問題は、$u = u(x, y)$ かつ $x = r \cos\theta$, $y = r \sin\theta$ であるとき、以下の問いに答える問題です。 (1) $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ を $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ で表しなさい。 (2) $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ を $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ で表しなさい。 (3) $\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2$ と $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ を $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$, $\frac{\partial^2 u}{\partial r^2}$, $\frac{\partial^2 u}{\partial \theta^2}$ で表しなさい。

はい、承知いたしました。画像にある数学の問題を解きます。

1. 問題の内容

問題は、$u = u(x, y)$ かつ $x = r \cos\theta$, $y = r \sin\theta$ であるとき、以下の問いに答える問題です。

(1) $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ を $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ で表しなさい。

(2) $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ を $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ で表しなさい。

(3) $\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2$ と $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ を $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$, $\frac{\partial^2 u}{\partial r^2}$, $\frac{\partial^2 u}{\partial \theta^2}$ で表しなさい。

2. 解き方の手順

(1) 合成関数の微分法より、

$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$

$\frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}$

$x = r \cos\theta$, $y = r \sin\theta$ より、

$\frac{\partial x}{\partial r} = \cos\theta$, $\frac{\partial y}{\partial r} = \sin\theta$, $\frac{\partial x}{\partial \theta} = -r\sin\theta$, $\frac{\partial y}{\partial \theta} = r\cos\theta$

したがって、

$\frac{\partial u}{\partial r} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta\frac{\partial u}{\partial y}$

$\frac{\partial u}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y}$

(2) (1)の結果を連立方程式と見て、$\frac{\partial u}{\partial x}$ と $\frac{\partial u}{\partial y}$ について解きます。

$\frac{\partial u}{\partial r} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta\frac{\partial u}{\partial y}$

$\frac{\partial u}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y}$

これを行列で表すと、

$\begin{pmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \end{pmatrix} = \begin{pmatrix} \frac{\partial u}{\partial r} \\ \frac{\partial u}{\partial \theta} \end{pmatrix}$

逆行列を左からかけると、

$\begin{pmatrix} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{pmatrix}^{-1} \begin{pmatrix} \frac{\partial u}{\partial r} \\ \frac{\partial u}{\partial \theta} \end{pmatrix}$

$\begin{pmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{pmatrix}$ の行列式は $r\cos^2\theta + r\sin^2\theta = r$ であるので、逆行列は

$\frac{1}{r}\begin{pmatrix} r\cos\theta & -\sin\theta \\ r\sin\theta & \cos\theta \end{pmatrix} = \begin{pmatrix} \cos\theta & -\frac{\sin\theta}{r} \\ \sin\theta & \frac{\cos\theta}{r} \end{pmatrix}$

したがって、

$\frac{\partial u}{\partial x} = \cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}$

$\frac{\partial u}{\partial y} = \sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta}$

(3)

$\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 = \left(\cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}\right)^2 + \left(\sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta}\right)^2$

$= \cos^2\theta \left(\frac{\partial u}{\partial r}\right)^2 - 2\cos\theta\sin\theta \frac{1}{r}\frac{\partial u}{\partial r}\frac{\partial u}{\partial \theta} + \frac{\sin^2\theta}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2 + \sin^2\theta \left(\frac{\partial u}{\partial r}\right)^2 + 2\sin\theta\cos\theta \frac{1}{r}\frac{\partial u}{\partial r}\frac{\partial u}{\partial \theta} + \frac{\cos^2\theta}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2$

$= \left(\cos^2\theta + \sin^2\theta\right)\left(\frac{\partial u}{\partial r}\right)^2 + \left(\frac{\sin^2\theta}{r^2} + \frac{\cos^2\theta}{r^2}\right)\left(\frac{\partial u}{\partial \theta}\right)^2$

$= \left(\frac{\partial u}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2$

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$

3. 最終的な答え

(1)

$\frac{\partial u}{\partial r} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta\frac{\partial u}{\partial y}$

$\frac{\partial u}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y}$

(2)

$\frac{\partial u}{\partial x} = \cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}$

$\frac{\partial u}{\partial y} = \sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta}$

(3)

$\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 = \left(\frac{\partial u}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2$

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$