練習問題メモ 03

練習問題メモ 03#

和達三樹. (2019). 微分積分. 第5章練習問題

[1]#

[1] 次の関数は原点 $(0, 0)$ を除いて連続である. 原点でも連続にできるか? 直線 $y$ $= m x$ に沿って原点に近づくとして調べよ.

(1) z = \frac{e^{x^{2} + y^{2}} - 1}{x^{2} + y^{2}}

z = \frac{e^{x^{2} + m^{2} x^{2}} - 1}{x^{2} + m^{2} x^{2}} = \frac{e^{(1 + m^{2}) x^{2}} - 1}{(1 + m^{2}) x^{2}}

指数関数のテイラー展開

e x p (a x) = 1 + a x + \frac{(a x)^{2}}{2!} + \frac{(a x)^{3}}{3!} + \dots

により、 $x$ が小さい時

e^{a x^{2}} = 1 + a x^{2}

となるため

z = \frac{e^{(1 + m^{2}) x^{2}} - 1}{(1 + m^{2}) x^{2}} = \frac{1 + (1 + m^{2}) x^{2} - 1}{(1 + m^{2}) x^{2}} = \frac{(1 + m^{2}) x^{2}}{(1 + m^{2}) x^{2}} = 1

となり、 $m$ の値の取り方（0への近づき方）によらず1に収束する

よって連続にできる

\begin{array}{r} z = {\begin{cases} \frac{e^{x^{2} + y^{2}} - 1}{x^{2} + y^{2}} & ((x, y) \neq (0, 0) の と き) \\ 1 & ((x, y) = (0, 0) の と き) \end{cases} \end{array}

../../../_images/f6f43f4d22418d9d52ba2c55719b4cba5ca9424ce3fa646b13101bb8f6d92990.png

(2) z = \frac{x^{2} y^{2}}{x^{4} + y^{4}}

lim_{x \to 0} z = lim_{x \to 0} \frac{x^{4} m^{2}}{x^{4} + m^{4} x^{4}} = lim_{x \to 0} \frac{m^{2}}{1 + m^{4}} = \frac{m^{2}}{1 + m^{4}}

$m$ の値によって異なる値をとるため、原点への近づけ方によって異なる値になってしまう。よって極限は存在せず、連続にできない。

[2]#

[2] 次の関数 $f (x, y)$ の $f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}, f_{y x}$ を計算せよ.

(1) $f (x, y) = x^{3} - x^{2} y + x y^{2} - y^{3} + 1$

(2) $f (x, y) = x^{2} \cos y - y^{2} \cos x$

(3) $f (x, y) = \arctan x^{2} y$

(1) $f (x, y) = x^{3} - x^{2} y + x y^{2} - y^{3} + 1$

\begin{array}{r} f_{x} = 3 x^{2} - 2 x y + y^{2} \\ f_{y} = 3 y^{2} + 2 x y - x^{2} \\ f_{x x} = 6 x - 2 y \\ f_{y y} = 6 y + 2 x \\ f_{x y} = - 2 x + 2 y \\ f_{y x} = - 2 x + 2 y \end{array}

(2) $f (x, y) = x^{2} \cos y - y^{2} \cos x$

\begin{array}{r} f_{x} = 2 x \cos y + y^{2} \sin x \\ f_{y} = - x^{2} \sin y - 2 y \cos x \\ f_{x x} = 2 \cos y + y^{2} \cos x \\ f_{y y} = - x^{2} \cos y - 2 \cos x \\ f_{x y} = - 2 x \sin y + 2 y \sin x \\ f_{y x} = - 2 x \sin y + 2 y \sin x \end{array}

(3) $f (x, y) = \arctan x^{2} y$

$(\arctan x)^{'} = 1 / (1 + x^{2})$ なので

\begin{array}{r} \begin{aligned} f_{x} & = \frac{1}{1 + x^{4} y^{2}} 2 x y \\ f_{y} & = \frac{1}{1 + x^{4} y^{2}} x^{2} \\ f_{x x} & = - \frac{(1 + x^{4} y^{2})^{'}}{(1 + x^{4} y^{2})^{2}} 2 x y + \frac{1}{1 + x^{4} y^{2}} (2 x y)^{'} (積 の 微 分 公 式 と 商 の 微 分 公 式 ） \\ = - \frac{4 x^{3} y^{2}}{(1 + x^{4} y^{2})^{2}} 2 x y + \frac{1}{1 + x^{4} y^{2}} 2 y \\ = - \frac{8 x^{4} y^{3}}{(1 + x^{4} y^{2})^{2}} + \frac{2 y}{1 + x^{4} y^{2}} \\ = - \frac{8 x^{4} y^{3}}{(1 + x^{4} y^{2})^{2}} + \frac{2 y (1 + x^{4} y^{2})}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 y + 2 x^{4} y^{2} - 8 x^{4} y^{3}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 y - 6 x^{4} y^{3}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 y (1 - 3 x^{4} y^{2})}{(1 + x^{4} y^{2})^{2}} \\ f_{y y} & = - \frac{(1 + x^{4} y^{2})^{'}}{(1 + x^{4} y^{2})^{2}} x^{2} + \frac{1}{1 + x^{4} y^{2}} (x^{2})^{'} (積 の 微 分 公 式 と 商 の 微 分 公 式 ） \\ = - \frac{2 x^{6} y}{(1 + x^{4} y^{2})^{2}} \\ f_{x y} & = \frac{(2 x y)^{'} (1 + x^{4} y^{2}) - 2 x y (1 + x^{4} y^{2})^{'}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x (1 + x^{4} y^{2}) - 4 x^{5} y^{2}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x + 2 x^{5} y^{2} - 4 x^{5} y^{2}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x - 2 x^{5} y^{2}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x (1 - x^{4} y^{2})}{(1 + x^{4} y^{2})^{2}} \\ f_{y x} & = \frac{(x^{2})^{'} (1 + x^{4} y^{2}) - x^{2} (1 + x^{4} y^{2})^{'}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x (1 + x^{4} y^{2}) - x^{2} \times 4 x^{3} y^{2}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x + 2 x^{5} y^{2} - 4 x^{5} y^{2}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x - 2 x^{5} y^{2}}{(1 + x^{4} y^{2})^{2}} \\ = \frac{2 x (1 - x^{4} y^{2})}{(1 + x^{4} y^{2})^{2}} \end{aligned} \end{array}

import sympy as sp
x, y = sp.symbols('x y')
f = sp.atan(x**2 * y)

df = sp.diff(f, x)
sp.diff(df, x)

- \frac{8 x^{4} y^{3}}{{(x^{4} y^{2} + 1)}^{2}} + \frac{2 y}{x^{4} y^{2} + 1}

import sympy as sp
x, y = sp.symbols('x y')
f = sp.atan(x**2 * y)

df = sp.diff(f, x)
sp.diff(df, x)

- \frac{8 x^{4} y^{3}}{{(x^{4} y^{2} + 1)}^{2}} + \frac{2 y}{x^{4} y^{2} + 1}

[3]#

[3] 次の関数の偏導関数 $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$ を求めよ.

(1) $z = x^{2} + 4 x y + y^{3}$

(2) $z = \frac{x}{y^{2}} - \frac{y}{x^{2}}$

(3) $z = \sin (2 x + 1) \cos (y^{2} + 4)$

(4) $3 x^{2} + 4 y^{2} - 5 z^{2} = 20$

(5) $\sin x y + \sin y z + \sin x z = 1$

(1) $z = x^{2} + 4 x y + y^{3}$

\begin{array}{r} \frac{\partial z}{\partial x} = 2 x + 4 y \\ \frac{\partial z}{\partial y} = 3 y^{2} + 4 x \end{array}

(2) $z = \frac{x}{y^{2}} - \frac{y}{x^{2}}$

\begin{array}{r} \begin{aligned} \frac{\partial z}{\partial x} & = \frac{1}{y^{2}} (x^{'}) - y {(\frac{1}{x^{2}})}^{'} \\ = \frac{1}{y^{2}} (x^{'}) + y \frac{(x^{2})^{'}}{(x^{2})^{2}} (商 の 微 分 公 式) \\ = \frac{1}{y^{2}} + \frac{2 y}{x^{3}} \\ \frac{\partial z}{\partial y} & = {(\frac{1}{y^{2}})}^{'} x - \frac{1}{x^{2}} (y)^{'} \\ = - \frac{2 y}{y^{4}} x - \frac{1}{x^{2}} \\ = - \frac{2 x}{y^{3}} - \frac{1}{x^{2}} \end{aligned} \end{array}

(3) $z = \sin (2 x + 1) \cos (y^{2} + 4)$

\begin{array}{r} \begin{aligned} \frac{\partial z}{\partial x} & = (\sin (2 x + 1))^{'} \cos (y^{2} + 4) \\ = 2 \cos (2 x + 1) \cos (y^{2} + 4) \\ \frac{\partial z}{\partial y} & = \sin (2 x + 1) (\cos (y^{2} + 4))^{'} \\ = - 2 y \sin (2 x + 1) \sin (y^{2} + 4) \end{aligned} \end{array}

import sympy as sp
x, y = sp.symbols('x y')
f = sp.sin(2*x + 1) * sp.cos(y**2 + 4)
sp.diff(f, y)

- 2 y \sin (2 x + 1) \sin (y^{2} + 4)

(4) $3 x^{2} + 4 y^{2} - 5 z^{2} = 20$

F (x, y, z) := 3 x^{2} + 4 y^{2} - 5 z^{2} - 20 = 0

とおく。

\frac{\partial F}{\partial x} = 6 x, \frac{\partial F}{\partial y} = 8 y, \frac{\partial F}{\partial z} = - 10 z

であり、 $\frac{\partial F}{\partial z} \neq 0$ のため、定理

\frac{\partial z}{\partial x} = - \frac{\partial F / \partial x}{\partial F / \partial z}, \frac{\partial z}{\partial y} = - \frac{\partial F / \partial y}{\partial F / \partial z}

より

\begin{array}{r} \begin{aligned} \frac{\partial z}{\partial x} & = \frac{\partial F / \partial x}{\partial F / \partial z} = \frac{6 x}{10 z} = \frac{3 x}{5 z} \\ \frac{\partial z}{\partial y} & = \frac{\partial F / \partial y}{\partial F / \partial z} = \frac{8 y}{10 z} = \frac{4 y}{5 z} \end{aligned} \end{array}

(5) $\sin x y + \sin y z + \sin x z = 1$

F (x, y, z) := \sin x y + \sin y z + \sin x z - 1 = 0

とおく。

\begin{array}{r} \begin{aligned} \frac{\partial F}{\partial x} & = y \cos x y + z \cos x z \\ \frac{\partial F}{\partial y} & = x \cos x y + z \cos y z \\ \frac{\partial F}{\partial z} & = y \cos y z + x \cos x z \end{aligned} \end{array}

であり、 $\frac{\partial F}{\partial z} \neq 0$ のため、定理より

\begin{array}{r} \begin{aligned} \frac{\partial z}{\partial x} & = - \frac{\partial F / \partial x}{\partial F / \partial z} = - \frac{y \cos x y + z \cos x z}{y \cos y z + x \cos x z} \\ \frac{\partial z}{\partial y} & = - \frac{\partial F / \partial y}{\partial F / \partial z} = - \frac{x \cos x y + z \cos y z}{y \cos y z + x \cos x z} \end{aligned} \end{array}

import sympy as sp
x, y, z = sp.symbols('x y z')
f = sp.sin(x*y) + sp.sin(y*z) + sp.sin(x*z) - 1
sp.diff(f, y)

x \cos (x y) + z \cos (y z)

[4]#

[4] $x = ρ \cos ϕ, y = ρ \sin ϕ$ のとき, 次のことを証明せよ.

(1) $x \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x} = 0$ ならば, $f (x, y)$ は $ρ$ だけの関数である.

(2) $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 0$ ならば, $f (x, y)$ は $ϕ$ だけの関数である.

(3) $\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = \frac{\partial^{2} f}{\partial ρ^{2}} + \frac{1}{ρ} \frac{\partial f}{\partial ρ} + \frac{1}{ρ^{2}} \frac{\partial^{2} f}{\partial ϕ^{2}}$

答例

(2) $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 0$ ならば, $f (x, y)$ は $ϕ$ だけの関数である.

$x = ρ \cos ϕ, y = ρ \sin ϕ$ 。

合成関数の微分（p.121）より

\frac{\partial f}{\partial ρ} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial ρ} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial ρ}

ここで

\frac{\partial x}{\partial ρ} = \cos ϕ = \frac{x}{ρ}, \frac{\partial y}{\partial ρ} = \sin ϕ = \frac{y}{ρ}

なので、

\frac{\partial f}{\partial ρ} = \frac{x}{ρ} \frac{\partial f}{\partial x} + \frac{y}{ρ} \frac{\partial f}{\partial y} = \frac{1}{ρ} (x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y})

もし

(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y})

がゼロならば $\frac{\partial f}{\partial ρ} = 0$ であり、導関数がゼロということはすなわち $f$ は $ρ$ に依存せず $ϕ$ のみの関数

$x = ρ \cos ϕ, y = ρ \sin ϕ$

(1) $x \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x} = 0$ ならば, $f (x, y)$ は $ρ$ だけの関数である.

定理より

\frac{\partial f}{\partial ϕ} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial ϕ} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial ϕ}

ここで

\frac{\partial x}{\partial ϕ} = - ρ \sin ϕ = - y, \frac{\partial y}{\partial ϕ} = ρ \cos ϕ = x

であるから

\frac{\partial f}{\partial ϕ} = x \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x}

となる

よって $x \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x} = 0$ なら、 $f$ は $ϕ$ で微分したとき変化がない定数ということなので、 $f$ は $ρ$ だけの関数

(3) $\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = \frac{\partial^{2} f}{\partial ρ^{2}} + \frac{1}{ρ} \frac{\partial f}{\partial ρ} + \frac{1}{ρ^{2}} \frac{\partial^{2} f}{\partial ϕ^{2}}$

考え方：2次の合成関数の偏導関数を求めていく

\begin{array}{r} \begin{aligned} \frac{\partial^{2} f}{\partial x^{2}} & = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) \\ = \frac{\partial}{\partial x} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x}) \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial x}) \cdot \frac{\partial ρ}{\partial x} + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial x}) \cdot \frac{\partial ϕ}{\partial x} \end{aligned} \end{array}

ここで $\frac{\partial ρ}{\partial x}$ や $\frac{\partial ϕ}{\partial x}$ を求めるために、 $ρ, ϕ$ の関数型とその導関数を求める必要がある

Step 1. $ρ, ϕ$ の関数型の同定：

\begin{array}{r} \begin{aligned} x^{2} + y^{2} & = ρ^{2} \cos^{2} ϕ + ρ^{2} \sin^{2} ϕ \\ = ρ^{2} \underset{= 1}{\underset{⏟}{(\cos^{2} ϕ + \sin^{2} ϕ)}} = ρ^{2} & \to ρ = \sqrt{x^{2} + y^{2}} \\ \frac{y}{x} & = \frac{ρ \sin ϕ}{ρ \cos ϕ} = \tan ϕ & \to ϕ = \arctan (\tan ϕ) = \arctan (\frac{y}{x}) \end{aligned} \end{array}

よって

ρ = \sqrt{x^{2} + y^{2}}, ϕ = \tan^{- 1} (\frac{y}{x})

Step 2. $ρ, ϕ$ の偏導関数

\frac{\partial ρ}{\partial x} = \frac{x}{\sqrt{x^{2} + y^{2}}} = \frac{x}{ρ} = \cos ϕ, \frac{\partial ρ}{\partial y} = \frac{y}{\sqrt{x^{2} + y^{2}}} = \frac{y}{ρ} = \sin ϕ

\frac{\partial ϕ}{\partial x} = - \frac{y}{x^{2} + y^{2}} = - \frac{\sin ϕ}{ρ}, \frac{\partial ϕ}{\partial y} = \frac{x}{x^{2} + y^{2}} = \frac{\cos ϕ}{ρ}

Step 3. $f$ の偏導関数

合成関数の偏微分

\begin{array}{r} \begin{array}{c} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x} \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial y} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial y} \end{array} \end{array}

に代入すると

\begin{array}{r} \begin{aligned} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial ρ} \cos ϕ - \frac{\partial f}{\partial ϕ} \frac{\sin ϕ}{ρ} \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial ρ} \sin ϕ + \frac{\partial f}{\partial ϕ} \frac{\cos ϕ}{ρ} \end{aligned} \end{array}

Step 4. $f$ の2次の偏導関数

\begin{array}{r} \begin{aligned} \frac{\partial^{2} f}{\partial x^{2}} & = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) \\ = \frac{\partial}{\partial x} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x}) \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial x}) \cdot \frac{\partial ρ}{\partial x} + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial x}) \cdot \frac{\partial ϕ}{\partial x} \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial x}) \cdot \cos ϕ + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial x}) \cdot (- \frac{\sin ϕ}{ρ}) \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x}) \cdot \cos ϕ + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x}) \cdot (- \frac{\sin ϕ}{ρ}) \\ = \frac{\partial}{\partial ρ} (\cos ϕ \frac{\partial f}{\partial ρ} - \frac{\sin ϕ}{ρ} \frac{\partial f}{\partial ϕ}) \cdot \cos ϕ + \frac{\partial}{\partial ϕ} (\cos ϕ \frac{\partial f}{\partial ρ} - \frac{\sin ϕ}{ρ} \frac{\partial f}{\partial ϕ}) (- \frac{\sin ϕ}{ρ}) \\ = \frac{\partial}{\partial ρ} (\cos^{2} ϕ \frac{\partial f}{\partial ρ}) - \frac{\partial}{\partial ρ} (\frac{\sin ϕ \cos ϕ}{ρ} \frac{\partial f}{\partial ϕ}) - \frac{\partial}{\partial ϕ} (\frac{\sin ϕ \cos ϕ}{ρ} \frac{\partial f}{\partial ρ}) + \frac{\partial}{\partial ϕ} (\frac{\sin^{2} ϕ}{ρ^{2}} \frac{\partial f}{\partial ϕ}) \\ = \cos^{2} ϕ \frac{\partial^{2} f}{\partial ρ^{2}} - \frac{\partial}{\partial ρ} (\frac{\sin ϕ \cos ϕ}{ρ} \frac{\partial f}{\partial ϕ}) - \frac{\partial}{\partial ϕ} (\frac{\sin ϕ \cos ϕ}{ρ} \frac{\partial f}{\partial ρ}) + \frac{\sin^{2} ϕ}{ρ^{2}} \frac{\partial^{2} f}{\partial ϕ^{2}} \\ = \cos^{2} ϕ \frac{\partial^{2} f}{\partial ρ^{2}} + \frac{2 \sin ϕ \cos ϕ}{ρ^{2}} \frac{\partial f}{\partial ϕ} - \frac{2 \sin ϕ \cos ϕ}{ρ} \frac{\partial^{2} f}{\partial ρ \partial ϕ} + \frac{\sin^{2} ϕ}{ρ} \frac{\partial f}{\partial ρ} + \frac{\sin^{2} ϕ}{ρ^{2}} \frac{\partial^{2} f}{\partial ϕ^{2}} \end{aligned} \end{array}

最後の2行がつながらないが、うまくやればいけそう？

\begin{array}{r} \begin{aligned} \frac{\partial^{2} f}{\partial y^{2}} & = \frac{\partial}{\partial y} (\frac{\partial f}{\partial y}) \\ = \frac{\partial}{\partial y} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial y} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial y}) \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial y}) \cdot \frac{\partial ρ}{\partial y} + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial y}) \cdot \frac{\partial ϕ}{\partial y} \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial y}) \cdot \sin ϕ + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial y}) \cdot \frac{\cos ϕ}{ρ} \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial y} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial y}) \cdot \sin ϕ + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial ρ} \frac{\partial ρ}{\partial y} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial y}) \cdot \frac{\cos ϕ}{ρ} \\ = \frac{\partial}{\partial ρ} (\frac{\partial f}{\partial ρ} \sin ϕ + \frac{\partial f}{\partial ϕ} \frac{\cos ϕ}{ρ}) \cdot \sin ϕ + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial ρ} \sin ϕ + \frac{\partial f}{\partial ϕ} \frac{\cos ϕ}{ρ}) \cdot \frac{\cos ϕ}{ρ} \\ = \sin^{2} ϕ \frac{\partial^{2} f}{\partial ρ^{2}} - \frac{2 \sin ϕ \cos ϕ}{ρ^{2}} \frac{\partial f}{\partial ϕ} + \frac{2 \sin ϕ \cos ϕ}{ρ} \frac{\partial^{2} f}{\partial ρ \partial ϕ} + \frac{\cos^{2} ϕ}{ρ} \frac{\partial f}{\partial ρ} + \frac{\cos^{2} ϕ}{ρ^{2}} \frac{\partial^{2} f}{\partial ϕ^{2}} \end{aligned} \end{array}

Step 5. $f$ の2次の偏導関数の和を求める

\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = \frac{\partial^{2} f}{\partial ρ^{2}} (\cos^{2} ϕ + \sin^{2} ϕ) + \frac{\partial^{2} f}{\partial ϕ^{2}} (\frac{\sin^{2} ϕ + \cos^{2} ϕ}{ρ^{2}}) + \frac{1}{ρ} \frac{\partial f}{\partial ρ}

$\cos^{2} ϕ + \sin^{2} ϕ = 1$ のため

\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = \frac{\partial^{2} f}{\partial ρ^{2}} + \frac{1}{ρ} \frac{\partial f}{\partial ρ} + \frac{1}{ρ^{2}} \frac{\partial^{2} f}{\partial ϕ^{2}}

ということらしい

[5]#

[5] 3 つの変数 $x, y, z$ の間に $F (x, y, z) = 0$ の関係がある. 次のことを示せ.

(1) ${(\frac{\partial y}{\partial x})}_{z} = 1 / {(\frac{\partial x}{\partial y})}_{z}$

(2) ${(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} {(\frac{\partial z}{\partial x})}_{y} = - 1$

(1) ${(\frac{\partial y}{\partial x})}_{z} = 1 / {(\frac{\partial x}{\partial y})}_{z}$

${(\frac{\partial y}{\partial x})}_{z}$ は $z$ を固定したもとでの偏導関数。よって $y$ は $x$ だけの関数であり、1変数の関数である。

逆関数の微分公式

\frac{d x}{d y} = 1 / \frac{d y}{d x} (\frac{d y}{d x} \neq 0 のとき)

より

{(\frac{\partial y}{\partial x})}_{z} = 1 / {(\frac{\partial x}{\partial y})}_{z}

(2) ${(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} {(\frac{\partial z}{\partial x})}_{y} = - 1$

なぜこうではないのか?：

定理より

\begin{array}{r} {(\frac{\partial x}{\partial y})}_{z} = - \frac{F_{y} (x, y, z)}{F_{x} (x, y, z)} \\ {(\frac{\partial y}{\partial z})}_{x} = - \frac{F_{z} (x, y, z)}{F_{y} (x, y, z)} \\ {(\frac{\partial z}{\partial x})}_{y} = - \frac{F_{x} (x, y, z)}{F_{z} (x, y, z)} \end{array}

問題の積は

\begin{array}{r} \begin{aligned} {(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} {(\frac{\partial z}{\partial x})}_{y} & = - \frac{F_{y} (x, y, z)}{F_{x} (x, y, z)} \times - \frac{F_{z} (x, y, z)}{F_{y} (x, y, z)} \times - \frac{F_{x} (x, y, z)}{F_{z} (x, y, z)} \\ = (- 1)^{3} \frac{F_{x} (x, y, z) F_{y} (x, y, z) F_{z} (x, y, z)}{F_{x} (x, y, z) F_{y} (x, y, z) F_{z} (x, y, z)} \\ = - 1 \end{aligned} \end{array}

[6]#

[6] $P (x, y) d x + Q (x, y) d y$ が全微分であるならば,

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

であることを示せ.

ある関数 $f (x, y)$ が存在して、

d f = P (x, y) d x + Q (x, y) d y

であるとする。

P (x, y) = \frac{\partial f}{\partial x} = f_{x}, Q (x, y) = \frac{\partial f}{\partial y} = f_{y}

であるため、

\frac{\partial P}{\partial y} = f_{x y}, \frac{\partial Q}{\partial x} = f_{y x}

を意味する。

定理より、

f_{x y} = f_{y x}

であるため

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

が成り立つ。

[7]#

[7] 関数 $F (x, y)$ がパラメータ $λ$ と定数 $p$ に対して,

F (λ x, λ y) = λ^{p} F (x, y)

をみたすとき, $p$ 次の同次関数(homogeneous function)であるという. このとき, 次式を証明せよ。

x \frac{\partial F}{\partial x} + y \frac{\partial F}{\partial y} = p F

Eulerの等式というらしい（参考）。

Eulerの等式

$F : R^{2} \to R$ が $p$ 次同次関数とする。このとき $F (x, y)$ について、

x \frac{\partial F}{\partial x} + y \frac{\partial F}{\partial y} = p F

が成立する。

F (λ x, λ y) = λ^{p} F (x, y)

の両辺を $λ$ で微分すると、

左辺は、 $F (λ x, λ y) = F (u (λ), v (λ))$ とおくと、全微分の合成関数の定理より

\frac{d F (u (λ), v (λ))}{d λ} = \frac{\partial F}{\partial u} \frac{d u}{d λ} + \frac{\partial F}{\partial v} \frac{d v}{d λ} = \frac{\partial F}{\partial u} x + \frac{\partial F}{\partial v} y

右辺は

\frac{d λ^{p} F (x, y)}{d λ} = p λ^{p - 1} F (x, y)

整理すると

\frac{\partial F}{\partial u} x + \frac{\partial F}{\partial v} y = p λ^{p - 1} F (x, y)

ここで $λ = 1$ とすると、 $u = x, v = y$ なので

\frac{\partial F}{\partial x} x + \frac{\partial F}{\partial y} y = p F (x, y)

[8]#

[8] $x = r \sin θ \cos ϕ, y = r \sin θ \sin ϕ, z = r \cos θ$ のとき, 次式を証明せよ.

\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}} = \frac{\partial^{2} f}{\partial r^{2}} + \frac{2}{r} \frac{\partial f}{\partial r} + \frac{1}{r^{2}} \frac{\partial^{2} f}{\partial θ^{2}} + \frac{1}{r^{2}} \cot θ \frac{\partial f}{\partial θ} + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} f}{\partial ϕ^{2}}

左辺は $f (x, y, z)$ （直交座標系）だが右辺は $f (r, θ, ϕ)$ （球面座標系）となっている。両者の関連を導く。

$x$ は $r, θ, ϕ$ の関数なので、

1階微分は

\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial f}{\partial θ} \frac{\partial θ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x}

2階微分は

\begin{array}{r} \begin{aligned} \frac{\partial^{2} f}{\partial^{2} x} & = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) \\ = \frac{\partial}{\partial x} (\frac{\partial f}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial f}{\partial θ} \frac{\partial θ}{\partial x} + \frac{\partial f}{\partial ϕ} \frac{\partial ϕ}{\partial x}) \\ = \frac{\partial}{\partial r} (\frac{\partial f}{\partial x}) \cdot \frac{\partial r}{\partial x} + \frac{\partial}{\partial θ} (\frac{\partial f}{\partial x}) \cdot \frac{\partial θ}{\partial x} + \frac{\partial}{\partial ϕ} (\frac{\partial f}{\partial x}) \cdot \frac{\partial ϕ}{\partial x} (お そ ら く 、 高 階 微 分 の 順 序 を 入 れ 替 え ら れ る た め) \end{aligned} \end{array}

$r, θ, ϕ$ が

r = \sqrt{x^{2} + y^{2} + z^{2}}, θ = \arctan (\sqrt{x^{2} + y^{2}} / z), ϕ = \arctan (y / x)

であることを知っていればまだ求められそう

（ $r$ の導関数）

\begin{array}{r} \begin{aligned} \frac{\partial r}{\partial x} & = \frac{1}{2} (x^{2} + y^{2} + z^{2})^{- 1 / 2} 2 x = \frac{x}{r} = \sin θ \cos ϕ \\ \frac{\partial r}{\partial y} & = \frac{y}{r} = \sin θ \sin ϕ \\ \frac{\partial r}{\partial z} & = \frac{z}{r} = \cos θ \end{aligned} \end{array}

$= \sin θ \cos ϕ$ とかはよくわからない

（ $θ$ の導関数）

公式より

$\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$

なので

$u = \sqrt{x^{2} + y^{2}} / z$ とおくと

\begin{array}{r} \begin{aligned} \frac{\partial θ}{\partial x} & = \frac{\partial \arctan (u)}{\partial u} \cdot \frac{\partial u}{\partial x} \\ = \frac{1}{1 + u^{2}} \cdot \frac{1}{z} \frac{1}{2} (x^{2} + y^{2})^{- 1 / 2} 2 x \\ = \frac{1}{1 + \frac{x^{2} + y^{2}}{z^{2}}} \cdot \frac{x}{z \sqrt{x^{2} + y^{2}}} \\ = \frac{1}{\frac{x^{2} + y^{2} + z^{2}}{z^{2}}} \cdot \frac{x}{z \sqrt{x^{2} + y^{2}}} \\ = \frac{z^{2}}{r^{2}} \cdot \frac{x}{z \sqrt{x^{2} + y^{2}}} \\ = \frac{z x}{r^{2} \sqrt{x^{2} + y^{2}}} \end{aligned} \end{array}

同様に

\begin{array}{r} \begin{aligned} \frac{\partial θ}{\partial y} & = \frac{z y}{r^{2} \sqrt{x^{2} + y^{2}}} \end{aligned} \end{array}

またzは

\begin{array}{r} \begin{aligned} \frac{\partial θ}{\partial z} & = \frac{\partial \arctan (u)}{\partial u} \cdot \frac{\partial u}{\partial z} \\ = \frac{1}{1 + u^{2}} \cdot - \frac{1}{z^{2}} \sqrt{x^{2} + y^{2}} \\ = \frac{z^{2}}{r^{2}} \cdot - \frac{\sqrt{x^{2} + y^{2}}}{z^{2}} \\ = - \frac{\sqrt{x^{2} + y^{2}}}{r^{2}} \end{aligned} \end{array}

まとめると

\frac{\partial θ}{\partial x} = \frac{z x}{r^{2} \sqrt{x^{2} + y^{2}}}, \frac{\partial θ}{\partial y} = \frac{z y}{r^{2} \sqrt{x^{2} + y^{2}}}, \frac{\partial θ}{\partial z} = - \frac{\sqrt{x^{2} + y^{2}}}{r^{2}}

（ $ϕ$ の導関数）