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

Lesson Quiz

Follow this link for lesson quiz.

Extra Links for Reading

  1. Article about the same from The Encyclopedia of Mathematics
  2. NPTEL Lectures about the section
  3. Previous article about Results on Resolvent Kernel

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