This is a repost concerning the same problem asked in Null Lagrangians and "Local Degree".
The problem is Evans' PDE exercise 8.5
Suppose $U$ is a bounded open subset on $\mathbb{R}^n$. Let $u: \mathbb{R}^n \to \mathbb{R}^n$ be smooth. Fix $x_0 \notin \partial U$, and choose a $C^1$ function $\eta: \mathbb{R}^n \to \mathbb{R}$ such that $supp(\eta) \subset B(x_0,r)$, where $r > 0$ is taken sufficiently small s.t. $B(x_0,r) \cap u(\partial U) = \emptyset$ and $\int_{B(x_0, r)} \eta =1$. Define the degree of $u$ relative to $x_0$ by $$\text{deg}(u,x_0) = \int_U \eta(u) \det(Du)dx.$$ Show that $\text{deg}(u,x_0)$ is an integer.
One observation (proved in Evans' exercise 8.4) is that $L(P,z,x) := \eta(z) \det(P)$ is a null lagrangian. So $\int_U \eta(u) \det(Du)dx$ depends only on $u|_{\partial(U)}$.
My questions:
- Using the fact that $L$ is null lagrangian, for each $u$, I think we could always define another function $w$ such that $w \equiv u$ on $\partial U$, and $w^{-1}(B(x_0,r)) \cap U = \emptyset$. In this case the degree should be identically $0$. I believe this is wrong, but I don't see which part is problematic;
- How should I approach this problem? I think we need to do change of variable somewhere because the form of the integral defining the degree; but here the sign of the $\det$ is changing.
Could you give some hint or suggestion? I appreciate it very much!