Iterated Kernels and Fredholm Theorems (IEL6)

Method of Successive Approximation

Solution Method 2 (Neumann Series Solution, Solution by Iterated Kernels)

For Fredholm IE

For the equation $g(s) = f(s) = \int K(s,t)g(s)ds$,

Put g(s) = f(s) or any appropriate function of s.

Find $g_m(s) = f(s) = \int K(s,t)g_{m-1}(s)ds$ iteratively.

$\Gamma(s,t;\lambda)=\sum_{m=i^\infty}\lambda^{m-1}K_m(s,t)$ where

$K_1(s,t) = K(s,t)$

$K_m(s,t) = \int K(s,x)K_{m-1}(x,t)dx$

and solution will be of the form $\phi(x)=f(s)+\lambda\int\Gamma(s,t;\lambda)f(t)dt$

For Volterra IE

$\Gamma(s,t;\lambda)=\sum_{m=i^\infty}\lambda^{m-1}K_m(s,t)$ where

$K_1(s,t) = K(s,t)$

$K_m(s,t) = \int_t^s K(s,x)K_{m-1}(x,t)dx$

and solution will be of the form $\phi(x)=f(s)+\lambda\int\Gamma(s,t;\lambda)f(t)dt$

Example

Solve the integral equation $\phi(x) = f(x)+\int_0^1 K(x,y)\phi(y)dy$ for $K(x,y)=xy^2$.

$K_1(x,y)=xy^2$

$K_2(x,y)=\int_0^1 xs^2\times sy^2 ds = xy^2\times (\frac {s^4}4)_0^1={xy^2\over4}$

Continuing, $K_m(x,y)= {xy^2\over4^{n-1}}$ . Hence, $R(x,t;\lambda)=\sum_{n=1}^\infty \lambda^{m-1}K_m(x,y)= \sum_{n=1}^\infty \lambda^{m-1} {xy^2\over4^{m-1}}$

$R(x,t;1)= \sum_{n=1}^\infty {xy^2\over4^{m-1}}=\frac1{ 1-1/4}=\frac43 xy^2$.

Hence, the solution, $\phi(x)=f(x)+ \frac43 x\int_0^1 y^2 f(y)dy$

Some Properties of Resolvent Kernels

1. $\Gamma(s,t;\lambda)$ is convergent absolutely and uniformly on $\latex |\lambda|<\frac1B$, where $B^2=\int_a^b \int_a^b |K(s,t)|^2dsdt$
2. $\Gamma(s,t;\lambda) =K(s,t)+\lambda\int K(s,x)\Gamma(x,t;\lambda)dx$
3. $\frac{\partial \Gamma(s,t;\lambda) }{ \partial \lambda}=\int\Gamma(s,x;\lambda) \Gamma(x,t;\lambda)dx$
4. $\int\Gamma(s,x;\lambda) \Gamma(x,t;\lambda)dx=\frac1{\lambda-\mu} \int\Gamma(s,x;\mu) \Gamma(x,t;\lambda)dx$

A property of iterated kernel: $K_m(s,t)=\int K_r(s,x)K_{m-r}(x,t)dx$

Fredholm’s Theorems

Solution Method 3 (Approximation)

For Fredholm IE with any integrable kernel, to find an approximate solution for $g(s)=f(s)+\lambda\int_a^b K(s,t)g(t)dt$

1. Choose n, and let $h={b-a\over n}$
2. Let $s_1=t_1=a,s_2=t_2=a+h,\cdots,s_n=t_n=a+(n-1)h$
3. Find $K_{ij}=K(s_i,t_j) \forall i,j;f_i=f(s_i),g_i=g(s_i)$
4. $g_i$ can be obtained by solving set of equations $g_i-\lambda h\sum_{j=1}^n K_{ij}g_j=f_i,\forall i$.
5. $\lambda$ can be approximated by putting $D(\lambda)=0$, where $d_ii=1-\lambda hK_{ii},d_{ij}=-\lambda hK_{ij},i\ne j$

Solution Method 4 (Using Fredholm’s First Theorem)

For Fredholm IE $g(s)=f(s)+\lambda\int_a^b K(s,t)g(t)dt$ , where f(s) and g(s) are integrable, $\Gamma(s,t;\lambda)={D(s,t;\lambda)\over D(\lambda)}$ where

$D(s,t;\lambda)=\sum_{p=0}^\infty {(-\lambda)^p\over p!}C_p(s,t)$ and $D(\lambda)=\sum_{p=0}^\infty {(-\lambda)^p\over p!}c_p$ where

$c_0=1,C+0(s,t)=K(s,t)$

$c_p=\int_a^b C_{p-1}(s,s)ds;$

$C_p(s,t)=c_pK(s,t)-p\int_a^bK(s,x)C_{p-1}(x,t)dx$

Example

For the linear IE $\phi(x)=x+\int_0^\frac12 \phi(\eta)d\eta$, find $R(x,\eta;1)$

$R(x,\eta;\lambda)={D(x,\eta;\lambda)\over D(\lambda)}$

$c_0=1,C_0(x,\eta)=1$

$c_1=\int_0^\frac12 C_0(x,x)dx= \int_0^\frac12 dx =\frac12$

$C_1(x,\eta)=\frac12\times 1)-1\int_0^\frac12 dx=\frac12 - \frac12 =0$

and $c_p=0,C_p(x,\eta)=0,\forall p>1$.

Hence, $R(x,\eta;\lambda)=\frac1{1-\lambda\frac12}\implies R(x,\eta;1) = 2$