Need to find Green Function to solve ODE

[romeo6]

Hi,

I have a basic ODE:

y''(x)+\frac{1}{4}y'(x)=f(x)

on 0<x<L

With Boundary conditions:

y(0)=y(L)=0

For which I would like to construct a Green Function.

Rather than just plain ask for help, I'll show you what I've been thinking and maybe someone wiser can help/correct me:

We must solve:

G&#039;&#039;(x,x&#039;)+\frac{1}{4}G&#039;(x,x&#039;&#039;)=\delta(x-x&#039;)

Since G vanishes at the boundaries we can expand as a Fourier sine series:

G(x,x&#039;)=\sum_{n=1}^\infty \gamma_n sin\frac{n\pi x}{L}

and:

\delta(x-x&#039;)=\sum_{n=1}^\infty A_n(x&#039;)sin\frac{n\pi x}{L}

Integrating the delta function I get:

A_n(x&#039;)=\frac{2}{L}sin\frac{n\pi x}{L}

I take the first and second derivative of the equation for G

and plug everything into the differential equation for G:

\sum_{n=1}^\infty \gamma_n (\frac{-n^2 \pi^2}{L^2})sin\frac{n\pi x}{L}+\frac{1}{4} \sum_{n=1}^\infty \gamma_n (\frac{n \pi}{L})cos\frac{n\pi x}{L}=\sum_{n=1}^\infty \frac{2}{L}sin\frac{n\pi x&#039;}{L}sin\frac{n\pi x}{L}

Now I am not sure what to do from here to get the required Green function...can someone guide me please?

 

[HallsofIvy]

It's not clear to me why you would use Fourier series. For all x except x', the Green's function satisfies G&quot;+ \frac{1}{4}G&#039;= 0 which has general solution
G(x)= A+ Be^{\frac{x}{4}}
The Green's function must be of the form
G(x, x&#039;)= \left{\begin{array}{c}A+ Be^{\frac{x}{4}} if 0\le x\le x&#039; \\C+ De^{\frac{x}{4}} if x&#039;\le x\le L\end{array}

A, B, C, D depend on x' and are chosen so that the boundary values are satisfied, the two "pieces" have the same value when x= x', and the difference in the derivative when x= x' (left side minus right) is 1: 4 conditions to determine 4 numbers.

 

[romeo6]

Ok, let me try and from some equations from the conditions:

Condition 1.

G(0,x&#039;)=0=A+Be^0
so A=-B

Conditon 2.

G(L,x&#039;)=0=C+De^{L/4}

soC=-De^{L/4}

Conditon 3:

Be^{x&#039;/4}=De^{1/4(x&#039;-L)}

Condition 4.

\frac{Bx&#039;(x-L)}{x-L}+B=1

Does this look about right??

Thanks!

 

[HallsofIvy]

If, by "condition 1", "condition 2", etc. you mean
"A, B, C, D depend on x' and are chosen so that the boundary values are satisfied, the two "pieces" have the same value when x= x', and the difference in the derivative when x= x' (left side minus right) is 1." that I mentioned before, then your equations for "condition 1" and "condition 2" are correct.
For the third one, having the same value when x= x',
A+ Be^{\frac{x&#039;}{4}}= C+ De^{\frac{x&#039;}{4}}

For the fourth, that the difference in left and right derivatives at x= x' be 1,
\frac{B}{4}e^{{x&#039;}{4}}- \frac{D}{4}e^{{x&#039;}{4}}= 1

 

[romeo6]

Thanks!

Yes, that's exactly what I meant by 4 conditions, but I was just borrowing your phrase '4 conditions to determine 4 numbers', so if I have used the phrase incorrectly I apologize.

I'll keep working on this.

In the last condition is that supposed to be 4e or is it a typo and supposed to be e/4?