Green function of ODE and checking uniqueness of solution

Question

I have solved the following ordinary differential equation
$$
u_{xx}=f(x) \,,\hspace{5mm}x\in[0,l] \\
u(0)=0,\,\hspace{5mm} u_x(l)=0
$$
and found the Green function to be
$$
G(x,\xi)=
\begin{cases}
-x & ,x<\xi \\
-\xi & ,x>\xi
\end{cases}
$$
I already know that $u(x)=\int_{0}^{l}f(\xi)G(\xi,x)\,d\xi$ and we know that Green function is symmetric i.e.
$$G(x,\xi)=G(\xi,x)$$
The problem is I cannot show that $u(x)$ I found by using Green function is equal to the solution when I solve above ode directly with known methods.

Answer 1

Let's find the Green's function together. It should have $\frac{\partial^2 G}{\partial \xi^2}=0$ except at $\xi=x$, and then at $\xi=x$ there should be a jump in $\frac{\partial G}{\partial \xi}$ of size $1$ while $G$ itself remains continuous. It should also vanish at $\xi=0$ and $\xi=l$. Ignoring that very last condition we get a form of

$$G(x,\xi)=\begin{cases} m(x) \xi & \xi<x \\ (m(x)+1)(\xi-x)+xm(x) & \xi>x \end{cases}.$$

Applying the last condition we get

\begin{align} (m(x)+1)(l-x)+xm(x)=0 & \Rightarrow l m(x)-xm(x)+l-x+xm(x)=0 \\ & \Rightarrow m(x)=\frac{x-l}{l}.\end{align}

Thus after simplifying you end up with

$$G(x,\xi)=\begin{cases} \frac{x}{l} \xi - \xi & \xi<x \\ \frac{x}{l} \xi-x & \xi>x \end{cases}.$$

Having fixed that you should be good.