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I know that

$$\begin{align*} \exp\left(\alpha\frac{d}{dx}\right)f(x)=f(x+\alpha)\,, \end{align*} \tag{1}$$

but I am looking for a definition for

$$\begin{align*} \exp\left(\alpha\frac{\partial}{\partial x}\frac{\partial}{\partial y}\right)f(x,y)\,, \end{align*} \tag{2}$$

so, the first question is: Is there a definition for the previous expression?.

Now, in the particular case where $f(x,y)$ is a product of two Gaussian functions centered in $y_{0}$ and $x_{0}$, that is $f(x,y)=\exp\left[-k_{2}(y-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right]$ (with $k_{1}$ and $k_{2}$ being real or complex constants), so, the second question is: Is it valid to apply equation (1) in order to displace first a Gaussian function?, I mean

$$\begin{align*} \exp\left(\alpha\frac{\partial}{\partial x}\frac{\partial}{\partial y}\right)\exp\left[-k_{2}(y-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right]\,, \end{align*} \tag{3}$$

taking $\hat{c} = \alpha\frac{\partial}{\partial x}$, the above expresión is $$\begin{align*} \exp\left(\hat{c}\frac{\partial}{\partial y}\right)\exp\left[-k_{2}(y-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right] \,, \end{align*} \tag{4}$$ Using equation (1) the $y$ variable is displaced $$\begin{align*} \exp\left[-k_{2}(\left[y+\hat{c}\right]-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right] \,, \end{align*} \tag{5}$$ or $$\begin{align*} =\exp\left[-k_{2}\left(\left[y+\alpha\frac{\partial}{\partial x}\right]-y_{0}\right)^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right] \ \end{align*} \tag{6}$$ then, the first exponential of above equation will become an operator that will act on the second exponential. I think that this procedure is wrong because if the exponentials of equation (3) are swapping, I would get

$$\begin{align*} \exp\left[-k_{1}(x-x_{0})^{2}\right] \exp\left[-k_{2}\left(\left[y+\alpha\frac{\partial}{\partial x}\right]-y_{0}\right)^{2}\right]\ \end{align*} \tag{7}$$

which clearly is not the same as the equation (6).

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    $\begingroup$ Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $\partial_{xy}$. Then your operator is just given by convolution with that fundamental solution. $\endgroup$
    – shalop
    Commented Dec 25, 2018 at 23:27
  • $\begingroup$ You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible. $\endgroup$
    – shalop
    Commented Dec 26, 2018 at 6:08

1 Answer 1

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Expressions like $\exp(\alpha T),$ where $T$ is some operator and $\alpha$ is a constant, are defined by the Maclaurin expansion: $$ \exp(\alpha T) = \sum_{n=0}^{\infty} \frac{1}{n!} \alpha^n T^n $$

Therefore, $$ \exp\left(\alpha\frac{\partial}{\partial x}\frac{\partial}{\partial y}\right) f(x,y) = \sum_{n=0}^{\infty} \frac{1}{n!} \alpha^n \left( \frac{\partial}{\partial x} \frac{\partial}{\partial y} \right)^n f(x,y) = \sum_{n=0}^{\infty} \frac{1}{n!} \alpha^n \left( \frac{\partial}{\partial x} \right)^n \left( \frac{\partial}{\partial y} \right)^n f(x,y), $$ where the last equality is valid since partial derivatives commute if $f$ is smooth.

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