How can I determine the Green's function $G:[0,1]\times [0,1] \to \mathbb R$ such that $u(x)=\int_0^1 G(x,\xi )f(\xi) \, d\xi $ is a solution of the boundary problem $$\begin{cases} u''(x)=f(x) \,, \,\,\,\,\,\,x \in (0,1) \\u(0)=0 \,, \,\,\,\,\, u(1)+u'(1)=0\end{cases}$$ and determine the solution for $f(x)=x$?
Help is much appreciated.
Remember that the Green's function obeys the same boundary conditions as the solution. So we look for a solutions to $G''(x, \xi) = \delta(x - \xi)$ which satisfies $G(0, \xi) = 0$, $G(1, \xi) + G'(1, \xi) = 0$.
For $x < \xi$, we have \begin{align*} G(x,\xi) = A_{<}x + B_{<} \end{align*} but since $G(0,\xi) = 0$, then $G(x, \xi) = A_{<}x$. For $x > \xi$ we have \begin{align*} G(x, \xi) = A_{>}x + B_{>} \end{align*} and the boundary condition $G(1, \xi) + G'(1, \xi) = 0$ implies that $B_{>} = -2A_{>}$, so that $G(x, \xi) = A_{>}(x-2)$. Continuity at $x = \xi$ implies \begin{align*} A_{<}\xi = A_{>}(\xi - 2) \end{align*} Letting $A_{<}\xi =: k$ gives \begin{align*} G(x, \xi) &= k\begin{cases} (\xi - 2)x & x < \xi \\ \xi(x -2) & x > \xi \end{cases} \end{align*} To determine the scaling, we use the definition $G''(x, \xi) = \delta(x-\xi)$ and integrate over the delta spike to obtain \begin{align*} \lim_{\varepsilon\to 0^{+}}\int_{\xi - \varepsilon}^{\xi + \varepsilon} G''(x, \xi) \, \mathrm{d}x = G'(\xi^{+}, \xi) - G'(\xi^{-}, \xi) = 2k = 1 \end{align*} so that \begin{align*} G(x, \xi) &= \frac{1}{2}\begin{cases} (\xi - 2)x & x < \xi \\ \xi(x -2) & x > \xi \end{cases} \end{align*}
This was a bit tedious so expect some typos.