Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.