- #1

sarrah1

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__General knowledge:__The homogeneous linear Fredholm integral equation

$\mu\ \varPsi(x)=\int_{a}^{b} \,k(x,s) \varPsi(s) ds$ (1)

has a nontrivial solution if and only if $\mu$ is an eigenvalue of the integral operator $K$. By multiplying (1) by $k(x,s)$ and integrating while changing the order of integration one obtains

${\mu}^{2}\varPsi(x)=\int_{a}^{b} \,{[k(x,s)]}^{(2)}\varPsi(s) ds$

where ${[k(x,s)]}^{(2)}=\int_{a}^{b} \,{[k(x,t)]}^{(1)}{[k(t,s)]}^{(1)}dt$ , where ${[k(x,s)]}^{(1)}=k(x,s)$

after $n$ iterations ${\mu}^{n} \varPsi(x)=\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds$ (2)

where ${[k(x,s)]}^{(n)}=\int_{a}^{b} \,{[k(x,t)]}^{(1)}{[k(t,s)]}^{(n-1)}dt$

__Now my Question:__

Solving the non-homogeneous equation

${\varPsi}(x)=f(x)+\lambda\int_{a}^{b} \,k(x,s) {\varPsi}(s) ds$ , (where $\lambda$ is a parameter)

by Picard successive approximations, results in the following equation after $n$ iterations

${\varPsi}_{n+1}(x)=f(x)+\lambda\int_{a}^{b} \,k(x,s) {\varPsi}_{n}(s) ds$

to prove convergence (I can use norms to reach the classical condition $|\lambda|<1/||K||$) however I want to reach a condition based on the spectral radius of the operator $K$ i.e. $\rho(K)$ . So one can show by imitating the above that after $n$ iterations

${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}\int_{a}^{b} \,{[k(x,s)]}^{(n)} [{\varPsi}_{0}(s)-{\varPsi}(s)] ds$ (3)

if $|\mu|<1$ then Eq. (2) gives $ \lim_{{n}\to{\infty}}\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds=0$ , $\lor x$

can this help me equally in Eq. (3) by saying that convergence occurs if

$|\lambda| < 1/\rho(K)$

grateful

Sarrah