I need to find a nontrivial function $f:\mathbb{R^2}\setminus\mathbb{D}\rightarrow \mathbb{R}$ ($\mathbb{D}$ denotes the unit disk) such that $\nabla^2f=0, \nabla f $ tends to zero as point $p$ goes to infinity, norm convergence, and $\nabla f $ is tangent to $\mathbb{S^1}$
I checked some book of fluid mechanics and have not found anything very useful, don't sure how to add the extra hypothesis.
A systematic way to proceed is to pass to polar coordinates, Laplace operator in two dimensions, so we have $$ f(x,y)=F(r,t)\ , $$ for a suitable function $F$, where there correspond $(x,y)=(r\cos t,r\sin t)$. The condition $\nabla^2 f=0$ becomes in polar coordinates $$ F_{rr}+\frac 1r F_r+\frac 1{r^2}F_{tt}=0\ . $$ Now we associate the helper function (by inversion) $$ H(r,t)=F\left(\frac 1r, t\right)\ , $$ and it can be shown that it satisfies also the harmonic condition, this time in the interior of the unit disk. Explicitly $$ \begin{aligned} &H_{rr}(r,t)+\frac 1r H_r(r,t)+\frac 1{r^2}H_{tt}(r,t) \\ &=F_{rr}(1/r,t)\cdot \frac 1{r^4} +F_{r}(1/r,t)\cdot \frac 2{r^3} +\frac 1r F_r(1/r,t)\cdot \frac {-1}{r^2}+\frac 1{r^2}F_{tt}(1/r,t) \\ &=\frac 1{r^4}\left[ F_{rr}+\frac 1{1/r} F_r+\frac 1{(1/r)^2}F_{tt} \right](1/r,t) \\ &=0\ . \end{aligned} $$ We obtain a harmonic function $H$ inside the punctured unit disk, with a removable singularity in $(0,0)$, where it can be continuously extended by zero, in general, boundedness is enough to remove the singularity. And on the margin we have the von Neumann condition of a zero radial derivative.
As far as i remember from my courses some 30 years ago, using Greens functions / kernels (existence and uniqueness of the solution) only the zero function satisfies the given conditions.