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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)(0,0) を除いて連続である. 原点でも連続にできるか? 直線 yy =mx=m x に沿って原点に近づくとして調べよ.

 (1) z=ex2+y2−1x2+y2\text { (1) } \quad z=\frac{e^{x^2+y^2}-1}{x^2+y^2}
(1)
z=ex2+m2x2−1x2+m2x2=e(1+m2)x2−1(1+m2)x2z=\frac{e^{x^2+m^2x^2}-1}{x^2+m^2x^2} =\frac{e^{(1+m^2)x^2}-1}{(1+m^2)x^2}
(2)

指数関数のテイラー展開

exp(ax)=1+ax+(ax)22!+(ax)33!+⋯exp(ax) = 1 + ax + \frac{(ax)^2}{2!} + \frac{(ax)^3}{3!} + \cdots
(3)

により、xxが小さい時

eax2=1+ax2e^{a x^2} = 1 + a x^2
(4)

となるため

z=e(1+m2)x2−1(1+m2)x2=1+(1+m2)x2−1(1+m2)x2=(1+m2)x2(1+m2)x2=1z = \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
(5)

となり、mm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

z={ex2+y2−1x2+y2((x,y)≠(0,0)のとき)1((x,y)=(0,0)のとき)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}
(6)
Source
import numpy as np
import matplotlib.pyplot as plt

a = 1
x = np.linspace(-1, 1, 100)
plt.plot(x, np.exp(a * x**2))
plt.xlabel("x")
plt.ylabel(f"$\exp(a x^2), a=1$")
plt.title(f"$\exp(a x^2), a=1$")
plt.show()
<Figure size 640x480 with 1 Axes>
 (2) z=x2y2x4+y4\text { (2) } \quad z=\frac{x^2 y^2}{x^4+y^4}
(7)
lim⁡x→0z=lim⁡x→0x4m2x4+m4x4=lim⁡x→0m21+m4=m21+m4\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}
(8)

mm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)f(x, y) の fx,fy,fxx,fyy,fxy,fyxf_x, f_y, f_{x x}, f_{y y}, f_{x y}, f_{y x} を計算せよ.

(1) f(x,y)=x3−x2y+xy2−y3+1f(x, y)=x^3-x^2 y+x y^2-y^3+1

(2) f(x,y)=x2cos⁡y−y2cos⁡xf(x, y)=x^2 \cos y-y^2 \cos x

(3) f(x,y)=arctan⁡x2yf(x, y)=\arctan x^2 y

(1) f(x,y)=x3−x2y+xy2−y3+1f(x, y)=x^3-x^2 y+x y^2-y^3+1

fx=3x2−2xy+y2fy=3y2+2xy−x2fxx=6x−2yfyy=6y+2xfxy=−2x+2yfyx=−2x+2yf_x = 3x^2 - 2xy + y^2\\ f_y = 3y^2 + 2xy - x^2\\ f_{xx} = 6x - 2y\\ f_{yy} = 6y + 2x\\ f_{xy} = - 2x + 2y\\ f_{yx} = - 2x + 2y\\
(9)

(2) f(x,y)=x2cos⁡y−y2cos⁡xf(x, y)=x^2 \cos y-y^2 \cos x

fx=2xcos⁡y+y2sin⁡xfy=−x2sin⁡y−2ycos⁡xfxx=2cos⁡y+y2cos⁡xfyy=−x2cos⁡y−2cos⁡xfxy=−2xsin⁡y+2ysin⁡xfyx=−2xsin⁡y+2ysin⁡xf_x = 2x \cos y + y^2 \sin x\\ f_y = - x^2 \sin y - 2y \cos x\\ f_{xx} = 2 \cos y + y^2 \cos x\\ f_{yy} = - x^2 \cos y - 2 \cos x\\ f_{xy} = - 2x \sin y + 2y \sin x\\ f_{yx} = - 2x \sin y + 2y \sin x\\
(10)

(3) f(x,y)=arctan⁡x2yf(x, y)=\arctan x^2 y

(arctan⁡x)′=1/(1+x2)(\arctan x)' = 1/(1+x^2)なので

fx=11+x4y22xyfy=11+x4y2x2fxx=−(1+x4y2)′(1+x4y2)22xy+11+x4y2(2xy)′(積の微分公式と商の微分公式）=−4x3y2(1+x4y2)22xy+11+x4y22y=−8x4y3(1+x4y2)2+2y1+x4y2=−8x4y3(1+x4y2)2+2y(1+x4y2)(1+x4y2)2=2y+2x4y2−8x4y3(1+x4y2)2=2y−6x4y3(1+x4y2)2=2y(1−3x4y2)(1+x4y2)2fyy=−(1+x4y2)′(1+x4y2)2x2+11+x4y2(x2)′(積の微分公式と商の微分公式）=−2x6y(1+x4y2)2fxy=(2xy)′(1+x4y2)−2xy(1+x4y2)′(1+x4y2)2=2x(1+x4y2)−4x5y2(1+x4y2)2=2x+2x5y2−4x5y2(1+x4y2)2=2x−2x5y2(1+x4y2)2=2x(1−x4y2)(1+x4y2)2fyx=(x2)′(1+x4y2)−x2(1+x4y2)′(1+x4y2)2=2x(1+x4y2)−x2×4x3y2(1+x4y2)2=2x+2x5y2−4x5y2(1+x4y2)2=2x−2x5y2(1+x4y2)2=2x(1−x4y2)(1+x4y2)2\begin{align} f_x &= \frac{1}{1+x^4 y^2} 2xy\\ f_y &= \frac{1}{1+x^4 y^2} x^2\\ f_{xx} &= -\frac{(1+x^4 y^2)'}{(1+x^4 y^2)^2} 2xy + \frac{1}{1+x^4 y^2} (2xy)' \quad (積の微分公式と商の微分公式）\\ &= -\frac{4x^3 y^2}{(1+x^4 y^2)^2} 2xy + \frac{1}{1+x^4 y^2} 2y \\ &= -\frac{8x^4 y^3}{(1+x^4 y^2)^2} + \frac{2y}{1+x^4 y^2} \\ &= -\frac{8x^4 y^3}{(1+x^4 y^2)^2} + \frac{2y(1+x^4 y^2)}{(1+x^4 y^2)^2} \\ &= \frac{2y+ 2x^4 y^2 - 8x^4 y^3}{(1+x^4 y^2)^2} \\ &= \frac{2y - 6x^4 y^3}{(1+x^4 y^2)^2} \\ &= \frac{2y(1 - 3x^4 y^2)}{(1+x^4 y^2)^2} \\ f_{yy} &=-\frac{(1+x^4 y^2)'}{(1+x^4 y^2)^2} x^2 + \frac{1}{1+x^4 y^2} (x^2)' \quad (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_{xy} &= \frac{(2xy)' (1 + x^4 y^2) - 2xy (1 + x^4 y^2)'}{(1 + x^4 y^2)^2}\\ &= \frac{2x (1 + x^4 y^2) - 4x^5 y^2}{(1 + x^4 y^2)^2}\\ &= \frac{2x + 2x^5 y^2 - 4x^5 y^2}{(1 + x^4 y^2)^2}\\ &= \frac{2x - 2x^5 y^2}{(1 + x^4 y^2)^2}\\ &= \frac{2x(1 - x^4 y^2)}{(1 + x^4 y^2)^2}\\ \\ f_{yx} &= \frac{(x^2)' (1 + x^4 y^2) - x^2 (1 + x^4 y^2)'}{(1 + x^4 y^2)^2}\\ &= \frac{2x (1 + x^4 y^2) - x^2 \times 4x^3 y^2}{(1 + x^4 y^2)^2}\\ &= \frac{2x + 2x^5 y^2 - 4x^5 y^2}{(1 + x^4 y^2)^2}\\ &= \frac{2x - 2x^5 y^2}{(1 + x^4 y^2)^2}\\ &= \frac{2x(1 - x^4 y^2)}{(1 + x^4 y^2)^2}\\ \end{align}
(11)
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)
Loading...
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)
Loading...

[3]

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

(1) z=x2+4xy+y3z=x^2+4 x y+y^3

(2) z=xy2−yx2z=\frac{x}{y^2}-\frac{y}{x^2}

(3) z=sin⁡(2x+1)cos⁡(y2+4)z=\sin (2 x+1) \cos \left(y^2+4\right)

(4) 3x2+4y2−5z2=203 x^2+4 y^2-5 z^2=20

(5) sin⁡xy+sin⁡yz+sin⁡xz=1\sin x y+\sin y z+\sin x z=1

(1) z=x2+4xy+y3z=x^2+4 x y+y^3

∂z∂x=2x+4y∂z∂y=3y2+4x\frac{\partial z}{\partial x} = 2x + 4y \\ \frac{\partial z}{\partial y} = 3y^2 + 4x
(14)

(2) z=xy2−yx2z=\frac{x}{y^2}-\frac{y}{x^2}

∂z∂x=1y2(x′)−y(1x2)′=1y2(x′)+y(x2)′(x2)2(商の微分公式)=1y2+2yx3∂z∂y=(1y2)′x−1x2(y)′=−2yy4x−1x2=−2xy3−1x2\begin{align} \frac{\partial z}{\partial x} &= \frac{1}{y^2} (x') - y \left(\frac{1}{x^2} \right)'\\ &= \frac{1}{y^2} (x') + y \frac{(x^2)'}{(x^2)^2} \quad (商の微分公式)\\ &= \frac{1}{y^2} + \frac{2y}{x^3}\\ \frac{\partial z}{\partial y} &= \left(\frac{1}{y^2}\right)' x - \frac{1}{x^2} (y)'\\ &= -\frac{2y}{y^4} x - \frac{1}{x^2}\\ &= -\frac{2x}{y^3} - \frac{1}{x^2}\\ \end{align}
(15)

(3) z=sin⁡(2x+1)cos⁡(y2+4)z=\sin (2 x+1) \cos \left(y^2+4\right)

∂z∂x=(sin⁡(2x+1))′cos⁡(y2+4)=2cos⁡(2x+1)cos⁡(y2+4)∂z∂y=sin⁡(2x+1)(cos⁡(y2+4))′=−2ysin⁡(2x+1)sin⁡(y2+4)\begin{align} \frac{\partial z}{\partial x} &= (\sin(2 x+1))' \cos(y^2+4)\\ &= 2 \cos(2x+1) \cos(y^2 + 4)\\ \\ \frac{\partial z}{\partial y} &= \sin(2 x+1) (\cos(y^2+4))'\\ &= -2y \sin(2x+1) \sin(y^2 + 4)\\ \end{align}
(16)
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)
Loading...

(4) 3x2+4y2−5z2=203 x^2+4 y^2-5 z^2=20

F(x,y,z):=3x2+4y2−5z2−20=0F(x,y,z) := 3x^2 + 4y^2 - 5z^2 - 20 = 0
(17)

とおく。

∂F∂x=6x,∂F∂y=8y,∂F∂z=−10z\frac{\partial F}{\partial x} = 6x, \quad \frac{\partial F}{\partial y} = 8y, \quad \frac{\partial F}{\partial z} = -10z
(18)

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

∂z∂x=−∂F/∂x∂F/∂z,∂z∂y=−∂F/∂y∂F/∂z\frac{\partial z}{\partial x} = -\frac{\partial F/\partial x}{\partial F / \partial z} , \quad \frac{\partial z}{\partial y} = -\frac{\partial F/\partial y}{\partial F / \partial z}
(19)

より

∂z∂x=∂F/∂x∂F/∂z=6x10z=3x5z∂z∂y=∂F/∂y∂F/∂z=8y10z=4y5z\begin{align} \frac{\partial z}{\partial x} &= \frac{\partial F/\partial x}{\partial F / \partial z} = \frac{6x}{10z} = \frac{3x}{5z} \\ \frac{\partial z}{\partial y} &= \frac{\partial F/\partial y}{\partial F / \partial z} = \frac{8y}{10z} = \frac{4y}{5z} \end{align}
(20)

(5) sin⁡xy+sin⁡yz+sin⁡xz=1\sin x y+\sin y z+\sin x z=1

F(x,y,z):=sin⁡xy+sin⁡yz+sin⁡xz−1=0F(x,y,z) := \sin x y+\sin y z+\sin x z - 1 = 0
(21)

とおく。

∂F∂x=ycos⁡xy+zcos⁡xz∂F∂y=xcos⁡xy+zcos⁡yz∂F∂z=ycos⁡yz+xcos⁡xz\begin{align} \frac{\partial F}{\partial x} &= y \cos xy + z \cos xz\\ \frac{\partial F}{\partial y} &= x \cos xy + z \cos yz\\ \frac{\partial F}{\partial z} &= y \cos yz + x \cos xz\\ \end{align}
(22)

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

∂z∂x=−∂F/∂x∂F/∂z=−ycos⁡xy+zcos⁡xzycos⁡yz+xcos⁡xz∂z∂y=−∂F/∂y∂F/∂z=−xcos⁡xy+zcos⁡yzycos⁡yz+xcos⁡xz\begin{align} \frac{\partial z}{\partial x} &= -\frac{\partial F/\partial x}{\partial F / \partial z} = -\frac{y \cos xy + z \cos xz}{y \cos yz + x \cos xz} \\ \frac{\partial z}{\partial y} &= -\frac{\partial F/\partial y}{\partial F / \partial z} = -\frac{x \cos xy + z \cos yz}{y \cos yz + x \cos xz} \end{align}
(23)
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)
Loading...

[4]

[4] x=ρcos⁡ϕ,y=ρsin⁡ϕx=\rho \cos \phi, y=\rho \sin \phi のとき, 次のことを証明せよ.

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

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

(3) ∂2f∂x2+∂2f∂y2=∂2f∂ρ2+1ρ∂f∂ρ+1ρ2∂2f∂ϕ2\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}

問題略解
ρ=x2+y2,ϕ=arctan⁡(y/x)∂ρ∂x=xρ=cos⁡ϕ,∂ρ∂y=yρ=sin⁡ϕ∂ϕ∂x=−yρ2=−sin⁡ϕρ,∂ϕ∂y=xρ2=cos⁡ϕρ∂f∂x=∂f∂ρ∂ρ∂x+∂f∂ϕ∂ϕ∂x=cos⁡ϕ∂f∂ρ−sin⁡ϕρ∂f∂ϕ∂f∂y=∂f∂ρ∂ρ∂y+∂f∂ϕ∂ϕ∂y=sin⁡ϕ∂f∂ρ+cos⁡ϕρ∂f∂ϕ\begin{aligned} & \rho=\sqrt{x^2+y^2}, \quad \phi = \arctan (y / x)\\ & \frac{\partial \rho}{\partial x} = \frac{x}{\rho}=\cos \phi, \quad \frac{\partial \rho}{\partial y} = \frac{y}{\rho}=\sin \phi \\ & \frac{\partial \phi}{\partial x} = -\frac{y}{\rho^2}= -\frac{\sin \phi}{\rho}, \quad \frac{\partial \phi}{\partial y}=\frac{x}{\rho^2}=\frac{\cos \phi}{\rho} \\ & \frac{\partial f}{\partial x}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}=\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi} \\ & \frac{\partial f}{\partial y}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y}=\sin \phi \frac{\partial f}{\partial \rho}+\frac{\cos \phi}{\rho} \frac{\partial f}{\partial \phi} \end{aligned}
(24)

(1) x(∂f/∂y)−y(∂f/∂x)=∂f/∂ϕ=0x(\partial f / \partial y)-y(\partial f / \partial x)=\partial f / \partial \phi=0. よって, ff は ϕ\phi によらず, ρ\rho だけの関数.

(2) x(∂f/∂x)+y(∂f/∂y)=ρ(∂f/∂ρ)=0x(\partial f / \partial x)+y(\partial f / \partial y)=\rho(\partial f / \partial \rho)=0. よって, ff は ρ\rho によらず, ϕ\phi だけの関数.

(3)

∂2f∂x2=∂∂x(∂f∂x)=∂∂ρ(∂f∂x)⋅∂ρ∂x+∂∂ϕ(∂f∂x)⋅∂ϕ∂x=∂∂ρ(cos⁡ϕ∂f∂ρ−sin⁡ϕρ∂f∂ϕ)⋅cos⁡ϕ+∂∂ϕ(cos⁡ϕ∂f∂ρ−sin⁡ϕρ∂f∂ϕ)(−sin⁡ϕρ)=cos⁡2ϕ∂2f∂ρ2+2sin⁡ϕcos⁡ϕρ2∂f∂ϕ−2sin⁡ϕcos⁡ϕρ∂2f∂ρ∂ϕ+sin⁡2ϕρ∂f∂ρ+sin⁡2ϕρ2∂2f∂ϕ2\begin{aligned} \frac{\partial^2 f}{\partial x^2}= & \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \\ = & \frac{\partial}{\partial \rho}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right) \cdot \cos \phi \\ & +\frac{\partial}{\partial \phi}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right)\left(-\frac{\sin \phi}{\rho}\right) \\ = & \cos ^2 \phi \frac{\partial^2 f}{\partial \rho^2}+\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}-\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi} \\ & +\frac{\sin ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\sin ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \end{aligned}
(25)

同様にして,

∂2f∂y2=sin⁡2ϕ∂2f∂ρ2−2sin⁡ϕcos⁡ϕρ2∂f∂ϕ+2sin⁡ϕcos⁡ϕρ∂2f∂ρ∂ϕ+cos⁡2ϕρ∂f∂ρ+cos⁡2ϕρ2∂2f∂ϕ2\begin{aligned} \frac{\partial^2 f}{\partial y^2}= & \sin ^2 \phi \frac{\partial^2 f}{\partial \rho^2}-\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}+\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi} \\ & +\frac{\cos ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\cos ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \end{aligned}
(26)

よって,

∂2f∂x2+∂2f∂y2=∂2f∂ρ2+1ρ∂f∂ρ+1ρ2∂2f∂ϕ2\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
(27)
合成関数の微分の定理（p.122）

z=f(x,y)z=f(x, y) において xxとyyは変数uuとvvに依存して、

x=g(u,v),y=h(u,v)x=g(u, v), \quad y=h(u, v)
(32)

とすると

dz=∂z∂xdx+∂z∂ydy,dx=∂x∂udu+∂x∂vdv,dy=∂y∂udu+∂y∂vdvd z=\frac{\partial z}{\partial x} d x+\frac{\partial z}{\partial y} d y, \quad d x=\frac{\partial x}{\partial u} d u+\frac{\partial x}{\partial v} d v, \quad d y=\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v
(33)

あとの2つの式を最初の式に代入して

dz=∂z∂x(∂x∂udu+∂x∂vdv)+∂z∂y(∂y∂udu+∂y∂vdv)=(∂z∂x∂x∂u+∂z∂y∂y∂u)du+(∂z∂x∂x∂v+∂z∂y∂y∂v)dv\begin{aligned} d z & =\frac{\partial z}{\partial x}\left(\frac{\partial x}{\partial u} d u+\frac{\partial x}{\partial v} d v\right)+\frac{\partial z}{\partial y}\left(\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v\right) \\ & =\left(\frac{\partial z}{\partial x} \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial u}\right) d u+\left(\frac{\partial z}{\partial x} \frac{\partial x}{\partial v}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial v}\right) d v \end{aligned}
(34)

zzはuuとvvの関数とみなせるから

dz=∂z∂udu+∂z∂vdvd z=\frac{\partial z}{\partial u} d u+\frac{\partial z}{\partial v} d v
(35)

duduとdvdvの係数（偏導関数の部分）をそれぞれ等しいとおくと

∂z∂u=∂z∂x∂x∂u+∂z∂y∂y∂u∂z∂v=∂z∂x∂x∂v+∂z∂y∂y∂v\begin{aligned} & \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \\ & \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial v}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \end{aligned}
(36)

x=ρcos⁡ϕ,y=ρsin⁡ϕx=\rho \cos \phi, y=\rho \sin \phi

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

定理より

∂f∂ϕ=∂f∂x∂x∂ϕ+∂f∂y∂y∂ϕ\frac{\partial f}{\partial \phi} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \phi} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \phi}
(37)

ここで

∂x∂ϕ=−ρsin⁡ϕ=−y,∂y∂ϕ=ρcos⁡ϕ=x\frac{\partial x}{\partial \phi} = - \rho \sin \phi = -y ,\quad \frac{\partial y}{\partial \phi} = \rho \cos \phi = x
(38)

であるから

∂f∂ϕ=x∂f∂y−y∂f∂x\frac{\partial f}{\partial \phi} = x \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x}
(39)

となる

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

(3) ∂2f∂x2+∂2f∂y2=∂2f∂ρ2+1ρ∂f∂ρ+1ρ2∂2f∂ϕ2\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}

考え方：2次の合成関数の偏導関数を求めていく

∂2f∂x2=∂∂x(∂f∂x)=∂∂x(∂f∂ρ∂ρ∂x+∂f∂ϕ∂ϕ∂x)=∂∂ρ(∂f∂x)⋅∂ρ∂x+∂∂ϕ(∂f∂x)⋅∂ϕ∂x\begin{aligned} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)\\ &= \frac{\partial}{\partial x}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right)\\ &= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \\ \end{aligned}
(40)

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

Step 1. ρ,ϕ\rho, \phiの関数型の同定：

x2+y2=ρ2cos⁡2ϕ+ρ2sin⁡2ϕ=ρ2(cos⁡2ϕ+sin⁡2ϕ)⏟=1=ρ2→ρ=x2+y2yx=ρsin⁡ϕρcos⁡ϕ=tan⁡ϕ→ϕ=arctan⁡(tan⁡ϕ)=arctan⁡(yx)\begin{aligned} x^2 + y^2 &= \rho^2 \cos^2 \phi + \rho^2 \sin^2 \phi\\ &= \rho^2 \underbrace{ (\cos^2 \phi + \sin^2 \phi) }_{=1} = \rho^2 \quad &\to \rho = \sqrt{x^2 + y^2} \\ \frac{y}{x} &= \frac{\rho \sin \phi}{\rho \cos \phi} = \tan \phi \quad &\to \phi = \arctan(\tan \phi) = \arctan \left(\frac{y}{x} \right) \end{aligned}
(41)

よって

ρ=x2+y2,ϕ=tan⁡−1(yx)\rho=\sqrt{x^2+y^2}, \quad \phi=\tan ^{-1}\left(\frac{y}{x}\right)
(42)

Step 2. ρ,ϕ\rho, \phiの偏導関数

∂ρ∂x=xx2+y2=xρ=cos⁡ϕ,∂ρ∂y=yx2+y2=yρ=sin⁡ϕ\frac{\partial \rho}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} = \frac{x}{\rho} =\cos \phi , \quad \frac{\partial \rho}{\partial y}=\frac{y}{\sqrt{x^2+y^2}} = \frac{y}{\rho} =\sin \phi
(43)
∂ϕ∂x=−yx2+y2=−sin⁡ϕρ,∂ϕ∂y=xx2+y2=cos⁡ϕρ\frac{\partial \phi}{\partial x}=-\frac{y}{x^2+y^2}=-\frac{\sin \phi}{\rho} , \quad \frac{\partial \phi}{\partial y}=\frac{x}{x^2+y^2}=\frac{\cos \phi}{\rho}
(44)
導出
∂ϕ∂x=11+(y/x)2⋅(−x′x2y)=−11+y2x2yx2=−1x2+y2x2yx2=−x2x2+y2yx2=−yx2+y2=−sin⁡ϕρ∂ϕ∂y=11+(y/x)2⋅1x=xx2+y2=cos⁡ϕρ\begin{aligned} \frac{\partial \phi}{\partial x} &= \frac{1}{1 + (y/x)^2}\cdot (-\frac{x'}{x^2} y)\\ &= -\frac{1}{1 + \frac{y^2}{x^2}} \frac{y}{x^2}\\ &= -\frac{1}{\frac{x^2 + y^2}{x^2}} \frac{y}{x^2}\\ &= -\frac{x^2}{x^2 + y^2} \frac{y}{x^2}\\ &= -\frac{y}{x^2+y^2}\\ &= -\frac{\sin \phi}{\rho}\\ \\ \frac{\partial \phi}{\partial y} &= \frac{1}{1 + (y/x)^2}\cdot \frac{1}{x}\\ &= \frac{x}{x^2 + y^2}\\ &= \frac{\cos \phi}{\rho}\\ \end{aligned}
(45)

Step 3. ffの偏導関数

合成関数の偏微分

∂f∂x=∂f∂ρ∂ρ∂x+∂f∂ϕ∂ϕ∂x∂f∂y=∂f∂ρ∂ρ∂y+∂f∂ϕ∂ϕ∂y\begin{gathered} \frac{\partial f}{\partial x}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \\ \frac{\partial f}{\partial y}=\frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \end{gathered}
(46)

に代入すると

∂f∂x=∂f∂ρcos⁡ϕ−∂f∂ϕsin⁡ϕρ∂f∂y=∂f∂ρsin⁡ϕ+∂f∂ϕcos⁡ϕρ\begin{aligned} & \frac{\partial f}{\partial x}=\frac{\partial f}{\partial \rho} \cos \phi-\frac{\partial f}{\partial \phi} \frac{\sin \phi}{\rho} \\ & \frac{\partial f}{\partial y}=\frac{\partial f}{\partial \rho} \sin \phi+\frac{\partial f}{\partial \phi} \frac{\cos \phi}{\rho} \end{aligned}
(47)

Step 4. ffの2次の偏導関数

∂2f∂x2=∂∂x(∂f∂x)=∂∂x(∂f∂ρ∂ρ∂x+∂f∂ϕ∂ϕ∂x)=∂∂ρ(∂f∂x)⋅∂ρ∂x+∂∂ϕ(∂f∂x)⋅∂ϕ∂x=∂∂ρ(∂f∂x)⋅cos⁡ϕ+∂∂ϕ(∂f∂x)⋅(−sin⁡ϕρ)=∂∂ρ(∂f∂ρ∂ρ∂x+∂f∂ϕ∂ϕ∂x)⋅cos⁡ϕ+∂∂ϕ(∂f∂ρ∂ρ∂x+∂f∂ϕ∂ϕ∂x)⋅(−sin⁡ϕρ)=∂∂ρ(cos⁡ϕ∂f∂ρ−sin⁡ϕρ∂f∂ϕ)⋅cos⁡ϕ+∂∂ϕ(cos⁡ϕ∂f∂ρ−sin⁡ϕρ∂f∂ϕ)(−sin⁡ϕρ)=∂∂ρ(cos⁡2ϕ∂f∂ρ)−∂∂ρ(sin⁡ϕcos⁡ϕρ∂f∂ϕ)−∂∂ϕ(sin⁡ϕcos⁡ϕρ∂f∂ρ)+∂∂ϕ(sin⁡2ϕρ2∂f∂ϕ)=cos⁡2ϕ∂2f∂ρ2−∂∂ρ(sin⁡ϕcos⁡ϕρ∂f∂ϕ)−∂∂ϕ(sin⁡ϕcos⁡ϕρ∂f∂ρ)+sin⁡2ϕρ2∂2f∂ϕ2=cos⁡2ϕ∂2f∂ρ2+2sin⁡ϕcos⁡ϕρ2∂f∂ϕ−2sin⁡ϕcos⁡ϕρ∂2f∂ρ∂ϕ+sin⁡2ϕρ∂f∂ρ+sin⁡2ϕρ2∂2f∂ϕ2\begin{aligned} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)\\ &= \frac{\partial}{\partial x}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right)\\ &= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \\ &= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial x}\right) \cdot \cos \phi + \frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \left( -\frac{\sin \phi}{\rho} \right) \\ &= \frac{\partial}{\partial \rho}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right) \cdot \cos \phi + \frac{\partial}{\partial \phi}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x} \right) \cdot \left( -\frac{\sin \phi}{\rho} \right) \\ &= \frac{\partial}{\partial \rho}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right) \cdot \cos \phi + \frac{\partial}{\partial \phi}\left(\cos \phi \frac{\partial f}{\partial \rho}-\frac{\sin \phi}{\rho} \frac{\partial f}{\partial \phi}\right)\left(-\frac{\sin \phi}{\rho}\right) \\ &= \frac{\partial}{\partial \rho}\left(\cos^2 \phi \frac{\partial f}{\partial \rho} \right) - \frac{\partial}{\partial \rho} \left( \frac{\sin \phi \cos \phi}{\rho} \frac{\partial f}{\partial \phi}\right) - \frac{\partial}{\partial \phi} \left( \frac{\sin \phi \cos \phi}{\rho} \frac{\partial f}{\partial \rho} \right) + \frac{\partial}{\partial \phi} \left( \frac{\sin^2 \phi}{\rho^2} \frac{\partial f}{\partial \phi}\right) \\ &= \cos^2 \phi \frac{\partial^2 f}{\partial \rho^2} - \frac{\partial}{\partial \rho} \left( \frac{\sin \phi \cos\phi}{\rho} \frac{\partial f}{\partial \phi}\right) - \frac{\partial}{\partial \phi} \left( \frac{\sin \phi \cos \phi}{\rho} \frac{\partial f}{\partial \rho} \right) + \frac{\sin^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \\ &= \cos ^2 \phi \frac{\partial^2 f}{\partial \rho^2}+\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}-\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi} +\frac{\sin ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\sin ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \\ \\ \end{aligned}
(48)

最後の2行がつながらないが、うまくやればいけそう？

∂2f∂y2=∂∂y(∂f∂y)=∂∂y(∂f∂ρ∂ρ∂y+∂f∂ϕ∂ϕ∂y)=∂∂ρ(∂f∂y)⋅∂ρ∂y+∂∂ϕ(∂f∂y)⋅∂ϕ∂y=∂∂ρ(∂f∂y)⋅sin⁡ϕ+∂∂ϕ(∂f∂y)⋅cos⁡ϕρ=∂∂ρ(∂f∂ρ∂ρ∂y+∂f∂ϕ∂ϕ∂y)⋅sin⁡ϕ+∂∂ϕ(∂f∂ρ∂ρ∂y+∂f∂ϕ∂ϕ∂y)⋅cos⁡ϕρ=∂∂ρ(∂f∂ρsin⁡ϕ+∂f∂ϕcos⁡ϕρ)⋅sin⁡ϕ+∂∂ϕ(∂f∂ρsin⁡ϕ+∂f∂ϕcos⁡ϕρ)⋅cos⁡ϕρ=sin⁡2ϕ∂2f∂ρ2−2sin⁡ϕcos⁡ϕρ2∂f∂ϕ+2sin⁡ϕcos⁡ϕρ∂2f∂ρ∂ϕ+cos⁡2ϕρ∂f∂ρ+cos⁡2ϕρ2∂2f∂ϕ2\begin{aligned} \frac{\partial^2 f}{\partial y^2} &= \frac{\partial }{\partial y} \left( \frac{\partial f}{\partial y} \right)\\ &= \frac{\partial }{\partial y} \left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \right)\\ &= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial y}\right) \cdot \frac{\partial \rho}{\partial y}+\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial y}\right) \cdot \frac{\partial \phi}{\partial y} \\ &= \frac{\partial}{\partial \rho}\left(\frac{\partial f}{\partial y}\right) \cdot \sin \phi + \frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial y}\right) \cdot \frac{\cos \phi}{\rho} \\ &= \frac{\partial}{\partial \rho}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \right) \cdot \sin \phi + \frac{\partial}{\partial \phi}\left( \frac{\partial f}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial y} \right) \cdot \frac{\cos \phi}{\rho} \\ &= \frac{\partial}{\partial \rho} \left( \frac{\partial f}{\partial \rho} \sin \phi +\frac{\partial f}{\partial \phi} \frac{\cos \phi}{\rho} \right) \cdot \sin \phi + \frac{\partial}{\partial \phi} \left( \frac{\partial f}{\partial \rho} \sin \phi +\frac{\partial f}{\partial \phi} \frac{\cos \phi}{\rho} \right) \cdot \frac{\cos \phi}{\rho} \\ &= \sin ^2 \phi \frac{\partial^2 f}{\partial \rho^2}-\frac{2 \sin \phi \cos \phi}{\rho^2} \frac{\partial f}{\partial \phi}+\frac{2 \sin \phi \cos \phi}{\rho} \frac{\partial^2 f}{\partial \rho \partial \phi} + \frac{\cos ^2 \phi}{\rho} \frac{\partial f}{\partial \rho}+\frac{\cos ^2 \phi}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} \end{aligned}
(49)

Step 5. ffの2次の偏導関数の和を求める

∂2f∂x2+∂2f∂y2=∂2f∂ρ2(cos⁡2ϕ+sin⁡2ϕ)+∂2f∂ϕ2(sin⁡2ϕ+cos⁡2ϕρ2)+1ρ∂f∂ρ\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}\left(\cos ^2 \phi+\sin ^2 \phi\right)+\frac{\partial^2 f}{\partial \phi^2}\left(\frac{\sin ^2 \phi+\cos ^2 \phi}{\rho^2}\right)+\frac{1}{\rho} \frac{\partial f}{\partial \rho}
(50)

cos⁡2ϕ+sin⁡2ϕ=1\cos ^2 \phi+\sin ^2 \phi=1のため

∂2f∂x2+∂2f∂y2=∂2f∂ρ2+1ρ∂f∂ρ+1ρ2∂2f∂ϕ2\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial f}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2}
(51)

ということらしい

[5]

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

(1) (∂y∂x)z=1/(∂x∂y)z\left(\frac{\partial y}{\partial x}\right)_z=1 /\left(\frac{\partial x}{\partial y}\right)_z

(2) (∂x∂y)z(∂y∂z)x(∂z∂x)y=−1\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1

(1) (∂y∂x)z=1/(∂x∂y)z\left(\frac{\partial y}{\partial x}\right)_z=1 /\left(\frac{\partial x}{\partial y}\right)_z

(∂y∂x)z\left(\frac{\partial y}{\partial x}\right)_zはzzを固定したもとでの偏導関数。よってyyはxx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

dxdy=1/dydx(dydx≠0 のとき )\frac{d x}{d y}=1 / \frac{d y}{d x} \quad\left(\frac{d y}{d x} \neq 0 \text { のとき }\right)
(52)

より

(∂y∂x)z=1/(∂x∂y)z\left(\frac{\partial y}{\partial x}\right)_z=1 /\left(\frac{\partial x}{\partial y}\right)_z
(53)

(2) (∂x∂y)z(∂y∂z)x(∂z∂x)y=−1\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1

略解

zz は x,yx, y の関数だから,

dz=(∂z∂x)ydx+(∂z∂y)xdyd z=\left(\frac{\partial z}{\partial x}\right)_y d x+\left(\frac{\partial z}{\partial y}\right)_x d y
(54)

ところで, (∂x/∂y)z(\partial x / \partial y)_z は, zz を一定, すなわち, dz=0d z=0 としたときの dx/dyd x / d y であるから,

上の式の両辺を dyd y で割って,

0=(∂z∂x)y(∂x∂y)z+(∂z∂y)x0=\left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial y}\right)_z+\left(\frac{\partial z}{\partial y}\right)_x
(55)

(1)により, (∂z/∂y)x=1/(∂y/∂z)x(\partial z / \partial y)_x=1 /(\partial y / \partial z)_x だから

(∂z∂x)y(∂x∂y)z(∂y∂z)x+1=0\left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x+1=0
(56)

よって，証明する式を得る.

なぜこうではないのか?：

定理より

(∂x∂y)z=−Fy(x,y,z)Fx(x,y,z)(∂y∂z)x=−Fz(x,y,z)Fy(x,y,z)(∂z∂x)y=−Fx(x,y,z)Fz(x,y,z)\left(\frac{\partial x}{\partial y}\right)_z = -\frac{F_y(x, y, z)}{F_x(x, y, z)}\\ \left(\frac{\partial y}{\partial z}\right)_x = -\frac{F_z(x, y, z)}{F_y(x, y, z)}\\ \left(\frac{\partial z}{\partial x}\right)_y = -\frac{F_x(x, y, z)}{F_z(x, y, z)}\\
(57)

問題の積は

(∂x∂y)z(∂y∂z)x(∂z∂x)y=−Fy(x,y,z)Fx(x,y,z)×−Fz(x,y,z)Fy(x,y,z)×−Fx(x,y,z)Fz(x,y,z)=(−1)3Fx(x,y,z)Fy(x,y,z)Fz(x,y,z)Fx(x,y,z)Fy(x,y,z)Fz(x,y,z)=−1\begin{aligned} \left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_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}
(58)

[6]

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

∂P∂y=∂Q∂x\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}
(59)

であることを示せ.

ある関数f(x,y)f(x,y)が存在して、

df=P(x,y)dx+Q(x,y)dydf = P(x, y) dx + Q(x, y) dy
(60)

であるとする。

P(x,y)=∂f∂x=fx,Q(x,y)=∂f∂y=fyP(x, y) = \frac{\partial f}{\partial x} = f_x ,\quad Q(x, y) = \frac{\partial f}{\partial y} = f_y
(61)

であるため、

∂P∂y=fxy,∂Q∂x=fyx\frac{\partial P}{\partial y} = f_{xy} ,\quad \frac{\partial Q}{\partial x} = f_{yx}
(62)

を意味する。

定理より、

fxy=fyxf_{xy} = f_{yx}
(63)

であるため

∂P∂y=∂Q∂x\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}
(64)

が成り立つ。

[7]

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

F(λx,λy)=λpF(x,y)F(\lambda x, \lambda y)=\lambda^p F(x, y)
(65)

をみたすとき, pp 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∂F∂x+y∂F∂y=pFx \frac{\partial F}{\partial x}+y \frac{\partial F}{\partial y}=p F
(66)

Eulerの等式というらしい（参考）。

F(λx,λy)=λpF(x,y)F(\lambda x, \lambda y) = \lambda^p F(x, y)
(68)

の両辺をλ\lambdaで微分すると、

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

dF(u(λ),v(λ))dλ=∂F∂ududλ+∂F∂vdvdλ=∂F∂ux+∂F∂vy\frac{dF(u(\lambda), v(\lambda))}{d\lambda} = \frac{\partial F}{\partial u} \frac{d u}{d \lambda} + \frac{\partial F}{\partial v} \frac{d v}{d \lambda} = \frac{\partial F}{\partial u} x + \frac{\partial F}{\partial v} y
(69)

右辺は

dλpF(x,y)dλ=pλp−1F(x,y)\frac{ d \lambda^p F(x, y) }{d \lambda} = p\lambda^{p-1} F(x,y)
(70)

整理すると

∂F∂ux+∂F∂vy=pλp−1F(x,y)\frac{\partial F}{\partial u} x + \frac{\partial F}{\partial v} y = p\lambda^{p-1} F(x,y)
(71)

ここでλ=1\lambda=1とすると、u=x,v=yu=x, v=yなので

∂F∂xx+∂F∂yy=pF(x,y)\frac{\partial F}{\partial x} x + \frac{\partial F}{\partial y} y = p F(x,y)
(72)

[8]

[8] x=rsin⁡θcos⁡ϕ,y=rsin⁡θsin⁡ϕ,z=rcos⁡θx=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi, z=r \cos \theta のとき, 次式を証明せよ.

∂2f∂x2+∂2f∂y2+∂2f∂z2=∂2f∂r2+2r∂f∂r+1r2∂2f∂θ2+1r2cot⁡θ∂f∂θ+1r2sin⁡2θ∂2f∂ϕ2\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 \theta^2}+\frac{1}{r^2} \cot \theta \frac{\partial f}{\partial \theta}+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 f}{\partial \phi^2}
(73)

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

xxはr,θ,ϕr,\theta,\phiの関数なので、

1階微分は

∂f∂x=∂f∂r∂r∂x+∂f∂θ∂θ∂x+∂f∂ϕ∂ϕ∂x\frac{\partial f}{\partial x} =\frac{\partial f}{\partial r} \frac{\partial r}{\partial x} +\frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} +\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}
(74)

2階微分は

∂2f∂2x=∂∂x(∂f∂x)=∂∂x(∂f∂r∂r∂x+∂f∂θ∂θ∂x+∂f∂ϕ∂ϕ∂x)=∂∂r(∂f∂x)⋅∂r∂x+∂∂θ(∂f∂x)⋅∂θ∂x+∂∂ϕ(∂f∂x)⋅∂ϕ∂x(おそらく、高階微分の順序を入れ替えられるため)\begin{aligned} \frac{\partial^2 f}{\partial^2 x} &=\frac{\partial }{\partial x} \left( \frac{\partial f}{\partial x} \right)\\ &=\frac{\partial }{\partial x} \left(\frac{\partial f}{\partial r} \frac{\partial r}{\partial x} +\frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} +\frac{\partial f}{\partial \phi} \frac{\partial \phi}{\partial x}\right) \\ &= \frac{\partial}{\partial r}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial r}{\partial x} +\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \theta}{\partial x} +\frac{\partial}{\partial \phi}\left(\frac{\partial f}{\partial x}\right) \cdot \frac{\partial \phi}{\partial x} \quad (おそらく、高階微分の順序を入れ替えられるため) \end{aligned}
(75)

r,θ,ϕr,\theta,\phiが

r=x2+y2+z2,θ=arctan⁡(x2+y2/z),ϕ=arctan⁡(y/x)r=\sqrt{x^2+y^2+z^2}, \quad \theta=\arctan(\sqrt{x^2+y^2} / z), \quad \phi=\arctan (y / x)
(76)

であることを知っていればまだ求められそう

（rrの導関数）

∂r∂x=12(x2+y2+z2)−1/22x=xr=sin⁡θcos⁡ϕ∂r∂y=yr=sin⁡θsin⁡ϕ∂r∂z=zr=cos⁡θ\begin{aligned} \frac{\partial r}{\partial x} &= \frac{1}{2} (x^2+y^2+z^2)^{-1/2} 2x =\frac{x}{r} =\sin \theta \cos \phi \\ \frac{\partial r}{\partial y} &=\frac{y}{r} = \sin \theta \sin \phi \\ \frac{\partial r}{\partial z} &=\frac{z}{r} = \cos \theta \end{aligned}
(77)

=sin⁡θcos⁡ϕ=\sin \theta \cos \phiとかはよくわからない

（θ\thetaの導関数）

公式より

ddxarctan⁡x=11+x2\frac{d}{d x} \arctan x = \frac{1}{1+x^2}

なので

u=x2+y2/zu=\sqrt{x^2+y^2} / zとおくと

∂θ∂x=∂arctan⁡(u)∂u⋅∂u∂x=11+u2⋅1z12(x2+y2)−1/22x=11+x2+y2z2⋅xzx2+y2=1x2+y2+z2z2⋅xzx2+y2=z2r2⋅xzx2+y2=zxr2x2+y2\begin{aligned} \frac{\partial \theta}{\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} 2x\\ &= \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{zx}{r^2 \sqrt{x^2+y^2}}\\ \end{aligned}
(78)

同様に

∂θ∂y=zyr2x2+y2\begin{aligned} \frac{\partial \theta}{\partial y} &= \frac{zy}{r^2 \sqrt{x^2+y^2}}\\ \end{aligned}
(79)

またzは

∂θ∂z=∂arctan⁡(u)∂u⋅∂u∂z=11+u2⋅−1z2x2+y2=z2r2⋅−x2+y2z2=−x2+y2r2\begin{aligned} \frac{\partial \theta}{\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}
(80)

まとめると

∂θ∂x=zxr2x2+y2,∂θ∂y=zyr2x2+y2,∂θ∂z=−x2+y2r2\frac{\partial \theta}{\partial x} = \frac{zx}{r^2 \sqrt{x^2+y^2}} ,\quad \frac{\partial \theta}{\partial y}= \frac{zy}{r^2 \sqrt{x^2+y^2}} ,\quad \frac{\partial \theta}{\partial z}= -\frac{\sqrt{x^2+y^2}}{r^2}
(81)

（ϕ\phiの導関数）

公式より

ddxarctan⁡x=11+x2\frac{d}{d x} \arctan x = \frac{1}{1+x^2}

なので

∂ϕ∂x=∂arctan⁡(y/x)∂(y/x)⋅∂(y/x)∂x=11+y2x2(−yx2)=−yx2+y2∂θ∂y=∂arctan⁡(y/x)∂(y/x)⋅∂(y/x)∂y=11+y2x2⋅1x=1x2+y2x2⋅1x=x2x2+y2⋅1x=xx2+y2∂θ∂z=0\begin{aligned} \frac{\partial \phi}{\partial x} &= \frac{\partial \arctan(y/x)}{\partial (y/x)} \cdot \frac{\partial (y/x)}{\partial x}\\ &= \frac{1}{1+\frac{y^2}{x^2}} (-\frac{y}{x^2})\\ &= -\frac{y}{x^2+y^2}\\ \\ \frac{\partial \theta}{\partial y} &= \frac{\partial \arctan(y/x)}{\partial (y/x)} \cdot \frac{\partial (y/x)}{\partial y}\\ &= \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{1}{x}\\ &= \frac{1}{\frac{x^2 + y^2}{x^2}} \cdot \frac{1}{x}\\ &= \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x}\\ &= \frac{x}{x^2 + y^2}\\ \\ \frac{\partial \theta}{\partial z} &=0 \end{aligned}
(82)

（複雑すぎてキャパオーバー

数学
練習問題メモ 02
数学
数理最適化