# On the Existence of Solutions of Nonlinear Fredholm Integral Equations from Kantorovich's Technique
## 4 Applications
### Application 1  
We first consider the following nonlinear Fredholm integral equation,  
\[x(s)=\sin(\pi s)+\frac{1}{5}\int_{0}^{1}\cos(\pi s)\sin(\pi t)x(t)^{3}\,dt, \tag{10}\]  
with \(s\in[0,1]\), that has been used by other authors as a numerical test [13; 21]. Observe that, in this case, kernel \(\mathfrak{K}(s,t)=\cos(\pi s)\sin(\pi t)\) is separable. Firstly, we apply Theorem 4 to obtain domains of existence and uniqueness of a solution. For this, we observe that the corresponding function \(\mathfrak{F}(x)\) defined in (2) and associated with (10) is defined in \(\Omega=\mathcal{C}[0,1]\). We then observe that condition (6) is required in Theorem 4. However, if we pay attention to the integral equation, we observe that the kernel is separable and we can then determine the corresponding operator \([\mathfrak{F}^{\prime}(x)]^{-1}\). For this, we write \([\mathfrak{F}^{\prime}(x)y](s)=z(s)\), so, if there exists \([\mathfrak{F}^{\prime}(x)]^{-1}\), we have  
\[[\mathfrak{F}^{\prime}(x)]^{-1}z(s)=y(s)=z(s)+\cos(\pi s)\left(\frac{3}{5}\int _{0}^{1}x(t)^{2}\sin(\pi t)y(t)dt\right).\]  
If we now denote \(\frac{3}{5}\int_{0}^{1}x(t)^{2}\sin(\pi t)y(t)dt=\mathcal{I}\), multiply next-to-last equality by \(\frac{3}{5}x(s)^{2}\sin(\pi s)\) and integrate it between \(0\) and \(1\), we obtain  
\[\mathcal{I}=\frac{\int_{0}^{1}x(s)^{2}\sin(\pi s)z(s)ds}{\frac{5}{3}-\int_{0}^ {1}x(s)^{2}\sin(\pi s)\cos(\pi s)ds}\]provided that  
\[\int_{0}^{1}x(s)^{2}\sin(\pi s)\cos(\pi s)ds\
eq\frac{5}{3}.\]  
Therefore,  
\[y(s)=[\mathfrak{F}^{\prime}(x)]^{-1}z(s)=z(s)+\frac{3}{5}\cos(\pi s)\frac{\int_{ 0}^{1}\sin(\pi t)x(t)^{2}z(t)dt}{1-\frac{3}{5}\int_{0}^{1}\sin(\pi t)\cos(\pi t )x(t)^{2}dt}.\]  
Now, as a consequence of the last formula, condition (6), that is required to prove the existence of the inverse operator \([\mathfrak{F}^{\prime}(x_{0})]^{-1}\), can be omitted, provided that  
\[\int_{0}^{1}\sin(\pi t)\cos(\pi t)x_{0}(t)^{2}dt\
eq\frac{5}{3}.\]  
Therefore, it is sufficient to choose some starting point \(x_{0}(s)\) for [PERSON]'s method such that the previous inequality holds. As \(x_{0}(s)=\sin(\pi s)\) is a reasonable choice as a starting point for [PERSON]'s method, as we can see in [12; 13; 14], the last inequality holds, since \(\int_{0}^{1}\sin(\pi t)\cos(\pi t)x_{0}(t)^{2}dt=0\), and condition (6) is omitted. After that, taking into account that \(\varpi=\frac{1}{5}\), \(S=\frac{2}{\pi}\), \(p=3\), \(\|x_{0}\|=1\) and \(\|F(x_{0})\|=\frac{3}{40}\), we construct the auxiliary scalar function  
\[\phi(t)=\frac{1}{40\pi}(16t^{3}+48t^{2}+8(6-5\pi)t+3\pi)\]  
and see that it has two positive real zeros \(r=0.1327\) and \(R=1.0589\). Therefore, the domains of existence and uniqueness of a solution of Equation (10) are respectively  
\[\left\{h\in\mathcal{C}[0,1]:\|h(s)-\sin(\pi s)\|\leq r=0.1327\right\},\]  
\[\left\{h\in\mathcal{C}[0,1]:\|h(s)-\sin(\pi s)\|<R=1.0588\right\}.\]  
On the other hand, we can write the function \(\phi(t)\) in the following way  
\[\phi(t)=(r-t)(R-t)\ell(t),\quad\ell(t)=\frac{2}{5\pi}(t+(4.19156))\]  
and obtain a priori error estimates from [PERSON]'s technique [19], that allow us to determine the number of iterations that we have to apply in [PERSON]'s method to reach a previously fixed precision. For this, we write \(\alpha_{n}=r-t_{n}\) and \(\gamma_{n}=R-t_{n}\), for all \(n\geq 0\).