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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.

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    It is probably easier to just solve the relation $u''=x$ by integrating twice. The result will contain two arbitrary constants that can be found by imposing boundary conditions. If you really need the Green's function, do the same procedure with an arbitrary $f$ in place of $x$. – 2017-02-06

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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.