Resolvent kernel of Fredholm integral equation.
$begingroup$
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.
integral-equations
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
add a comment |
$begingroup$
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.
integral-equations
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
add a comment |
$begingroup$
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.
integral-equations
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
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.
integral-equations
integral-equations
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
asked May 24 '18 at 6:41
ASHWINI SANKHEASHWINI SANKHE
12710
12710
add a comment |
add a comment |
1 Answer
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$begingroup$
$k(x,t)=k_1(x,t)=1$,
$k_2(x,t)=int_{0}^{1/2} k(x,z)k_1(z,t) dz=frac {1}{2}$
$k_3(x,t)=int_{0}^{1/2} k(x,z)k_2(z,t) dz=frac {1}{2^2}$......
$k_n(x,t)=int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=frac {1}{2^{n-1}}$
Hence $R(x,t,lambda)=sum_{i=0}^infty lambda^i k_{i+1}(x,t)$
$R(x,t,1)=frac{1}{2}+frac {1}{2^2}+frac {1}{2^3}+....=frac{1}{1-(1/2)}=2$
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$k(x,t)=k_1(x,t)=1$,
$k_2(x,t)=int_{0}^{1/2} k(x,z)k_1(z,t) dz=frac {1}{2}$
$k_3(x,t)=int_{0}^{1/2} k(x,z)k_2(z,t) dz=frac {1}{2^2}$......
$k_n(x,t)=int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=frac {1}{2^{n-1}}$
Hence $R(x,t,lambda)=sum_{i=0}^infty lambda^i k_{i+1}(x,t)$
$R(x,t,1)=frac{1}{2}+frac {1}{2^2}+frac {1}{2^3}+....=frac{1}{1-(1/2)}=2$
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
add a comment |
$begingroup$
$k(x,t)=k_1(x,t)=1$,
$k_2(x,t)=int_{0}^{1/2} k(x,z)k_1(z,t) dz=frac {1}{2}$
$k_3(x,t)=int_{0}^{1/2} k(x,z)k_2(z,t) dz=frac {1}{2^2}$......
$k_n(x,t)=int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=frac {1}{2^{n-1}}$
Hence $R(x,t,lambda)=sum_{i=0}^infty lambda^i k_{i+1}(x,t)$
$R(x,t,1)=frac{1}{2}+frac {1}{2^2}+frac {1}{2^3}+....=frac{1}{1-(1/2)}=2$
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
add a comment |
$begingroup$
$k(x,t)=k_1(x,t)=1$,
$k_2(x,t)=int_{0}^{1/2} k(x,z)k_1(z,t) dz=frac {1}{2}$
$k_3(x,t)=int_{0}^{1/2} k(x,z)k_2(z,t) dz=frac {1}{2^2}$......
$k_n(x,t)=int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=frac {1}{2^{n-1}}$
Hence $R(x,t,lambda)=sum_{i=0}^infty lambda^i k_{i+1}(x,t)$
$R(x,t,1)=frac{1}{2}+frac {1}{2^2}+frac {1}{2^3}+....=frac{1}{1-(1/2)}=2$
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
$endgroup$
$k(x,t)=k_1(x,t)=1$,
$k_2(x,t)=int_{0}^{1/2} k(x,z)k_1(z,t) dz=frac {1}{2}$
$k_3(x,t)=int_{0}^{1/2} k(x,z)k_2(z,t) dz=frac {1}{2^2}$......
$k_n(x,t)=int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=frac {1}{2^{n-1}}$
Hence $R(x,t,lambda)=sum_{i=0}^infty lambda^i k_{i+1}(x,t)$
$R(x,t,1)=frac{1}{2}+frac {1}{2^2}+frac {1}{2^3}+....=frac{1}{1-(1/2)}=2$
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
answered Dec 24 '18 at 8:27
ASHWINI SANKHEASHWINI SANKHE
12710
12710
add a comment |
add a comment |
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