必要十分条件の証明

■必要条件の証明

$3$\displaystyle{Pdx+Qdy=0}$ が完全微分方程式であるとすれば， $3$\displaystyle{du=Pdx+Qdy}$ となる関数 $3$\displaystyle{u $ x,y $ }$ が存在する．

$3$\displaystyle{du=\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}dx+\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}dy}$

より

$3$\displaystyle{P=\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ ， $3$\displaystyle{Q=\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$

である．

$3$\displaystyle{\frac{\hspace{2} {\partial}P \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ =\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}}$ ･･････(1)

$3$\displaystyle{\frac{\hspace{2} {\partial}Q \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ =\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$ ･･････(2)

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}u \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$ ･･････(3)　　(偏微分の順序交換より)

(1)，(2)，(3)より

$3$\displaystyle{\frac{\hspace{2} {\partial}P \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2} {\partial}Q \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$

となる．

以上より

$3$\displaystyle{Pdx+Qdy=0}$ が完全微分方程式であれば $3$\displaystyle{\frac{\hspace{2} {\partial}P \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2} {\partial}Q \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ が成り立つ．

■十分条件の証明

$3$\displaystyle{\frac{\hspace{2} {\partial}P \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2} {\partial}Q \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ が成り立つとする．

$3$\displaystyle{F $ x,y $ =\displaystyle{ \int _{a}^{x} P $ t,y $ dt }}$

$3$\displaystyle{G $y$ =\displaystyle{ \int _{a}^{y} Q $ a,s $ ds }}$

とし

$3$\displaystyle{u $ x,y $ =F $ x,y $ +G $y$ }$

とおく．

$3$\displaystyle{\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} \{ F $ x,y $ +G $y$ \} }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}F $ x,y $ }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\displaystyle{ \int _{a}^{x} P $ t,y $ dt }}$

$3$\displaystyle{=P $ x,y $ }$

$3$\displaystyle{\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} \{ F $ x,y $ +G $y$ \} }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}F $ x,y $ +\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}G $y$ }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\displaystyle{ \int _{a}^{x} P $ t,y $ dt }+\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\displaystyle{ \int _{a}^{y} Q $ a,s $ ds }}$

$3$\displaystyle{=\displaystyle{ \int _{a}^{x} \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}P $ t,y $ dt }+Q $ a,y $ }$ 　　(微分と積分の順序交換より)

$3$\displaystyle{ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}P\left({ t,y }\right)=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}t \hspace{2}}Q\left({ t,y }\right) }$ より

$3$\displaystyle{=\displaystyle{ \int _{a}^{x} \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}t \hspace{2}}Q $ t,y $ dt }+Q $ a,y $ }$

$3$\displaystyle{= \[ Q $ t,y $ \] _{a}^{x}+Q $ a,y $ }$

$3$\displaystyle{=Q $ x,y $ - Q $ a,y $ +Q $ a,y $ }$

$3$\displaystyle{=Q $ x,y $ }$

よって

$3$\displaystyle{Pdx+Qdy=\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}dx+\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}du=du}$

したがって $3$\displaystyle{Pdx+Qdy=0}$ は完全微分方程式である．

以上より

$3$\displaystyle{\frac{\hspace{2} {\partial}P \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2} {\partial}Q \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ が成り立つならば $3$\displaystyle{Pdx+Qdy=0}$ は完全微分方程式である．

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　最終更新日： 2024年5月17日