# Green's Function - modified operator

1. Feb 28, 2015

### RightFresh

Hi, I'm stuck with a question from one of my examples sheets from uni.

The question is as follows:

If G(x,x') is a greens function for the linear operator L, then what is the corresponding greens function for the linear operator L'=f(x)L, where f(x) =/=0?

So I've started by writing L'G'(x,x')= delta(x-x') =f(x)LG'(x,x') from the definition of the greens function.

Also, LG(x,x')=delta(x-x').

Therefore, LG=f(x)LG'.

I'm now stuck! I don't know where to go from here. I've tried to integrate over a certain range x=[a,b] and use the delta function properties but I've got nowhere!

2. Feb 28, 2015

### RUber

What do you get if you apply G to L'? Since f is not zero, you can divide by it.

3. Feb 28, 2015

### RightFresh

I get L'G=f(x)LG=f(x) delta(x-x')

I also have LG'=delta(x-x')/f(x)

Does any of that seem right?

4. Feb 28, 2015

### RUber

I think the goal is to write G' in terms of G and f. I think you should have to tools to do that now.

5. Feb 28, 2015

### RightFresh

I know that, but I can't seem to remove the dependence on either an L or L' operator. Do you have a hint on how to proceed?

6. Feb 28, 2015

### RightFresh

So I've tried integrating these over an arbitrary range to remove the delta function, is there anything I can do with the integral of LG' or L'G ??

Thanks

7. Feb 28, 2015

### RUber

Is there any reason it wouldn't just be G/f?

8. Feb 28, 2015

### RightFresh

I did initially think that, but when you do L'(G/f), won't L' also need to act on f ?

9. Mar 1, 2015

### RUber

What if we were to multiply by a function of x' which, assuming L acts on x, will commute right through, like $G'(x,x') = G(x,x') \frac 1 {f(x')}$?

10. Mar 1, 2015

### RightFresh

Oh! I see. Thanks a lot for your help!