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For the linear integral equation
$ y(x)=x+\int_{0}^{1/2} y(t) dt$.
Find Resolvent kernel $R(x,t,1)$.

I tried to find resolvent kernel of Volterra integral equation by taking kernel as 1.Then I got $R(x,t,1)=e^{(x-t)}$.But I don't know how to find resolvent kernel of nonhomogeneous Fredholm integral equation of second kind.please guide me.Thanks a lot.

ASHWINI SANKHE
  • 480

1 Answers1

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$k(x,t)=k_1(x,t)=1$, $k_2(x,t)=\int_{0}^{1/2} k(x,z)k_1(z,t) dz=\frac {1}{2}$
$k_3(x,t)=\int_{0}^{1/2} k(x,z)k_2(z,t) dz=\frac {1}{2^2}$...... $k_n(x,t)=\int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=\frac {1}{2^{n-1}}$
Hence $R(x,t,\lambda)=\sum_{i=0}^\infty \lambda^i k_{i+1}(x,t)$
$R(x,t,1)=\frac{1}{2}+\frac {1}{2^2}+\frac {1}{2^3}+....=\frac{1}{1-(1/2)}=2$

ASHWINI SANKHE
  • 480