Can a Neumann Series Be Proposed for the Resolvent Kernel in Iterative Methods?

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SUMMARY

The discussion centers on the proposal of a Neumann series for the resolvent kernel in the context of solving the integral equation f(x)=λ ∫₀¹ K(x,y)f(y) dy, where λ is a small eigenvalue and K is a known kernel. The iterative scheme f_{n+1}(x)=λ ∫₀¹ K(x,y)f_n(y) dy is suggested for solving the equation. The main inquiry is whether the Neumann series R(x,y) = ∑ₙ₌₀^∞ λⁿ Kⁿ can be proposed for the resolvent kernel, given that the equation is homogeneous and all functions except the kernel are unknown.

PREREQUISITES
  • Understanding of integral equations, specifically the form f(x)=λ ∫₀¹ K(x,y)f(y) dy.
  • Familiarity with eigenvalues and their role in iterative methods.
  • Knowledge of Neumann series and their application in functional analysis.
  • Experience with iterative schemes in numerical methods.
NEXT STEPS
  • Research the properties and applications of Neumann series in functional analysis.
  • Study the role of eigenvalues in iterative methods for solving integral equations.
  • Explore advanced techniques for analyzing the convergence of iterative schemes.
  • Investigate the implications of homogeneous integral equations in numerical solutions.
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Mathematicians, researchers in functional analysis, and anyone involved in numerical methods for solving integral equations will benefit from this discussion.

eljose
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let be the integral equation:

f(x)=\lambda \int_{0}^{1}dyK(x,y)f(y)

where the Kernel is known and "lambada" is an small eigenvalue... the question is if i want to solve i propose the iterative scheme:

f_{n+1} (x)=\lambda \int_{0}^{1}dyK(x,y)f_n(y)

My question is if in this case i can propose a "Neumann series" for the resolvent Kernel R(x,y) in the form...

R= \sum_{n=0}^{\infty} \lambda ^ n K^{n}

for K^n the n-th iterated kernel since the equation is HOmogeneus (all the functions involved except the Kernel are unknown) or if we have enough with the iteration procedure to solve the equation..thanks.
 
Physics news on Phys.org
"lambda" is an eigenvalue. "Lambada" is a dance!
 
HallsofIvy said:
"lambda" is an eigenvalue. "Lambada" is a dance!
:smile: :smile: :smile:
 
- Ooooh...¡what a wonderful and marvelous joke¡¡..I ask myself daily why you are here wasting your time when you could be a "millionaire" Hollywood comediant like Eddie Murphy or appear on "Saturday Night LIfe"...
 
I keep asking myself that!
 

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