## 3 Solving VFIE of Convolution Type  
The method described in the previous section requires the existence and construction in closed form of the inverse operator \(\mathcal{J}^{-1}\) of the linear Volterra operator of the second kind \(\mathcal{J}\). Finding the exact inverse is not an easy task, and in many cases it is not possible. [PERSON] and [PERSON] [104] provide the closed-form solution for several classes of linear Volterra integral equations (VIE) of the second kind. Here, we distinguish and deal with the special case of the VIE of convolution type that can be explicitly solved by applying the Laplace transform method. It is well known that the Laplace transform of a function \(f(x)\) defined for all \(x\geq 0\) is a function \(F(s)\) of the variable \(s\) defined by  
\[F(s)=\mathcal{L}(f)=\int_{0}^{\infty}e^{-sx}f(x)dx,\]  
where \(\mathcal{L}\) denotes the Laplace transform operator, which is linear and injective. Suppose \(f(x)\) is continuous and of exponential order; that is, there exist real constants \(\gamma\) and \(M\) such that \(|f(x)|\leq Me^{\gamma x}\) for all \(x\geq 0\). Then the Laplace transform of \(f\) exists for all \(s>\gamma\). The inverse of the Laplace transform is \(f(x)=\mathcal{L}^{-1}(F)\). The convolution of two functions of exponential order \(f,g\) is defined by  
\[(f*g)(x)=\int_{0}^{x}f(x-t)g(t)dt,\]  
and \(\mathcal{L}(f*g)=\mathcal{L}(f)\mathcal{L}(g)\) (see for example [109; 110]). Let the linear Volterra integral operator \(\mathcal{K}\) in (3) be of convolution type, i.e.,  
\[\mathcal{K}y(x)=\sum_{i=1}^{n}\int_{0}^{x}k_{i}(x-t)y(t)dt=\sum_{i=1}^{n}(k_{i} *y)(x), \tag{23}\]and let the convolution-type Volterra integral equation (CVIE) of the second kind  
\[\mathcal{J}y(x)=y(x)-\mathcal{K}y(x)=y(x)-\sum_{i=1}^{n}(k_{i}*y)(x)=r(x),\quad r \in X. \tag{24}\]  
By applying the Laplace transform operator on both sides of (24) and utilizing the convolution theorem, we get  
\[\mathcal{L}\{\mathcal{J}y\} = \mathcal{L}\Bigg{\{}y-\sum_{i=1}^{n}(k_{i}*y)\Bigg{\}}\] \[= Y(s)-Y(s)\sum_{i=1}^{n}K_{i}(s)=R(s),\]  
where \(Y(s)=\mathcal{L}\{y\}\), \(K_{i}(s)=\mathcal{L}\{k_{i}\}\), \(i=1,2,\ldots,m\), and \(R(s)=\mathcal{L}\{r\}\). Multiplication by the transfer function  
\[Q(s)=\frac{1}{1-(K_{1}(s)+K_{2}(s)+\cdots+K_{n}(s))},\]  
yields  
\[Y(s)=R(s)Q(s).\]  
Taking the inverse Laplace transform, we obtain the solution of (24), namely  
\[y=\mathcal{J}^{-1}r=\mathcal{L}^{-1}\{R(s)Q(s)\}. \tag{25}\]  
Since the Equation (25) holds for every \(r(x)\in X\), it also implies that the operator \(\mathcal{J}\) is bijective. To summarize and facilitate the implementation of the methods presented in the previous section and here, we provide the following algorithm in Algorithm 1. ```
input \(k_{1}(x),k_{2}(x),\ldots,k_{n}(x)\), \(g_{1}(x),g_{2}(x),\ldots,g_{m}(x)\), \(h_{1}(x),h_{2}(x),\ldots,h_{m}(x)\), \(b\), \(f(x)\) compute \(K_{i}(s):=\mathcal{L}\{k_{i}\},\quad i=1,2,\ldots,n\) \(Q(s):=\dfrac{1}{1-(K_{1}(s)+K_{2}(s)+\cdots+K_{n}(s))}\) \(G_{j}(s):=\mathcal{L}\big{\{}g_{j}\big{\}},\quad j=1,2,\ldots,m\) \(\mathcal{J}^{-1}g_{j}(x):=\mathcal{L}^{-1}\big{\{}G_{j}(s)Q(s)\big{\}},\quad j= 1,2,\ldots,m\) \(\mathcal{J}^{-1}g:=\big{(}\ \mathcal{J}^{-1}g_{1}(x)\quad\cdots\quad\mathcal{J}^{-1}g_{m}(x)\ \ \big{)}\) \(F(s):=\mathcal{L}\{f\}\) \(\mathcal{J}^{-1}f:=\mathcal{L}^{-1}\{F(s)Q(s)\}\) linear case \(\big{(}\tilde{\varphi}_{j}(t,y(t))\equiv y(t)\big{)}\): \(\Phi_{i}\big{(}\mathcal{J}^{-1}g_{j}\big{)}:=\int_{a}^{b}h_{i}(t)\mathcal{J}^ {-1}g_{j}(t)dt,\quad i,j=1,2,\ldots,m\) \(\[PERSON]\big{(}\mathcal{J}^{-1}g\big{)}:=\left( \begin{array}{ccc}\[PERSON]_{1}\big{(} \mathcal{J}^{-1}g_{1}\big{)}&\cdots&\[PERSON]_{1}\big{(}\mathcal{J}^{-1}g_{m}\big{)} \\ \vdots&\ddots&\vdots\\ \[PERSON]_{m}\big{(}\mathcal{J}^{-1}g_{1}\big{)}&\cdots&\[PERSON]_{m}\big{(}\mathcal{J}^ {-1}g_{m}\big{)}\end{array}