Derivation of the Fourier Transform of a PDE Given the following PDE:
$$
-\nabla \cdot (\sigma(x,z)\nabla \phi (x,y,z))=\frac{\partial\rho}{\partial t}\delta(x_s)\delta(y_s)\delta(z_s) \tag 1
$$
A paper I am reading Fourier transforms this equation in the $y$ direction and obtains:
$$
-\nabla \cdot (\sigma(x,z)\nabla \hat{\phi} (x,K_y,z))+K_y^2\sigma(x,z)\hat{\phi}(x,K_y,z)=\frac{\partial\rho}{\partial t}\delta(x_s)\delta(z_s) \tag 2
$$
I have done Fourier transform with functions, but never with equations.
Could you explain to me how to go from $(1)$ to $(2)$?
 A: To make things easier to see, writing out $(1)$ explicitly yields
\begin{align}
- \nabla \cdot [\sigma(x,z) \nabla \phi(x,y,z)] &= - [(\sigma(x,z) \phi_x(x,y,z))_x + (\sigma(x,z) \phi_y(x,y,z))_y + (\sigma(x,z) \phi_z(x,y,z))_z] \\
&= \delta(x_s) \delta(y_s) \delta(z_s) p_t \\
\end{align}
Denoting the Fourier transform with respect to $y$ by $\mathcal{F_y}$ and using linearity of the integral, we get
\begin{align}
- \mathcal{F_y} \{ [\sigma(x,z) \phi_x(x,y,z)]_x \} - \mathcal{F_y} \{ [\sigma(x,z) \phi_y(x,y,z)]_y \} - \mathcal{F_y} \{ [\sigma(x,z) \phi_z(x,y,z)]_z \}&= \mathcal{F_y} \{ \delta(x_s) \delta(y_s) \delta(z_s) p_{t} \} \quad (*) \\
\end{align}
Now, $\mathcal F_y : y \to K_y$ and hence, for functions with no derivatives, we have (where $(\cdot)'$ denotes the Fourier transform of $(\cdot)$)
\begin{align}
- \mathcal{F_y} \{ [\sigma(x,z) \phi_x(x,y,z)]_x \} &= - \sigma(x,z) \mathcal{F_y} \{ \phi_{xx}(x,y,z) \} - \sigma_x(x,z) \mathcal{F_y} \{ \phi_{x}(x,y,z) \} \\
&= - \sigma(x,z) \phi'_{xx}(x,K_y,z) - \sigma_x(x,z) \phi'_{x}(x,K_y,z) \\
&= - (\sigma(x,z) \phi'_x(x,K_y,z))_x \\
- \mathcal{F_y} \{ [\sigma(x,z) \phi_z(x,y,z)]_z \} &= - \sigma(x,z) \mathcal{F_y} \{ \phi_{zz}(x,y,z) \} - \sigma_z(x,z) \mathcal{F_y} \{ \phi_{z}(x,y,z) \} \\
&= - \sigma(x,z) \phi'_{zz}(x,K_y,z) - \sigma_z(x,z) \phi'_{z}(x,K_y,z) \\
&= - (\sigma(x,z) \phi'_z(x,K_y,z))_z
\end{align}
We can write the sum of these two terms above as
$$- \nabla \cdot (\sigma(x,z) \nabla \phi'(x, K_y, z))$$
so $(*)$ becomes
\begin{align}
- \nabla \cdot (\sigma(x,z) \nabla \phi'(x, K_y, z)) - \mathcal{F_y} \{ [\sigma(x,z) \phi_y(x,y,z)]_y \} &= \mathcal{F_y} \{ \delta(x_s) \delta(y_s) \delta(z_s) p_{t} \} \quad (**) \\
\end{align}
As $\sigma$ is independent of $y$, we have that $[\sigma(x,z) \phi_y(x,y,z)]_y = \sigma(x,z) \phi_{yy}(x,y,z)$, so
\begin{align}
- \mathcal{F_y} \{ [\sigma(x,z) \phi_y(x,y,z)]_y \} &= - \sigma(x,z) \mathcal{F_y} \{ \phi_{yy}(x,y,z) \} \\
&= - \sigma(x,z) \cdot [- K_y^2 \phi'(x,K_y,z)] \\
&= K_y^2 \sigma(x,z) \phi'(x,K_y,z)
\end{align}
Replacing in $(**)$ yields
\begin{align}
- \nabla \cdot (\sigma(x,z) \nabla \phi'(x, K_y, z)) + K_y^2 \sigma(x,z)\phi'(x,K_y,z) &= \mathcal{F_y} \{ \delta(x_s) \delta(y_s) \delta(z_s) p_{t} \} \\
\end{align}
Finally, using that $\mathcal{F_y} \{ \delta(y_s) \} = 1$, we have
\begin{align}
- \nabla \cdot (\sigma(x,z) \nabla \phi'(x, K_y, z)) + K_y^2 \sigma(x,z)\phi'(x,K_y,z) &= \mathcal{F_y} \{ \delta(x_s) \delta(y_s) \delta(z_s) p_{t} \} \\
&= \delta(x_s) \mathcal{F_y} \{ \delta(y_s) \} \delta(z_s) p_{t} \\
&= \delta(x_s) \cdot 1 \cdot \delta(z_s) p_{t} \\
&= \delta(x_s) \delta(z_s) p_{t}
\end{align}
as was to be shown.
