# Iterative Matrix Techniques Based on Averages
## 5 Solution of a Fredholm Integral Equation of the First Kind by Means of Orthogonal Polynomials  
This section is devoted to solving integral equations of the Fredholm type of the first kind, given by the following expression:  
\[\int_{a}^{b}k(x,y)v(y)dy=u(x), \tag{3}\]  
assuming that \(u,v\in\mathcal{L}^{2}([a,b])\), \(k\in\mathcal{L}^{2}([a,b]\times[a,b])\), and the function \(u\) is known. The notation \(\mathcal{L}^{2}([a,b])\) means the space of square-integrable real functions defined on a compact real interval \([a,b]\). We must notice that the solution of this equation is not unique, since if \(v^{*}\) is a solution of (3) and \(h\) belongs to the null space of the operator \(K\) defined as follows:  
\[Kf(x):=\int_{a}^{b}k(x,y)f(y)dy,\]  
then \(v^{*}+h\) is a solution as well. The map \(k(x,y)\) is called the kernel of the Fredholm operator \(K\). The following are interesting particular cases of \(k\):  
* If \(k(x,y)=e^{-ixy}\), \(Kf\) is the Fourier Transform of \(f\). * For \(k(x,y)=1/(x-y)\), we have the Hilbert Transform of \(f\). * When \(k(x,y)=e^{-xy}\), the Laplace Transform of \(f\) is obtained. We consider in this section an arbitrary kernel \(k\in\mathcal{L}^{2}([a,b]\times[a,b])\). Assuming that \([a,b]=[-1,1]\), we consider the expansions of \(u\) and \(v\) in terms of an orthogonal basis of the space \(\mathcal{L}^{2}\) in order to find a solution of (3). In this case we have chosen the polynomials of Legendre, which compose an orthogonal basis of \(\mathcal{L}^{2}([-1,1])\). They are defined for \(n=0,1,\ldots\) as  
\[P_{n}(x)=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}^{2}(x+1)^{n-k}(x-1)^{k}.\]  
Legendre polynomials satisfy the following equalities:  
\[\int_{-1}^{1}P_{n}(x)P_{m}(x)dx= \begin{cases}0&n\
eq\,m\\ \frac{2}{2k+1}&n=\,m.\end{cases} \]  
Figure 2: Successive distances of the approximate solutions of a system. We have considered here orthonormal Legendre polynomials \(p_{n}\), that is to say,  
\[p_{n}(x):=P_{n}(x)/||P_{n}(x)||_{2}=\sqrt{\frac{2n+1}{2}}P_{n}(x),\]  
in order to facilitate the calculus of the coefficients of a function with respect to the basis. In this way, for \(f\in\mathcal{L}^{2}([-1,1])\),  
\[f(x)=\sum_{m=0}^{\infty}c_{m}p_{m}(x),\]  
where  
\[c_{m}=<f,p_{m}>=\int_{-1}^{1}f(x)p_{m}(x)dx.\]  
The convergence of the sum holds with respect to the norm \(||\cdot||_{2}\) in \(\mathcal{L}^{2}([-1,1])\). If the domain of the integral of the Equation (3) is a different interval, a simple change of variable transforms \((P_{m})\) into a system of orthogonal polynomials defined on it. To solve the Equation (3), let us consider Legendre expansions of the maps \(u,v\) up to the \(q\)-th term:  
\[u(x)\simeq\sum_{m=0}^{q}u_{m}p_{m}(x),\]  
\[v(x)\simeq\sum_{m=0}^{q}v_{m}p_{m}(x).\]  
The discretization of the integral Equation (3) becomes  
\[u(x)=\sum_{m=0}^{q}u_{m}p_{m}(x)=\int_{a}^{b}k(x,y)\left(\sum_{j=1}^{q}v_{j}p_ {j}(y)\right)dy=\sum_{j=0}^{q}v_{j}\int_{a}^{b}k(x,y)p_{j}(y)dy.\]  
Thus,  
\[u_{m}=<u,p_{m}>=\sum_{j=0}^{q}v_{j}<\int_{a}^{b}k(x,y)p_{j}(y)dy,p_{m}>=\sum_{ j=0}^{q}v_{j}\int_{a}^{b}\int_{a}^{b}k(x,y)p_{m}(x)p_{j}(y)dxdy.\]  
Denoting \(c_{mj}:=\int_{a}^{b}\int_{a}^{b}k(x,y)p_{m}(x)p_{j}(y)dxdy\), we obtain a system of linear equations whose unknowns are the coefficients \((v_{j})\) of the sought function \(v\) :  
\[u_{m}=\sum_{j=0}^{q}c_{mj}v_{j},\]  
for \(m,j=0,1,2,\ldots,q\). The linear system can be written as  
\[U=CV,\]  
where \(C=(c_{mj})\), \(U=(u_{0},u_{1},u_{2},\ldots,u_{q})^{T}\) and \(V=(v_{0},v_{1},v_{2},\ldots,v_{q})^{T}\). The system can be transformed into the fixed-point problem  
\[p_{n}(x):=P_{n}(x)/||P_{n}(x)||_{2}=\sqrt{\frac{2n+1}{2}}P_{n}(x),\]0  
as explained in the previous section. This equation can be solved using the described method whenever \(||I-C||\leq 1\) for some subordinate norm of the matrix \(I-C\). **Example 8**.