I currently read a paper about solving Fredholm integral equations. There is a theorem given to solve them, but I cant seem to solve the given example.
Theorem: The solution can be calculated by:
$$h(x) = Q(L)G \textrm{ where } G(x) = \begin{cases} g(x)+v(x) \textrm{ in D} \\ u(x) \textrm{ in }\mathbb{R}^n \setminus D \end{cases}$$
where g is a solution to $$P(L)g = f$$ and $u(x)$ and $v(x)$ are the unique solutions to: $$Q(L)u = 0 \textrm{ in } \mathbb{R}^n \setminus D$$ $$P(L)v = 0 \textrm{ in } D$$ $$\partial_N^ju = \partial_N^j(v+g) \textrm{ on } \partial D \textrm{ for } 0 \leq j \leq \alpha$$
Now I have an example where $P(\lambda) = 1, Q(\lambda) = (\lambda^2+1)/2, \alpha=0, D=[-1,1], L = -i\frac{d}{dx}$
The solution to this is: $$h(x) = \frac{-f''+f}{2} + \frac{f'(1)+f(1)}{2}\delta(x-1) + \frac{-f'(-1)+f(-1)}{2}\delta(x+1)$$
We immediately get from the theorem, since $P(L)=1$:
$g=f$ and $v=0$ in $[-1,1]$ which would give us for the solution in D: $$h(x) = Q(L)G = Q(L)g = \frac{-\frac{d^2}{dx^2}+1}{2} (f) = \frac{-f''+f}{2}$$
But $$Q(L)u = 0 \textrm{ in } \mathbb{R}^n \setminus D$$ should make the other term $0$, so I dont understand where the part with the delta function comes from?? And also should it be the second derivative of f instead of the first one in the alpha terms?? ( In the paper its the first, but I think thats a mistake)
I need this for a presentation I have soon, so I would really appreciate help!!