Chain rule in airy case

Question

I have the following differential operator \begin{equation} L_1:=\hbar\partial^{1}-\frac{1}{2} Ax_{2}x_{3}-\hbar -\frac{1}{2}C \partial^{2}\partial^{3}-\hbar D, \end{equation} where $\partial^{c}=\frac{\partial }{\partial x_c}$, A,B,C, D are constant. $L_1$ annihilates a function say $f(x_1, x_2 , x_3) $. I want to define a new function say $$F(x_1, x_2 , x_3 ,s)= f(\frac{x_1}{s}, \frac{x_2}{s},\frac{x_3}{s} )$$ So I claim that by chain rule there exist an operator in $\partial^c,\frac{\partial }{\partial s}$ that annihilates $F(x_1, x_2 , x_3 ,s)$. My question how to get the annihilator in this particular case ?

Answer 1

It looks simple, let $x_i = s y_i$. Then $f(y_1, y_2, y_3) = F(s y_1, s y_2, s y_3, s)$. We use \begin{equation} \hbar\partial^{1}f(y_1, y_2, y_3)-\frac{1}{2} Ay_{2}y_{3}f(y_1, y_2, y_3)-\hbar f(y_1, y_2, y_3) -\frac{1}{2}C \partial^{2}\partial^{3}f(y_1, y_2, y_3)-\hbar D f(y_1, y_2, y_3) = 0 \end{equation} Then use that $\partial^i f(y_1, y_2, y_3) = s \partial^i F(s y_1, s y_2, s y_3, s)$ and $\partial^i\partial^j f(y_1, y_2, y_3) = s^2\partial^i\partial^j F(s y_1, s y_2, s y_3, s)$. Finally, replace $s y_i$ by $x_i$ and we are left with \begin{equation} \hbar s \partial^{1}F-\frac{1}{2} A \frac{x_{2}x_{3}}{s^2}F-\hbar F -\frac{1}{2}C s^2 \partial^{2}\partial^{3}F-\hbar D F = 0 \end{equation}

Also note that being homogeneous of degree $0$, $F$ anihilates another operator: \begin{equation} x_1 \partial^1 + x_2 \partial^2 + x_3 \partial^3 + s \partial^4 \end{equation}