# 調和関数

## ■問題

$f\left(x,y\right)={\mathrm{tan}}^{-1}\frac{y}{x}$

## ■ヒント

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

がなりたつことを示す．

## ■答

${f}_{x}=\frac{1}{1+{\left(\frac{y}{x}\right)}^{2}}×\left(-\frac{y}{{x}^{2}}\right)$$=\frac{{x}^{2}}{{x}^{2}+{y}^{2}}×\left(-\frac{y}{{x}^{2}}\right)$$=-\frac{y}{{x}^{2}+{y}^{2}}$

${f}_{xx}=-\frac{-y\cdot 2x}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$$=\frac{2xy}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$

${f}_{y}=\frac{1}{1+{\left(\frac{y}{x}\right)}^{2}}×\left(\frac{1}{x}\right)$$=\frac{{x}^{2}}{{x}^{2}+{y}^{2}}×\left(\frac{1}{x}\right)$$=\frac{x}{{x}^{2}+{y}^{2}}$

${f}_{yy}=-\frac{-x\cdot 2y}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$$=-\frac{2xy}{{\left({x}^{2}+{y}^{2}\right)}^{2}}$

$\therefore \Delta f={f}_{xx}+{f}_{yy}$$=\frac{2xy}{\left({x}^{2}+{y}^{2}\right)}+\left\{-\frac{2xy}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\right\}$$=0$

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