I'm seeing this equality that I'm having a a bit of trouble reproducing. If we have a couple of Functions defined as
$$G(\vec x,y) = (-\nabla^2)^{-1}F(\vec x,y)$$
where the inverse Laplacian has the Fourier transform:
$$(-\nabla^2)^{-1} \rightarrow (\vec q^2 + k^2),$$
where $\vec q$ is the wavevector in the $\vec x$ subspace, and $k$ is the wavevector in the $y$ direction. I have to show that Fourier transforming in the $x$ subspace gives:
$$G(q,y) = \frac{1}{2|q|}\int_{-\infty}^\infty dy' e^{-|q| |y-y'|}F(q,y')$$
I have tried to either start with $-\nabla^2G(x,y) = F(x,y)$, and plug in the definition of the fourier transform, or to start with $G(q,y) = \int e^{i k y}G(q,k)$ and use the full Fourier transform definition, but I can't get the integral over the auxiliary variable $y'$. Any hints?
Thanks!
To avoid confusing notation, we let $g(q,y)$ to be the Fourier transform of $G(x,y)$ in the $x$ variables (with $y$ fixed), and similarly with $f$.
Taking Fourier transforms of $-\nabla^2G =F$ gives $$ (|q|^2 - \partial_y^2) g(q,y)= f(q,y),\quad \text{or}\quad \partial_y^2 g(q,y)= |q|^2 g(q,y)- f(q,y), $$ which is an ODE in $y$ for each fixed $q$. Note that the function $u(q,t)=A(q)e^{-|q||y|}$ solves the homogeneous equation and has good decay as $y\to\pm\infty$ (there's a hiccup at $y=0$ that will come into play in a bit*); now to solve the inhomogeneous problem we use the variation of parameters/Duhamel's formula: \begin{equation}\tag{1} v(q,t)= \int_{-\infty}^{\infty} A(q)e^{-|q||y-y'|}f(y')\, dy'. \end{equation} It's easy to check, by breaking up the integral into the $\int_{-\infty}^y$ and $\int_y^\infty$ parts, \begin{equation} \begin{split} \partial_yv(q,y) &= \int_{-\infty}^{\infty} |q|A(q)e^{|q||y-y'|}f(q,y')\, dy',\\ \partial_y^2 v(q,y) &= -2|q| A(q)f(q,y) + \int_{-\infty}^\infty |q|^2A(q)e^{-|q||y-y'|}f(q,y')\, dy'. \end{split} \end{equation} Therefore, the choice of $A(q)=\frac{1}{2|q|}$ gives $$ \partial_y^2 v(q,y)= |q|^2 v(q,y) - f(q,y). $$ By uniqueness for the ODE, we see $g(q,y)=v(q,y)$, and combined with $(1)$, is exactly the identity we want.
*: Perhaps the right way to phrase this is that the function $u$ is the fundamental solution of the operator $-\partial_y^2+|q|^2$, so we're not really solving the homogeneous equation, but rather the equation with a Dirac delta as the forcing term; then the "Duhamel formula" is simply the fact that the convolution satisfies $f*\delta_y = f(y)$.