Days ago and I'm trying to understand this equation
please help
If the linear differential operator $$C= C_1 \frac{\partial}{\partial x}+C_2 \frac{\partial}{\partial y}+C_0$$
and $\phi(x,y)$ is any solution of $C[\phi(x,y)]=0$
then $$\phi^{-1}C\phi = C_1 \frac{\partial}{\partial x}+C_2 \frac{\partial}{\partial y}$$
If we let $ E=\phi^{-1}C\phi $, then $C=\phi E \phi^{-1} $
By substituting for the operator $C$ and simplifying , we get
$$e^{c C}f(x,y)=e^{c\phi E \phi^{-1}}f(x,y)= \phi (x,y) e^{cE}[\phi^{-1}(x,y)f(x,y)]$$
Is this statement is true, and why?
$e^{c\phi E \phi^{-1}}f(x,y)= \phi (x,y) e^{cE}[\phi^{-1}(x,y)f(x,y)]$
Thanks for participating
Update:
It seems that $\;\phi^{-1} = 1/\phi\;$ and that the product operator (i.e function) happens to be at the same time a solution. Then everything on that page seems indeed to be correct and: $$ e^{c\phi E \phi^{-1}}f(x,y)= \left[ \sum_{n=0}^\infty \left(\phi\, cE \,\phi^{-1}\right)^n / n! \right] f(x,y) $$ Where: $$ \left(\phi\,cE\,\phi^{-1}\right)^n = \left(\phi\,cE\,\phi^{-1}\right)\left(\phi \,cE\,\phi^{-1}\right)\left(\phi\,cE\,\phi^{-1}\right)\cdots =\\ \phi\,cE\cdot 1/\phi\ \cdot\phi\,cE\cdot 1/\phi \cdot \phi\,cE\cdot 1/\phi\cdots = \phi\left(cE\right)^n\phi^{-1} $$ Because $\,1/\phi\cdot \phi = 1$ . Hence the formula follows: $$ e^{c\phi E \phi^{-1}}f(x,y)= \left[ \sum_{n=0}^\infty \phi\left(cE\right)^n \,\phi^{-1}/n! \right]f(x,y) = \phi \left[ \sum_{n=0}^\infty \left(cE\right)^n /n! \right] \phi^{-1} f(x,y) \\ = \phi (x,y) e^{cE}[\phi^{-1}(x,y)f(x,y)] $$