I have the problem of finding the Green's function for the problem $$ u''(x)=f(x), $$ with $f \in C[0,1]$, $u \in C^2[0,1]$, and $u(0)=u'(1)=0$.
First of all I can easily see that the operator is self-adjoint.
So I need to solve the problem $$ G''(x,y)=\delta (x-y). $$
I want to solve this using the Laplace transform.
Taking the transform on both sides and solving for $G$ I get $$G(s) =\frac{e^{-ys}+G'(0)}{s^2}. $$
Taking the inverse Laplace transform I get $$G(x,y)=G'(0)x+xH(x-y)-yH(x-y)$$ where $H$ is the Heaviside function.
The problem is that I wasn't able to find $G'(0)$ since the boundary conditions are $u(0)=u'(1)=0$, there is no condition for $u'(0)$.
However I know that $G'(0)$ must be zero, since I also computed the Green's function using a different way and got as solution $$ G(x,y)=xH(x-y)-yH(x-y). $$
So my question is, how can I find $G'(0)$?