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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.

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    This is a Neumann boundary condition, named after Carl Neumann. John von Neumann was a different person. – 2017-02-11
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    Cool, I'll edit that. – 2017-02-12

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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, $$ u_{\xi}(x) = \left\{\begin{array}{cc} (1-\xi)/c, & 0 < x < \xi \\ (1-x)/c, & \xi < x < 1 \end{array} \right. $$

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    At the step where you define $u_\xi (x)$ piecewise how do you know to shift x by $(1-x)$? – 2017-02-12