Green's function

Remarks

Under the assumption, the Green's function exists, and the solution above is the only one. For fixed $s$, the function $x \mapsto G(x, s)$ solves $L y = 0$ on $[a, s)$ and on $(s, b]$ and satisfies the boundary conditions. Its derivatives of order up to $n - 2$ are continuous at $x = s$, and its derivative of order $n - 1$ jumps by $1$ there. Formally, $L_x G(x, s) = \delta(x - s)$.

Example: for $L y = y''$ on $[0, 1]$ with $y(0) = y(1) = 0$, take $G(x, s) = x (s - 1)$ for $x \le s$ and $G(x, s) = s (x - 1)$ for $x \ge s$. It vanishes at $x = 0$ and $x = 1$, it is linear in $x$ on each side of $s$, and $\partial_x G$ jumps from $s - 1$ to $s$, a jump of $1$. For $f = 1$ it gives $y(x) = \frac{x (x - 1)}{2}$, and indeed $y'' = 1$ and $y(0) = y(1) = 0$.

Non-example: for $L y = y''$ on $[0, 1]$ with $y'(0) = y'(1) = 0$, the constants solve the homogeneous problem, so the assumption fails. There is no Green's function: $y'' = 1$ has no solution with these conditions, since $y'(1) - y'(0) = \int_0^1 y'' = 1$.