二階偏微分の連鎖律(chain rule)

定理

2変数関数 f(u,v)f(u,v) について,u=u(x,y),v=v(x,y)u = u(x,y), v = v(x,y) により x,yx,y 変数に置換したとき,2階偏微分は次のようになる。

2fx2=2fu2(ux)2+22fuvuxvx+2fv2(vx)2+fu2ux2+fv2vx2\begin{aligned} \dfrac{\partial^2 f}{\partial x^2} &= \dfrac{\partial^2 f}{\partial u^2} \left( \dfrac{\partial u}{\partial x} \right)^2 + 2 \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial x} \dfrac{\partial v}{\partial x} + \dfrac{\partial^2 f}{\partial v^2} \left( \dfrac{\partial v}{\partial x} \right)^2\\ &\quad + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial x^2} \end{aligned}

2fxy=2fu2uxuy+2fuvuyvx+2fuvuxvy+2fv2vxvy+fu2uxy+fv2vxy\begin{aligned} \dfrac{\partial^2 f}{\partial x \partial y} &= \dfrac{\partial^2 f}{\partial u^2} \dfrac{\partial u}{\partial x} \dfrac{\partial u}{\partial y} + \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial y} \dfrac{\partial v}{\partial x} + \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial x} \dfrac{\partial v}{\partial y}\\ &\quad + \dfrac{\partial^2 f}{\partial v^2} \dfrac{\partial v}{\partial x} \dfrac{\partial v}{\partial y} + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial x \partial y} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial x \partial y} \end{aligned}

2fy2=2fu2(uy)2+22fuvuyvy+2fv2(vy)2+fu2uy2+fv2vy2\begin{aligned} \dfrac{\partial^2 f}{\partial y^2} &= \dfrac{\partial^2 f}{\partial u^2} \left( \dfrac{\partial u}{\partial y} \right)^2 + 2 \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial y} \dfrac{\partial v}{\partial y} + \dfrac{\partial^2 f}{\partial v^2} \left( \dfrac{\partial v}{\partial y} \right)^2\\ &\quad + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial y^2} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial y^2} \end{aligned}

なお,f,u,vf,u,v はどれも C2C^2 級関数としている。

この記事では2変数の合成関数の2階偏微分公式を証明します。

連鎖律(チェーンルール)を使い倒します。復習するときはこちらからどうぞ。→ 連鎖律(チェインルール)~多変数関数の合成関数の微分

証明

証明

気合で計算します

  1. 2fx2\dfrac{\partial^2 f}{\partial x^2}

2fx2=x(fuux+fvvx)\begin{aligned} \dfrac{\partial^2 f}{\partial x^2} &= \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial x} + \dfrac{\partial f}{\partial v} \dfrac{\partial v}{\partial x} \right) \end{aligned}

積の微分公式より ()=x(fu)ux+fux(ux)+x(fv)vx+fvx(vx)\begin{aligned} (\text{式}) &= \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial u} \right) \dfrac{\partial u}{\partial x} + \dfrac{\partial f}{\partial u} \dfrac{\partial}{\partial x} \left( \dfrac{\partial u}{\partial x} \right) \\ &\quad + \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial v} \right) \dfrac{\partial v}{\partial x} + \dfrac{\partial f}{\partial v} \dfrac{\partial}{\partial x} \left( \dfrac{\partial v}{\partial x} \right)\\ \end{aligned} である。

チェーンルールをもう一度用いて整理すると ()=(2fu2ux+2fuvvx)ux+fu2ux2+(2fuvux+2f2vvx)vx+fv2vx2=2fu2(ux)2+22fuvuxvx+2fv2(vx)2+fu2ux2+fv2vx2\begin{aligned} (\text{式}) &= \left( \dfrac{\partial^2 f}{\partial u^2} \dfrac{\partial u}{\partial x} + \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial v}{\partial x} \right) \dfrac{\partial u}{\partial x} + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial x^2} \\ &\quad + \left( \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial x} + \dfrac{\partial^2 f}{\partial^2 v} \dfrac{\partial v}{\partial x} \right) \dfrac{\partial v}{\partial x} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial x^2} \\ &= \dfrac{\partial^2 f}{\partial u^2} \left( \dfrac{\partial u}{\partial x} \right)^2 + 2 \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial x} \dfrac{\partial v}{\partial x} + \dfrac{\partial^2 f}{\partial v^2} \left( \dfrac{\partial v}{\partial x} \right)^2\\ &\quad + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial x^2} \end{aligned} となる。

  1. 2fxy\dfrac{\partial^2 f}{\partial x \partial y}

同様に計算する。 2fxy=x(fuuy+fvvy)\begin{aligned} \dfrac{\partial^2 f}{\partial x \partial y} &= \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial y} + \dfrac{\partial f}{\partial v} \dfrac{\partial v}{\partial y} \right) \\ \end{aligned}

積の微分公式を用いると ()=x(fu)uy+fux(uy)+x(fv)vy+fvx(vy)\begin{aligned} (\text{式}) &= \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial u} \right) \dfrac{\partial u}{\partial y} + \dfrac{\partial f}{\partial u} \dfrac{\partial}{\partial x} \left( \dfrac{\partial u}{\partial y} \right) \\ &\quad + \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial v} \right) \dfrac{\partial v}{\partial y} + \dfrac{\partial f}{\partial v} \dfrac{\partial}{\partial x} \left( \dfrac{\partial v}{\partial y} \right)\\ \end{aligned} である。

チェーンルールをもう一度用いて整理すると ()=(2fu2ux+2fuvvx)uy+fu2uxy+(2fuvux+2f2vvx)vy+fv2vxy=2fu2uxuy+2fuvuyvx+2fuvuxvy+2fv2vxvy+fu2uxy+fv2vxy\begin{aligned} (\text{式}) &= \left( \dfrac{\partial^2 f}{\partial u^2} \dfrac{\partial u}{\partial x} + \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial v}{\partial x} \right) \dfrac{\partial u}{\partial y} + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial x \partial y} \\ &\quad + \left( \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial x} + \dfrac{\partial^2 f}{\partial^2 v} \dfrac{\partial v}{\partial x} \right) \dfrac{\partial v}{\partial y} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial x \partial y} \\ &= \dfrac{\partial^2 f}{\partial u^2} \dfrac{\partial u}{\partial x} \dfrac{\partial u}{\partial y} + \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial y} \dfrac{\partial v}{\partial x} + \dfrac{\partial^2 f}{\partial u \partial v} \dfrac{\partial u}{\partial x} \dfrac{\partial v}{\partial y}\\ &\quad + \dfrac{\partial^2 f}{\partial v^2} \dfrac{\partial v}{\partial x} \dfrac{\partial v}{\partial y} + \dfrac{\partial f}{\partial u} \dfrac{\partial^2 u}{\partial x \partial y} + \dfrac{\partial f}{\partial v} \dfrac{\partial^2 v}{\partial x \partial y} \end{aligned} となる。

  1. 2fy2\dfrac{\partial^2 f}{\partial y^2}

1番と同じなので省略。

計算上の注意点

uxuy2uxy\dfrac{\partial u}{\partial x} \dfrac{\partial u}{\partial y} \neq \dfrac{\partial^2 u}{\partial x \partial y} です。しばしばこのように間違えてしまうことで,計算をミスしてしまう人がいます。

計算がたいへんですね。