Green's function for periodic boundary condition How to find a explicit Green's function for the problem $$-u''(x)+q(x)u(x)=g(x),$$ $x\in (0,1)$ with conditions $u(0)=u(1)$ and $u'(0)=u'(1)$? Here $q$ has whatever property you want except being constant.
All I found about it is for separated end-point conditions saying "Yes, there is a green's function". 
 A: A Green's function $G(x,s)$ for your problem is a solution of $-u''(x) + q(x) u(x) = \delta(x-s)$
with your boundary conditions.  That is:
(1) $-u''(x) + q(x) u(x) = 0$ for $x \ne s$;
(2) $\lim_{x \to s+} u'(x) - \lim_{x \to s-} u'(x) = -1$; and
(3) $u$ continuous at $x=s$.
If you want an explicit solution, you first of all will need a fundamental set of solutions $y_1(x)$, $y_2(x)$ of the homogeneous equation
$$-u''(x) + q(x) u(x) = 0.$$  Then, you want to have
$$G(x,s) =
\begin{cases}
 a_1(s) y_1(x) + a_2(s) y_2(x), &\text{for}~ -\infty < x \le s< +\infty 
\\
 b_1(s) y_1(x) + b_2(s) y_2(x), &\text{for}~ -\infty < s \le x< +\infty. 
\end{cases}
$$
Here we introduced four, as yet unknown, constants.  The four constants are determined by the following system of four equations.
$$ \eqalign{a_1(s)\; y_1(0) + a_2(s)\; y_2(0) &= b_1(s)\; y_1(1) + b_2(s)\; y_2(1)\cr
            a_1(s)\; y_1'(0) + a_2(s)\; y_2'(0) &= b_1(s)\; y_1'(1) + b_2(s)\; y_2'(1)\cr
            a_1(s)\; y_1(s) + a_2(s)\; y_2(s) &= b_1(s)\; y_1(s) + b_2(s)\; y_2(s)\cr
            a_1(s)\; y_1'(s) + a_2(s)\; y_2'(s) &= 1 + b_1(s)\; y_1'(s) + b_2(s) \;y_2'(s)\cr}$$
So you must solve this set of four linear equations for the unknowns $a_1(s), a_2(s), b_1(s), b_2(s)$.
A: I attempt to find a explicit Green's function for the problem
$$-u''(x)+\underbrace{\left[-m^2\right]}_{q(x)}\,u(x)=g(x),
\tag{1}$$
where $x\in (0,1)$ with conditions $u(0)=u(1)$ and $u'(0)=u'(1)$? Here $q$ is, in fact, a constant.
A fundamental solution is
$$u(x) = \sin(m\,x) + \cos(m\,x).$$
Then,
$$G{(x,s)}
= 
\begin{cases}
a_1(s)\,\sin{(m\,x)} + a_2(s)\,\cos{(m\,x)}
,&~\text{for} 0< x<x'<2\,\pi
\\
b_1(s)\,\sin{(m\,x)} + b_2(s)\,\cos{(m\,x)}
,&~\text{for} 0< x'<x<2\,\pi
\end{cases}
$$
I now attempt to solve for the four unknowns using the following four equations
\begin{align}
a_1(s)\; \sin(m\,0) + a_2(s)\; \cos(m\,0) &= b_1(s)\; \sin(m\,1) + b_2(s)\; \cos(m\,1)
\\
m\,a_1(s)\; \cos(m\,0) - m\,a_2(s)\; \sin'(m\,0) &= m\,b_1(s)\; \cos(m\,1) - m\,b_2(s)\; \sin(m\,1)
\\
a_1(s)\; \sin(m\,s) + a_2(s)\; \cos(m\,s) &= b_1(s)\;\sin(m\,s) + b_2(s)\; \cos(m\,s)
\\
m\,a_1(s)\; \cos(m\,s) - m\,a_2(s)\; \sin(m\,s) &= 1 + m\,b_1(s)\; \cos(m\,s) -m\,b_2(s) \;\sin(m\,s)
\end{align}
Thus
\begin{align}
a_2(s)  &= b_1(s)\; \sin(m ) + b_2(s)\; \cos(m )
\\
 a_1(s)   &=  b_1(s)\; \cos(m ) -  b_2(s)\; \sin(m )
\\
a_1(s)\; \sin(m\,s) + a_2(s)\; \cos(m\,s) &= b_1(s)\;\sin(m\,s) + b_2(s)\; \cos(m\,s)
\\
m\,a_1(s)\; \cos(m\,s) - m\,a_2(s)\; \sin(m\,s) &= 1 + m\,b_1(s)\; \cos(m\,s) -m\,b_2(s) \;\sin(m\,s)
\end{align}
From the third equation in the equations array  above, I find $a_1 = b_1$ and $a_2 = b_2$. Substituting these results into the fourth equation in the equaiton array, I find an inconsistency, namely that
\begin{align}
0 &= 1.  
\end{align}
How come I got an inconsistent solution? Are Cauchy boundary conditions generally inconsistent with the governing differential equation equation in (1)?
