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Could someone please tell me why this is true ?

Let $$g=g(x,z)$$ $$f(x)=\exp(ikx)\left(1+i{g \over k}-{g_z \over k^2}\right)\bigg|_{z=x}-{1\over k^2}\int_x^\infty g_{zz}\exp(ikz)\,\,dz$$where $g,g_z\to 0 $ as $z\to\infty$.

Also, let $$g_{xx}=g_{zz}+ug$$ when $z>x$ and where $u=-2(g_x+g_z)\big|_{z=x}$.

Then $$\left\{-\partial_x^2-2(g_x+g_z)\big|_{z=x}\right\}f=k^2f$$

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From the definition of $f$, it's easy to prove that $$f(x) = e^{ikx} + \int_0^\infty g(x,z) e^{ikz} dz.$$ I've been unable to derive the equality from here. Are you sure you've stated the problem correctly and completely? – Pragabhava Nov 9 '12 at 20:58
@Pragabhava , firstly, thanks for the edit! Secondly, you, and anyone else who attempted to help, have my sincere apologies. There is indeed a bit about $u$ from further up the page that I missed. Now edited in. Does it work now? – Terry Nov 9 '12 at 21:33
Thats more likely! Check my answer. – Pragabhava Nov 9 '12 at 22:18
up vote 0 down vote accepted

Given $$ f(x) = e^{ikx} + \int_x^\infty g(x,z) e^{ikz} dz $$ we have $$ \partial_x f(x) = e^{ikx}\big(ik - g(x,x)\big) + \int_x^\infty g_x(x,z) e^{ikz}dz $$ and $$ \partial_{xx} f(x) = -e^{ikx}\big(k^2 + ik g(x,x) + g_z(x,x) + 2g_x(x,x)\big) + \int_0^\infty g_{xx}(x,z) e^{ikx}. $$ Using $g_{xx} = g_{zz} + ug$, \begin{multline} \partial_{xx} f(x) = -e^{ikx}\big(k^2 + ik g(x,x) + g_z(x,x) + 2g_x(x,x)\big) \\ + \int_0^\infty g_{zz}(x,z) e^{ikx} + \int_0^\infty u g(x,z) e^{ikz} dz. \end{multline}

Substitute the integral involving $g_{zz}$ and use the fact that $u(x) = - 2 \big(g_z(x,x) + 2g_x(x,x)\big)$ and you'll be able to finish the proof.

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