Let $u: D\_1 \to \mathbf{R}$ be a smooth function defined on the unit disk $D\_1 \subset \mathbf{R}^2$ which describes the minimal graph $G$. Suppose that at the origin $G$ is tangent to the horizontal plane $\{ x^{3} = 0 \}$, that is: $u(0) = 0$ and $Du(0) = 0$. Suppose that in a small disk $D\_r$ around the origin, the intersection of $G$ and the plane decomposes like
\begin{equation}
\{ u = 0 \} \cap D\_r \setminus \{ 0 \} = \gamma\_1 \cup \cdots \cup \gamma\_4.
\end{equation}
The $\gamma\_i$ are smooth curves meeting at right angles at the origin, their common endpoint. After rotation, we may assume that $\gamma\_1'(0) = e\_1$, $\gamma\_2'(0) = e\_2$, $\gamma\_3'(0) = -e\_1$ and $\gamma\_4'(0) = -e\_2$.



> 
> Is it possible for $\gamma\_1 \cup \gamma\_2$ to form a graph over the $x^1$-axis, that is can there be $\varphi \in C^0(-r,r)$ so that $\gamma\_1 \cup \gamma\_2 = \{ (t,\varphi(t),0) \mid t \in (-r,r) \}$?
> 
> 
> 



'Degenerate' tangent point of a minimal graph
Yes, this is possible. Consider the intersection of the helicoid $z = \tan^{-1}(y/x)$ with its tangent plane $z = y$ at $(1,0,0)$. The projection of the intersection curves to $\{z = 0\}$ consists of the line $y = 0$ and the curve $x = y/\tan(y) \sim 1 - y^2/3$. After a rigid motion so that the tangent plane is horizontal and the tangent point is the origin, the union of $\gamma\_1$ and $\gamma\_2$ will be a $C^{1/2}$ graph $\{(t,\,\varphi(t),\,0)\}$ where $\varphi(t) = 0$ for $t \geq 0$ and $\varphi(t) \sim (-6t)^{1/2}$ for $t < 0$.


One can generate many more examples using Cauchy-Kovalevskaya, by choosing appropriate Cauchy data on a line segment through 0 and solving the minimal surface equation in a neighborhood of the origin. For example, taking $u = 0$ and $u\_y = x + x^2/2$ on $\{y = 0\}$ gives
$$u = xy + \frac{1}{2}x^2y - \frac{1}{6}y^3 + O((x^2+y^2)^2),$$
hence $\{u = 0\}$ locally resembles the line $y = 0$ and the curve $y^2 = 6x$.


