Derivation of the Fourier Transform of a PDE

Question

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 $(2)$?

Answer 1

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.