Fredholm alternative question for Green's functions

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During my course in mathematical methods for physics we were studying spectral decomposition on eigenfunctions for linear differential and integral operators.

When dealing with a problem like

$$ \begin{align} &\hat L g = f & f,g\in L^2(\Omega) \end{align} $$

where $\hat L$ is a linear differential or integral operator, we said that if we know the spectral properties of the operator

$$ \hat L \phi_n =\lambda_n\phi_n $$

where, and this is the important part, $\lambda_n\neq0$, then we could write our solution as a linear combination of the eigenfunctions, mainly

$$ \begin{align} &f(x) = \sum_n a_n\phi_n&a_n = \langle\phi_n| f\rangle = \int_\Omega\overline{\phi_n}(x')\;f(x')\,\mathrm dx'\tag1\\ &g(x) = \sum_nb_n\phi_n \end{align} $$

From this we could find the $b_n$ as

$$ \hat L g = \hat L\left(\sum_n b_n\phi_n\right) =\sum_nb_n\lambda_n\phi_n = \sum_na_n\phi_n\\ \sum_n(b_n\lambda_n-a_n)\phi_n=0 \Leftrightarrow b_n=\frac{a_n}{\lambda_n} $$

In this case

$$ g(x) = \sum_n \frac{a_n}{\lambda_n}\phi_n \overset{\text{From }(1)}{=} \int_\Omega \left[\sum_n \frac{\overline{\phi}_n(x')\phi_n(x)}{\lambda_n}\right] f'(x)\,\mathrm dx' $$

from that we can define the Green's function for this problem as

$$ G(x,x') = \sum_n \frac{\overline{\phi}_n(x')\phi_n(x)}{\lambda_n} $$

Then our professor said the following statement without going into details:

If one of the eigenvalues $\lambda_n$ is zero, then we invoke the Fredholm alternative theorem

Serching online I found lot's of definitions for this theorem, seems to me that the theorem is stated for finite dimensional linear operators and then was adapted for infinite dimensional ones. But I cannot wrap my head around it.

What does the statement means? How can I use the Fredholm alternative? What does the Fredholm alternative theorem say about infinite dimensional operators? How can I apply it to the Green's function problem? How does the solution to the problem change when $\lambda_n=0$ for one, or even more, $n$'s?

I know that probably this are a lot of questions but I think that can be answered all together. I want to know more about it because it seems very interesting and my professor didn't go in more details.

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What does the statement means? How can I use the Fredholm alternative? What does the Fredholm alternative theorem say about infinite dimensional operators? How can I apply it to the Green's function problem? How does the solution to the problem change when λn=0 for one, or even more, n's?

My book has a section on this, which is talking about the same exact thing as you.

If $\lambda =0 $ is an eigenvalue then the Green's function doesn't exist. In order to understand this we re-examine the original non-homogeneous problem $$ L(u) = f(x) \tag{1} $$

where it is subject to homogeneous boundary conditions

As you approached it with the method of eigenfunctions expansion. We have

$$ u = \sum_{n=1}^{\infty} a_{n} \phi_{n}(x) \tag{2} $$

which yields the following by orthogonality

$$ -a_{n}\lambda_{n} = \frac{\int_{a}^{b}f(x) \phi_{n}(x) dx}{\int_{a}^{b} \phi_{n}^{2} \sigma(x) dx } \tag{3} $$

which means that if $\lambda_{n} = 0$ there may not be any solutions to the non-homogeneous case. It actually defines Generalized Green's functions later in the chapter.

$$ L(u) = f \tag{4}$$

subject to homogeneous boundary conditions when $\lambda = 0$

$$L[ G(x,x_{0}) ] = \delta(x-x_{0}) \tag{5} $$

we're going to analyze the case where $\lambda=0$

$$ \int_{a}^{b} f(x) \phi_{h}(x) dx =0 \tag{6}$$

however this function may not exist for all $x_{0}$ since $\delta(x-x_{0})$ is not orthogonal to $\phi_{h}(x)$

The solution becomes

$$ \delta(x-x_{0}) +c\phi_{h}(x) \tag{7}$$

If $c$ is chosen properly.

$$ \int_{a}^{b} \phi_{h}(x) [\delta(x-x_{0}) +c\phi_{h}(x) ] dx = \phi_{h}(x_{0}) +c\int_{a}^{b} \phi_{h}^{2}(x) dx \tag{8} $$

which allows us to introduce the Generalized Greens function $G_{m}(x,x_{0})$

$$ L[G_{m}(x,x_{0})] = \delta(x,x_{0}) -\frac{\phi_{h}(x)\phi_{h}(x_{0})}{\int_{a}^{b}\phi_{h}^{2}(x) dx} \tag{9}$$

I was going to include an example but the post is rather long.