(1)

$$\boldsymbol \sigma_u = \begin{pmatrix} 1 \\ 0 \\ f_u \end{pmatrix} \quad \boldsymbol \sigma_v = \begin{pmatrix} 0 \\ 1 \\ f_v \end{pmatrix} \quad \boldsymbol n = \frac{\boldsymbol \sigma_u \times \boldsymbol \sigma_v}{\|\boldsymbol \sigma_u \times \boldsymbol \sigma_v\|} = \frac{1}{\sqrt{1 + f_u^2 + f_v^2}} \begin{pmatrix} -f_u \\ -f_v \\ 1 \end{pmatrix}$$

(2)

$$E = \boldsymbol \sigma_u \cdot \boldsymbol \sigma_u = 1 + f_u^2 \\ F = \boldsymbol \sigma_u \cdot \boldsymbol \sigma_v = f_u f_v \\ G = \boldsymbol \sigma_v \cdot \boldsymbol \sigma_v = 1 + f_v^2$$

$$L = \boldsymbol \sigma_{uu} \cdot \boldsymbol n = \frac{f_{uu}}{\sqrt{1 + f_u^2 + f_v^2}} \\ M = \boldsymbol \sigma_{uv} \cdot \boldsymbol n = \frac{f_{uv}}{\sqrt{1 + f_u^2 + f_v^2}} \\ N = \boldsymbol \sigma_{vv} \cdot \boldsymbol n = \frac{f_{vv}}{\sqrt{1 + f_u^2 + f_v^2}}$$

(3)

$$K(\boldsymbol p) = \frac{LN - M^2}{EG - F^2} = \frac{f_{uu}f_{vv} - f_{uv}^2}{(1 + f_u^2 + f_v^2)^2} \\ \quad \\ H(\boldsymbol p) = \frac{GL - 2FM + EN}{2(EG - F^2)} = \frac{(1+f_v^2)f_{uu} - 2f_uf_vf_{uv} + (1+f_u^2)f_{vv}}{2(1+f_u^2+f_v^2)^{3/2}}$$

一定要注意定义域是开区间，这意味着环面上有两条线上的点是无法用这个参数表示的（实际上换个参数表示很可能也表示不了）。

红线为 $T_{R,r}\backslash \boldsymbol \sigma(D)$

所以我们考虑计算形算子 $\Sigma_{\boldsymbol p}$ 的表现行列来得到。首先我们需要得到切空间，所以计算曲面方程的梯度来获取法向量场，注意取正方向。

$$f(x,y,z) := (\sqrt{x^2 + y^2} - R)^2 + z^2 - r^2$$

$$\nabla f = \begin{pmatrix} 2(\sqrt{x^2 + y^2} - R)\frac{x}{\sqrt{x^2 + y^2}} \\ 2(\sqrt{x^2 + y^2} - R)\frac{y}{\sqrt{x^2 + y^2}} \\ 2z \end{pmatrix}$$

$$\begin{aligned} \boldsymbol n = -\frac{\nabla f}{\|\nabla f\|} &= -\frac{1}{\sqrt{(\sqrt{x^2 + y^2} - R)^2 + z^2}} \begin{pmatrix} (\sqrt{x^2 + y^2} - R)\frac{x}{\sqrt{x^2 + y^2}} \\ (\sqrt{x^2 + y^2} - R)\frac{y}{\sqrt{x^2 + y^2}} \\ z \end{pmatrix} \\ &= -\frac{1}{r}\begin{pmatrix} (\sqrt{x^2 + y^2} - R)\frac{x}{\sqrt{x^2 + y^2}} \\ (\sqrt{x^2 + y^2} - R)\frac{y}{\sqrt{x^2 + y^2}} \\ z \end{pmatrix} \end{aligned}$$

接着构造切空间的基底。这是最考验观察力的部分。如果这里的基选取不好，会极大程度加重后续计算难度，出于这里是展示算法，直接给出良好性质的第一个基底

$$\boldsymbol u_1 = \begin{pmatrix} -y \\ x \\ 0 \end{pmatrix}$$

注意基本的几何事实：$\boldsymbol u_1$ 位于切空间 $T_{\boldsymbol p}(T_{R,r})$ 中，而法向量垂直于切空间，所以可以利用向量积来获取另外一个正交基，长度不纳入考虑，可以移除系数

$$\boldsymbol u_2 = \boldsymbol n \times \boldsymbol u_1 = -\frac{1}{r} \begin{pmatrix} -xz \\ -yz \\ (\sqrt{x^2 + y^2} - R)\sqrt{x^2 + y^2} \end{pmatrix} \longrightarrow \begin{pmatrix} -xz \\ -yz \\ (\sqrt{x^2 + y^2} - R)\sqrt{x^2 + y^2} \end{pmatrix}$$

那么此时显然 ($\boldsymbol u_1, \boldsymbol u_2$) 线性独立并成为 $T_{\boldsymbol p}(T_{R,r})$ 内的基底。

令 $\boldsymbol v = {}^t(v_1, v_2, v_3) \in T_{\boldsymbol p}(T_{R,r})$，则对于满足 $\boldsymbol \gamma(0) = \boldsymbol p, \frac{d{\boldsymbol \gamma}}{dt}(0) = \boldsymbol v$ 的 $C^\infty$ 曲线

$$\boldsymbol \gamma(t) = \begin{pmatrix} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{pmatrix}$$

代入可得

$$\boldsymbol n(\boldsymbol \gamma) = -\frac{1}{r} \left( \begin{pmatrix} \gamma_1 \\ \gamma_2 \\ \gamma_3 \end{pmatrix} - \frac{R}{\sqrt{\gamma_1^2 + \gamma_2^2}} \begin{pmatrix} \gamma_1 \\ \gamma_2 \\ 0 \end{pmatrix} \right)$$

进而

$$\begin{aligned} \Sigma_{\boldsymbol p}(\boldsymbol v) = -\frac{d\bold n}{dt}(\boldsymbol \gamma(t)) \bigg |_{t=0} &= \frac{1}{r} \left( \begin{pmatrix} \gamma_1' \\ \gamma_2' \\ \gamma_3' \end{pmatrix} + \frac{R(\gamma_1 \gamma_1' + \gamma_2 \gamma_2')}{(\gamma_1^2 + \gamma_2^2)^{3/2}} \begin{pmatrix} \gamma_1 \\ \gamma_2 \\ 0 \end{pmatrix} - \frac{R}{\sqrt{\gamma_1^2 + \gamma_2^2}} \begin{pmatrix} \gamma_1' \\ \gamma_2' \\ 0 \end{pmatrix} \right) \bigg |_{t=0} \\ &= \frac{1}{r} \left( \boldsymbol v + \frac{R(x v_1 + y v_2)}{(x^2 + y^2)^{3/2}} \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} - \frac{R}{\sqrt{x^2 + y^2}} \begin{pmatrix} v_1 \\ v_2 \\ 0 \end{pmatrix} \right) \end{aligned}$$

代入基底计算

$$\Sigma_{\boldsymbol p}(\boldsymbol u_1) = \frac{1}{r} (1 - \frac{R}{\sqrt{x^2 + y^2}}) \boldsymbol u_1$$

$$\Sigma_{\boldsymbol p}(\boldsymbol u_2) = \frac{1}{r} \boldsymbol u_2$$

所以可得表现行列

$$\Sigma_{\boldsymbol p} = \frac{1}{r} \begin{pmatrix} 1 - \frac{R}{\sqrt{x^2 + y^2}} & 0 \\ 0 & 1 \end{pmatrix}$$

注意此处因为我们挑选的第一个基底性质足够好，才使得后面计算出来的表现行列直接就是对角行列，不然的话还要对角化一遍。最后得到主曲率（特征值）和高斯曲率，平均曲率

$$\kappa_1 = \frac{1}{r}, \quad \kappa_2 = \frac{1}{r}(1 - \frac{R}{\sqrt{x^2 + y^2}})$$

$$K(\boldsymbol p) = \kappa_1 \kappa_2 = \frac{1}{r^2}(1 - \frac{R}{\sqrt{x^2 + y^2}})$$

$$H(\boldsymbol p) = \frac{\kappa_1 + \kappa_2}{2} = \frac{1}{2r}(2 - \frac{R}{\sqrt{x^2 + y^2}})$$