# 偏微分の順序交換

$z=f\left(x,y\right)$  について，$\frac{{\partial }^{2}z}{\partial y\partial x}$$\frac{{\partial }^{2}z}{\partial x\partial y}$ が共に連続ならば

$\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}$

${f}_{xy}\left(x,y\right)={f}_{yx}\left(x,y\right)$

が成り立つ．

## ■証明

${f}_{x}\left(x,y\right)=\frac{\partial f}{\partial x}$$=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h,y\right)-f\left(x,y\right)}{h}$

${f}_{y}\left(x,y\right)=\frac{\partial f}{\partial y}$$=\underset{k\to 0}{\mathrm{lim}}\frac{f\left(x,y+k\right)-f\left(x,y\right)}{k}$

また

${f}_{xy}=\frac{\partial {f}_{x}}{\partial y}=\underset{k\to 0}{\mathrm{lim}}\frac{{f}_{x}\left(x,y+k\right)-{f}_{x}\left(x,y\right)}{k}$

$=\underset{k\to 0}{\mathrm{lim}}\frac{\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h,y+k\right)-f\left(x,y+k\right)}{h}-\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h,y\right)-f\left(x,y\right)}{h}}{k}$

$=\underset{k\to 0}{\mathrm{lim}}\left[\frac{1}{k}\left\{\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h,y+k\right)-f\left(x,y+k\right)-f\left(x+h,y\right)+f\left(x,y\right)}{h}\right\}\right]$

$=\underset{k\to 0}{\mathrm{lim}}\left[\underset{h\to 0}{\mathrm{lim}}\frac{1}{hk}\left\{f\left(x+h,y+k\right)-f\left(x,y+k\right)-f\left(x+h,y\right)+f\left(x,y\right)\right\}\right]$  ･･････(1)

${f}_{yx}=\frac{\partial {f}_{y}}{\partial x}=\underset{h\to 0}{\mathrm{lim}}\frac{{f}_{y}\left(x+h,y\right)-{f}_{y}\left(x,y\right)}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{\underset{k\to 0}{\mathrm{lim}}\frac{f\left(x+h,y+k\right)-f\left(x+h,y\right)}{k}-\underset{k\to 0}{\mathrm{lim}}\frac{f\left(x,y+k\right)-f\left(x,y\right)}{k}}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\left[\frac{1}{h}\left\{\underset{k\to 0}{\mathrm{lim}}\frac{f\left(x+h,y+k\right)-f\left(x+h,y\right)}{k}-\underset{k\to 0}{\mathrm{lim}}\frac{f\left(x,y+k\right)-f\left(x,y\right)}{k}\right\}\right]$

$=\underset{h\to 0}{\mathrm{lim}}\left[\underset{k\to 0}{\mathrm{lim}}\frac{1}{hk}\left\{f\left(x+h,y+k\right)-f\left(x,y+k\right)-f\left(x+h,y\right)+f\left(x,y\right)\right\}\right]$  ･･････(2)

ここで

$\Delta =f\left(x+h,y+k\right)-f\left(x,y+k\right)$$-f\left(x+h,y\right)$$+f\left(x,y\right)$  ･･････(3)

とおき，

${g}_{1}\left(x\right)=f\left(x,y+k\right)-f\left(x,y\right)$

とおくと，(3)は

$\Delta ={g}_{1}\left(x+h\right)-{g}_{1}\left(x\right)$

$\Delta ={{g}^{\prime }}_{1}\left(x+{\theta }_{1}h\right)h$     ($0<{\theta }_{1}<1$ )

ここで${{g}^{\prime }}_{1}\left(x\right)$

${{g}^{\prime }}_{1}\left(x\right)=\frac{d}{dx}{g}_{1}\left(x\right)$

$=\left\{\frac{d}{dx}{g}_{1}\left(x+{\theta }_{1}h\right)\right\}h$

$=\frac{\partial }{\partial x}\left\{f\left(x,y+k\right)-f\left(x,y\right)\right\}$

$={f}_{x}\left(x,y+k\right)-{f}_{x}\left(x,y\right)$

よって

$\Delta =\left\{{f}_{x}\left(x+{\theta }_{1}h,y+k\right)-{f}_{x}\left(x+{\theta }_{1}h,y\right)\right\}h$  ･･････(4)

${g}_{2}\left(y\right)={f}_{x}\left(x+{\theta }_{1}h,y\right)$  とおくと(4)は，

$\Delta =\left\{{g}_{2}\left(y+k\right)-{g}_{2}\left(y\right)\right\}h$

${g}_{2}\left(y+k\right)-{g}_{2}\left(y\right)={{g}^{\prime }}_{2}\left(y+{\theta }_{2}k\right)k$

となるので

$\Delta =\left\{{{g}^{\prime }}_{2}\left(y+{\theta }_{2}k\right)k\right\}h$     ( $0<{\theta }_{2}<1$ )

$=\left\{{{g}^{\prime }}_{2}\left(y+{\theta }_{2}k\right)\right\}hk$

ここで${{g}^{\prime }}_{2}\left(y\right)$

${{g}^{\prime }}_{2}\left(y\right)=\frac{d}{dy}{g}_{2}\left(y\right)$

$=\frac{\partial }{\partial y}{f}_{x}\left(x+{\theta }_{1}h,y\right)$

$={f}_{xy}\left(x+{\theta }_{1}h,y\right)$

よって

$\Delta ={f}_{xy}\left(x+{\theta }_{1}h,y+{\theta }_{2}k\right)hk$  ･･････(5)

$\Delta ={h}_{1}\left(y+k\right)-{h}_{1}\left(y\right)$

$\Delta ={{h}^{\prime }}_{1}\left(y+{\varphi }_{1}k\right)k$     ($0<{\varphi }_{1}<1$ )

ここで${{h}^{\prime }}_{1}\left(y\right)$

${{h}^{\prime }}_{1}\left(y\right)=\frac{d}{dy}{h}_{1}\left(y\right)$

$=\frac{\partial }{\partial y}\left\{f\left(x+h,y\right)-f\left(x,y\right)\right\}$

$={f}_{y}\left(x+h,y\right)-{f}_{y}\left(x,y\right)$

よって

$\Delta =\left\{{f}_{y}\left(x+h,y+{\varphi }_{1}k\right)-{f}_{y}\left(x,y+{\varphi }_{1}k\right)\right\}k$  ･･････(6)

${h}_{2}\left(x\right)={f}_{y}\left(x,y+{\varphi }_{1}k\right)$  とおくと(6)は

$\Delta =\left\{{h}_{2}\left(x+h\right)-{h}_{2}\left(x\right)\right\}k$

${h}_{2}\left(x+h\right)-{h}_{2}\left(x\right)={{h}^{\prime }}_{2}\left(x+{\varphi }_{2}h\right)h$

となるので

$\Delta =\left\{{{h}^{\prime }}_{2}\left(x+{\varphi }_{2}h\right)h\right\}k$     ( $0<{\varphi }_{2}<1$ )

$=\left\{{{h}^{\prime }}_{2}\left(x+{\varphi }_{2}h\right)\right\}hk$

ここで${{h}^{\prime }}_{2}\left(x\right)$

${{h}^{\prime }}_{2}\left(x\right)=\frac{d}{dx}{h}_{2}\left(x\right)$

$=\frac{\partial }{\partial x}{f}_{y}\left(x,y+{\varphi }_{1}k\right)$

$={f}_{yx}\left(x,y+{\varphi }_{1}k\right)$

よって

$\Delta ={f}_{yx}\left(x+{\varphi }_{2}h,y+{\varphi }_{1}k\right)hk$  ･･････(7)

(1)，(5)より

$\underset{\begin{array}{l}h\to 0\\ k\to 0\end{array}}{\mathrm{lim}}\frac{\Delta }{hk}=\underset{\begin{array}{l}h\to 0\\ k\to 0\end{array}}{\mathrm{lim}}{f}_{xy}\left(x+{\theta }_{1}h,y+{\theta }_{2}k\right)$$={f}_{xy}\left(x,y\right)$  ･･････(8)

(2)，(7)より

$\underset{\begin{array}{l}h\to 0\\ k\to 0\end{array}}{\mathrm{lim}}\frac{\Delta }{hk}=\underset{\begin{array}{l}h\to 0\\ k\to 0\end{array}}{\mathrm{lim}}{f}_{yx}\left(x+{\varphi }_{2}h,y+{\varphi }_{1}k\right)$$={f}_{yx}\left(x,y\right)$  ･･････(9)

(8)，(9)より

${f}_{xy}\left(x,y\right)={f}_{yx}\left(x,y\right)$

ホーム>>カテゴリー別分類>>微分>>偏微分>>偏微分の順序交換