Identidad de expansión potencial newtoniano

Question

Preliminares

Considerar el potencial newtoniano <span class="math-container">$$\frac{1}{|\vec x - \vec y|}$$ with <span class="math-container">$ \vec{x}, \vec{y} \in \mathbb{R}^3$</span> and <span class="math-container">$|\vec{x}| = x > y = |\vec{y}|$</span>. Its Taylor expansion is given by <span class="math-container">$$\frac{1}{|\vec x - \vec y|} = \sum{n=0}^\infty \frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}\vec{x})^n \frac{1}{x},$$</span> which can also be written in terms of Legendre polynomials as <span class="math-container">$$\frac{1}{|\vec x - \vec y|} = \sum_{n = 0}^\infty Pn (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}}.$$</span> The key identity here is <span class="math-container">$$\frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}\vec{x})^n \frac{1}{x} = Pn (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}}$$</span> Using the key identity twice, we have <span class="math-container">$$\frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}\vec{x})^n \frac{1}{x} = P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}} = \left(\frac{y}{x}\right)^{2n+1} Pn (\hat{x} \cdot \hat{y}) \frac{x^n}{y^{n+1}} = \left(\frac{y}{x}\right)^{2n+1} \frac{(-1)^n}{n!} (\vec{x} \cdot\vec{\nabla}\vec{y})^n \frac{1}{y}.$$</span></span>

La pregunta

¿Hay una manera de probar la identidad <span class="math-container">$$\boxed{(\vec{y} \cdot\vec{\nabla}\vec{x})^n \frac{1}{x} = \left(\frac{y}{x}\right)^{2n+1} (\vec{x} \cdot\vec{\nabla}\vec{y})^n \frac{1}{y}}$ $</span> directamente, es decir, sin recurrir a polinomios de Legendre?

Answer 1

$\color{brown}{\textbf{Edition of 18.12.2018}}$

Permítanos presentar la cuestión de la identidad $$(\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = \left(\frac{y}{x}\right)^{2n+1} (\vec{x} \cdot\vec{\nabla}_\vec{y})^n \frac{1}{y}\tag1$$ en el formulario de $$f_n(x,y)=f_n(y,x),\tag2$$ donde $$\begin{align} f_n(x,y)=\left(x_1^2+x_2^2+x_3^2\right)^{(2n+1)/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)^{(n)} \left(x_1^2+x_2^2+x_3^2\right)^{(-1/2)} \end{align}.\tag3$$ Entonces $$\begin{align} &f_1(x,y) = \left(x_1^2+x_2^2+x_3^2\right)^{3/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(x_1^2+x_2^2+x_3^2\right)^{-1/2}\\[4pt] &= - \left(x_1^2+x_2^2+x_3^2\right)^{3/2}\left(y_1 x_1 + y_2 x_2+y_3 x_3\right) \left(x_1^2+x_2^2+x_3^2\right)^{-3/2}, \end{align}$$ $$f_1(x,y) = -\left(y_1x_1 + y_2x_2 + y_3 x_3\right),\tag4$$

$$\begin{align} &f_{m+1}(x,y)=\left(x_1^2+x_2^2+x_3^2\right)^{(2m+3)/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)^{m+1} \left(x_1^2+x_2^2+x_3^2\right)^{(-1/2)}\\[4pt] &=\left(x_1^2+x_2^2+x_3^2\right)^{(2m+3)/2}\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(f_m(x,y)\left(x_1^2+x_2^2+x_3^2\right)^{-(2m+1)/2}\right), \end{align}$$ $$\begin{align} &f_{m+1}(x,y)=\left(x_1^2+x_2^2+x_3^2\right)\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)f_m(x,y)\\[4pt] &\quad -(2m+1)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)f_m(x,y). \end{align}\tag5$$

El uso de $(4)-(5),$ uno puede conseguir $$\begin{align} &f_2(x,y)=-\left(x_1^2+x_2^2+x_3^2\right)\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad +3\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2, \end{align}$$ $$f_2(x,y) = - \left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2 + y_2^2 + y_3^2\right) + 3\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2,\tag6$$

$$\begin{align} &f_3(x,y)=-\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\\[4pt] &\qquad\times\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(x_1^2 + x_2^2 + x_3^2\right)\\[4pt] &\quad +3\left(x_1^2+x_2^2+x_3^2\right)\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2\\[4pt] &\quad +5\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)-15\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^3\\[4pt] &=-2\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad +6\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad +5\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_3^2\right)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)-15\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^3, \end{align}$$ $$f_3(x,y) = 9\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2 + y_2^2 + y_3^2\right) + 3\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^2.\tag7$$

A partir de las fórmulas de $(4),(6)-(7)$ debe $$f_1(x,y)=f_1(y,x)\in \mathbb F,\quad f_2(x,y)=f_2(y,x)\in \mathbb F,\quad f_3(x,y)=f_3(y,x)\in \mathbb F,\tag8$$ donde $$\mathbb F = \left\{\sum_{k=0}^Q C_k\left(x_1^2+x_2^2+x_3^2\right)^{A_k} \left(y_1^2 + y_2^2 + y_3^2\right)^{A_k} \left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k}\right\},$$

Deje que nosotros resultó $$f_m(x,y)\in \mathbb F,$$ entonces $$f_m(x,y)=f_m(y,x).\tag9$$

Tomando en cuenta $(3)-(4)$, uno puede conseguir $$\begin{align} &f_{m+1}(x,y)=\left(x_1^2+x_2^2+x_3^2\right) \sum_{k=0}^{Q(m)}C_k\left(y_1^2 + y_2^2 +y_3^2\right)^{A_k}\\[4pt] &\qquad\times\left(y_1\dfrac{\partial}{\partial x_1}+y_2\dfrac{\partial}{\partial x_2}+y_3\dfrac{\partial}{\partial x_3}\right)\Big(\left(x_1^2+x_2^2+x_3^2\right)^{A_k} \left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k}\Big)\\[4pt] &\quad -(2m+1)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)\\[4pt] &\quad\times \sum_{k=0}^{Q(m)}C_k\left(x_1^2+x_2^2+x_3^2\right)^{A_k}\left(y_1^2 + y_2^2 + y_3^2\right)^{A_k}\left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k}\\[4pt] &=\sum_{k=0}^{Q(m)}C_k \left(x_1^2+x_2^2+x_3^2\right)^{A_k}\left(y_1^2 + y_2^2 +y_3^2\right)^{A_k}\Big(2A_k\left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k+1}\\[4pt] &\quad + B_k\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2 + y_2^2 +y_3^2\right) \left(x_1y_1+x_2y_2+x_3y_3\right)^{B_k-1}\\[4pt] &\quad -(2m+1)\left(y_1x_1 + y_2x_2 + y_3 x_3\right)^{B_k+1}\Big)\in\mathbb F, \end{align}$$ $$f_{m+1}(x,y) = f_{m+1}(y,x).\tag{10}$$ Por lo tanto, la identidad de $(2)$ es válido para $n=m+1$ y, teniendo en cuenta $(8)-(10),$ por inducción para todos los $n>1.$

$\color{brown}{\textbf{Proved.}}$