Let $\mathbb D^2$ be the closed unit disk in $\mathbb R^2$. Let $f:\mathbb D^2 \to \mathbb{R}^2$ be a real-analytic orientation preserving immersion, and let $\omega:\mathbb D^2 \to \mathbb{R}^2$ be the unique harmonic map satisfying $\omega|\_{\partial \mathbb D^2}=f|\_{\partial \mathbb D^2}$



> 
> Does $d\omega \neq 0 $ everywhere on $\mathbb D^2$? 
> 
> 
> 


I have two observations:



> 
> 1. There is an open neighbourhood of $\partial \mathbb D^2$ where $d\omega \neq 0 $ .
> 2. $d\omega$ is invertible outside a set of Hausdorff dimension $\le 1$.
> 
> 
> 


Claim $(1)$ follows from the fact that for $p \in \partial \mathbb D^2$, we have  

$$ \text{rank}(d\omega\_p)\ge \text{rank}\big(d(\omega|\_{\partial \mathbb D^2})\_p\big)= \text{rank}\big(d(f|\_{\partial \mathbb D^2})\_p\big)=1.$$


For point $2$, note that $d\omega$ cannot be singular everywhere, since $$ \int\_{\mathbb D^2} \det d\omega = \int\_{\mathbb D^2} \det df>0.$$


Thus, $\big(\det(d\omega)\big)^{-1}(0)$ is the zero-set of a real-analytic function which is not identically zero, which implies dimension $\le 1$.




---



Can harmonic maps with immersive boundary conditions have singular points?
No, this is not so.


I'll explain how to construct a counter-example, though will leave some details in the form of exercises. I will also assume that we consider just smooth maps from the disk since smooth maps can be $C^{\infty}$ approximated by analytic ones, it will be obvious from the construction that there is no difference.


Let us parametrise the boundary of $\mathbb D^2$ by angle $t$, $t\in [0,2\pi]$. Then it will be enough to find an immersion $f=(f\_1,f\_2)$ such that when we restrict $f\_1$ and $f\_2$ to the unit circle, the following equalities hold:


$$\int\_{0}^{2\pi} f\_1(t)\cos(t)=\int\_{0}^{2\pi} f\_1(t)\sin(t)=\int\_{0}^{2\pi} f\_2(t)\cos(t)=\int\_{0}^{2\pi} f\_2(t)\sin(t)=0.$$


Indeed, if we construct such an immersion then the corresponding harmonic functions $(\omega\_1, \omega\_2)=\omega$ will satisfy $d\omega(0,0)=0$.


The existence of such $f$ is quite obvious, plenty of ways to construct it, I'll indicate one way.


First, we start with a simple exercise:


*Exercise 1.* Suppose we have a finite number of distinct points $(x\_1,y\_1),\ldots, (x\_n, y\_n)\in\mathbb R^2$ then for any $0<t\_1<t\_2<\ldots <t\_n<2\pi$ there always exists an immersion (even an embedding) from a disk $\tilde f:\mathbb D\to \mathbb R^2$ such $\tilde f(t\_i)=(x\_i,y\_i)$.


*Exercise 2.* There exists $n$, distinct $t\_i$'s and distinct $(x\_i,y\_i)$'s such that


$$\sum\_i \cos(t\_i)x\_i=\sum\_i \sin(t\_i)x\_i=\sum\_i \cos(t\_i)y\_i=\sum\_i \sin(t\_i)y\_i=0.$$


There is a huge amount of flexibility in finding such $t\_i, x\_i, y\_i$.


Finally, we consider an immersion $\tilde f$ from Exercise 1, so that the boundary of $\mathbb D$ passes through $(x\_i,y\_i)$ at time $t\_i$, but reparametrise it in such a way that the circle $\tilde f(t)$ spends time $2\pi(\frac{1}{n}-\epsilon)$ very close to each point $(x\_i, y\_i)$ and then in time $2\pi\epsilon$ runs to the point $(x\_{i+1},x\_{i+1})$.
In such case the second 4-tuple of equalities on sums will approximately guarantee the first 4 -tuple of equalities on integrals. After this a simple perturbation argument solves the problem. 


**Proof of $d\omega(0,0)=0$.** Recall that every harmonic function $h$ can be decomposed as $h=a\_0+a\_1 Re(z)+b\_1 Im(z)+ a\_2 Re(z^2)+b\_2 Im(z^2)+...$. Such a function has zero derivative at $(0,0)$ if an only if $a\_1=b\_1=0$, but the integrals written above give exactly the same conditions, since $\sin(mt)$, $\cos(mt)$ are orthogonal to each other for different $m$. 


