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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 $\gamma_1 \cup \gamma_2 = \{ (t,\varphi(t),0) \mid t \in (-r,r) \}$?

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  • $\begingroup$ I'm a bit confused by the setup since how can the union of two curves be a single graph. But have you considered ennepers surface? If you intersect with the right tangent plane you'll get two straight lines (usually this is the xy plane through the origin) $\endgroup$ Commented May 12, 2023 at 13:21
  • $\begingroup$ @OtisChodosh Ultimately I was hoping one could say that $\gamma_1$ intersects the first coordinate axis (to which it is tangent at the origin). For the Enneper surface, I think $\gamma_1$ would coincide with it. I phrased the question like this because I thought it might be easier to exclude this extreme case. So for example, to say that $\gamma_2$ cannot be the graph of $x_1 \mapsto \sqrt{-x_1}$. $\endgroup$
    – Leo Moos
    Commented May 12, 2023 at 14:15

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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 $y^2 = 6x$.

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  • $\begingroup$ This is so cool, thanks! Did your reasoning go something as follows? You saw that there examples that are harmonic functions, and then the helicoid was a natural guess, and the calculations worked out at the point? And as for 'why' there 'should' be harmonic examples, it seems like it's got something to do with $u$ being decomposed into different frequencies, with them interacting at intermediate scales - is there an explanation along these lines? $\endgroup$
    – Leo Moos
    Commented May 12, 2023 at 17:58
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    $\begingroup$ Sure! Yes, playing with harmonic functions (expansions of minimal surfaces over their tangent planes) was the key. The helicoid is one of my favorite examples because it's harmonic and infinity-harmonic, thus minimal (since the MSE is Laplace - (Infinity Laplace)/(1+grad^2) = 0). $\endgroup$ Commented May 12, 2023 at 19:10

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