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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".

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up vote 4 down vote accepted

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, $-u''(x) + q(x) u(x) = 0$ for $x \ne s$, with $\lim_{x \to s+} u'(x) - \lim_{x \to s-} u'(x) = -1$ but $u$ continuous at $s$.

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 $G(x,s) = a_1(s) y_1(x) + a_2(s) y_2(x)$ for $x \le s$ and $b_1(s) y_1(x) + b_2(s) y_2(x)$ for $x > s$, such that $$ \eqalign{a_1(s)\; y_1(0) + a_2(s)\; y_2(0) &= b_1(s)\; y_1(1) + b_2(s)\; y_2(1)\cr a_1(s)\; y_1'(0) + a_2(s)\; y_2'(0) &= b_1(s)\; y_1'(1) + b_2(s)\; y_2'(1)\cr a_1(s)\; y_1(s) + a_2(s)\; y_2(s) &= b_1(s)\; y_1(s) + b_2(s)\; y_2(s)\cr a_1(s)\; y_1'(s) + a_2(s)\; y_2'(s) &= 1 + b_1(s)\; y_1'(s) + b_2(s) \;y_2'(s)\cr}$$ So you must solve this set of four linear equations for the unknowns $a_1(s), a_2(s), b_1(s), b_2(s)$.

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