Homogeneous Fredholm equation of the second kind

[yiorgos]

Hi,
during the analysis of a problem in my phd thesis
I have resulted in the following equation.

\varphi(x)= \int_a^b K(x,t)\varphi(t)dt

which is clearly a homogeneous Fredholm equation of the second kind

The problem is that I can't find in any text any way of solving it.
Solutions are provided only for special cases like when the kernel K
is symmetric

K(x,t)=K(x,t)
or when it is separable which are both not my case.

The particular form of the equation I am dealing with is
\varphi(x)= \int_a^b \Lambda(x,t)g(x)\varphi(t)dt

where \Lambda(x,t) is symmetric and g(x) a known function involving logarithm.

Any ideas of how to deal with this kind of form?
Thank you in advance

 

[Anthony]

Your question very much depends on whether the integral operator defined by:

K\varphi = \int_a^b K(x,t) \varphi(t)\, \mathrm{d} t

is compact. In this case, you can apply Fredholm theory (for instance, your equation can only have finitely many solutions). Alternatively, if you can show \|K\|<1 then you can construct a convergent (in the operator norm) Neumann series to show the only solution is \varphi=0.

 

[yiorgos]

Your question very much depends on whether the integral operator defined by:

K\varphi = \int_a^b K(x,t) \varphi(t)\, \mathrm{d} t

is compact. In this case, you can apply Fredholm theory (for instance, your equation can only have finitely many solutions). Alternatively, if you can show \|K\|<1 then you can construct a convergent (in the operator norm) Neumann series to show the only solution is \varphi=0.

Thank you for the reply.
So, you say that if |K|<1 then \varphi vanishes?

One more question. Since my kernel is not of a specific form,
is it more convenient to take h(t)=g(x)*\varphi(t)
and translate the initial equation to the form

\varphi(x)= \int_a^b K(x,t)h(t)dt

which is a Fredholm equation of the second kind?

Is this form easier to be solved or it will make things worse?

PS: Do you have any good book to suggest?
Every book I have searched treats only the trivial cases of kernels (separable etc.)

EDIT: The solution phi=0 has no physical meaning in my case, so it should be considered as unacceptable.

 

[Anthony]

Thank you for the reply.
So, you say that if |K|<1 then \varphi vanishes?

Yes, if the operator norm is less than one. I.e. if K:X\rightarrow Y and X is a normed space, then:

\| K\| = \sup_{\|\varphi\|=1} \| A\varphi \| \

So if \|K\| &lt;1, then the following Neumann series converges (in the operator norm):

S = \sum_{n=0}^\infty K^n

and you can check S (I-K) = (I-K)S = I, i.e. S = (I-K)^{-1}.

One more question. Since my kernel is not of a specific form,
is it more convenient to take h(t)=g(x)*\varphi(t)
and translate the initial equation to the form

\varphi(x)= \int_a^b K(x,t)h(t)dt

which is a Fredholm equation of the second kind?

Certainly not - your equation is still of the 1st kind.

PS: Do you have any good book to suggest?
Every book I have searched treats only the trivial cases of kernels (separable etc.)

Kress has a good book which is fairly accessible.

 

[yiorgos]

I forgot to mention that I know for \phi(x)
that it is defined only in [a,b] and I'm interesting particularly for a domain of the form [-a,a].
Additionally, I expect \phi(x) to be continuous and symmetric about zero.
Would these properties help by any means?

 

[Anthony]

I have no idea what \phi(x) is.