Green Function

[logistic_guy]

Determine Green’s Function and the solution for the following initial value problem:

$\displaystyle u'' + \frac{1}{x}u' - \frac{1}{x^2}u = f(x), \ \ \ u(1) = 0, \ \ \ u'(1) = 1$

 

[logistic_guy]

Let us first enjoy the taste of Green's Function structure!

$\displaystyle g(x,s) = \frac{\Delta(x,s)}{a_0(s)W(s)}$

 

[BeansNRice]

Congratulations, you have killed this board. I bet you feel pretty proud of yourself.

For your next feat you can go beat up a little girl. Go you.

 

[logistic_guy]

Congratulations, you have killed this board. I bet you feel pretty proud of yourself.

For your next feat you can go beat up a little girl. Go you.

What have you done so far?

Don't panic BeansNRice, we can solve this if you show us exactly where you're stuck at. I am not saying it's easy, but it's solvable!

Let us start with $\displaystyle \Delta(x,s)$

Since this Delta function depends on the homogeneous solution of the original differential equation,

we have to solve this first:

$\displaystyle u'' + \frac{1}{x}u' - \frac{1}{x^2}u = 0$

This is called Euler's differential equation. Have you solved this type of equations before?

The more information you tell us about what you know the better we can help you!

 

[logistic_guy]

I have shown in other posts the concept of how to solve such equation in details. The solution to this homogeneous differential equation is:

$\displaystyle u(x) = c_1x + \frac{c_2}{x}$

 

[logistic_guy]

$\displaystyle \Delta(x,s) = \left| \begin{matrix}s & \frac{1}{s} \\[5 pt] x & \frac{1}{x}\end{matrix} \right| = \frac{s}{x} - \frac{x}{s} = \frac{s^2 - x^2}{xs}$

 

[logistic_guy]

The first coefficient in the original differential equation is $\displaystyle a_0(s)$.

Then,

$\displaystyle a_0(s) = 1$

 

[logistic_guy]

The Wronskian is:

$\displaystyle W(s) = \left| \begin{matrix}s & \frac{1}{s} \\[5 pt] 1 & -\frac{1}{s^2}\end{matrix} \right| = -\frac{1}{s} - \frac{1}{s} = -\frac{2}{s}$