's boundary. So you can probably apply the divergence theorem to write the desired quantity as $\mathrm{div}\,F(u) + G(u)$ where $\int_U G(u)$ is independent of $u$ and corresponds to some topological quantity. However such an approach would also require some form of differential topology along the way.
