Finding a Green Function for y''+1/24y=f(x) with Bounded Conditions | Math Help

[tomboy67]

Hey,

I am trying to find a GF for the function:

$$y''+\frac{1}{24}y=f(x)$$

The function is bounded by:

$$y(0)=y(\pi)=0$$

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

$$y''+k^2y=f(x)$$

and have found a nice looking general solution:

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

for x<x'

and

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

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!

 

[Clausius2]

I...I don't see how can diverge. It's a multiplication of trigonometric functions isn't it?. How can that integral diverge?.

 

[tomboy67]

hmmmmm...maybe its just my formula table...but when you integrate there is a factor of nx on the denominator and when you plug in the limits the zero gives you infinty.

 

[HallsofIvy]

What formula is that? If you are integrating sin(x/4)f(x), you certainly won't get any "x" in the denominator from the sine!

 

[tomboy67]

Sorry, I figured it now...you are right. I carried an x' through into the denominator, but it should only have been the coefficient of x. That was the origin of the infinity. Thanks, for your assistance!