I'm trying to solve the BVP $f''(x)=\delta(x-a)$ where $0<a<1$ and $f(0)=f(1)=1$ but I'm really not sure where to start. I tried taking the Laplace Transform of the equation, to get (after applying $f(0)=0$):
$p^2\bar{f}(p)-f'(0)=e^{-ap}$
But I'm not sure how to proceed. I need to somehow eliminate the $f'(0)$ term, and I'm sure I have to do this via the other condition $f(1)=0$, but not sure how to apply it. Then I'd invert $\bar{f}(p)$.
Some ideas: Maybe a change of variables $x\to 1-x$? and appeal to symmetry of delta function. Any help would be great. Thanks.
Do a first of integration "by hand":
$$f'(t)=Y(t-a)+k$$
where $Y$ is Heaviside function, and $k$ a constant to be obtained later (by Initial Conditions).
Then only, apply Laplace Transform :
$$sF(s)-\underbrace{f(0)}_1=\frac{e^{-sa}}{s}+ks$$
$$F(s)=\dfrac{e^{-sa}}{s^2}+\dfrac{1}{s}+k$$
Can you proceed from here ?