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Homogeneous Fredholm equation of the second kind

  1. Jul 9, 2010 #1
    during the analysis of a problem in my phd thesis
    I have resulted in the following equation.

    [tex]\varphi(x)= \int_a^b K(x,t)\varphi(t)dt[/tex]

    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

    or when it is separable which are both not my case.

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

    where [tex]\Lambda(x,t)[/tex] 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
  2. jcsd
  3. Jul 9, 2010 #2
    Your question very much depends on whether the integral operator defined by:

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

    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 [tex]\|K\|<1[/tex] then you can construct a convergent (in the operator norm) Neumann series to show the only solution is [tex]\varphi=0[/tex].
  4. Jul 9, 2010 #3
    Thank you for the reply.
    So, you say that if |K|<1 then [tex]\varphi[/tex] vanishes?

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

    [tex]\varphi(x)= \int_a^b K(x,t)h(t)dt[/tex]

    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.
  5. Jul 9, 2010 #4
    Yes, if the operator norm is less than one. I.e. if [tex]K:X\rightarrow Y[/tex] and [tex]X[/tex] is a normed space, then:

    [tex] \| K\| = \sup_{\|\varphi\|=1} \| A\varphi \| \[/tex]

    So if [tex]\|K\| <1[/tex], then the following Neumann series converges (in the operator norm):

    [tex] S = \sum_{n=0}^\infty K^n [/tex]

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

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

    Kress has a good book which is fairly accessible.
  6. Jul 9, 2010 #5
    I forgot to mention that I know for [tex]\phi(x)[/tex]
    that it is defined only in [a,b] and I'm interesting particularly for a domain of the form [-a,a].
    Additionally, I expect [tex]\phi(x)[/tex] to be continuous and symmetric about zero.
    Would these properties help by any means?
  7. Jul 9, 2010 #6
    I have no idea what [tex]\phi(x)[/tex] is.
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