I tried to solve my homework, but unfortunately got stuck at some point. Could you please help me to get through it?
For a given boundary value problem: $$ Ly =y'''+y''\\ y(0)+y'(0) = y''(0)=y(1) = 0 $$ Establish whether there is a Green's function and if so, construct it.
I start with $$G'''(x,s) + G''(x,s) = \delta(x-s)\\\text{If } x \neq s:G(x,s) = c_1e^{-x}+c_2x+c_3$$
$$For \ x < s : \begin{cases}
a_2+a_3 = 0\\
a_1 = 0
\end{cases}
(\text{the boundary condition at } x=0)\\
For \ x > s : \frac{b_1}{e}+b_2+b_3 = 0\ (\text{the boundary condition at } x=1)
$$
$$
G(x,s) = \begin{cases}
G_1(x,s)=a_2x+a_3 \ \ \ \text{for } xs
\end{cases}
$$
Added:
Continuity of the Green’s function permits
$$
G_1(s,s) =G_2(s,s)\\
a_2s+a_3 = b_1e^{-s}+b_2s+b_3\\
s(a_2-b_2)+(a_3-b_3) =b_1e^{-s}
$$
While the jump discontinuity condition gives
$$
\begin{cases}
\lim_{\epsilon \rightarrow 0}\left.\frac{dG(x,s)}{dx}\right|_{s-\epsilon}^{s+\epsilon}= 1
\\
\lim_{\epsilon \rightarrow 0}\left.\frac{d^2G(x,s)}{dx^2}\right|_{s-\epsilon}^{s+\epsilon}= 1
\end{cases}
\\
\begin{cases}
\frac{dG_2(s,s)}{dx}-\frac{dG_1(s,s)}{dx}=1
\\
\frac{d^2G_2(s,s)}{dx^2}-\frac{d^2G_1(s,s)}{dx^2}=1
\end{cases}\\
\begin{cases}
b_2-b_1e^{-s}-a_2=1
\\
b_1e^{-s}=1
\end{cases}
\\
\begin{cases}
b_1 = e^{s}
\\
b_2-a_2 = 2
\end{cases}
$$
Now:
$$
\begin{cases}
a_1 =0
\\
a_2 = -a_3
\\
b_1 = e^{s}
\\
b_2-a_2 = 2
\\
s(a_2-b_2)+(a_3-b_3) = 1
\\
e^{s-1} +b_2 +b_3 =0
\end{cases}
\\
\left\{\begin{matrix}
a_1 = 0\\
a_2 = -\frac{2es+e+e^s}{2es}\\
a_3 =\frac{2es+e+e^s}{2es}\\
b_1 = e^s\\
b_2 = \frac{2es-e-e^s}{2es}\\
b_3 = -\frac{(e^s+e)(2s-1)}{2es}
\end{matrix}\right.
\\
G(x,s) = \begin{cases}
\frac{2es+e+e^s}{2es}(1-x)\ \ \ \text{for } xs
\end{cases}
$$