How to transform this periodic boundary value problem to an integral equation? While reading this paper, I find this passage interesting, but I dont know how to prove it exactly(I know that It uses Green's function..)

"We study the existence of a solution of the following periodic system:
\begin{align}
& u^{\prime}+\lambda_{1} u-\lambda_{2} v=f(t, u)+g(t, v)+\lambda_{1} u-\lambda_{2} v \\
& v^{\prime}+\lambda_{1} v-\lambda_{2} u=f(t, v)+g(t, u)+\lambda_{1} v-\lambda_{2} u
\end{align}
together with the periodicity conditions,
$$
u(0)=u(T) \quad \text { and } \quad v(0)=v(T)
$$
This problem is equivalent to the integral equations:
$$
\begin{align}
u(t)=\int_0^T & G_1(t, s)\left[f(s, u)+g(s, v)+\lambda_1 u-\lambda_2 v\right] \\
& {}+G_2(t, s)\left[f(s, v)+g(s, u)+\lambda_1 v-\lambda_2 u\right] \mathrm{d} s \\
v(t)=\int_0^T & G_{1}(t, s)\left[f(s, v)+g(s, u)+\lambda_1 v-\lambda_2 u\right] \\
& {}+G_2(t, s)\left[f(s, u)+g(s, v)+\lambda_1 u-\lambda_2 v\right] \mathrm{d} s
\end{align}
$$
where $$G_{1}(t, s)=\left\{\begin{array}{ll}
\frac{1}{2}\left[\frac{e^{\sigma_{1}(t-s)}}{1-e^{\sigma_{1} T}}+\frac{e^{\sigma_{2}(t-s)}}{1-e^{\sigma_{2} T}}\right] & 0 \leq s<t \leq T \\
\frac{1}{2}\left[\frac{e^{\sigma_{1}(t+T-s)}}{1-e^{\sigma_{1} T}} + \frac{e^{\sigma_{2}(t+T-s)}}{1-e^{\sigma_{2} T}}\right] & 0 \leq t<s \leq T
\end{array}\right.$$
and
$$G_{2}(t, s)=\left\{\begin{array}{ll}
\frac{1}{2}\left[\frac{e^{\sigma_{2}(t-s)}}{1-e^{\sigma_{2} T}}-\frac{e^{\sigma_{1}(t-s)}}{1-e^{\sigma_{1} T}}\right] & 0 \leq s<t \leq T \\
\frac{1}{2}\left[\frac{e^{\sigma_{2}(t+T-s)}}{1-e^{\sigma_{2} T}}-\frac{e^{\sigma_{1}(t+T-s)}}{1-e^{\sigma_{1} T}}\right] & 0 \leq t<s \leq T
\end{array}\right.$$
Here, $ \sigma_{1}=-\left(\lambda_{1}+\lambda_{2}\right)$  and $ \sigma_{2}=\left(\lambda_{2}-\lambda_{1}\right) .$"

 A: $\def\d{\delta}
\def\l{\lambda} 
\def\s{\sigma}$We wish to solve
\begin{align*}
u'+\l_1 u-\l_2 v &= f_1(t,u,v) \\ 
v'+\l_1 v -\l_2 u &= f_2(t,u,v).
\end{align*}
This is equivalent to solving
\begin{align*}
a'-\s_2 a &= f_1+f_2 \\
b'-\s_1 b &= f_1-f_2,
\end{align*}
where
\begin{align*}
a &= u+v \\
b &= u-v.
\end{align*}
Note that $a(T)=a(0)$ and $b(T)=b(0)$.
We have decoupled the original system and solve
\begin{align*}
G_a'-\s_2 G_a &= \d(t-s) \\
G_b'-\s_1 G_b &= \d(t-s)
\end{align*}
with the appropriate boundary conditions.
It is a straightforward exercise to show that
\begin{align*}
G_a(t,s) &= \begin{cases}
\displaystyle \frac{e^{\s_2(t-s)}}{1-e^{\s_2 T}}, 
    & 0\le s<t\le T \\
\displaystyle \frac{e^{\s_2(t+T-s)}}{1-e^{\s_2 T}}, 
    & 0\le t<s\le T.
\end{cases}
\end{align*}
A similar result holds for $G_b$.
Thus,
\begin{align*}
a &= \int(f_1+f_2)G_a ds \\
b &= \int(f_1-f_2)G_b ds
\end{align*}
and so
\begin{align*}
u &= \frac12(a+b) \\ 
&= \int\left[ 
\frac12 (f_1+f_2)G_a + \frac12(f_1-f_2)G_b\right]ds \\
&= \int\left[\frac12(G_a+G_b)f_1 + \frac12(G_a-G_b)f_2\right]ds. 
\end{align*}
A similar result holds for $v$.
Lastly, one can verify that
\begin{align*}
G_1&=\frac12(G_a+G_b) \\
G_2&=\frac12(G_a-G_b).
\end{align*}
