# Can we perturb the Dirichlet boundary conditions to make harmonic maps locally invertible?

While analyzing a variational problem, I came to the following question:

Let $$\mathbb D^n \subseteq \mathbb{R}^n$$ be the closed $$n$$-dimensional unit ball, and let $$f: \mathbb D^n \to \mathbb{R}^n$$ be a smooth orientation-preserving immersion. Denote by $$\omega_f :\mathbb D^n \to \mathbb{R}^n$$ the unique harmonic map satisfying $$\omega_f|_{\partial \mathbb D^n}=f|_{\partial \mathbb D^n}$$.

$$d\omega_f$$ must be invertible outside a set of measure zero in $$\mathbb D^n$$. Indeed, $$\omega_f$$ is real-analytic, and so is $$\det d\omega_f$$, which is not identically zero, since $$\int_{\mathbb D^n} \det d\omega_f = \int_{\mathbb D^n} \det df>0.$$

Now, the zero-set of a real-analytic function which is not identically zero has measure zero.

Question: Do there exist $$f_k \in C^{\infty}(\mathbb D^n, \mathbb{R}^n)$$ such that $$d\omega_{f_k} \in \text{GL}^+$$ are everywhere and $$f_k \to f$$ in $$W^{1,2}$$?

($$\omega_{f_k}$$ is the harmonic map corresponding to the Dirichlet problem imposed by $$f_k$$.)

Note that even though $$d\omega_f \in \text{GL}$$ a.e., it can "spend time" in both $$\text{GL}^+$$ and $$\text{GL}^-$$. Here is an example:

Let $$f : \mathbb D^2 \to \mathbb R^2$$ be defined by $$f(x,y) = (x-2y^2,y).$$ We have $$df=\left(\begin{matrix}1 & -4y \\ 0 & 1\end{matrix}\right)$$

and thus $$f$$ is an orientation-preserving immersion.

The solution to the Dirichlet problem in this case is $$\omega_f(x,y) = (x^2 - y^2 + x - 1,y)$$, so $$d\omega_f=\left(\begin{matrix}1+2x & -2y \\ 0 & 1\end{matrix}\right)$$ and $$\det(d\omega_f)=1+2x>0 \iff x>-\frac{1}{2}$$.

• In the highlighted Question, do you mean $f_k\to f$? Aug 24 '19 at 10:44
• Yes, thank you. This was a typo. Aug 24 '19 at 11:41

It seems that the answer is negative for dimension $$n=2$$ . I am not sure if higher dimensions can be reduced to the $$2D$$ case.

Here is the argument for $$n=2$$:

Suppose that there exist $$f_k \in C^{\infty}(\mathbb D^2, \mathbb{R}^2)$$ such that $$d\omega_{f_k} \in \text{GL}^+$$ everywhere and $$f_k \to f$$ in $$W^{1,2}$$.

The convergence $$f_k \to f$$ in $$W^{1,2}$$ implies that $$d\omega_{f_k} \to d\omega_f$$ in $$L^2$$. (This follows from the fact that the trace operator is a continuous surjection onto the fractional Sobolev space $$W^{1/2,2}(\partial \Omega)$$, see here for details).

This implies that $$\det(d\omega_{f_k}) \to \det(d\omega_f)$$ in $$L^1$$ (here use the fact that the dimension is $$2$$), so up to passing to a subsequence, $$\det(d\omega_{f_k})$$ converges pointwise a.e. to $$\det(d\omega_f)$$.

By our assumption, $$\det(d\omega_{f_k}) > 0$$ everywhere, so this implies $$\det(d\omega_f) \ge 0$$ a.e.

Now, taking any $$f$$ whose $$\omega_f$$ does not satisfy this, we get a contradiction. (e.g. the example described in the question).