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dg.differential geometry – Non-linear hyperbolic PDE Answer

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dg.differential geometry – Non-linear hyperbolic PDE

I’ve the next PDE in two dimensions

$$
2partial_xpartial_ysqrt{1-u^2}+left(partial^2_x-partial^2_y privilege)u=0,
$$

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

$$
2partial_xpartial_ysin2theta(x,y)+left(partial^2_x-partial^2_y privilege)cos2theta(x,y)=0,
$$

with actual $theta(x,y)simtheta(x,y)+2pi$, on some province of the airplane. Now, numerically I can secure the options very snappily specifying some province and an preliminary Cauchy line (because the equation hyperbolic), however I want to have a deeper judgement of the options, so I’d love to behold if there is a route to secure analytic options. For occasion, I do know that $u=cos(2arctan(y/x))$ and $theta(x,y)=arctan(y/x)pm1/2arccos(c_1+c_2/(x^2+y^2))$, with $c_1, c_2$ some reals constants, are analytic, explicit options, which strongly means that some common resolution with capricious constants is believable.

The downside is encountered within the context of elasticity of skinny sheets. A so-called director bailiwick is imprinted on a skinny elastic sheet, and it generates curvature upon a course of referred to as activation [1]. The director bailiwick $theta(x,y)$ will induce a Riemannian metric on the fresh, 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 $2times2$ rotation matrix and $lambda_1,lambda_2$ some optimistic, identified constants. Now, the aforementioned metric has a Gaussian curvature proportional to the equation written earlier than, and the query I’m addressing is, for which $theta(x,y)$s the generated curvature is zero ?, aside from presumably remoted factors the place it might stray. Now, the options I wrote earlier than coincide to cones, however there ought to breathe extra analytic options.

Any concepts ? Have you seen this equation or somebody comparable earlier than ?

Thank you so mighty.

[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 abstract of the answer from Robert Bryant’s noble respond, in a language extra close to physicist.

Consider the (in common muliply-)related province $mathscr{{W}}subseteqmathbb{R}^{2}$, with Cartesian coordinates $u$ and $v$, and the duty $f:mathscr{{W}}rightarrowmathbb{R}$ that solves $frac{partial^{2}f(u,v)}{partial upartial v}=f(u,v)$,
with $f(u,v)$ and $partial f(u,v)/partial v$ non-vanishing. This is a linear hyperbolic equation, so it is Cauchy downside is at all times nicely outlined on $mathscr{{W}}$, and the house of options is at all times non-empty.

The 1-forms

$$
alpha_{1} equiv fcosleft(u-vright)mathrm{d} u+frac{partial f}{partial v}sinleft(u-vright)mathrm{d}v
,::alpha_{2} equiv fsinleft(u-vright)mathrm{d} u-frac{partial f}{partial v}cosleft(u-vright)mathrm{d}v,
$$

are closed. Therefore we are able to write domestically $mathrm{d} x =alpha_{1},:
mathrm{d} y =alpha_{2},$
for some features $x$ and $y$ on $mathscr{{W}}$. We outline the duty $(x,y):mathscr{{W}}rightarrowmathbb{R}^{2}$ and the province $mathscr{{Z}mathbb{subseteq R}}^{2}$ because the picture of $(x,y)$, i.e., $(x,y)left(mathscr{{W}}privilege)=mathscr{{Z}}$. We can employ $x$ and $y$ as coordinates of $mathscr{{Z}}$, as by definition they mask the latter fully. The duty $u-v:mathscr{{W}}rightarrowmathbb{R}$ can breathe pulled via $mathscr{{Z}}$, that’s $u-v=thetacirc(x,y)$, with $theta:mathscr{{Z}}rightarrowmathbb{R}$ a duty outlined by the earlier relation.

Inverting the definitions for $u$ and $v$ as features of $x$ and $y$ we are able to write

$$frac{partial}{partial x} =frac{1}{f}cosleft(u-vright)frac{partial}{partial u}+frac{1}{frac{partial f}{partial v}}sinleft(u-vright)
frac{partial}{partial v}
frac{partial}{partial y} =frac{1}{f}sinleft(u-vright)frac{partial}{partial u}-frac{1}{frac{partial f}{partial v}}cosleft(u-vright)frac{partial}{partial v},
$$

and it is only a signify of long-suffering to confirm that

$$
commence{align}
&2frac{partial^{2}}{partial xpartial y}sin2theta+left(frac{partial^{2}}{partial x^{2}}-frac{partial^{2}}{partial y^{2}}privilege)cos2theta
&=Bigg{ left[frac{1}{f}cosleft(u-vright)frac{partial}{partial u}+frac{1}{frac{partial f}{partial v}}sinleft(u-vright)frac{partial}{partial v}right]^{2}
&::::-left[frac{1}{f}sinleft(u-vright)frac{partial}{partial u}-frac{1}{frac{partial f}{partial v}}cosleft(u-vright)frac{partial}{partial v}right]^{2}Bigg} cos2left(u-vright)
&+2left[frac{1}{f}cosleft(u-vright)frac{partial}{partial u}+frac{1}{frac{partial f}{partial v}}sinleft(u-vright)frac{partial}{partial v}right] &:::timesleft[frac{1}{f}sinleft(u-vright)frac{partial}{partial u}-frac{1}{frac{partial f}{partial v}}cosleft(u-vright)frac{partial}{partial v}right]sin2left(u-vright)
&=frac{4cos^{2}(u-v)}{fleft(frac{partial f}{partial v}privilege)^{2}}left(f(u,v)-frac{partial f}{partial upartial v}privilege)
&=0.
aim{align}
$$

The explicit options talked about earlier than are obtained with $f(u,v)=e^{alpha u+v/alpha}$, with some ceaseless $alpha$. But of passage some other $f(u,v)$ will generate an answer. The most common actual, separable resolution is

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

for some capricious kernel $C(rho)$. So one can classify arbitrarily many options by $C(rho)$. From a calculative standpoint, as soon as chosen some $f(u,v)$ the issue is that in common it is troublesome to decipher the algebraic system to put in writing down explicitly $u(x,y)$ and $v(x,y)$, so one can say that the non-linear differential equation in two variables was remodeled into an issue of two non-linear algebraic equations.

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