# Exponential of two different derivatives

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).

• 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. Commented Dec 25, 2018 at 23:27
• 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. Commented Dec 26, 2018 at 6:08

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.