## 2 Formulation and Solution of Nonlinear VFIDEs
: By multiplying the nonlinear equation  
\[Tu=Vu-g\Psi(u)=f,\]  
by inverse operator \(V^{-1}\), we obtain  
\[u=V^{-1}f+V^{-1}g\Psi(u). \tag{15}\]  
By applying vector \(\Psi\) on both sides, we have  
\[\Psi(u)=\Psi\Big{(}V^{-1}f+V^{-1}g\Psi(u)\Big{)}.\]  
We set \(\mathbf{c}=\Psi(u)\) and write  
\[\mathbf{c}=\Psi\Big{(}V^{-1}f+V^{-1}g\mathbf{c}\Big{)},\]  
which is a nonlinear algebraic (transcendental) system with vector \(\mathbf{c}\) unknown. Let \(\mathbf{c}^{*}\) be a solution of this system. By substituting \(\mathbf{c}^{*}\) into (15), we obtain the solution in (13). **3. Solution of Linear VFIDEs**  
In the case in which \(\varphi_{j}\Big{(}t,u(t),u^{\prime}(t),\ldots,u^{(n)}(t)\Big{)}\) is a linear function in \(u(t),u^{\prime}(t),\ldots,\)\(u^{(n)}(t)\), namely,  
\[\varphi_{j}\Big{(}t,u(t),u^{\prime}(t),\ldots,u^{(n)}(t)\Big{)}=\sum_{i=0}^{n }\varphi_{ji}(t)u^{(n-i)}(t),\quad j=1,2,\ldots,m, \tag{16}\]  
where \(\varphi_{ji}(t)\in X\), \(j=1,2,\ldots,m\), \(i=1,2,\ldots,n\), Equation (8) is simplified as  
\[\bar{K}u=\sum_{j=1}^{m}g_{j}(x)\int_{a}^{b}h_{j}(t)\sum_{i=0}^{n}\varphi_{ji}( t)u^{(n-i)}(t)dt,\]  
and Equation (9) is written as  
\[\Psi(u)=\left( \begin{array}{c}\Psi_{1}(u)\\ \Psi_{2}(u)\\ \vdots\\ \Psi_{m}(u)\end{array} \right),\quad\Psi_{j}(u)=\int_{a}^{b}h_{j}(t)\sum_{i=0}^ {n}\varphi_{ji}(t)u^{(n-i)}(t)dt,\]  
where now \(\Psi_{j}\), \(j=1,2,\ldots,m\), are linear bounded functionals defined on \(C^{n}[a,b]\) (see [35] (Theorem 3, p. 480)). We introduce the \(m\times m\) matrix  
\[\Psi(g)=\left( \begin{array}{ccc}\Psi_{1}(g_{1})&\cdots&\Psi_{1}(g_{m})\\ \vdots&\ddots&\vdots\\ \Psi_{m}(g_{1})&\cdots&\Psi_{m}(g_{m})\end{array} \right),\]  
where element \(\Psi_{i}(g_{j})\) is the value of functional \(\Psi_{i}\) on element \(g_{j}\). It is noted that for a \(m\times k\) constant matrix \(\mathbf{C}\),  
\[\Psi(g\mathbf{C})=\Psi(g)\mathbf{C}. \tag{17}\]  
Define the linear Volterra-Fredholm integro-differential operator \(S:X\to X\) as  
\[Su = Vu-g\Psi(u),\] \[\mathcal{D}(S) = \mathcal{D}(V)=\{u:u\in C^{n}[a,b],\ \Phi(u)-\mathbf{N}\Theta(u)= \mathbf{0}\},\]  
and write the linear boundary value problem (3), (4) and (16) in the symbolic form \[Su=f,\quad f\in X.\]  
We now give the following theorem, where we use the notation \(I_{m}\) to denote the \(m\times m\) identity matrix. Theorem 2: _If linear operator \(V\) is bijective on \(X\), then linear operator \(S\) is injective if and only if_  
\[\det\tilde{\mathbf{W}}=\det\Bigl{[}I_{m}-\Psi\Bigl{(}V^{-1}g\Bigr{)}\Bigr{]} \
eq 0,\]  
_and the unique solution of the boundary value problem_  
\[Su=f,\quad\text{for all}\quad f\in X, \tag{18}\]  
_is given by_  
\[u=V^{-1}f+V^{-1}g\tilde{\mathbf{W}}^{-1}\Psi\Bigl{(}V^{-1}f\Bigr{)}. \tag{19}\]  
Proof: Suppose that \(\det\tilde{\mathbf{W}}\
eq 0\) and \(u\in\ker S\). Then, \(Su=Vu-g\Psi(u)=0\); hence,  
\[u=V^{-1}g\Psi(u). \tag{20}\]  
By applying vector \(\Psi\) on both sides and using (17), we have  
\[\Psi(u)=\Psi\Bigl{(}V^{-1}g\Psi(u)\Bigr{)}=\Psi\Bigl{(}V^{-1}g\Bigr{)}\Psi(u),\]  
and thus  
\[\Bigl{[}I_{m}-\Psi\Bigl{(}V^{-1}g\Bigr{)}\Bigr{]}\Psi(u)=\tilde{\mathbf{W}} \Psi(u)=\mathbf{0},\]  
from where it follows that \(\Psi(u)=\mathbf{0}\). Then, from (20), it is implied that \(u=0\), i.e., \(\ker S\ =\ \{0\}\); therefore, operator \(S\) is injective. Conversely, we assume that \(S\) is injective, and we will show that \(\det\tilde{\mathbf{W}}\
eq 0\), or equivalently, we suppose that \(\det\tilde{\mathbf{W}}=0\), and we will show that \(S\) is not injective. Let \(\mathbf{c}\) be a nonzero constant vector \(\mathbf{c}\) such that \(\tilde{\mathbf{W}}\mathbf{c}=\mathbf{0}\), and let element \(v=V^{-1}g\mathbf{c}\in X\), which is different from zero; otherwise,  
\[Su = Vu-g\Psi(u),\]0  
which is a contradiction. Then,  
\[Su = Vu-g\Psi(u),\]1  
which means that \(\ker S\
eq\{0\}\); therefore, \(S\) is not injective.