## 3 Main Results
_If_ \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]2 (10)_then the operator \(B\) can be factorized into \(Bu=B_{2}B_{1}u\) where the two linear integro-differential operators \(B_{1},B_{2}:X\to X\) are defined by_  
\[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]3 \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]4 (12) _where_ \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]5 (13) _(ii) If there exists a vector \(P=(p_{1},\ldots,p_{m})\in[D(A_{2})]^{m}\) satisfying the equation_ \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]6 (14) _then the operator \(B\) is bijective if and only if_ \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]7 (15) _In this case, the unique solution to the boundary value problem_ \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]8 (16) _is given by_ \[A_{2}P-G\Psi(A_{2}P)=S, \tag{5}\]9 (17) \[Bu=B_{2}B_{1}u,\]0 \[Bu=B_{2}B_{1}u,\]1  
Proof. : (i) If Equation (10) holds, then from Lemma 2 it is implied that the operator \(B_{2}\) defined in (12) is bijective and that the unique solution of the boundary value problem \(B_{2}P=A_{2}P-G\Psi(A_{2}P)=S\) is given in (13). Then, from Lemma 1 it follows that the operator \(B\) can be decomposed into \(Bu=B_{2}B_{1}u\) where the operators \(B_{1}\) and \(B_{2}\) are given in (11) and (12), respectively. (ii) From Lemma (1) it is deduced that \(Bu=B_{2}B_{1}u\) with \(B_{1}\) and \(B_{2}\) given by (11) and (12), respectively, and \(B_{2}P=S\). Substituting into (8), we have  
\[Bu=B_{2}B_{1}u,\]2 \[Bu=B_{2}B_{1}u,\]3 \[= Au-\big{(} \begin{array}{cc}B_{2}P&G\end{array} \big{)}\bigg{(} \begin{array}{c}\Phi(A_{1}u)\\ \Psi(Au)\end{array} \bigg{)}\] \[= Au-\big{(} \begin{array}{cc}B_{2}P&G\end{array} \big{)}\bigg{(} \begin{array}{c}\Phi(A_{2}^{-1}A_{2}A_{1}u)\\ \Psi(Au)\end{array} \bigg{)}\] \[= Au-\big{(} \begin{array}{cc}B_{2}P&G\end{array} \big{)}\bigg{(} \begin{array}{c}\[PERSON](A_{2}^{-1}Au)\\ \Psi(Au)\end{array} \bigg{)}\] \[= Au-\hat{G}\hat{\Psi}(Au),\]  
where  
\[\hat{G}=\big{(} \begin{array}{cc}B_{2}P&G\end{array} \big{)},\quad\hat{\Psi} (Au)=\bigg{(} \begin{array}{c}\[PERSON](A_{2}^{-1}Au)\\ \Psi(Au)\end{array} \bigg{)}.\]By applying Theorem 1, the operator \(B\) is bijective if and only if  
\[\det\hat{W} = \det\bigl{[}I_{2m}-\hat{\Psi}(\hat{G})\bigr{]}\] \[= \det\biggl{[}\left[ \begin{array}{cc}I_{m}&0_{m}\\ 0_{m}&I_{m}\end{array} \right]-\left[ \begin{array}{cc}\[PERSON](A_{2}^{-1}B_{2}P)& \[PERSON](A_{2}^{-1}G)\\ \[PERSON](B_{2}P)&\Psi(G)\end{array} \right]\biggr{]}\] \[= \det\biggl{[} \begin{array}{cc}I_{m}-\[PERSON](A_{2}^{-1}B_{2}P)&- \[PERSON](A_{2}^{-1}G)\\ -\[PERSON](B_{2}P)&I_{m}-\[PERSON](G)\end{array} \biggr{]}\] \[= \det\biggl{[} \begin{array}{cc}I_{m}-\[PERSON](P)+\[PERSON](A_{2}^{-1}G) \[PERSON](A_{2}P)&-\[PERSON](A_{2}^{-1}G)\\ -\[PERSON](A_{2}P)+\[PERSON](G)\[PERSON](A_{2}P)&I_{m}-\[PERSON](G)\end{array} \biggr{]}\
eq 0.\]  
Multiplying from the right the elements of the second column by \(\[PERSON](A_{2}P)\) and adding to the corresponding elements of the first column, we obtain  
\[\det\hat{W} = \det\biggl{[} \begin{array}{cc}I_{m}-\Phi(P)&-\Phi(A_{2}^{-1}G) \\ 0_{m}&I_{m}-\Psi(G)\end{array} \biggr{]}\] \[= \det V\det W\
eq 0.\]  
So we proved that the operator \(B\) is bijective if and only if the two conditions in (15) are fulfilled. By means of Lemma 1 the boundary value problem (16) degenerates to \(B_{2}B_{1}u=f\) where the operators \(B_{1}\) and \(B_{2}\) are given in (11) and (12), respectively. We set \(B_{1}u=v\) and factor the problem to two simpler boundary value problems  
\[B_{1}u = v,\quad D(B_{1})=D(A_{1}), \tag{18}\] \[B_{2}v = f,\quad D(B_{2})=D(A_{2}). \tag{19}\]  
From Lemma 2 it follows that both operators \(B_{1}\) and \(B_{2}\) are bijective.