In the article : "Third order semipositone boundary value problems"
They assume that $w(t)$ is nondecreasing and $w(t)>0$ on $(q,1]$ .
and they found the Green's function $$G(t,s)=-t(p-s)\chi_{[0,p]}(s)+\frac{(t-s)^2}{2} \chi_{[0,t]}(s)+\frac{t(2p-1)}{2}W(s)\chi_{[q,1]}(s)+\frac{t(2p-t)}{2}\chi_{[0,q]}(s)$$
Where $$W(s)=(\int_0^1 w(v)dv)^{-1}\int w(v) dv, s\in [q,1]$$
we have that $G(t,s)>0 $ for $t,s \in (0,1)$
And what i dont unserstand is this :
For the function $e(t)$ given in Eq (1.1), define $$\gamma(t)=\int_0^1 G(t,s)e(s) ds, t\in [0,1]$$
Since $e\in L(0,1)$, we have that $\gamma \in C^2[0,1]\cap C^3(0,1)$ and moreover $\gamma(t)$ is the unique solution of the BVP consisting of the equation $$u'''=e(t), t\in (0,1)$$ and BC(1.2)