Integral Equations - Green's Function

1. Mar 14, 2015

Gyges

1. The problem statement, all variables and given/known data

I'm trying to show that,

$\phi(x')=b\frac{\sin kx'}{k}+a\cos kx'+\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx$

is the solution of,

$\frac{d^{2}}{dx'^{2}}\phi(x')+k^{2}\phi(x')=f(x')dx$ where $0 \leq x'<\infty$.
2. Relevant equations

N/A

3. The attempt at a solution

If we consider one dimensional Schrodinger equation with potential U(x)

$(\frac{d^{2}}{dx^{2}} + k^{2})\phi(x)=U(x)\phi(x)$

Supposing that $\phi(0)=a~~ and~~ \phi'(0)=b$ are given

Solution for $x>0$

$L\phi(x)=f(x)~~ with~~ L=\frac{d^{2}}{dx^{2}}+k^{2}$

$\int_{0}^{\infty} g(x,x')L\phi(x)=\int_{0}^{\infty} g(x,x')f(x)d(x)$

$\int_{0}^{\infty} (Lg(x,x'))\phi(x)dx+g(x,x')\phi'(x)|_{x=0}^{x=\infty}-\frac{dg(x,x')}{d(x)}\phi(x)|_{x=0}^{x=\infty}=\int_{0}^{\infty}g(x,x')f(x)dx$

$g(\infty,x')=0 ~~and~~ \frac{dg}{dx}(\infty,x')=0$

$Lg(x,x')=\delta(x-x')$

$\phi(x')=bg(0,x')-a\frac{dg}{dx}(0,x')+\int_{0}^{\infty} g(x,x')f(x)dx$

Boundry conditions are given by

$(\frac{d^{2}}{dx^{2}} + k^{2})g(x,x')=\delta(x-x') ~~on~~ x\in(0,\infty)~~ with~~ x'(\in0,\infty)$

Boundry Condition 1

$g(\infty,x')=0$

Boundry Condition 2

$\frac{dg}{dx}(\infty,x')=0$

For $x<x’$

$g(x,x')=A\sin kx + B\cos kx$

For $x>x’$

$g(x,x')=C\sin kx + D\cos kx$

If we apply boundry conditions then

C=D=0

so

$g(x,x')=0~~for x>x'$

$\frac{dg}{dx}(x'+\xi,x')-\frac{dg}{dx}(x'-\xi,x')=1$

$g(x'+\xi,x')=g(x'-\xi,x')$

$\xi\rightarrow0$

$\begin{cases}A\sin kx'+B\cos kx'=0\\-A\cos kx'+B\sin kx'=\frac{1}{k}\end{cases}$

so

$A=\frac{-\cos kx'}{k}$

$B=\frac{\sin kx'}{k}$

Then Green Function found

$g(x,x')=\begin{cases}\frac{\sin k(x-x')}{k} ~ , ~x<x' \\0 ~,~ x>x'\end{cases}$

Then

$\phi(x')=bg(0,x')-a\frac{dg}{dx}(0,x')+\int_{0}^{\infty} g(x,x')f(x)dx$

becomes

$\phi(x')=b\frac{\sin kx'}{k}+a\cos kx'+\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx$

Now finally I should show that this is the solution of

$\frac{d^{2}}{dx'^{2}}\phi(x')+k^{2}\phi(x')=f(x')dx ~~ where ~~ 0 \leq x'<\infty$

2. Mar 14, 2015

Ray Vickson

You have "derived" the solution, starting from the DE. Congratulations on that, but that was not what the question asked! It asked you to just verify that the given formula solves the DE.

Anyway, I think the formula has an error in it. When I apply the operator $\frac{d^2}{d x'^2} + k^2$ to your $\phi(x')$ I get $-f(x')$ instead of $+f(x')$. That can be remedied by a small, rather obvious, change in the formula for $\phi$.

3. Mar 14, 2015

Gyges

Thank your for reply, this was a part of question, and I wanted to share the part I completed. So you could comment if it is wrong and if it is correct maybe it can be helpful for someone else.

I checked again and it seems that I didn't make any typing error, question maybe be wrong and I will ask again.

If we come back the part I asked, can you write the steps that shows how you applied operator $\frac{d^2}{d x'^2} + k^2$ . Did you use Leibniz rule, if so can you show the method please.

4. Mar 14, 2015

Ray Vickson

I just let the computer algebra package Maple do all the work. It knows all the standard methods and applies them as needed.

5. Mar 14, 2015

Gyges

I checked it now, which bundle or bundles you use? And if it shows the steps can you please post it?
Finally, as I have not seen bundle output yet, it will be highly appreciated, if someone can show me that,

$\phi(x')=b\frac{\sin kx'}{k}+a\cos kx'+\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx$

is the solution of,

$\frac{d^{2}}{dx'^{2}}\phi(x')+k^{2}\phi(x')=f(x')dx$ where $0 \leq x'<\infty$.

as this part is the last thing, I have to complete after other calculations, I showed above.

6. Mar 14, 2015

vela

Staff Emeritus
It's against the forum rules to do this. It's your job to do the problem. Take a shot at it and post your attempt.

7. Mar 15, 2015

Gyges

$\phi(x')=b\frac{\sin kx'}{k}+a\cos kx'+\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx$

I take the first and second derivative of RHS as follows

1st derivative
$\frac{d}{dx'}(b\frac{\sin kx'}{k}+a\cos kx'+\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx)=b\cos (kx')-ak\sin(kx')+\frac{d}{dx}(\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx)$

2nd derivative
$\frac{d^{2}}{dx'^{2}}(b\frac{\sin kx'}{k}+a\cos kx'+\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx)=-k^{2}a\cos(kx')-kb\sin(kx')\frac{d^{2}}{dx'^{2}}(\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx)$

But then how can I take the derivative of integral part $\frac{d}{dx'}(\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx)$ and $\frac{d^{2}}{dx'^{2}}(\int_{0}^{x'}\frac{\sin k(x-x')}{k} f(x)dx)$
Thanks

8. Mar 15, 2015

vela

Staff Emeritus
Like you said, use the Leibniz rule. I can tell you're reluctant to try it. I just can't figure out why.

9. Mar 15, 2015

Gyges

I am confused because derivative is $\frac{d}{dx'}$ but we have $f(x)dx$ and $(x-x')$ in integral, can you show me an example. Maybe not directly my problem but another example similar that can help me to understand. Thanks.

10. Mar 15, 2015

Ray Vickson

Do you know the actual statement of Leibniz' Rule? If you do, just sit down and apply it term-by-term, applying exactly what the formula tells you to do.

If you do not know the actual statement of Leibniz' Rule (but somehow know the name, anyway), you can find many discussions of it on-line, including step-by-step examples worked out in detail. Others have already done that for you, so there is no reason why we should repeat those efforts.