# Homogeneous Fredholm equation of the second kind

1. Jul 9, 2010

### 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?

2. Jul 9, 2010

### 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 $$\varphi=0$$.

3. Jul 9, 2010

### yiorgos

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.

4. Jul 9, 2010

### Anthony

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 $$\|K\| <1$$, 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. $$S = (I-K)^{-1}$$.

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

Kress has a good book which is fairly accessible.

5. Jul 9, 2010

### 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?

6. Jul 9, 2010

### Anthony

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