在探討拉普拉斯算子的旋轉不變性之前，先來說說方向導數，定義如下：
考慮函數 f ( x , y ) \small f(x,y) f ( x , y ) f \small f f P 0 ( x 0 , y 0 ) \small P_0(x_0,y_0) P 0 ( x 0 , y 0 ) U ( P 0 ) \small U(P_0) U ( P 0 ) l = ( cos ⟨ l , x ⟩ , sin ⟨ l , x ⟩ ) \small l=(\cos\langle l,x\rangle,\sin\langle l,x\rangle) l = ( cos ⟨ l , x ⟩ , sin ⟨ l , x ⟩ ) ⟨ l , x ⟩ \langle l,x\rangle ⟨ l , x ⟩ l l l x x x P \small P P P 0 P → = ρ l \small \overrightarrow{P_0P}=\rho\,l P 0 P = ρ l ρ > 0 \rho>0 ρ > 0 P \small P P P 0 \small P_0 P 0 lim ρ → 0 + f ( P ) − f ( P 0 ) ρ
\lim_{\rho\to0^+}\frac{f(P)-f(P_0)}{\rho}
ρ → 0 + lim ρ f ( P ) − f ( P 0 )
存在，則稱此極限爲函數 f \small f f P 0 \small P_0 P 0 l l l 方向導數 ，記作∂ f ∂ l ∣ P 0 o r f l ( P 0 )
\frac{\partial f}{\partial l}\Big|_{P_0} \,\,or\,\,f_l(P_0)
∂ l ∂ f ∣ ∣ ∣ P 0 o r f l ( P 0 )
直覺性較強的讀者應該能夠看出：若 l l l x x x f x ( P 0 ) \small f_x(P_0) f x ( P 0 ) l l l y y y f y ( P 0 ) \small f_y(P_0) f y ( P 0 )
那能否通過偏導數 f x ( P 0 ) , f y ( P 0 ) \small f_x(P_0),f_y(P_0) f x ( P 0 ) , f y ( P 0 )
答案是肯定的，有如下定理：
設 f \small f f P 0 ( x 0 , y 0 ) \small P_0(x_0,y_0) P 0 ( x 0 , y 0 ) f \small f f P 0 \small P_0 P 0 l = ( cos ⟨ l , x ⟩ , sin ⟨ l , x ⟩ ) \small l=(\cos\langle l,x\rangle,\sin\langle l,x\rangle) l = ( cos ⟨ l , x ⟩ , sin ⟨ l , x ⟩ ) ∂ f ∂ l = ∂ f ∂ x cos ⟨ l , x ⟩ + ∂ f ∂ y sin ⟨ l , x ⟩
\frac{\partial f}{\partial l}=\frac{\partial f}{\partial x}\cos\langle l,x\rangle+\frac{\partial f}{\partial y}\sin\langle l,x\rangle
∂ l ∂ f = ∂ x ∂ f cos ⟨ l , x ⟩ + ∂ y ∂ f sin ⟨ l , x ⟩
這個定理的證明有兩種思路，不知客官喜歡哪一種？
證法 1： 根據定義
設 P ( x , y ) \small P(x,y) P ( x , y ) P 0 P → = ρ l , ρ > 0 \small \overrightarrow{P_0P}=\rho\,l,\,\rho>0 P 0 P = ρ l , ρ > 0 { x − x 0 = Δ x = ρ cos ⟨ l , x ⟩ y − y 0 = Δ y = ρ sin ⟨ l , x ⟩ (*)
\begin{cases}
x-x_0=\Delta x=\rho\cos\langle l,x \rangle\\
y-y_0=\Delta y=\rho\sin\langle l,x \rangle \tag{*}
\end{cases}
{ x − x 0 = Δ x = ρ cos ⟨ l , x ⟩ y − y 0 = Δ y = ρ sin ⟨ l , x ⟩ ( * )
由 f f f P 0 \small P_0 P 0
f ( P ) − f ( P 0 ) = f x ( P 0 ) Δ x + f y ( P 0 ) Δ y + o ( ρ )
f(P)-f(P_0)=f_x(P_0)\Delta x+f_y(P_0)\Delta y+o(\rho)
f ( P ) − f ( P 0 ) = f x ( P 0 ) Δ x + f y ( P 0 ) Δ y + o ( ρ )
兩邊同除 ρ \rho ρ ( ∗ ) (*) ( ∗ )
f ( P ) − f ( P 0 ) ρ = f x ( P 0 ) Δ x ρ + f y ( P 0 ) Δ y ρ + o ( ρ ) ρ = f x ( P 0 ) cos ⟨ l , x ⟩ + f y ( P 0 ) sin ⟨ l , x ⟩ + o ( ρ ) ρ
\begin{aligned}
\frac{f(P)-f(P_0)}{\rho}&=
f_x(P_0)\frac{\Delta x}{\rho}+f_y(P_0)\frac{\Delta y}{\rho}+\frac{o(\rho)}{\rho}\\&=
f_x(P_0)\cos\langle l,x \rangle+f_y(P_0)\sin\langle l,x \rangle+\frac{o(\rho)}{\rho}
\end{aligned}
ρ f ( P ) − f ( P 0 ) = f x ( P 0 ) ρ Δ x + f y ( P 0 ) ρ Δ y + ρ o ( ρ ) = f x ( P 0 ) cos ⟨ l , x ⟩ + f y ( P 0 ) sin ⟨ l , x ⟩ + ρ o ( ρ )
兩邊取極限 ρ → 0 + \rho\to 0^+ ρ → 0 + f l ( P 0 ) = lim ρ → 0 + f ( P ) − f ( P 0 ) ρ = f x ( P 0 ) cos ⟨ l , x ⟩ + f y ( P 0 ) sin ⟨ l , x ⟩
f_l(P_0)=\lim_{\rho\to0^+}\frac{f(P)-f(P_0)}{\rho}=f_x(P_0)\cos\langle l,x \rangle+f_y(P_0)\sin\langle l,x \rangle
f l ( P 0 ) = ρ → 0 + lim ρ f ( P ) − f ( P 0 ) = f x ( P 0 ) cos ⟨ l , x ⟩ + f y ( P 0 ) sin ⟨ l , x ⟩ 證法 2： 根據座標旋轉變換
將原座標系 x O y \small xOy x O y ϕ \phi ϕ x ′ O y ′ \small x'Oy' x ′ O y ′ P \small P P O P → \small \overrightarrow{OP} O P r r r x , x ′ x,x' x , x ′ θ , θ ′ \small \theta,\theta' θ , θ ′ θ ′ = θ − ϕ
\theta'=\theta-\phi
θ ′ = θ − ϕ
則點 P \small P P { x = r cos θ y = r sin θ
\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}
{ x = r cos θ y = r sin θ
在新座標系下的座標爲{ x ′ = r cos θ ′ = r cos ( θ − ϕ ) = r ( cos θ cos ϕ + sin θ sin ϕ ) = x cos ϕ + y sin ϕ y ′ = r sin θ ′ = r sin ( θ − ϕ ) = r ( sin θ cos ϕ − cos θ sin ϕ ) = − x sin ϕ + y cos ϕ
\begin{cases}
\begin{aligned}
x'=r\cos\theta'=&\,\,r\cos(\theta-\phi)=r(\cos\theta\cos\phi+\sin\theta\sin\phi)=x\cos\phi+y\sin\phi\\
y'=r\sin\theta'=&\,\,r\sin(\theta-\phi)=r(\sin\theta\cos\phi-\cos\theta\sin\phi)=-x\sin\phi+y\cos\phi
\end{aligned}
\end{cases}
{ x ′ = r cos θ ′ = y ′ = r sin θ ′ = r cos ( θ − ϕ ) = r ( cos θ cos ϕ + sin θ sin ϕ ) = x cos ϕ + y sin ϕ r sin ( θ − ϕ ) = r ( sin θ cos ϕ − cos θ sin ϕ ) = − x sin ϕ + y cos ϕ
化爲矩陣形式( x ′ y ′ ) = ( cos ϕ sin ϕ − sin ϕ cos ϕ ) ( x y )
\begin{pmatrix}x'\\y'\end{pmatrix}=
\begin{pmatrix}
\cos\phi & \sin\phi\\
-\sin\phi & \cos\phi
\end{pmatrix}
\begin{pmatrix}x\\y\end{pmatrix}
( x ′ y ′ ) = ( cos ϕ − sin ϕ sin ϕ cos ϕ ) ( x y )
同理可推出(或者兩邊左乘旋轉矩陣的逆)，可得( x y ) = ( cos ϕ − sin ϕ sin ϕ cos ϕ ) ( x ′ y ′ )
\begin{pmatrix}x\\y\end{pmatrix}=
\begin{pmatrix}
\cos\phi & -\sin\phi\\
\sin\phi & \cos\phi
\end{pmatrix}
\begin{pmatrix}x'\\y'\end{pmatrix}
( x y ) = ( cos ϕ sin ϕ − sin ϕ cos ϕ ) ( x ′ y ′ )
所以∂ x ∂ x ′ = cos ϕ , ∂ y ∂ x ′ = sin ϕ
\frac{\partial x}{\partial x'}=\cos\phi,\,\,\frac{\partial y}{\partial x'}=\sin\phi
∂ x ′ ∂ x = cos ϕ , ∂ x ′ ∂ y = sin ϕ f ( x , y ) \small f(x,y) f ( x , y ) x ′ x' x ′ ∂ f ∂ x ′ = ∂ f ∂ x ∂ x ∂ x ′ + ∂ f ∂ y ∂ y ∂ x ′ = ∂ f ∂ x cos ϕ + ∂ f ∂ y sin ϕ
\begin{aligned}
\frac{\partial f}{\partial x'}&=
\frac{\partial f}{\partial x}\frac{\partial x}{\partial x'}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x'}\\
&=\frac{\partial f}{\partial x}\cos\phi+\frac{\partial f}{\partial y}\sin\phi
\end{aligned}
∂ x ′ ∂ f = ∂ x ∂ f ∂ x ′ ∂ x + ∂ y ∂ f ∂ x ′ ∂ y = ∂ x ∂ f cos ϕ + ∂ y ∂ f sin ϕ f \small f f y ′ y' y ′ ∂ f ∂ y ′ = − ∂ f ∂ x sin ϕ + ∂ f ∂ y cos ϕ
\frac{\partial f}{\partial y'}=-\frac{\partial f}{\partial x}\sin\phi+\frac{\partial f}{\partial y}\cos\phi
∂ y ′ ∂ f = − ∂ x ∂ f sin ϕ + ∂ y ∂ f cos ϕ
對 ∂ f / ∂ y ′ \partial f/\partial y' ∂ f / ∂ y ′ ∂ f ∂ y ′ = ∂ f ∂ x cos ( ϕ + π / 2 ) + ∂ f ∂ y sin ( ϕ + π / 2 )
\frac{\partial f}{\partial y'}=\frac{\partial f}{\partial x}\cos(\phi+\pi/2)+\frac{\partial f}{\partial y}\sin(\phi+\pi/2)
∂ y ′ ∂ f = ∂ x ∂ f cos ( ϕ + π / 2 ) + ∂ y ∂ f sin ( ϕ + π / 2 ) ( ϕ + π / 2 ) (\phi+\pi/2) ( ϕ + π / 2 ) y ′ y' y ′ x x x
有人會有這樣的疑問：假如 x ′ , y ′ x',y' x ′ , y ′
答案是肯定的，因爲d ( x ′ + a ) d x ′ = 1 = d ( y ′ + b ) y ′
\frac{d(x'+a)}{dx'}=1=\frac{d(y'+b)}{y'}
d x ′ d ( x ′ + a ) = 1 = y ′ d ( y ′ + b ) a , b a,b a , b
拉普拉斯算子∇ f = Δ 2 f = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2
\nabla f= \Delta^2 f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}
∇ f = Δ 2 f = ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f 拉普拉斯算子具有旋轉不變性，emm，這該怎麼理解呢？
先說下我的看法：
將原座標系 x O y \small xOy x O y x ′ O y ′ \small x'Oy' x ′ O y ′ ∂ 2 f ∂ x ′ 2 + ∂ 2 f ∂ y ′ 2 = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2
\frac{\partial^2f}{\partial x'^2}+\frac{\partial^2f}{\partial y'^2} =
\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}
∂ x ′ 2 ∂ 2 f + ∂ y ′ 2 ∂ 2 f = ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f 旋轉不變性 .
其實一般算子的旋轉不變性也是這樣定義的，比如我們可以自己造一個算子，叫作模長算子，容易驗證，它也具有旋轉不變性.
就算寫到這裏，我對拉普拉斯算子的 旋轉不變性 仍表示懷疑，因爲還沒有自己把它證出來，下面就讓我們試着證一下！
已知∂ g ∂ x ′ = ∂ g ∂ x cos ϕ + ∂ g ∂ y sin ϕ (1)
\frac{\partial g}{\partial x'}=
\frac{\partial g}{\partial x}\cos\phi+\frac{\partial g} {\partial y}\sin\phi \tag{1}
∂ x ′ ∂ g = ∂ x ∂ g cos ϕ + ∂ y ∂ g sin ϕ ( 1 ) ∂ g ∂ y ′ = − ∂ g ∂ x sin ϕ + ∂ g ∂ y cos ϕ (2)
\frac{\partial g}{\partial y'}=
-\frac{\partial g}{\partial x}\sin\phi+\frac{\partial g}{\partial y}\cos\phi \tag{2}
∂ y ′ ∂ g = − ∂ x ∂ g sin ϕ + ∂ y ∂ g cos ϕ ( 2 )
將 g = ∂ f ∂ x ′ \displaystyle g=\frac{\partial f}{\partial x'} g = ∂ x ′ ∂ f ( 1 ) \small (1) ( 1 )
∂ 2 f ∂ x ′ 2 = ∂ ∂ x ′ ( ∂ f ∂ x ′ ) = ∂ ∂ x ( ∂ f ∂ x ′ ) cos ϕ + ∂ ∂ y ( ∂ f ∂ x ′ ) sin ϕ
\frac{\partial^2 f}{\partial x'^2}=
\frac{\partial}{\partial x'}(\frac{\partial f}{\partial x'})=
\frac{\partial}{\partial x}(\frac{\partial f}{\partial x'})\cos\phi+\frac{\partial}{\partial y}(\frac{\partial f}{\partial x'})\sin\phi
∂ x ′ 2 ∂ 2 f = ∂ x ′ ∂ ( ∂ x ′ ∂ f ) = ∂ x ∂ ( ∂ x ′ ∂ f ) cos ϕ + ∂ y ∂ ( ∂ x ′ ∂ f ) sin ϕ
將 ( 1 ) \small (1) ( 1 ) f f f g g g ∂ 2 f ∂ x ′ 2 = ∂ ∂ x ( ∂ f ∂ x cos ϕ + ∂ f ∂ y sin ϕ ) cos ϕ + ∂ ∂ y ( ∂ f ∂ x cos ϕ + ∂ f ∂ y sin ϕ ) sin ϕ = ∂ 2 f ∂ x 2 cos 2 ϕ + 2 ∂ 2 f ∂ x ∂ y sin ϕ cos ϕ + ∂ 2 f ∂ y 2 sin 2 ϕ
\begin{aligned}
\frac{\partial^2 f}{\partial x'^2}&=
\frac{\partial}{\partial x}\big(\frac{\partial f}{\partial x}\cos\phi+\frac{\partial f}{\partial y}\sin\phi\big)\cos\phi+
\frac{\partial}{\partial y}\big(\frac{\partial f}{\partial x}\cos\phi+\frac{\partial f}{\partial y}\sin\phi\big)\sin\phi\\&=
\frac{\partial^2f}{\partial x^2}\cos^2\phi+2\frac{\partial^2f}{\partial x\partial y}\sin\phi\cos\phi+\frac{\partial^2f}{\partial y^2}\sin^2\phi
\end{aligned}
∂ x ′ 2 ∂ 2 f = ∂ x ∂ ( ∂ x ∂ f cos ϕ + ∂ y ∂ f sin ϕ ) cos ϕ + ∂ y ∂ ( ∂ x ∂ f cos ϕ + ∂ y ∂ f sin ϕ ) sin ϕ = ∂ x 2 ∂ 2 f cos 2 ϕ + 2 ∂ x ∂ y ∂ 2 f sin ϕ cos ϕ + ∂ y 2 ∂ 2 f sin 2 ϕ ∂ 2 f ∂ y ′ 2 = ∂ ∂ y ′ ( ∂ f ∂ y ′ ) = − ∂ ∂ x ( ∂ f ∂ y ′ ) sin ϕ + ∂ ∂ y ( ∂ f ∂ y ′ ) cos ϕ = − ∂ ∂ x ( − ∂ f ∂ x sin ϕ + ∂ f ∂ y cos ϕ ) sin ϕ + ∂ ∂ y ( − ∂ f ∂ x sin ϕ + ∂ f ∂ y cos ϕ ) cos ϕ = ∂ 2 f ∂ x 2 sin 2 ϕ − 2 ∂ 2 f ∂ x ∂ y sin ϕ cos ϕ + ∂ 2 f ∂ y 2 cos 2 ϕ
\begin{aligned}
\frac{\partial^2 f}{\partial y'^2}&=
\frac{\partial}{\partial y'}(\frac{\partial f}{\partial y'})\\&=
-\frac{\partial}{\partial x}(\frac{\partial f}{\partial y'})\sin\phi+\frac{\partial}{\partial y}(\frac{\partial f}{\partial y'})\cos\phi\\&=
-\frac{\partial}{\partial x}\big(-\frac{\partial f}{\partial x}\sin\phi+\frac{\partial f}{\partial y}\cos\phi\big)\sin\phi+\frac{\partial}{\partial y}\big(-\frac{\partial f}{\partial x}\sin\phi+\frac{\partial f}{\partial y}\cos\phi\big)\cos\phi\\&=
\frac{\partial^2f}{\partial x^2}\sin^2\phi-2\frac{\partial^2f}{\partial x\partial y}\sin\phi\cos\phi+\frac{\partial^2f}{\partial y^2}\cos^2\phi
\end{aligned}
∂ y ′ 2 ∂ 2 f = ∂ y ′ ∂ ( ∂ y ′ ∂ f ) = − ∂ x ∂ ( ∂ y ′ ∂ f ) sin ϕ + ∂ y ∂ ( ∂ y ′ ∂ f ) cos ϕ = − ∂ x ∂ ( − ∂ x ∂ f sin ϕ + ∂ y ∂ f cos ϕ ) sin ϕ + ∂ y ∂ ( − ∂ x ∂ f sin ϕ + ∂ y ∂ f cos ϕ ) cos ϕ = ∂ x 2 ∂ 2 f sin 2 ϕ − 2 ∂ x ∂ y ∂ 2 f sin ϕ cos ϕ + ∂ y 2 ∂ 2 f cos 2 ϕ
所以∂ 2 f ∂ x ′ 2 + ∂ 2 f ∂ y ′ 2 = ∂ 2 f ∂ x 2 ( cos 2 ϕ + sin 2 ϕ ) + ∂ 2 f ∂ y 2 ( sin 2 ϕ + cos 2 ϕ ) = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2
\begin{aligned}
\frac{\partial^2f}{\partial x'^2}+\frac{\partial^2f}{\partial y'^2} &=
\frac{\partial^2f}{\partial x^2}(\cos^2\phi+\sin^2\phi)+\frac{\partial^2f}{\partial y^2}(\sin^2\phi+\cos^2\phi) \\&=
\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}
\end{aligned}
∂ x ′ 2 ∂ 2 f + ∂ y ′ 2 ∂ 2 f = ∂ x 2 ∂ 2 f ( cos 2 ϕ + sin 2 ϕ ) + ∂ y 2 ∂ 2 f ( sin 2 ϕ + cos 2 ϕ ) = ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f
哈哈，這下我信了！
參考文獻 ：華東師範大學數學系. 數學分析(下冊)[M]. 第四版. 北京：高等教育出版社，2010.
