How to Find Green Function for y'' + 1/4y = f(x) with Boundary Conditions?

[spock0149]

Homework Statement

Hey folks,

I need to find a Green function for the equation:

y'' +1/4y = f(x)

With boundary conditions y(0)=y(pi) = 0

The Attempt at a Solution

I tried some combination of solutions that look like sin(kx) and sin(k-pi)
and looked at the strum liouville equation too and meassed with a Wronkskian.

I was just wandering if there was an easier way to do this?

 

[spock0149]

anyone?
:)

 

[Tom1992]

pde problems are usually messy and tedious, which i have so far avoided and even prof. mathwonk himself admitted he is lacking in. perhaps you should post in the differential equations forum where there are some pde's experts to assist you.

 

[HallsofIvy]

Homework Statement

Hey folks,

I need to find a Green function for the equation:

y'' +1/4y = f(x)

With boundary conditions y(0)=y(pi) = 0

The Attempt at a Solution

I tried some combination of solutions that look like sin(kx) and sin(k-pi)
and looked at the strum liouville equation too and meassed with a Wronkskian.

I was just wandering if there was an easier way to do this?

Do you mean y"+ (1/4)y ? (Not y"+ 1/(4y).)

What is the DEFINITION of Green's function?

The Green's function for this problem, G(x,t), must satisfy:

G_xx+ (1/4)G= 0 for all $x\ne t$.
G(0,t)= 0, G($\pi$,t)= 0
G is continuous at x= t.
Derivative of G at t, from the right, minus derivative of G at t,from the left, must equal 1.

The general solution of y"+ (1/4)y= 0 is A cos((1/2)x+ B sin((1/2)x) so Green's function must be of the form
[tex]G(x,t)= \left{\begin{array}{c}A cos((1/2)x)+ B sin((1/2)x if x< t \\C cos((1/2)x)+ D sin((1/2)x) if x> t \end{array}\right[/itex]<br /> <br /> G(0,t)= A= 0, G($\pi$,t)= D= 0<br /> <br /> G(t,t)= A cos((1/2)t)+ B sin((1/2)t= C cos((1/2)t)+ D sin((1/2)t)<br /> <br /> -(1/2)C sin((1/2)t)+ (1/2)D cos((1/2)t)+ (1/2)A sin((1/2)t)- (1/2)B cos((1/2)t)= 1<br /> <br /> Solve for A, B, C, D.[/tex]