A How Does Green's Function Solve Nonlinear Boundary Value Problems?

[Krikri]

I have this BVP $$u''+u' =f(x)-\lambda |u(x)| $$, $x\in [0,1]$ we BC $u(0)=u(1)=0$.
Following an ''algorithm'' for calculating the green's function I got something like $$g(x,t)=\Theta(x-t)(1+e^{t-x}) + \frac{e^{t}-e}{e-1} +\frac{e-e^{t}}{e-1}e^{-x}$$. At some point there is this integral $j(x)=- \int_{0}^{1}g(x,t)dt$ and since $j(0)=j(1)=0 , j'' + j'=-1$ which leads to $j(x)= \frac{e}{e-1} -x -\frac{e}{e-1}e^{-x}$
Can someone show me how it goes from defining the integral to find this form of $j(x)$. I mean for the conditions I see there is a straight connection with our initial form BC. For the second I understand that the differential operator acts on j(x) but why it gives -1 and how j(x) takes the final form?

 

[Delta2]

Which function is f(x)? if f(x) is a known function and g(x,t) is the Green's function then the solution is $u(x)=\int\limits_{D(f)} g(x,t)f(t)dt$.

 

[jasonRF]

Since the ode has the absolute value term, it isn't linear in general. So I wouldn't expect the solution to be a nice linear combination like the integral of a Green's function. Perhaps if $g$ and $u$ do not change sign at all over the interval it is fine?