問題4.3.4
$z=f(x,y)$、$x=r \cos \theta$、$x=r \sin \theta$ のとき、次の関係式を示せ。$$z_{xx}+z_{yy}=z_{rr}+\dfrac{1}{r}z_r+\dfrac{1}{r^2}z_{\theta \theta}$$
《ポイント》
本問はラプラシアン$\varDelta$を極座標表示された関数に対して適用するときの関係式の証明問題です。この式を暗記するというよりは導出方法を身に付けるべきでしょう。
《解答例》
$$\begin{align}\dfrac{\partial z}{\partial r} &= \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial r} \\ &= \dfrac{\partial z}{\partial x}\cos \theta+\dfrac{\partial z}{\partial y}\sin \theta \end{align}$$となるから、$$\begin{align}\dfrac{\partial^2 z}{\partial r^2} &= \dfrac{\partial}{\partial r}\dfrac{\partial z}{\partial x}\cos \theta+\dfrac{\partial}{\partial r}\dfrac{\partial z}{\partial y}\sin \theta \\ &= \dfrac{\partial}{\partial x}\dfrac{\partial z}{\partial r}\cos \theta+\dfrac{\partial}{\partial y}\dfrac{\partial z}{\partial r}\sin \theta \\ &= \dfrac{\partial}{\partial x}\left(\dfrac{\partial z}{\partial x}\cos \theta+\dfrac{\partial z}{\partial y}\sin \theta\right)\cos \theta \\ & \ \ \ \ +\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial x}\cos \theta+\dfrac{\partial z}{\partial y}\sin \theta\right)\sin \theta \\ &= \dfrac{\partial^2 z}{\partial x^2}\cos^2 \theta+\dfrac{\partial^2 z}{\partial x \partial y}\sin \theta \cos \theta+\dfrac{\partial^2 z}{\partial y^2}\sin^2 \theta \end{align}$$となる。
また、$$\begin{align}\dfrac{\partial z}{\partial \theta} &= \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial \theta}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial \theta} \\ &= \dfrac{\partial z}{\partial x}(-r\sin \theta)+\dfrac{\partial z}{\partial y}r\cos \theta \end{align}$$となるから、$$\begin{align}\dfrac{\partial^2 z}{\partial \theta^2} &= \dfrac{\partial}{\partial \theta}\left\{\dfrac{\partial z}{\partial x}(-r\sin \theta)\right\}+\dfrac{\partial}{\partial \theta}\left\{\dfrac{\partial z}{\partial y}(r\cos \theta)\right\} \\
&= \left\{\dfrac{\partial}{\partial \theta}\left(\dfrac{\partial z}{\partial x}\right)\right\}(-r\sin \theta)+\dfrac{\partial z}{\partial x}(-r\cos \theta) \\ &\ \ \ \ +\left\{\dfrac{\partial}{\partial \theta}\left(\dfrac{\partial z}{\partial y}\right)\right\}(r\cos \theta)+\dfrac{\partial z}{\partial x}(-r\sin \theta) \\
&= \left\{\dfrac{\partial}{\partial x}\left(\dfrac{\partial z}{\partial \theta}\right)\right\}(-r\sin \theta)+\dfrac{\partial z}{\partial x}(-r\cos \theta) \\ &\ \ \ \ +\left\{\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial \theta}\right)\right\}(r\cos \theta)+\dfrac{\partial z}{\partial x}(-r\sin \theta) \\
&= \dfrac{\partial^2 z}{\partial x^2}(r^2\sin^2 \theta)-2\dfrac{\partial^2 z}{\partial x \partial y}(r^2\sin \theta \cos \theta) \\ & \ \ \ \ +\dfrac{\partial^2 z}{\partial y^2}(r^2 \cos^2 \theta)+\dfrac{\partial z}{\partial x}(-r\cos \theta)+\dfrac{\partial z}{\partial y}(-r\sin \theta) \end{align}$$となる。
題意の関係式の右辺 $\dfrac{\partial^2 z}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial z}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 z}{\partial \theta^2}$ を計算すると $\dfrac{\partial^2 z}{\partial x^2}+\dfrac{\partial^2 z}{\partial y^2}$ に一致するから、$$z_{xx}+z_{yy}=z_{rr}+\dfrac{1}{r}z_r+\dfrac{1}{r^2}z_{\theta \theta}$$が成立することが示された。
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