# [SOLVED]eigenvalue (and function) of integral equation

#### dwsmith

##### Well-known member
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
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 $$f_{1/4}(x)=x$$.

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#### dwsmith

##### Well-known member
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
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
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:

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

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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.