調和関数に関する問題

■問題

次の関数は調和関数であることを示せ．

$3$\displaystyle{ f $ x,y $ ={ \tan }^{ - 1 }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} }$

■ヒント

$3$\displaystyle{ {\Delta}f=\left({ \frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}} }\right)f }$ $3$\displaystyle{ =\frac{\hspace{2} {{\partial}}^{2}f \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}f \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}} }$ $3$\displaystyle{ =\hspace{1}{f}_{ xx }+\hspace{1}{f}_{ yy } }$ $3$\displaystyle{ =0 }$

がなりたつことを示す．

■答

$3$\displaystyle{\hspace{1}{f}_{x}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{ $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }^{2} \hspace{2}}\times $ - \frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} $ }$ $3$\displaystyle{=\frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2} {x}^{2}+{y}^{2} \hspace{2}}\times $ - \frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} $ }$ $3$\displaystyle{=- \frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{2}+{y}^{2} \hspace{2}}}$

$3$\displaystyle{\hspace{1}{f}_{ xx }=- \frac{\hspace{2} - y\cdot 2x \hspace{2}}{\hspace{2} { $ {x}^{2}+{y}^{2} $ }^{2} \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2} 2xy \hspace{2}}{\hspace{2} { $ {x}^{2}+{y}^{2} $ }^{2} \hspace{2}}}$

$3$\displaystyle{\hspace{1}{f}_{y}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{ $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }^{2} \hspace{2}}\times $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} $ }$ $3$\displaystyle{=\frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2} {x}^{2}+{y}^{2} \hspace{2}}\times $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} $ }$ $3$\displaystyle{=\frac{\hspace{2}x\hspace{2}}{\hspace{2} {x}^{2}+{y}^{2} \hspace{2}}}$

$3$\displaystyle{\hspace{1}{f}_{ yy }=- \frac{\hspace{2} - x\cdot 2y \hspace{2}}{\hspace{2} { $ {x}^{2}+{y}^{2} $ }^{2} \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} 2xy \hspace{2}}{\hspace{2} { $ {x}^{2}+{y}^{2} $ }^{2} \hspace{2}}}$

$3$\displaystyle{{}_\bullet\limits{ }^\bullet {}_\bullet {\Delta}f=\hspace{1}{f}_{ xx }+\hspace{1}{f}_{ yy }}$ $3$\displaystyle{=\frac{\hspace{2} 2xy \hspace{2}}{\hspace{2} $ {x}^{2}+{y}^{2} $ \hspace{2}}+ \{ - \frac{\hspace{2} 2xy \hspace{2}}{\hspace{2} { $ {x}^{2}+{y}^{2} $ }^{2} \hspace{2}} \} }$ $3$\displaystyle{=0}$

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最終更新日： 2024年5月28日