A Green's Function ODE BVP

1. Jan 3, 2018

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?

2. Jan 8, 2018

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jan 9, 2018

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 $u(x)=\int\limits_{D(f)} g(x,t)f(t)dt$.

4. Jan 9, 2018

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?