Green's Functions solution to PDE with Neumann Boundary Condition I am trying to find the Green's function for $-cu''=f(x)$ with the boundary condition that $u'(0)=0=u(1)$.
So far I've started by supposing $f(x)=\delta_\xi(x)$ and then considering $ \frac{d^2G}{dx^2}=\delta_\xi(x)$. And then I consider the homogenous equation, giving $G=ax+b$, with  $a,b\in \mathbb{R}$ but that doesn't seem right and think I am making a mistake. Where should I proceed from here? I know how to solve for coefficients and such, but I get stuck in the actual set up of the problem. 
Thanks.
 A: You want a solution of
$$
                   u_{\xi}''(x)=0,\;\;\; 0 < x < \xi,\;\; \xi < x < 1, \\
                           u_{\xi}'(0)=0,\;\; u_{\xi}(1)=0, \\
                           u_{\xi}(\xi+0)=u_{\xi}(\xi-0), \\
                          -cu_{\xi}'(\xi+0)+cu_{\xi}'(\xi-0)=1.
$$
The last condition gives the $\delta_{\xi}$ property for $-cu''$. The solution to the left of $\xi$ is a constant. The solution to the right of $\xi$ is a constant times $1-x$. Thus,
$$
          u_{\xi}(x) = \left\{\begin{array}{cc}
                          A, & 0 < x < \xi \\
                          B(1-x), & \xi < x < 1
                        \end{array} \right.
$$
where $A$ and $B$ are determined by the last two conditions. Continuity gives $A=B(1-\xi)$ and the jump condition gives $-cB(-1)-0=1$, or $B=1/c$. So $A=(1-\xi)/c$. Therefore,
$$          u_{\xi}(x) = \left\{\begin{array}{cc}
                          (1-\xi)/c, & 0 < x < \xi \\
                          (1-x)/c, & \xi < x < 1
                        \end{array} \right.
$$
