Homogeneous Fredholm equation of the second kind

**yiorgos** · July 9th 2010, 08:22 AM

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 depends only in the difference x-t, $\Lambda(x,t)=\Lambda(x-t)$ , and g(x) a known function involving logarithm.

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

**Jose27** · July 12th 2010, 05:39 PM

I don't really know about integral equations, but it is an easy consequence of Banach's fixed point theorem that if $1> \| K \| _{\infty} (b-a)$ then your equation only has $\varphi (x)=0$ as solution. In general the eigenvalue problem $\lambda \varphi (x) = \int_a^b K(x,t)\varphi (t)dt$ has only the trivial solution if $|\lambda | > \| K\| _{\infty} (b-a)$ where $K: [a,b]\times [a,b] \rightarrow \mathbb{R}$ is continous and we consider the mapping $F(f)(x)= \int_a^b K(x,t)f(t)dt$ from $C[a,b]$ to itself.

**Jose27** · July 12th 2010, 08:55 PM

Hey, I've been thinking about this problem and here's some things that I came up with: We can consider the same operator $F$ , only this time defined in other spaces.

1.- $F: L^1[a,b] \rightarrow L^1[a,b]$ and $g\in L^1[a,b] \cap L^{\infty} [a,b]$ and we then have (noting that $K(x,t)=K(x-t)$ ) $\| F(\varphi ) \| _1 \leq \| g\| _{\infty } \|K\| _{1 } \| \varphi \| _1$ and so if $\| g\| _{\infty } \|K\| _{1 } \leq \alpha \in (0,1)$ we get only the trivial solution. What I did here was note that $\int_a^b K(x,t)\varphi (t)dt$ is just a convolution, and use said operation's properties.

2.- $F:L^2[a,b] \rightarrow L^2[a,b]$ and $K \in L^1 \left( [a,b]\times [a,b] \right) \cap L^{\infty } \left( [a,b] \times [a,b] \right)$ we have then $\| F(\varphi ) \| _2 \leq c\| \varphi \| _2 \| \overline{K} g \| _2$ where $\overline{K} (x)= \mbox{ess} \sup_t \ K(x,t)$ . This one's a little harder to get (I'm still not sure I have all the calculations right) and uses Jensen's inequality (if you want the details just ask).

If the equation does (or should) have a non-trivial solution well...

**yiorgos** · July 13th 2010, 08:07 AM

Thank you for your reply.
I think that the trivial solution always exists as a consequence of the Fredholm theory.

Anyway, the trivial solution is definitely not what I am looking for.
The φ(x) function corresponds to a physical quantity in my problem
and it should not be zero.

In fact I expect a family of solutions depending on some parameter.

**Jose27** · July 13th 2010, 08:15 AM

Originally Posted by yiorgos

Thank you for your reply.
I think that the trivial solution always exists as a consequence of the Fredholm theory.

Anyway, the trivial solution is definitely not what I am looking for.
The φ(x) function corresponds to a physical quantity in my problem
and it should not be zero.

In fact I expect a family of solutions depending on some parameter.

Well, the trivial solution exists simply because F is a linear operator.

If you want a non-trivial solution first check that none of the above inequalities hold, and maybe if K is smooth (or $C^1$ ) you could differentiate the equation to get some sort of differential equation for $\varphi$ . If all else fails maybe a qualitative approach suffices for you.

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