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## Main Question or Discussion Point

Hey,

I am trying to find a GF for the function:

[tex] y''+\frac{1}{24}y=f(x)[/tex]

The function is bounded by:

[tex]y(0)=y(\pi)=0[/tex]

I have followed a math textbook that goes through the exact process for the function:

[tex] y''+k^2y=f(x)[/tex]

and have found a nice looking general solution:

[tex]G(x,x')=-\frac{sin(x/4)sin(\frac{1}{4}[\pi-x'])}{\frac{1}{4}sin(\pi/4)}[/tex]

for x<x'

and

[tex]G(x,x')=-\frac{sin(x'/4)sin(\frac{1}{4}[\pi-x])}{\frac{1}{4}sin(\pi/4)}[/tex]

for x>x'

Now, here is my problem:

I need to find y(x) for f(x)=sin(x)

This sounds easy right, just use the Green function, multiply by f(x) and integrate over the boundary 0 to pi.

The problem is that the integral only converges for k^2 is some integer.

For non integer k^2 it seems to diverge...not good as I have k^2=1/24

I was just wandering if anyone has any suggestions????

Thanks!

I am trying to find a GF for the function:

[tex] y''+\frac{1}{24}y=f(x)[/tex]

The function is bounded by:

[tex]y(0)=y(\pi)=0[/tex]

I have followed a math textbook that goes through the exact process for the function:

[tex] y''+k^2y=f(x)[/tex]

and have found a nice looking general solution:

[tex]G(x,x')=-\frac{sin(x/4)sin(\frac{1}{4}[\pi-x'])}{\frac{1}{4}sin(\pi/4)}[/tex]

for x<x'

and

[tex]G(x,x')=-\frac{sin(x'/4)sin(\frac{1}{4}[\pi-x])}{\frac{1}{4}sin(\pi/4)}[/tex]

for x>x'

Now, here is my problem:

I need to find y(x) for f(x)=sin(x)

This sounds easy right, just use the Green function, multiply by f(x) and integrate over the boundary 0 to pi.

The problem is that the integral only converges for k^2 is some integer.

For non integer k^2 it seems to diverge...not good as I have k^2=1/24

I was just wandering if anyone has any suggestions????

Thanks!