I've found lots of resources deriving the Green's function for the 1D Poisson Equation, but no examples actually solving a DE with them. As a result, I'm very confused setting up the integrals for different forcing functions.
Consider the ODE with fixed boundary conditions $u(0)=0$ and $u(L)=L$: $$\frac{d^2u}{dx^2}=\frac{-f(x)}{T}$$ It can be shown that the Green's function for this problem is $$ G(x|\xi) = \begin{cases} \frac{(L-\xi)}{LT}x, & \text{for $x<\xi$} \\ \frac{(L-x)}{LT}\xi, & \text{for $x>\xi$} \end{cases}$$ Hence $$u(x)=\int_0^LG(x|\xi)f(\xi) \ d\xi=\int_0^xG(x|\xi)f(\xi) \ d\xi+\int_x^LG(x|\xi)f(\xi) \ d\xi$$ Now consider the forcing function $$f(x) = \begin{cases} f_0, & \text{for $0<x<\frac{L}{2}$} \\ -f_0, & \text{for $\frac{L}{2}<x<L$} \end{cases}$$ find $u(x)$
So far, I've broken the problem into 2 regions and set up $u(x)$ separately in both regions. I get a reasonable answer except it's discontinuous at $\frac{L}{2}$. When deriving the Green's function, I thought we used continuity of the string as a condition.
It's a lot to type out, so if anyone is interested in seeing the details, I'll link to a photo and an interactive graph of the solution.
Green's function
Your Green's function only makes sense if you reformulate the differential equation as $$ -Tu''(x)=f(x). $$ Then indeed the solution to $-T\frac{\partial^2 G(x|\xi)}{\partial\xi^2}=\delta(\xi-x)$, $0=G(x|0)=G(x|L)$ is $$ G(x|\xi)=\begin{cases} \frac{(L-x)\xi}{LT} &\text{ for }\xi\le x, \\ \frac{(L-\xi)x}{LT} &\text{ for }\xi\ge x. \end{cases} $$
Homogeneous boundaries
To apply the Green function, you need homogeneous boundary conditions. To that end, select a function $u_0$ satisfying the boundary conditions and construct the equation for the difference. Here $u_0(x)=x$ and $v=u-u_0$ satisfies the same differential equation, as $v''=u''$.
Integrating the solution
The function $f$ that you insert is not a constant. It is a piecewise constant. This you have to include in your interval division.
Thus for $0\le x\le L/2\le L$ you get \begin{align} u(x)&=x+\int_0^L G(x|\xi)f(\xi)\,d\xi \\ &=x+\int_0^x \frac{(L-x)\xi}{LT}f_0\,d\xi+\int_x^{L/2} \frac{(L-\xi)x}{LT}f_0\,d\xi+\int_{L/2}^L \frac{(L-\xi)x}{LT}(-f_0)\,d\xi \\ &=x+\frac{f_0(L-x)x^2}{2LT}+\frac{f_0[-(L/2)^2+(L-x)^2]x}{2LT}-\frac{f_0(L/2)^2x}{2LT} \\ &=x+\frac{f_0x(L-2x)}{4T} \end{align} and similarly for $L/2\le x\le L$ \begin{align} u(x)&=x+\int_0^L G(x|\xi)f(\xi)\,d\xi \\ &=x+\int_0^{L/2} \frac{(L-x)\xi}{LT}f_0\,d\xi+\int_{L/2}^x \frac{(L-x)\xi}{LT}(-f_0)\,d\xi+\int_x^L \frac{(L-\xi)x}{LT}(-f_0)\,d\xi \\ &=x+\frac{f_0(L-x)(L/2)^2}{2LT}-\frac{f_0(L-x)[x^2-(L/2)^2]}{2LT}-\frac{f_0(L-x)^2x}{2LT} \\ &=x+\frac{f_0(L-x)(L-2x)}{4T} \end{align}