曲线浸入曲面

This is a note for Nadir’s work, but I have not fully understood it.

概述

我们从生长模型的角度看这个问题，对于沿 $t$ 方向生长，有： $ \frac{\partial \mathbf{X}}{\partial t}=\rm{U}\mathbf{n} $

定义曲面的度规：

$ g_{11}=\frac{\partial \mathbf{X}}{\partial \sigma}\cdot\frac{\partial \mathbf{X}}{\partial \sigma}=\frac{\partial\mathbf{X}}{\partial s}\cdot\frac{\partial\mathbf{X}}{\partial s}*(\frac{\partial s}{\partial \sigma})^2=(\frac{\partial s}{\partial \sigma})^2 $

$ g_{22}=\frac{\partial \mathbf{X}}{\partial t}\cdot\frac{\partial \mathbf{X}}{\partial t}=\rm{U^2} $

$ g_{12}=0 $

度规的导数

$ \frac{\partial g}{\partial \sigma}=2g\sqrt{g}\frac{\partial \mathbf{X}_s}{\partial \sigma}\mathbf{X}_s=0 $

$ \frac{\partial \sqrt{g}}{\partial t}=2g\mathrm{U}\frac{\partial \mathbf{n}}{\partial s}\mathbf{T}=-\sqrt{g}\kappa _g\mathrm{U} $

曲线标架的变换关系

${\rm{T},\rm{n},\rm{N}}$ 描述了曲面上曲线的运动，沿着${s,t}$ 两个方向由于活动标架的正交归一性，必须满足对应的运动方程。

s方向

$s$方向，Darboux标架满足： $ \partial _s\left( \begin{array}{c} \mathbf{T}
\mathbf{n}
\mathbf{N}
\end{array} \right) =\left( \begin{matrix} 0& \kappa _g& \kappa _N
-\kappa _g& 0& \tau _g
-\kappa _N& -\tau _g& 0
\end{matrix} \right) \left( \begin{array}{c} \mathbf{T}
\mathbf{n}
\mathbf{N}
\end{array} \right) $

$ \kappa_n=\frac{\partial^2\mathbf{X}}{\partial s^2}\cdot\mathbf{N} $

$ \kappa_g=\frac{\partial^2\mathbf{X}}{\partial s^2}\cdot\mathbf{n} $

t方向

首先推导偏导数的对易关系： $ \partial _t\partial _s=\partial _t\left( \frac{1}{\sqrt{g}}\partial _{\sigma} \right) =\partial _s\partial _t+\partial _t\left( \frac{1}{\sqrt{g}} \right) \partial _{\sigma}=\partial _s\partial _t+\kappa _g\mathrm{U}\partial _s $ $t$方向，Darboux标架满足: $ \partial _t\left( \begin{array}{c} \mathbf{T}
\mathbf{n}
\mathbf{N}
\end{array} \right) =\left( \begin{matrix} 0& \frac{\partial \mathrm{U}}{\partial s}& \mathrm{U}\tau _g
-\frac{\partial \mathrm{U}}{\partial s}& 0& \mathrm{U}\kappa _{N,2}
-\mathrm{U}\tau _g& -\mathrm{U}\kappa _{N,2}& 0
\end{matrix} \right) \left( \begin{array}{c} \mathbf{T}
\mathbf{n}
\mathbf{N}
\end{array} \right) $

曲面本构条件

首先得到第一第二基本型系数为： $ \mathrm{E}=(\frac{\partial s}{\partial \sigma})^2=g;\mathrm{F}=0;\mathrm{G}=\mathrm{U}^2 $

$ \mathrm{L}=\frac{\partial ^2\mathbf{X}}{\partial \sigma ^2}\cdot \mathbf{N}=\sqrt{g}\frac{\partial}{\partial s}\left( \mathbf{T}\sqrt{g} \right) \cdot \mathbf{N}=g\kappa _N $

$ \mathrm{M}=\frac{\partial ^2\mathbf{X}}{\partial \sigma \partial t}\cdot \mathbf{N}=\frac{\partial s}{\partial \sigma}\frac{\partial}{\partial s}\left( \mathrm{U}\mathbf{n} \right) \cdot \mathbf{N}=\sqrt{g}\mathrm{U}\tau _g $

$ \mathrm{N}=\frac{\partial ^2\mathbf{X}}{\partial t^2}\cdot \mathbf{N}=\frac{\partial}{\partial t}\left( \mathrm{U}\mathbf{n} \right) \cdot \mathbf{N}=\mathrm{U}\frac{\partial \mathbf{n}}{\partial t}\cdot \mathbf{N}=\mathrm{U}^2\frac{\partial \mathbf{n}}{\mathrm{U}\partial t}\cdot \mathbf{N}=\mathrm{U}^2\kappa _{N,2} $

由(8) ~ (11)计算得到Christoffel曲面的符号为： $ \Gamma _{11}^{1}=\frac{\mathrm{E}_u}{2\mathrm{E}}=\frac{\partial \sqrt{g}}{\partial s}; \Gamma _{12}^{1}=\frac{\mathrm{E}_v}{2\mathrm{E}}=-\mathrm{U}\kappa _g; \Gamma _{22}^{1}=-\frac{\mathrm{G}_u}{2\mathrm{E}}=-\frac{\mathrm{U}}{\sqrt{g}}\frac{\partial \mathrm{U}}{\partial s} $

$ \Gamma _{22}^{2}=\frac{\mathrm{G}_v}{2\mathrm{G}}=\frac{1}{\mathrm{U}}\frac{\partial \mathrm{U}}{\partial t}; \Gamma _{12}^{2}=\frac{\mathrm{G}_u}{2\mathrm{G}}=\frac{\sqrt{g}}{\mathrm{U}}\frac{\partial \mathrm{U}}{\partial s}; \Gamma _{11}^{2}=-\frac{\mathrm{E}_v}{2\mathrm{G}}=\frac{g}{\mathrm{U}}\kappa _g $

Codazzi-Mainardi 方程

$ \Gamma _{12}^{1}\mathrm{L}+\left( \Gamma _{12}^{2}-\Gamma _{11}^{1} \right) \mathrm{M}-\Gamma _{11}^{2}\mathrm{N}=\mathrm{L}_v-\mathrm{M}_u $

$ \Gamma _{22}^{1}\mathrm{L}+\left( \Gamma _{22}^{2}-\Gamma _{21}^{1} \right) \mathrm{M}-\Gamma _{21}^{2}\mathrm{N}=\mathrm{M}_v-\mathrm{N}_u $

带入Mainardi方程得到： $ -\mathrm{Ug}\kappa _N\kappa _g+\left( \frac{\sqrt{g}}{\mathrm{U}}\frac{\partial \mathrm{U}}{\partial s}-\frac{\partial \sqrt{g}}{\partial s} \right) \sqrt{\mathrm{g}}\mathrm{U}\tau _g-\frac{g}{\mathrm{U}}\kappa _g\mathrm{U}^2\kappa _{N,2}=\frac{\partial \mathrm{g}\kappa _N}{\partial t}-\frac{\partial \left( \sqrt{\mathrm{g}}\mathrm{U}\tau _g \right)}{\partial \sigma} $

$ -\frac{\mathrm{U}}{\sqrt{g}}\frac{\partial \mathrm{U}}{\partial s}\mathrm{g}\kappa _N+\left( \mathrm{U}\kappa _g+\frac{1}{\mathrm{U}}\frac{\partial \mathrm{U}}{\partial t} \right) \sqrt{\mathrm{g}}\mathrm{U}\tau _g-\frac{\sqrt{g}}{\mathrm{U}}\frac{\partial \mathrm{U}}{\partial s}\mathrm{U}^2\kappa _{N,2}=\frac{\partial \left( \sqrt{\mathrm{g}}\mathrm{U}\tau _g \right)}{\partial t}-\frac{\partial \left( \mathrm{U}^2\kappa _{N,2} \right)}{\partial \sigma} $

化简得到： $ \frac{\partial \kappa _N}{\partial t}=\frac{\partial \left( \mathrm{U}\tau _g \right)}{\partial s}+\frac{\partial \mathrm{U}}{\partial s}\tau _g+\mathrm{U}\kappa _g\left( \kappa _N-\kappa _{N,2} \right) $

$ \frac{\partial \tau _g}{\partial t}=\frac{\partial \left( \mathrm{U}\kappa _{N,2} \right)}{\partial s}-\frac{\partial \mathrm{U}}{\partial s}\kappa _N+2\mathrm{U}\kappa _g\tau _g $

Gauss-Bonnet 定理