# Not understand online notes on the Green function

1. Feb 22, 2012

### fluidistic

1. The problem statement, all variables and given/known data
I'm trying to self study Green function and I can't follow the very last step of a demonstration in an online notes (that I attach in this post). Page 7 to 8.
Basically he says that from $G_{tt}(t,t')+\omega G(t,t')=0$ for all $t>t'$ with the conditions $G(t,t'+\varepsilon)=0$ and $G_{t} (t,t'+\varepsilon)=1$ when $\varepsilon$ tends to 0. One can "easily" find out that $G(t,t')=\frac{1}{\omega }\sin [\omega (t-t')]$.
My question is how do you find this out? I have no idea.

2. Relevant equations
No idea!

3. The attempt at a solution
100% clueless.

#### Attached Files:

• ###### Green.pdf
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2. Feb 22, 2012

### MathWarrior

3. Feb 22, 2012

### vela

Staff Emeritus
There's a typo in his notes. The differential equation should be $y''(t) + \omega^2 y(t) = 0$. Considering what I've seen in your other threads, I find it hard to believe you don't know how to solve this ordinary differential equation with the given initial conditions.

Perhaps it's the notation that's throwing you off. Just think of t' as a parameter.

4. Feb 22, 2012

### fluidistic

Ok thanks, this makes the 3rd typo on this same page (I emailed him for the 2 others I saw and he replied with a thank you message).
I realize I misunderstood the initial conditions and consequently miswrote them here.

He uses the notation (for simplicity only) $G(t,t')=y(t)$. So I am lead to think that when he writes $y(t') = 0$ he means $G(t',t')=0$? Shouldn't it be $G(0,t')=0$?
I'm totally confused.

5. Feb 22, 2012

### vela

Staff Emeritus
Just consider t' to be fixed. Its value partitions the number line into two regions: t<t' and t>t'. What he did was solve the homogeneous equation to find a solution y1(t), valid for t<t', and a solution y2(t), valid for t>t'. There were initial conditions given for t<t', namely y(0)=0 and y'(0)=0, so he used those to pin down what y1(t) equaled for t<t'. Then by integrating the differential equation across the boundary, he derived what the initial conditions are for y2(t) — in other words, what y2(t') and y'2(t') equal.

Earlier, he wrote y2(t) = B sin ω(t-t1). Here, B and t1 are the arbitrary constants you get when you solve a second-order differential equation. You want to find B and t1 so that y2(t') = 0 and y'2(t') = 1.

6. Feb 22, 2012

### fluidistic

Thank you very much vela. I now solved the problem using his notation and also keeping $G(t,t')$ instead of y(t).
I reach the result $G(t,t')=0$ for $t<t'$ and $G(t,t')=\frac{\sin [\omega (t-t')] }{\omega }$ for $t>t'$ as it should.