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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。$$z_{xx}+z_{yy}=z_{rr}+{\displaystyle \frac{1}{r}}{z}_r+{\displaystyle \frac{1}{r^2}}{z}_{\theta \theta }$$

 

《ポイント》

本問はラプラシアン$\Delta$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。

 

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《解答例》

$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial r}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial r}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial r}}\\ & ={\displaystyle \frac{\partial z}{\partial x}}\cos \theta +{\displaystyle \frac{\partial z}{\partial y}}\sin \theta \end{array}$$となるから、$$\begin{array}{rl}{\displaystyle \frac{{\partial }^{2}z}{\partial r^2}}& ={\displaystyle \frac{\partial }{\partial r}}{\displaystyle \frac{\partial z}{\partial x}}\cos \theta +{\displaystyle \frac{\partial }{\partial r}}{\displaystyle \frac{\partial z}{\partial y}}\sin \theta \\ & ={\displaystyle \frac{\partial }{\partial x}}{\displaystyle \frac{\partial z}{\partial r}}\cos \theta +{\displaystyle \frac{\partial }{\partial y}}{\displaystyle \frac{\partial z}{\partial r}}\sin \theta \\ & ={\displaystyle \frac{\partial }{\partial x}}({\displaystyle \frac{\partial z}{\partial x}}\cos \theta +{\displaystyle \frac{\partial z}{\partial y}}\sin \theta )\cos \theta \\ & +{\displaystyle \frac{\partial }{\partial y}}({\displaystyle \frac{\partial z}{\partial x}}\cos \theta +{\displaystyle \frac{\partial z}{\partial y}}\sin \theta )\sin \theta \\ & ={\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}{\cos }^2\theta +{\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y}}\sin \theta \cos \theta +{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}}{\sin }^2\theta \end{array}$$となる。

また、$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial \theta }}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial \theta }}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial \theta }}\\ & ={\displaystyle \frac{\partial z}{\partial x}}(-r\sin \theta )+{\displaystyle \frac{\partial z}{\partial y}}r\cos \theta \end{array}$$となるから、$$\begin{array}{rl}{\displaystyle \frac{{\partial }^{2}z}{\partial {\theta }^2}}& ={\displaystyle \frac{\partial }{\partial \theta }}\{{\displaystyle \frac{\partial z}{\partial x}}(-r\sin \theta )\}+{\displaystyle \frac{\partial }{\partial \theta }}\{{\displaystyle \frac{\partial z}{\partial y}}(r\cos \theta )\}\\ & =\left\{{\displaystyle \frac{\partial }{\partial \theta }}\left({\displaystyle \frac{\partial z}{\partial x}}\right)\right\}(-r\sin \theta )+{\displaystyle \frac{\partial z}{\partial x}}(-r\cos \theta )\\ & +\left\{{\displaystyle \frac{\partial }{\partial \theta }}\left({\displaystyle \frac{\partial z}{\partial y}}\right)\right\}(r\cos \theta )+{\displaystyle \frac{\partial z}{\partial x}}(-r\sin \theta )\\ & =\left\{{\displaystyle \frac{\partial }{\partial x}}\left({\displaystyle \frac{\partial z}{\partial \theta }}\right)\right\}(-r\sin \theta )+{\displaystyle \frac{\partial z}{\partial x}}(-r\cos \theta )\\ & +\left\{{\displaystyle \frac{\partial }{\partial y}}\left({\displaystyle \frac{\partial z}{\partial \theta }}\right)\right\}(r\cos \theta )+{\displaystyle \frac{\partial z}{\partial x}}(-r\sin \theta )\\ & ={\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}(r^2{\sin }^2\theta )-2{\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y}}(r^2\sin \theta \cos \theta )\\ & +{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}}(r^2{\cos }^2\theta )+{\displaystyle \frac{\partial z}{\partial x}}(-r\cos \theta )+{\displaystyle \frac{\partial z}{\partial y}}(-r\sin \theta )\end{array}$$となる。

題意の関係式の右辺 ${\displaystyle \frac{{\partial }^{2}z}{\partial r^2}}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial z}{\partial r}}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{{\partial }^{2}z}{\partial {\theta }^2}}$ を計算すると ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}+{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}}$ に一致するから、$$z_{xx}+z_{yy}=z_{rr}+{\displaystyle \frac{1}{r}}{z}_r+{\displaystyle \frac{1}{r^2}}{z}_{\theta \theta }$$が成立することが示された。

 

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