$$(*)X'=A(t)X - system$$ $$(*)PX(\alpha)+QX(\beta)=0.$$-border conditions, where P,Q constant square matrices $n \times n $. Let $Y(t)$ be the fundamental matrix for the system $(*)$ normed for$ t= \alpha$ aka: $Y(\alpha)=E$
$Definition:$ We call the $G(t,s)$ which acts from : $\Omega=\{(t,s): t \in [\alpha, \beta], s\in [\alpha, \beta], s \neq t\}$ to square matrices $n \times n$ Green's function for the border assignment $(*)$ if the following are satisfied: $1.) \forall t \in [\alpha, s] \cup [s, \beta] $: $\frac{dG}{dt} =A(t)G.$
$2.)\forall s \in [\alpha, \beta]: PG(\alpha, s)+ Q(\beta,s)=0$
$3.) G(s+0)-G(s-0)=E,E-identity\ matrix$\
Now this is the part where it becomes tough, the conclusions arrive from the above typed, could anyone explain the steps how these conclusions are made? Here they go (I don't understand from &&&&& on..)
From condition 1.) of definition, and the fact that the product of fundamental matrix $Y(t)$ and some regular constant matrix $n \times n$ is once again a fundamental matrix for the system, we have : &&&&&&&&&&&&&&&&&&&&&&& $$G(t,s)= \begin{cases} Y(t)S(s), \alpha \leq t <s \\ Y(t)T(s), s<t \leq \beta \end{cases}$$ where S and T, for now, unknown square matrices $ n \times n$ Using 2.) and 3.) (this I really dont get and would very much like to) $$PS+QY(\beta)T=0, \ \ \ Y(S-T)=-E$$ From here: (the following I get)$$S-T=-X^{-1}$$ but not: $$S(s)=-[P+QY(\beta)]^{-1}QY(\beta) Y^{-1}(s)\\ T(s)=\{E-[P+QY(\beta)]^{-1}QY(\beta)\}Y^{-1}(s)$$ From here: $$G(t,s)=\begin{cases} -Y(t)[P+QY(\beta)]^{-1}QY(\beta)Y^{-1}(s), \alpha \leq t < s \\ X(t)\{ E-[P-QY(\beta)]^{-1}Q Y(\beta) \}Y^{-1}(s), s<t \leq \beta \end{cases}$$ This I would love to understand aswell: From this it's not hard to see: (I cant see it !): $$\frac{dG}{ds}=GA(t), G(t, t-0)-G(t,t+0)=E$$
This is Greenes function, have a read more about it. There are many links about said on the internet, and the concept is quite straight forward. Best of luck.