Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $y=u(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.