# Null Lagrangians and “Local Degree”

Let $u: U\subset\mathbb R^n \rightarrow \mathbb R^n$ be a smooth function, $U$ bounded. Let $x_0$ and $r$ be such that $B_r(x_0)$ is disjoint from $\partial U$. Let $\eta$ be a smooth bump function supported in $B_r(x_0)$ with total integral one. Define the local degree of $u$ at $x_0$ as

$$\int_U \eta(u) \det(Du) dx.$$

I am trying to prove that this integral is an integer. This is an exercise in Chapter 8 of Evans' "Partial Differential Equations." The preceding exercise has us show that $\eta(u) \det(Du)$ is a null lagrangian, so the integral depends only on the restriction of $u$ to $\partial U$. So, it seems that we need to use this fact somehow, and look at a function whose boundary values agree with those of $u$. But, I do not see a way to do this. Any suggestions?

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This is not an easy exercise (at least for me). I would do the following: (1) the integral does not depend on $\eta$ as long as $\eta$ satisfies the constraints of the problem. (Try to differentiate the integral with $t\eta_1+(1-t)\eta_2$ with respect to $t$, and see that the derivative is $0$). (2) by Sard's lemma, we can make sure that the support of $\eta$ does not contain any critical values of $u$ and fits within a neighborhood which $u$ covers nicely (as a covering map) (3) change the variables to get a finite sum of integrals of the kind $\pm \int \eta(x)\,dx$. – user53153 Dec 25 '12 at 6:31

This question need the change variables formula, the degree is actually the cardinal of $u^{-1}(x_0)$, so we have to discuss whether $Du$ is invertible near each of this point. Then change variable to $y=u(x)$.