# Help With Difficult Proof

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?

-
You could begin by labelling with (1), (2), and (3) the equations you refer to by those labels. Then you could tell us what all those $\delta$-symbols are - are they supposed to be $\partial$, partial derivatives? And what's $\Delta$? And is $i$ the same as $j$ the same as $\sqrt{-1}$? And if all of this comes from some book, you could give the name and page number, or a link to a scan, or something. –  Gerry Myerson Sep 30 '12 at 5:03
Another question - where do the $u$ and $v$ in your target come from? They are nowhere to be found in the other equations. –  Gerry Myerson Sep 30 '12 at 5:05
All valid criticisms. I have addressed most of them now, I think. Thanks. –  John Roberts Sep 30 '12 at 14:02
Good. What's $r$? (As in $A(r)$, when $A$ has been defined as a function of 3 variables, not 1). My advice would be to give it to a good computer algebra package. –  Gerry Myerson Sep 30 '12 at 23:01
I have fixed this up as well, and added a clarification to the Laplace operator definition. Which computer algebra package would you recommend? –  John Roberts Oct 1 '12 at 13:25

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 $$Y(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+X(\partial_{yy}Y-\frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0.$$
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.
Hey man, thanks for the answer. Just a quick question - Should the line: "$Y(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+X(\partial_{yy}Y- \frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0$" actually be "$\frac{1}{X}(\partial_{xx}X-\frac{2ikx}{q}\partial_xX-2ik\partial_zX)+\frac{1}{‌​Y}(\partial_{yy}Y-\frac{2iky}{q}\partial_yY-2ik\partial_zY)+2kXY\partial_zZ = 0$", or am I mistaken? –  John Roberts Oct 2 '12 at 18:43
No, the terms $1/X(...)$, $1/Y(...)$ you get after you divide the equation by $XY$, which is in the next step. –  Ondřej Černotík Oct 2 '12 at 18:58
Ah, I see now. But how do we then end up getting rid of the $\frac{W^{2}}{2}$ term that appears outside the $1/X$ and $1/Y$ functions? –  John Roberts Oct 2 '12 at 19:07
Those cancel with the terms you get when you express the derivative with respect to $x$ as derivative with respect to $u$. –  Ondřej Černotík Oct 2 '12 at 19:34