How to find the poles of a green function? I am trying to construct a green function for $y''+\alpha^2u=f(x), u(0)=u(1), u'(0)=u'(1)$. For that I am trying to follow the procedure described here:(Construct the Green s function for the equation)
I was not able to know how to find "$a$". 
 A: For fixed $t \in (0,1)$, the Green function $G(x,t)$ is the solution of
$$
         G_{xx}(x,t)+\alpha^{2}G(x,t) = \delta_{t}(x) \\
           G(0,t)=G(1,t),\\
           G_{x}(0,t)=G_{x}(1,t)
$$
Let $Lf=f''+\alpha^{2}f$. The above means $LG(x,t)=0$ for $x < t$ and for $x > t$. So you have to stitch together solutions of $Lf=0$ so that they (a) come together continuously at $t$ (b) have a unit jump in the derivative at $t$ and (c) are periodic along with the first derivative.
To the left of $t$ the solution will be $Ae^{-i\alpha x}+Be^{i\alpha x}$ and to the right it will be $Ce^{-i\alpha x}+De^{i\alpha x}$. To enforce periodicity,
$$
            A +B= Ce^{-i\alpha}+De^{i\alpha},\\
            -A+B = -Ce^{-i\alpha}+De^{i\alpha}
$$
Continuity at $x = t$ gives
$$
                  Ae^{-i\alpha t}+Be^{i\alpha t} = Ce^{-i\alpha t}+De^{i\alpha t}.
$$
A jump in the derivative of $1$ as $x$ crosses $t$ gives
$$
   (-iCe^{-i\alpha t}+iDe^{i\alpha t})-(-iAe^{-i\alpha t}+iD^{i\alpha t}) = 1.
$$
Four equations, four unknowns. You'll get a unique solution unless $e^{i\alpha}=1$, which is the case where $Lf=0$ has non-trivial periodic solutions on $[0,1]$.
A: Start by writing solutions for $x<x'$ and $x>x'$ as 
$$\begin{align}
&g(x,x')=A(x')\sin \alpha x+B(x')\cos \alpha x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdots x<x' \tag1\\\\
&g(x,x')=C(x')\sin \alpha(x-1)+D(x')\cos \alpha(x-1)\,\cdots x>x'\tag2
\end{align}$$
Applying the boundary conditions $g(0,x')=g(1,x')$ and $g_1(0,x')=g_1(1,x')$ into $(1)$ and $(2)$ reveals that $A(x')=C(x')$ and $B(x')=D(x')$.
We also enforce continuity of $g$ at $x=x'$ along with the condition $g_1(x,x_>')-g_1(x,x_<')=1$ to find that 
$$A(x')=\frac{\cos \alpha(x'-1)-\cos \alpha x'}{4\alpha\sin^2(\alpha/2)}$$
$$B(x')=-\frac{\sin \alpha(x'-1)-\sin \alpha x'}{4\alpha\sin^2(\alpha/2)}$$
and thus
$$\begin{align}
&g(x,x')= \frac{\sin \alpha(x-x'+1)-\sin \alpha(x-x')}{4\alpha\sin^2(\alpha/2)}\,\,\cdots x<x' \\\\
&g(x,x')=\frac{\sin \alpha(x-x')-\sin \alpha(x-x'-1)}{4\alpha\sin^2(\alpha/2)}\,\cdots x>x'
\end{align}$$
which is a valid solution for all $\alpha \ne 2m\pi$, $m$ an integer.
