## 2 Direct Operator Method for Solving Linear and Nonlinear VFIE
### Linear VFIE  
First, we consider the case of linear VFIE where without loss of generality we assume that \(\bar{\varphi}_{j}(t,y(t))\equiv y(t)\), \(j=1,2,\ldots,m\). In this case, the Fredholm type functionals \(\Phi_{j}(y)\) in (6) become  
\[\Phi_{j}(y)=\int_{0}^{b}h_{j}(t)y(t)dt,\quad j=1,2,\ldots,m, \tag{7}\]  
which are bounded linear functionals, and the vector \(\Phi\in[X^{*}]^{m}\) where \(X^{*}\) is the space of all bounded linear functionals on \(X\). Moreover, for a vector of functions \(g\), such as in (5), and a constant vector \(\mathbf{c}\), the following relation holds  
\[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]  
**Theorem 1** (Linear VFIE). : _Let the linear Volterra-Fredholm integral operator of the second kind \(\mathcal{I}:X\to X\) be defined by_  
\[\mathcal{I}y=\mathcal{J}y-g\Phi(y),\quad y=y(x)\in X, \tag{9}\]  
_where the Volterra-type integral operator of the second kind \(\mathcal{J}:X\to X\) is defined as_  
\[\mathcal{J}y=y-\mathcal{K}y,\quad y=y(x)\in X, \tag{10}\]  
_and the operator \(\mathcal{K}\) as in (3), the vector of functions \(g\) as in (5) and \(\Phi\) is a vector of linear bounded functionals of the kind (7). If the integral operator \(\mathcal{J}\) is bijective on \(X\), and its inverse is \(\mathcal{J}^{-1}\), then the operator \(\mathcal{I}\) is injective if and only if_  
\[\det W=\det\Bigl{[}I_{m}-\Phi\Bigl{(}\mathcal{J}^{-1}g\Bigr{)}\Bigr{]}\
eq 0. \tag{11}\]  
_Furthermore, the unique solution of the linear VFIE_  
\[\mathcal{I}y=f,\quad\text{for all}\quad f=f(x)\in X, \tag{12}\]  
_is given by_  
\[y=\mathcal{I}^{-1}f=\mathcal{J}^{-1}f+\mathcal{J}^{-1}g\mathcal{W}^{-1}\Phi \Bigl{(}\mathcal{J}^{-1}f\Bigr{)}. \tag{13}\]  
Proof. : Let us assume that \(\det W\
eq 0\) and take an element \(y\in\ker\mathcal{I}\). Then  
\[\mathcal{I}y=\mathcal{J}y-g\Phi(y)=0,\]  
and since the operator \(\mathcal{J}\) is bijective, we have  
\[y=\mathcal{J}^{-1}g\Phi(y). \tag{14}\]  
Acting by the vector \(\Phi\) on both sides of (14) and by using (8), we get  
\[\Phi(y)=\Phi\Bigl{(}\mathcal{J}^{-1}g\Phi(y)\Bigr{)}=\Phi\Bigl{(}\mathcal{J}^ {-1}g\Bigr{)}\Phi(y),\]  
and hence  
\[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]0  
Since \(\det W\
eq 0\), we have \(\Phi(y)=\mathbf{0}\). Then, from (14), it follows that \(y=0\), i.e., \(\ker\mathcal{I}=\{0\}\), which implies that the operator \(\mathcal{I}\) is injective. Conversely, we assume \(\mathcal{I}\) is injective, and we will show that \(\det W\
eq 0\), or, equivalently, we assume \(\det W=0\), and wewill prove that \(\mathcal{I}\) is not injective. In this case, there exists a nonzero constant vector \(\mathbf{c}\) such that \(W\mathbf{c}=\mathbf{0}\). Let the element \(y_{0}=\mathcal{J}^{-1}g\mathbf{c}\in X\). Note that \(y_{0}\
eq 0\) since \(y_{0}=0\), we have  
\[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]1  
which contradicts the assumption that \(\mathbf{c}\) is nonzero. Then  
\[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]2 \[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]3 \[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]4 \[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]5 \[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]6  
This means that \(\ker\mathcal{I}\
eq\{0\}\), and, therefore, \(\mathcal{I}\) is not injective. Let now the VFIE in (12), viz. \[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]7  
By applyingthe operator \(\mathcal{J}^{-1}\) on (15), we get  
\[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]8  
Acting as aboveby the vector \(\Phi\), we have  
\[\Phi(g\mathbf{c})=\Phi(g)\mathbf{c}. \tag{8}\]9  
and hence  
\[\mathcal{I}y=\mathcal{J}y-g\Phi(y),\quad y=y(x)\in X, \tag{9}\]0 \[\mathcal{I}y=\mathcal{J}y-g\Phi(y),\quad y=y(x)\in X, \tag{9}\]1  
Substitution of (17) into (16) yields  
\[\mathcal{I}y=\mathcal{J}y-g\Phi(y),\quad y=y(x)\in X, \tag{9}\]2  
which is the solution of the linear VFIE in (12) for every \(f\in X\).