Mathematical Methods for Physics and Engineering - Matematica.NET

23.5 NEUMANN SERIES23.5 Neumann seriesAs mentioned above, most integral equations met in practice will not be of thesimple forms discussed in the last section and so, in general, it is not possible tofind closed-form solutions. In such cases, we might try to obtain a solution in theform of an infinite series, as we did for differential equations (see chapter 16).Let us consider the equationy(x) =f(x)+λ∫ baK(x, z)y(z) dz, (23.34)where either both integration limits are constants (for a Fredholm equation) orthe upper limit is variable (for a Volterra equation). Clearly, if λ were small thena crude (but reasonable) approximation to the solution would bey(x) ≈ y 0 (x) =f(x),where y 0 (x) stands for our ‘zeroth-order’ approximation to the solution (and isnot to be confused with an eigenfunction).Substituting this crude guess under the integral sign in the original equation,we obtain what should be a better approximation:y 1 (x) =f(x)+λ∫ baK(x, z)y 0 (z) dz = f(x)+λ∫ baK(x, z)f(z) dz,which is first order in λ. Repeating the procedure once more results in thesecond-order approximationy 2 (x) =f(x)+λ= f(x)+λ∫ ba∫ baK(x, z)y 1 (z) dz∫ b ∫ bK(x, z 1 )f(z 1 ) dz 1 + λ 2 dz 1 K(x, z 1 )K(z 1 ,z 2 )f(z 2 ) dz 2 .It is clear that we may continue this process to obtain progressively higher-orderapproximations to the solution. Introducing the functionsK 1 (x, z) =K(x, z),K 2 (x, z) =K 3 (x, z) =∫ ba∫ bK(x, z 1 )K(z 1 ,z) dz 1 ,and so on, which obey the recurrence relationaK n (x, z) =a∫ bdz 1 K(x, z 1 )K(z 1 ,z 2 )K(z 2 ,z) dz 2 ,a∫ baK(x, z 1 )K n−1 (z 1 ,z) dz 1 ,813a