偏微分を含む証明

■問題

$z=\frac{1}{x}f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}$ ならば $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=0\text{\hspace{0.17em}}$ である．

■ヒント

$z\text{\hspace{0.17em}}$$x\text{\hspace{0.17em}},\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ でそれぞれ偏微分し，２式を連立させる． このとき，積の微分合成関数の微分を用いる．

■解説

$u=\frac{y}{x}\text{\hspace{0.17em}}$ とおくと

$z=\frac{1}{x}f\left(u\right)$$\frac{\partial u}{\partial x}=-\frac{y}{{x}^{2}}$$\frac{\partial u}{\partial y}=\frac{1}{x}$

$\frac{\partial z}{\partial x}\text{\hspace{0.17em}}=\left\{\frac{\partial }{\partial x}\left(\frac{1}{x}\right)\right\}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}f\left(u\right)+\frac{1}{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial }{\partial x}f\left(u\right)$

$=\left\{\frac{\partial }{\partial x}\left(\frac{1}{x}\right)\right\}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}f\left(u\right)+\frac{1}{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{f}^{\prime }\left(u\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial u}{\partial x}$

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

$=-\frac{1}{{x}^{2}}·f\left(\frac{y}{x}\right)-\frac{y}{{x}^{3}}·{f}^{\prime }\left(\frac{y}{x}\right)$

より

$-\frac{\partial z}{\partial x}-\frac{1}{{x}^{2}}f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}=\frac{y}{{x}^{3}}{f}^{\prime }\left(\frac{y}{x}\right)$

$-\frac{{x}^{3}}{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial x}-\frac{x}{y}f\left(\frac{y}{x}\right)={f}^{\prime }\left(\frac{y}{x}\right)$

$\frac{\partial z}{\partial y}\text{\hspace{0.17em}}=\left\{\frac{\partial }{\partial y}\left(\frac{1}{x}\right)\right\}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}f\left(u\right)+\frac{1}{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial }{\partial y}f\left(u\right)\text{\hspace{0.17em}}$

$\frac{\partial z}{\partial y}\text{\hspace{0.17em}}=\left\{\frac{\partial }{\partial y}\left(\frac{1}{x}\right)\right\}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}f\left(u\right)+\frac{1}{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{f}^{\prime }\left(u\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial u}{\partial y}$

$=0·f\left(u\right)+\frac{1}{x}·{f}^{\prime }\left(\frac{y}{x}\right)·\frac{1}{x}$

$=\frac{1}{{x}^{2}}·{f}^{\prime }\left(\frac{y}{x}\right)$

より

${x}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial y}={f}^{\prime }\left(\frac{y}{x}\right)$

$-\frac{{x}^{3}}{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial x}-\frac{x}{y}f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}={x}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial y}$

$\frac{{x}^{3}}{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial x}+{x}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial y}+\frac{x}{y}f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}=0$

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+\frac{1}{x}f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}=0$

$z=\frac{1}{x}f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}$ を代入する．

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z\text{\hspace{0.17em}}=0$

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