Green's Function fo a Boundary Value Problem

[maxtor101]

Homework Statement

$$L[y] = \frac{d^2y}{dx^2}$$

Show that the Green's function for the boundary value problem with $y(-1) = 0$ and $y(1) = 0$ is given by

$G(x,y) = \frac{1}{2}(1-x)(1+y) for -1\leq y \leq x \leq 1\$


$G(x,y) = \frac{1}{2}(1+x)(1-y) for -1\leq x \leq y \leq 1\$

Homework Equations

The Attempt at a Solution

Well in class we had defined $L[y] = (py')' + qy$ as the Strum-Liouville self-adjoint operator

So that gives me:

$L[y] = (py')' + qy = \frac{d^2y}{dx^2}$


Do I treat this problem like other Strum-Liouville Boundary Problems by writting it as :

$L[y] = (py')' + qy = f(x)$

Where $f(x) = \frac{d^2y}{dx^2}$

And continue on as I usually would?

Any help on this would be greatly appreciated!

 

[mighty2000]

it is important to understand and be able to solve boundary value problems in order to study and analyze various physical phenomena. In this case, we are given a boundary value problem with a self-adjoint operator L[y] = \frac{d^2y}{dx^2} and boundary conditions y(-1) = 0 and y(1) = 0. Our goal is to find the Green's function for this problem, which will help us solve for the solution y(x) for any given source function f(x).

To start, we can rewrite the boundary value problem as:

L[y] = \frac{d^2y}{dx^2} = f(x), with boundary conditions y(-1) = 0 and y(1) = 0.

Next, we can use the definition of the Green's function for a self-adjoint operator L[y] to write:

L[G(x,y)] = \delta(x-y), where \delta(x-y) is the Dirac delta function.

Substituting L[y] = \frac{d^2y}{dx^2} and using the boundary conditions, we get:

\frac{d^2G(x,y)}{dx^2} = \delta(x-y), with boundary conditions G(x,y)|_{x=-1} = 0 and G(x,y)|_{x=1} = 0.

Integrating both sides with respect to x, we get:

\frac{dG(x,y)}{dx} = H(x-y) + C_1, where H(x) is the Heaviside step function and C_1 is a constant of integration.

Integrating again and applying the boundary conditions, we get:

G(x,y) = \frac{1}{2}(1-x)(1+y), for -1\leq y \leq x \leq 1 and

G(x,y) = \frac{1}{2}(1+x)(1-y), for -1\leq x \leq y \leq 1.

Thus, the Green's function for the given boundary value problem is given by:

G(x,y) = \frac{1}{2}(1-x)(1+y), for -1\leq y \leq x \leq 1 and

G(x,y) = \