I've been scratching my head for quite some time on how one could find the surface embedded in 3D flat space that induces a given 2x2 metric. Usually one asks the opposite question: given a surface, find the corresponding metric (e.g. Finding the metric of a surface embedded in $\mathbb{R}^3$). But I'm interested in the inverse problem. To my surprise, almost nothing can be found online about this problem.
I can write down the partial differential equations that the parametrization of the surface should satisfy, but I am missing the boundary conditions. Can anyone help me with this?
Here is my reasoning: first suppose you have a 2x2 metric, so that $ds^2 = m_{\alpha\alpha}(\alpha,\beta) d\alpha^2 + m_{\alpha\beta}(\alpha,\beta) d\alpha d\beta + m_{\beta\beta}(\alpha,\beta) d\beta^2$. Now we claim that there exists a surface in flat 3D space that would induce this metric. Such a surface is parametrized by \begin{align} x &= f_x(\alpha,\beta)~,\\ y &= f_y(\alpha,\beta)~, \\ z &= f_z(\alpha,\beta)~. \end{align} From the constraints above it is quite direct to find \begin{align} dx &= \frac{\partial f_x(\alpha,\beta)}{\partial \alpha}d\alpha + \frac{\partial f_x(\alpha,\beta)}{\partial \beta}\partial \beta~,\\ dy &= \frac{\partial f_y(\alpha,\beta)}{\partial \alpha}d\alpha + \frac{\partial f_y(\alpha,\beta)}{\partial \beta}d\beta~, \\ dz &= \frac{\partial f_z(\alpha,\beta)}{\partial \alpha}d\alpha + \frac{\partial f_z(\alpha,\beta)}{\partial \beta}d\beta~. \end{align} Since we are in euclidean $\mathbb{R}^3$ we have $ds^2 = dx^2 + dy^2 + dz^2$. By inserting these constraints in the length element we find a set of 3 differential equations for 3 unknown functions so that the parametrization above induces the metric we had at the start \begin{align} \left(\frac{\partial f_x(\alpha,\beta)}{\partial \alpha}\right)^2 + \left(\frac{\partial f_y(\alpha,\beta)}{\partial \alpha}\right)^2 + \left(\frac{\partial f_z(\alpha,\beta)}{\partial \alpha}\right)^2 &= m_{\alpha\alpha}(\alpha,\beta)~,\\ \frac{\partial f_x(\alpha,\beta)}{\partial \alpha}\frac{\partial f_x(\alpha,\beta)}{\partial \beta} + \frac{\partial f_y(\alpha,\beta)}{\partial \alpha}\frac{\partial f_y(\alpha,\beta)}{\partial \beta} + \frac{\partial f_z(\alpha,\beta)}{\partial \alpha}\frac{\partial f_z(\alpha,\beta)}{\partial \beta} &= m_{\alpha\beta}(\alpha,\beta)~,\\ \left(\frac{\partial f_x(\alpha,\beta)}{\partial \beta}\right)^2 + \left(\frac{\partial f_y(\alpha,\beta)}{\partial \beta}\right)^2 + \left(\frac{\partial f_z(\alpha,\beta)}{\partial \beta}\right)^2 &= m_{\beta\beta}(\alpha,\beta)~. \end{align}
To solve these PDEs one would need boundary conditions of the type $f_x(\alpha,0)=g_x(\alpha)$, but I don't know how I should find this $g_x$ that defines the boundary condition for $f_x$ (of course, the same is true for the other boundary conditions).
Any help in how to continue this problem is greatly appreciated, same for any correction or different approach to solve this.