Help With Difficult Proof

Question

Suppose we have the following equation 1: $$\tag{1} A_G(x,y,z) = \frac{A_1}{q(z)} e^{-ik \frac{x^2 + y^2}{2q(z)}} $$ where $$ q(z) = z+iz_0 $$ and $i$ is equal to $\sqrt{-1}$.

Suppose we have another equation 2 (where $X(.)$, $Y(.)$, and $Z(z)$ are real functions): $$\tag{2} A(x,y,z) = X( \sqrt{2} \frac{x}{W(z)})Y(\sqrt{2} \frac{y}{W(z)})e^{iZ(z)}A_G(x,y,z) $$

Lastly, we have equation 3: $$ \Delta _T A(x,y,z) - 2ik \frac{ \partial A}{ \partial z} =0 \tag{3}$$

where $\Delta _T=\partial_{xx}+\partial_{yy}$ is the transverse Laplacian operator.

I need to show that substituting equation (2) into equation (3), given that equation (1) is also a solution of equation (3), will produce the following equation: $$ \frac{1}{X} ( \frac{\partial ^{2}X}{\partial u^{2}} - 2u \frac{\partial X}{\partial u}) + \frac{1}{Y} ( \frac{\partial ^{2}Y}{\partial v^{2}} - 2v \frac{\partial Y}{\partial v}) + kW^{2}(z) \frac{\partial Z}{\partial z}=0 $$ where $$ u=\frac{\sqrt2x}{W(z)} $$ and $$ v=\frac{\sqrt2y}{W(z)} $$ Can somebody give me some pointers as to how to begin here?

Answer 1

This is not as difficult as it might look at first. The calculation is long, but when you don't forget anything, it works fine.

First, you need to find all the derivatives in Eq. (3). It's important to remember, that also $X$ and $Y$ are functions of $z$, so that the $z$ derivative gives $i XYe^{iZ}A_G\partial_z Z + XYe^{iZ}\partial_z A_G + Ye^{iZ}A_G\partial_z X + Xe^{iZ}A_G \partial_z Y$. Second derivatives with respect to $x$ and $y$ each give three terms, proportional to $\partial_{xx}X$, $\partial_x X\partial_x A_G$, $\partial_{xx}A_G$ (similar for $y$). Plugging all these into (3) and using that $A_G$ fulfilles (3) as well, you can get rid of terms with $\partial_{xx}A_G$, $\partial_{yy}A_G$ and $\partial_zA_G$. Next, using $\partial_xA_G = -ikxA_G/q$, you can divide the equation by $A_G$ and $e^{iZ}$, so that you should arrive at \begin{equation} Y(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+X(\partial_{yy}Y-\frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0. \end{equation}

Multiplying this by $W^2/(2XY)$ you directly get the $Z$ term. To obtain the terms with $X$ and $Y$, you need to explicitly make the derivatives, so that \begin{eqnarray} \partial_z X(\sqrt{2}x/W(z)) &=& -X'\frac{\sqrt{2}x}{W^2}\partial_zW \\ \partial_x X(\sqrt{2}x/W(z)) &=& \frac{\sqrt{2}}{W} X' \\ \partial_{xx} X(\sqrt{2}x/W(z)) &=& \frac{2}{W^2} X'', \end{eqnarray} denoting $X'$, $X''$ first and second derivatives with respect to $u = \sqrt{2}x/W$. If you now express the derivative $\partial_z W$ and use the relations that bind the parameters of the Gaussian beam, you can arrive at the final equation.