東京大学 新領域創成科学研究科 複雑理工学専攻 2018年8月実施 専門基礎科目 第1問

标签：

• Tokyo-University
• Laplacian-in-Polar-Coordinates
• Helmholtz-Equation
• Separation-of-Variables-for-Partial-Differential-Equation
• Bessel-Equation
• Power-Series-Solution-and-Coefficient-Recurrence

Author

之遥

Description

$f(x,y)(x,y \in \mathbb{R},(x,y) \neq (0,0))$ を実関数とし，微分方程式

$$\begin{align} \frac{\partial ^2 f}{\partial x^2} + \frac{\partial ^2 f}{\partial y^2} + f = 0 \end{align}$$

を極座標 $(r,\theta)(r,\theta \in \mathbb{R},r > 0)$ を用いて考える。ここで, $x = r\cos\theta,y = r\sin\theta$ とする。$g(r,\theta) = f(x(r,\theta),y(r,\theta))$ として, 以下の問に答えよ。

(問 1)

$r$ を $x,y$ を用いて表せ。

(問 2)

$\frac{\partial f}{\partial x} = A(r,\theta)\frac{\partial g}{\partial r} + B(r,\theta)\frac{\partial g}{\partial \theta}$ と表されるとき, $A(r,\theta),B(r,\theta)$ を求めよ。$\frac{\partial \theta}{\partial x} = -\frac{\sin\theta}{r}$ を用いてよい。

(問 3)

$\frac{\partial ^2 f}{\partial x^2} = D(r,\theta)\frac{\partial ^2 g}{\partial r^2} + E(r,\theta)\frac{\partial g}{\partial r} + F(r,\theta)\frac{\partial ^2 g}{\partial \theta^2} + G(r,\theta)\frac{\partial g}{\partial \theta} + H(r,\theta)\frac{\partial ^2g}{\partial r\partial \theta}$ と表されるとき, $D(r,\theta),E(r,\theta),F(r,\theta),G(r,\theta),H(r,\theta)$ を求めよ。

(問 4)

式(1)の微分方程式において, 変数を極座標に変換して $g(r,\theta)$ に関する微分方程式を求めよ。$\frac{\partial \theta}{\partial y} = \frac{\cos\theta}{r}$ を用いてよい。

(問 5)

(問 4)で求めた微分方程式の解が, $g(r,\theta) = R(r)\sin\theta$ と書けると仮定する。 $R(r)$ は微分方程式

$$\begin{align} \frac{\text{d}^2R}{\text{d}r^2} + \frac{1}{r}\frac{\text{d}R}{\text{d}r} + \big(1 - \frac{1}{r^2}\big)R = 0 \end{align}$$

を満たすことを示せ。

(問 6)

式(2)の微分方程式の解が, 項別微分可能な級数 $R(r) = r + \sum_{m=1}^{\infty}C_{m}r^{m+1}$ と書けると仮定する。

(i) $C_1,C_2$ の値を求めよ。

(ii) 整数 $n \ge 1$ に対して, $C_{n+2}$ を $C_n$ を用いて表せ。

题目描述

设 $f(x,y)$ 是定义在 $\mathbb R^2\setminus\{(0,0)\}$ 上的实函数，满足二维 Helmholtz 方程

$$\frac{\partial^2f}{\partial x^2} +\frac{\partial^2f}{\partial y^2}+f=0. \tag{1}$$

采用极坐标

$$x=r\cos\theta,\qquad y=r\sin\theta,\qquad r>0,$$

并定义 $g(r,\theta)=f(x(r,\theta),y(r,\theta))$。回答：

1. 用 $x,y$ 表示 $r$。
2. 若
   $$\frac{\partial f}{\partial x} =A(r,\theta)\frac{\partial g}{\partial r} +B(r,\theta)\frac{\partial g}{\partial\theta},$$
   求 $A,B$；可用 $\partial\theta/\partial x=-\sin\theta/r$。
3. 若
   $$\frac{\partial^2f}{\partial x^2} =Dg_{rr}+Eg_r+Fg_{\theta\theta}+Gg_\theta+Hg_{r\theta},$$
   求 $D(r,\theta),E(r,\theta),F(r,\theta),G(r,\theta),H(r,\theta)$。
4. 将式 (1) 完整变换成关于 $g(r,\theta)$ 的极坐标偏微分方程；可用 $\partial\theta/\partial y=\cos\theta/r$。
5. 假设解可分离为
   $$g(r,\theta)=R(r)\sin\theta,$$
   证明径向函数满足
   $$R''+\frac1rR'+\left(1-\frac1{r^2}\right)R=0. \tag{2}$$
6. 再假设式 (2) 有可逐项微分的级数解
   $$R(r)=r+\sum_{m=1}^{\infty}C_mr^{m+1}.$$
   1. 求 $C_1,C_2$；
   2. 对整数 $n\ge1$，用 $C_n$ 表示 $C_{n+2}$。

Kai

(問 1)

$$r = \sqrt{x^2 + y^2}$$

(問 2)

$$\begin{aligned} &\frac{\partial r}{\partial x} = \frac{2x}{2\sqrt{x^2 + y^2}} = \frac{r\cos\theta}{r} = \cos\theta \\ &\frac{\partial f(x,y)}{\partial x} = \frac{\partial f(x(r,\theta),y(r,\theta))}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f(x(r,\theta),y(r,\theta))}{\partial \theta}\frac{\partial \theta}{\partial x} \\ &= \frac{\partial g}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial g}{\partial \theta}\frac{\partial \theta}{\partial x} = \cos\theta\frac{\partial g}{\partial r} + \big(-\frac{\sin\theta}{r}\big)\frac{\partial g}{\partial \theta} \\ &\therefore A(r,\theta) = \cos\theta,B(r,\theta) = -\frac{\sin\theta}{r} \end{aligned}$$

(問 3)

$$\begin{aligned} &\frac{\partial ^2f}{\partial x^2} = \frac{\partial }{\partial x}(\frac{\partial f}{\partial x}) = \frac{\partial }{\partial r}(\frac{\partial f}{\partial x})\cdot\frac{\partial r}{\partial x} + \frac{\partial }{\partial \theta}(\frac{\partial f}{\partial x})\cdot\frac{\partial \theta}{\partial x} \\ &= \frac{\partial }{\partial r}\bigg[\cos\theta\frac{\partial g}{\partial r} + (-\frac{\sin\theta}{r})\frac{\partial g}{\partial \theta}\bigg] \cdot \frac{\partial r}{\partial x} + \frac{\partial }{\partial \theta}\bigg[\cos\theta\frac{\partial g}{\partial r} + (-\frac{\sin\theta}{r})\frac{\partial g}{\partial \theta}\bigg] \cdot \frac{\partial \theta}{\partial x} \\ &= \bigg[\cos\theta\frac{\partial ^2g}{\partial r^2} + \frac{\sin\theta}{r^2}\frac{\partial g}{\partial \theta} + (-\frac{\sin\theta}{r})\frac{\partial ^2g}{\partial \theta\partial r}\bigg]\cos\theta \\ &\qquad + \bigg[-\sin\theta\frac{\partial g}{\partial r} + \cos\theta\frac{\partial g}{\partial r\partial \theta} + (-\frac{\cos\theta}{r})\frac{\partial g}{\partial \theta} +(-\frac{\sin\theta}{r})\frac{\partial ^2g}{\partial \theta^2}\bigg](-\frac{\sin\theta}{r}) \\ &\therefore D(r,\theta) = \cos^2\theta,E(r,\theta) = \frac{\sin^2\theta}{r} ,F(r,\theta) = \frac{\sin^2\theta}{r^2},G(r,\theta) = \frac{2\sin\theta\cos\theta}{r^2} ,\\ &\quad H(r,\theta) = - \frac{2\sin\theta\cos\theta}{r^2} \end{aligned}$$

(問 4)

$$\begin{aligned} &\frac{\partial f}{\partial y} = \frac{\partial g}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial g}{\partial \theta}\frac{\partial \theta}{\partial y} = \sin\theta\frac{\partial g}{\partial r} + \frac{\cos\theta}{r}\frac{\partial g}{\partial \theta} \\ &\frac{\partial ^2f}{\partial y^2} = \bigg[\sin\theta\frac{\partial ^2g}{\partial r^2}+ (-\frac{\cos\theta}{r^2})\frac{\partial g}{\partial \theta} + \frac{\cos\theta}{r}\frac{\partial ^2g}{\partial \theta\partial r}\bigg]\sin\theta \\ &\qquad + \bigg[-\cos\theta\frac{\partial g}{\partial r} + \sin\theta\frac{\partial g}{\partial r\partial \theta} + (-\frac{\sin\theta}{r})\frac{\partial g}{\partial \theta} + \frac{\cos\theta}{r}\frac{\partial ^2g}{\partial \theta^2}\bigg]\frac{\cos\theta}{r} \\ &\text{Since }\frac{\partial ^2f}{\partial x^2} + \frac{\partial ^2f}{\partial y^2} + f = 0 \\ &\frac{\partial ^2g}{\partial r^2} + \frac{\partial g}{\partial r}\frac{1}{r} + \frac{\partial ^2g}{\partial \theta^2} \frac{1}{r^2} + g = 0 \end{aligned}$$

(問 5)

$$\begin{aligned} &\frac{\partial g}{\partial r} = \frac{\text{d}R}{\text{d}r}\sin\theta,\frac{\partial ^2g}{\partial r^2} = \frac{\text{d}^2R}{\text{d}r^2}\sin\theta \\ &\frac{\partial g}{\partial \theta} = R\cos\theta,\frac{\partial ^2g}{\partial \theta^2} = -R\sin\theta \\ &\therefore \frac{\text{d}^2R}{\text{d}r^2}\sin\theta + \frac{\text{d}R}{\text{d}r}\sin\theta\frac{1}{r} + (-R\sin\theta)\frac{1}{r^2} + R\sin\theta = 0 \\ &\frac{\text{d}^2R}{\text{d}r^2} + \frac{1}{r}\frac{\text{d}R}{\text{d}r} + (1 - \frac{1}{r^2})R = 0 \end{aligned}$$

(問 6)

(i)

$$\begin{aligned} &\frac{\text{d}R}{\text{d}r} = 1 + \sum_{m=1}^{\infty}(m+1)C_{m}r^{m},\frac{\text{d}^2R}{\text{d}r^2} = \sum_{m = 1}^{\infty}m(m+1)C_{m}r^{m-1} \\ &\therefore\frac{\text{d}^2R}{\text{d}r^2} + \frac{1}{r}\frac{\text{d}R}{\text{d}r} + (1 - \frac{1}{r^2})R \\ &= \sum_{m=1}^{\infty}m(m+1)C_{m}r^{m-1} + \frac{1}{r} + \sum_{m=1}^{\infty}(m+1)C_{m}r^{m-1} + r - \frac{1}{r} + \sum_{m=1}^{\infty}C_{m}r^{m+1} - \sum_{m=1}^{\infty}C_{m}r^{m-1} \\ &= r + \sum_{m=1}^{\infty}(m^2 + m + m + 1 - 1)C_{m}r^{m-1} + \sum_{m=1}^{\infty}C_{m}r^{m+1} \\ &= r + \sum_{m+1}^{\infty}(m^2 + 2m)C_{m}r^{m-1} + \sum_{m=1}^{\infty}C_{m}r^{m+1} \\ &= r + 3C_1 + 8C_2r + \sum_{m=3}^{\infty}m(m+2)C_{m}r^{m-1} + \sum_{m=1}^{\infty}C_{m}r^{m+1} \\ &= 3C_1 + (1 + 8C_2)r + \sum_{m=1}^{\infty}(m + 2)(m + 4)C_{m+2}r^{m+1} + \sum_{m=1}^{\infty}C_{m}r^{m+1} = 0 \\ \end{aligned}$$

$$\therefore \left\{ \begin{aligned} &3C_1 = 0 \\ &1 + 8C_2 = 0 \\ &(m + 2)(m + 4)C_{m+2} + C_m = 0 \end{aligned} \right. , \left\{ \begin{aligned} &C_1 = 0 \\ &C_2 = -\frac{1}{8} \\ &C_{m+2} = -\frac{C_m}{(m+2)(m+4)} \end{aligned} \right.$$

(ii)

$$C_{n+2} = -\frac{C_n}{(n+2)(n+4)}$$