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[SOLVED] eigenvalue (and function) of integral equation

dwsmith

Well-known member
Feb 1, 2012
1,673
Given
\[
f(x) = \lambda\int_0^1xy^2f(y)dy
\]
I am trying to determine the eigenvalues and eigenfunction. I know that the \(\frac{1}{\lambda}\) are the eigenvalues.
We can write \(f(x) = xA\) and \(A = \lambda\int_0^1y^2f(y)dy\).
\[
A\Bigg(1 - \lambda\int_0^1y^3dy\Bigg) = 0\quad (*)
\]
So is the eigenvalue only one value which would be what I get when I solve for 1 over lambda?
\[
\frac{1}{\lambda} = \frac{1}{4}
\]
Is \((*)\) the eigenfunction? If not, how do I find the eigenfunction?
 

zzephod

Well-known member
Feb 3, 2013
134
Re: eigenvalue (and function) of integral eq

Given
\[
f(x) = \lambda\int_0^1xy^2f(y)dy
\]
I am trying to determine the eigenvalues and eigenfunction. I know that the \(\frac{1}{\lambda}\) are the eigenvalues.
We can write \(f(x) = xA\) and \(A = \lambda\int_0^1y^2f(y)dy\).
\[
A\Bigg(1 - \lambda\int_0^1y^3dy\Bigg) = 0\quad (*)
\]
So is the eigenvalue only one value which would be what I get when I solve for 1 over lambda?
\[
\frac{1}{\lambda} = \frac{1}{4}
\]
Is \((*)\) the eigenfunction? If not, how do I find the eigenfunction?
I presume you want to find the eigen values and functions of the operator defined by: \(Lf(x) = \int_0^1 x y^2 f(y)\ dy\).

We observe that this is a linear operator over the space of functions it is defined on, as we require, and that we are looking for an eigen value \(\lambda\) and function \(f_{\lambda}(x)\) such that:

\[Lf_{\lambda}(x)=\lambda f_{\lambda}(x)\]

As \(L\) maps any function in its domain to a multiple of \(x\) and any multiple of an eigen function is also an eigen function we may as well take the eigen function to be \(f(x)=x\), then \(Lf(x)=\int_0^1 x y^2 f(y)\ dy=\frac{1}{4}x\).

Hence the eigen value of \(L\) is \(1/4\) and a corresponding eigen function is \(f_{1/4}(x)=x\).

.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Re: eigenvalue (and function) of integral eq

I presume you want to find the eigen values and functions of the operator defined by: \(Lf(x) = \int_0^1 x y^2 f(y)\ dy\).

We observe that this is a linear operator over the space of functions it is defined on, as we require, and that we are looking for an eigen value \(\lambda\) and function \(f_{\lambda}(x)\) such that:

\[Lf_{\lambda}(x)=\lambda f_{\lambda}(x)\]

As \(L\) maps any function in its domain to a multiple of \(x\) and any multiple of an eigen function is also an eigen function we may as well take the eigen function to be \(f(x)=x\), then \(Lf(x)=\int_0^1 x y^2 f(y)\ dy=\frac{1}{4}x\).

Hence the eigen value of \(L\) is \(1/4\) and a corresponding eigen function is \(f_{1/4}(x)=x\).

.
From your working, how did you obtain the 1/4? Is the eigenfunction always \(f(x) = \text{x terms}\)?
 

zzephod

Well-known member
Feb 3, 2013
134
Re: eigenvalue (and function) of integral eq

From your working, how did you obtain the 1/4? Is the eigenfunction always \(f(x) = \text{x terms}\)?
As the image of any function under \(L\) is a multiple of \(x\) the eigen function must be a multiple of \(x\) for this operator.

The \(1/4\) when \(f(x)=x\) comes from:

\[ x \int_0^1 y^2f(y)\ dx=x \int_0^1 y^3\ dx=x\ \left[ \frac{y^4}{4}\right]_0^1\]

.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Re: eigenvalue (and function) of integral eq

As the image of any function under \(L\) is a multiple of \(x\) the eigen function must be a multiple of \(x\) for this operator.

The \(1/4\) when \(f(x)=x\) comes from:

\[ x \int_0^1 y^2f(y)\ dx=x \int_0^1 y^3\ dx=x\ \left[ \frac{y^4}{4}\right]_0^1\]

.
How would I do this problem using the Fredholm series? I know how to use the Fredholm series to solve an integrable equation but not how to find the eigenvalue(s) or function.