I have the following PDE in two dimensions

$$ 2\partial_x\partial_y\sqrt{1-u^2}+\left(\partial^2_x-\partial^2_y \right)u=0, $$

with $u=u(x,y)$ with values between $-1$ and $1$, or alternatively

$$ 2\partial_x\partial_y\sin2\theta(x,y)+\left(\partial^2_x-\partial^2_y \right)\cos2\theta(x,y)=0, $$

with real $\theta(x,y)\sim\theta(x,y)+2\pi$, on some domain of the plane. Now, numerically I can obtain the solutions very quickly specifying some domain and an initial Cauchy line (as the equation hyperbolic), but I wish to have a deeper understanding of the solutions, so I'd like to see if there's a way to obtain analytic solutions. For instance, I know that $u=\cos(2\arctan(y/x))$ and $\theta(x,y)=\arctan(y/x)\pm1/2\arccos(c_1+c_2/(x^2+y^2))$, with $c_1, c_2$ some reals constants, are analytic, particular solutions, which strongly suggests that some general solution with arbitrary constants is plausible.

The problem is encountered in the context of elasticity of thin sheets. A so-called director field is imprinted on a thin elastic sheet, and it generates curvature upon a process called activation [1]. The director field $\theta(x,y)$ will induce a Riemannian metric on the new, curved sheet

$$
g(x,y)=R[\theta(x,y)]diag(\lambda_1,\lambda_2)R[\theta(x,y)]^T,
$$
with $R[\theta(x,y)]$ a $2\times2$ rotation matrix and $\lambda_1,\lambda_2$ some positive, known constants. Now, the aforementioned metric has a Gaussian curvature proportional to the equation written before, and the question I'm addressing is, *for which $\theta(x,y)$s the generated curvature is zero* ?, except for possibly isolated points where it may diverge. Now, the solutions I wrote before correspond to cones, but there should be more analytic solutions.

Any ideas ? Have you seen this equation or someone similar before ?

Thank you so much.

[1] *Mostajeran, Cyrus; Warner, Mark; Ware, Taylor H.; White, Timothy J.*, **Encoding Gaussian curvature in glassy and elastomeric liquid crystal solids**, Proc. R. Soc. Lond., A, Math. Phys. Eng. Sci. 472, No. 2189, Article ID 20160112, 16 p. (2016). ZBL1371.82141.

**Edit:**

Here is a summary of the solution from Robert Bryant's great answer, in a language more familiar to physicist.

Consider the (in general muliply-)connected domain $\mathscr{{W}}\subseteq\mathbb{R}^{2}$, with Cartesian coordinates $u$ and $v$, and the function $f:\mathscr{{W}}\rightarrow\mathbb{R}$ that solves $\frac{\partial^{2}f(u,v)}{\partial u\partial v}=f(u,v)$, with $f(u,v)$ and $\partial f(u,v)/\partial v$ non-vanishing. This is a linear hyperbolic equation, so it's Cauchy problem is always well defined on $\mathscr{{W}}$, and the space of solutions is always non-empty.

The 1-forms

$$ \alpha_{1} \equiv f\cos\left(u-v\right)\mathrm{d} u+\frac{\partial f}{\partial v}\sin\left(u-v\right)\mathrm{d}v ,\:\:\alpha_{2} \equiv f\sin\left(u-v\right)\mathrm{d} u-\frac{\partial f}{\partial v}\cos\left(u-v\right)\mathrm{d}v, $$

are closed. Therefore we can write locally $\mathrm{d} x =\alpha_{1},\: \mathrm{d} y =\alpha_{2},$ for some functions $x$ and $y$ on $\mathscr{{W}}$. We define the function $(x,y):\mathscr{{W}}\rightarrow\mathbb{R}^{2}$ and the domain $\mathscr{{Z}\mathbb{\subseteq R}}^{2}$ as the image of $(x,y)$, i.e., $(x,y)\left(\mathscr{{W}}\right)=\mathscr{{Z}}$. We can use $x$ and $y$ as coordinates of $\mathscr{{Z}}$, as by definition they cover the latter completely. The function $u-v:\mathscr{{W}}\rightarrow\mathbb{R}$ can be pulled through $\mathscr{{Z}}$, that is $u-v=\theta\circ(x,y)$, with $\theta:\mathscr{{Z}}\rightarrow\mathbb{R}$ a function defined by the previous relation.

Inverting the definitions for $u$ and $v$ as functions of $x$ and $y$ we can write

$$\frac{\partial}{\partial x} =\frac{1}{f}\cos\left(u-v\right)\frac{\partial}{\partial u}+\frac{1}{\frac{\partial f}{\partial v}}\sin\left(u-v\right) \frac{\partial}{\partial v}\\ \frac{\partial}{\partial y} =\frac{1}{f}\sin\left(u-v\right)\frac{\partial}{\partial u}-\frac{1}{\frac{\partial f}{\partial v}}\cos\left(u-v\right)\frac{\partial}{\partial v}, $$

and it's just a matter of patience to verify that

$$ \begin{align} &2\frac{\partial^{2}}{\partial x\partial y}\sin2\theta+\left(\frac{\partial^{2}}{\partial x^{2}}-\frac{\partial^{2}}{\partial y^{2}}\right)\cos2\theta\\ &=\Bigg\{ \left[\frac{1}{f}\cos\left(u-v\right)\frac{\partial}{\partial u}+\frac{1}{\frac{\partial f}{\partial v}}\sin\left(u-v\right)\frac{\partial}{\partial v}\right]^{2}\\ &\:\:\:\:-\left[\frac{1}{f}\sin\left(u-v\right)\frac{\partial}{\partial u}-\frac{1}{\frac{\partial f}{\partial v}}\cos\left(u-v\right)\frac{\partial}{\partial v}\right]^{2}\Bigg\} \cos2\left(u-v\right)\\ &+2\left[\frac{1}{f}\cos\left(u-v\right)\frac{\partial}{\partial u}+\frac{1}{\frac{\partial f}{\partial v}}\sin\left(u-v\right)\frac{\partial}{\partial v}\right]\\ &\:\:\:\times\left[\frac{1}{f}\sin\left(u-v\right)\frac{\partial}{\partial u}-\frac{1}{\frac{\partial f}{\partial v}}\cos\left(u-v\right)\frac{\partial}{\partial v}\right]\sin2\left(u-v\right)\\ &=\frac{4\cos^{2}(u-v)}{f\left(\frac{\partial f}{\partial v}\right)^{2}}\left(f(u,v)-\frac{\partial f}{\partial u\partial v}\right)\\ &=0. \end{align} $$

The particular solutions mentioned before are obtained with $f(u,v)=e^{\alpha u+v/\alpha}$, with some constant $\alpha$. But of course any other $f(u,v)$ will generate a solution. The most general real, separable solution is

$$ f(u,v)=\int_{-\infty}^{\infty}\mathrm{d}\rho\:C(\rho)\:e^{\rho u+\rho^{-1}v}, $$

for some arbitrary kernel $C(\rho)$. So one can classify arbitrarily many solutions by $C(\rho)$. From a calculative standpoint, once chosen some $f(u,v)$ the problem is that in general it's difficult to solve the algebraic system to write down explicitly $u(x,y)$ and $v(x,y)$, so one can say that the non-linear differential equation in two variables was transformed into a problem of two non-linear algebraic equations.