Charlier Polynomials Method, we develop a new algorithm for solving linear and nonlinear integral equations using Galerkin weighted residual numerical method with Charlier’s polynomials [1] is given as
3.7.1 Methodology
Definition 3.5. 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 
.For this, we assume that
where 
are Charlier polynomials of degree defined in equation (3.137) and 
are unknown parameters, to be determined. Substituting (3.139) into (3.138)
we get,
Then the Galerkin equations are obtained by multiplying both sides of (3.140) 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 , 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 ,k=0,1,2,…,n.
Finally (3.141) represents the system of n+1 linear equations in n+1 unknowns, are given by
where
Now the unknown parameters are determined by solving the system of equations (3.142) and substituting these values of parameters in (3.139), we get the approximate solution of the integral equation (3.138).
Definition 3.6. 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) and u(x) 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 (3.143) and substituting these values of parameters in (3.139), we get the approximate solution of the integral equation (3.144).
3.7.2 Applications of Charlier Polynomials Method
Example 3.28 Consider the Volterra Integral equation of the first kind [2]
The exact solution of Eq. (3.145)
According to the proposed technique, consider the trail solution
Consider 1st order Charlier’s Polynomials, i.e. for n=1, we have Eq. (3.147) is
Substituting Eq. (3.148) into Eq. (3.145)
Now multiply Eq. (3.149) by ,j=0,1 and integrating both side from 0 to 1
if β=1,then the matrix form of Eq. (3.150) is
After solving we get 
and 
Consequently we have the approximate solution
Table 3.19 Comparison of the Exact Solution and Approximate Solutions of Eq. (3.145) for n=1, using Charlier’s Polynomials Method
Fig 3.19 Comparison of exact and approximate Solutions of Eq. (3.145) for n=1.
Consider 2nd order Charlier’s Polynomials, i.e. for n=2, we have Eq. (3.147) is
Substituting Eq. (3.151) into Eq. (3.145)
Now multiply Eq. (3.152) by ,j=0,1,,2 and integrating both side from 0 to 1, we have
If β=1,then the matrix form of Eq. (3.153) is
After solving we get 
and 
Consequently we have the exact solution is
u(x) = x + x²
Fig 3.20 Comparison of Exact and Approximate Solutions of Eq. (3.145) for n=2.
Table 3.20 Comparison of the Exact Solution and Approximate Solutions of Eq. (3.145) for n=2 using Charlier’s Polynomials Method
Example 3.29 Consider the linear Fredholm Integral equation [2] as
The exact solution of Eq. (3.154) is
According to the proposed technique, consider the trail solution
Consider 5th order Charlier’s Polynomials, i.e. for n=5, we have Eq. (3.156) is
Substituting Eq. (3.157) into Eq. (3.154)
Now multiply Eq. (3.158) by where j=0,1,2,3,4,5 and integrating both side from 0 to π/4, we have
If β=1,then the matrix form of Eq. (3.159) is
After solving we get,
and 
Consequently we have the approximate solution is
Fig 3.21 Comparison of Exact and Approximate Solutions of Eq. (3.154) for n=5.
Table 3.21 Comparison of the Exact Solution and Approximate Solutions of Eq. (3.154) for obtained using Charlier’s Polynomials Method
Consider 10th order Charlier’s Polynomials, i.e. for n=10 we have Eq. (3.156) is
Substituting Eq. (3.160) into Eq. (3.154)
Now multiply Eq. (3.161) by where j=0,1,2,…,10 and integrating both side from 0 to π/4, we have
If β=1,then the matrix form of Eq. (3.162) is
After solving we get 
Consequently we have the exact solution is
Fig 3.22 Comparison of Exact and Approximate Solutions of Eq. (3.154) for n=10.
Table 3.22 Comparison of the Exact Solution and Approximate Solutions of Eq. (3.154) for n=5, using Charlier Polynomials Method
Example 3.30 Consider the linear Fredholm Integral equation [2]
The exact solution of Eq. (3.163) is
According to the proposed technique, consider the trail solution
Consider 2nd order Charlier’s Polynomials, i.e. for n=2, we have Eq. (3.165) is
or
Substituting Eq. (3.166) into Eq. (3.163)
Now multiply Eq. (3.167) by 
j=0,1,2 and integrating both side from -1 to 1,we have
If β=1,then the matrix form of Eq. (3.168) is
after solving we get,
Consequently we have the approximate solution is
Table 3.23 Comparison of Exact and Approximate Solutions of Eq. (3.163) for obtained from Charlier’s Polynomials Method
Fig 3.23 Comparison of Exact and Approximate Solutions of Eq. (3.163) for n=3.
Consider 5th order Charlier’s Polynomials, i.e. for n=5, we have Eq. (3.165) is
or
substituting Eq. (3.170) into Eq. (3.163)
Now multiply Eq. (3.171) by j=0,1,2,3,4,5 and integrating both side from -1 to 1, we have the matrix form of Eq. (5.523) is
after solving we get,
Consequently we have the approximate solution is
Table 3.24 Comparison of the Exact Solution and Approximate Solutions of Eq. (3.163) for obtained using Charlier’s Polynomials Method
Fig 3.24 Comparison of Exact and Approximate Solutions of Eq. (3.163) for n=5.
Example 3.31 Consider the weakly-singular integral equation [2]
The exact solution of Eq. (3.174)
According to the proposed technique, consider the trail solution
Consider 1st order Charlier’s Polynomials, i.e. for n=1, we have Eq. (3.175) is
Substituting Eq. (3.176) into Eq. (3.174)
Now multiply Eq. (33) by 
j=0,1 and integrating both side from 0 to 1, we have
if β=1,then the matrix form of Eq. (3.178) is
After solving we get
Consequently we have the approximate solution is
Fig 3.25 Comparison of Exact and Approximate Solutions of Eq. (3.174) for n=1.
Table 3.25 Comparison of the Exact Solution and Approximate Solutions of Eq.(3.174) for using Charlier Polynomials Method
Consider 2nd order Charlier’s Polynomials, i.e. for n=2,we have Eq. (3.175) is
Substituting Eq. (3.179) into Eq. (3.174)
Now multiply Eq. (3.180) by 
j=0,1,2 and integrating both side from 0 to 1, we have
if β=1,then the matrix form of Eq. (3.181) is
After solving we get
Consequently we have the exact solution is
u(x) = 1+ x²
Fig 3.26 Comparison of Approximate and Exact Solutions of Eq. (3.174) for n=3.
Table 3.26 Comparison of the Exact Solution and Approximate Solutions of Eq. (3.174)
for using Charlier’s Polynomials Method
References
[1]T.M. Dunster, Uniform asymptotic expansions for Charlier polynomials, Journal of approximation theory 112(2001), 93–133.
[2]A.M.Wazwaz, Linear and Nonlinear Integral Equations Method and Applications, Springer Heidelberg Dordrecht London, New York, 2011.