Fibonacci Polynomials Method develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Fibonacci polynomials [1]Fibonacci polynomials is given as
4.1.1 Methodology
Definition 4.1. Consider the integral equation of the 1st kind is given as
where u (t) is the unknown function, to be determined, , k (x,t), the kernel, is a continuous or discontinuous and square integrable function f(x), being the known function.
Now we use the technique of Galerkin method, [Lewis, 2], to find an approximate solution ū(x) of Eq. (4.9). For this, we assume that
where 
are Fibonacci polynomials of degree k defined in equation and 
are unknown parameters, to be determined. Substituting (4.10) into (4.9), we get
Then the Galerkin equations are obtained by multiplying both sides of Eq. (4.11) by 
and then integrating with respect to x from a to b , we have
Since in each equation, there are two integrals. The inner integrand of the left side is a function of x, and t , and is integrated with respect to t from a to x . As a result the outer integrand becomes a function of x only and integration with respect to x from a to b yields a constant. Thus for each j= 0,1,2,…,n we have a linear equation with n+1 unknowns and k= 0,1,2,…,n
Finally Eq. (4.12) represents the system of n+1 linear equations in n+1 unknowns, are given by
Now the unknown parameters are determined by solving the system of equations Eq. (4.13) and substituting these values of parameters in Eq. (4.10), we get the approximate solution ū(x) of the integral equation Eq. (4.9).
Definition 4.2. Consider the integral equation of the 2nd kind is
where u(t) is the unknown function, to be determined,k(x,t) , the kernel, is a continuous or discontinuous and square integrable function,f(x), u(x) and being the known function and λ is a constant. Proceeding as before
Where
Now the unknown parameters are determined by solving the system of equations Eq. (4.15) and substituting these values of parameters in Eq. (4.10), we get the approximate solution ū(x) of the integral equation Eq. (4.12).
4.1.2 Applications of Fibonacci Polynomials Method
Example 4.3 Consider the Fredholm Integral equation [2]
The exact solution of Eq. (4.16) is
Consider 3rd order Fibonicca Polynomials, i.e. for n=3 we have Eq. (4.16) is
Substituting Eq. (4.18) into Eq. (4.16)
Now multiply Eq. (4.19) by 
and integrating both side from -1 to 1, we have
the matrix form of Eq. (4.20) is
After solving we get
Consequently we have the approximate solution is
TABLE 4.1 Comparison of the Exact Solution and Approximate Solutions of Eq. (4.16) for n=3 obtained from Fibonacci Polynomials Method
Fig 4.1 Comparison of Exact and Approximate Solutions of Eq. (4.16) for n=3.
Consider 5th order Fibonacci Polynomials, i.e. for we have Eq. (4.16) is
Substituting Eq. (4.22) into Eq. (4.16)
Now multiply Eq. (4.23) by and integrating both side from -1 to 1, we have
after solving we get
Consequently we have approximate solution as
TABLE 4.2 Comparison of the Exact and Approximate Solutions of Eq. (4.16) for n=5 obtained from Fibonacci Polynomials Method
Fig 4.2 Comparison of Exact and Approximate Solutions of Eq. (4.16) for n= 5
Consider 6th order Fibonacci Polynomials, i.e. for n= 6 we have Eq. (4.16) is
Substituting Eq. (4.26) into Eq. (4.16)
Now multiply Eq. (4.27) by 
and integrating both side from -1 to 1, we have
Now equation (4.28) can be written in the matrix form as,
After solving we get
Consequently we have exact solution as
TABLE 4.3 Comparison of the Exact Solution and Approximate Solutions of Eq. (4.16) for n=6 obtained from Fibonacci Polynomials Method
Fig 4.3 Comparison of Exact and Approximate Solutions of Eq. (4.16) for n=6
Example 4.4 Consider the Volterra Integral equation [2]
The exact solution of Eq. (4.30) is
u (x) = 6(x)
According to the proposed technique, consider the trail solution
Consider 1st order Fibonacci Polynomials, i.e. for n=1 we have Eq. (4.32) is
Substituting Eq. (4.32) into Eq. (4.29)
Now multiply Eq. (27) by 
,where j =0,1 and integrating both side from 0 to 1, we have
now by solving Eq. (4.35) we get,
1.5000
=4.000
after solving we get
=1.800.
Consequently we have the approximate solution is
u(x) = 3.6667
Table 4.4 Comparison of the Exact Solution and Approximate Solutions of Eq. (4.40) for n=1, obtained from Fibonacci Polynomials Method
Fig 4.4 Comparison of Exact and Approximate Solutions of Eq. (4.40) for n=1.
Consider 2nd order Fibonacci Polynomials, i.e. for n=2, we have Eq. (4.42) is
Substituting Eq. (4.37) into Eq. (4.28)
Now multiply Eq. (3.45) by where j=0,1,2 and integrating both side from 0 to 1, we have
the matrix form of Eq. (4.39) is
After solving we get
Consequently we have the exact solution is
u(x)=6x (4.40)
Table 4.5 Comparison of the Exact Solution and Approximate Solutions of Eq. (4.40) for n=2 obtained from Fibonacci Polynomials Method
Fig 4.5 Comparison of Exact and Approximate Solutions of Eq. (4.40) for n=3.
Example 4.5 Consider the Abel’s Integral equation [2]
The exact solution of Eq. (4.41) is
u(x) = sin(x+3) 
The transformation
According to the proposed technique, consider the trail solution
Consider 1st order Fibonacci Polynomials, i.e. for n=1 ,we have Eq. (4.44) is
Substituting Eq. (4.45) into Eq. (4.41)
Now multiply Eq. (4.46) by 
where j=0,1 and integrating both side from 0 to 1,
we have
Eq. (4.47) can be written as,
after solving we get 
consequently we have the approximate solution is
now by putting back into Eq. (4.43) we get
Which is approximate solution.
TABLE 4.6 Comparison of the Exact Solution and Approximate Solutions of Eq. (4.41) for n=1, obtained from Fibonacci Polynomials Method
Fig 4.6 Comparison of Exact and Approximate Solutions of Eq. (4.41) for n=1.
Consider 2nd order Fibonacci Polynomials, i.e. for n=2, we have Eq. (4.44) is
Substituting Eq. (4.49) into Eq. (4.44)
Now multiply Eq. (4.50) by where j=0,1 and integrating both side from 0 to 1, we have
the matrix form of Eq. (4.51) is
after solving we get 
Consequently we have solution is
This gives exact solution by
obtained by using Eq. (4.43).
TABLE 4.7 Comparison of the Exact Solution and Approximate Solutions of Eq.(4.41) for n=3, obtained from Fibonacci Polynomials Method
Fig 4.7 Comparison of Exact and Approximate Solutions of Eq. (4.41) for n=3.
References
[1]. Benjamin, Arthur T.; Quinn, Jennifer J. “9.4 Fibonacci and Lucas Polynomial”.Proofs that Really Count. MAA. . ISBN 0-88385-333-7. (141)(2003).
[2]. A.M. Wazwaz, Linear and Nonlinear Integral Equations Method and Applications, Springer Heidelberg Dordrecht London, New York, 2011.
[3]. A.M. Wazwaz, Partial Differential Equations and Solitary Waves Theory, HEP and Springer, Beijing and Berlin, (2009).
[4]. J.Y. Xiao, L. H. Wen, D. Zhang, Solving second kind integral equations by periodic wavelet Galerkin method, Appl. Math. Comput. 175(2006), 508-518.