Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Solving this integral equation

  1. Nov 3, 2017 #1
    I have the following expression :

    $$ y_{E} = \int_{0}^{\infty} 0.5 * [E_{1}(µ(E)*r) - E_{1}(\frac{µ(E)*r}{cos \alpha})] * f(r) dr $$

    where :
    - $y_{E}$ has been measured for some E (something like 5 different $E_{i}$, to give you an idea)
    - µ(E) is retrieved from a table in the litterature (basically that means I have no analytical expression for µ(E) but I can easily get access for its value). If you're curious about the shape of that thing in function of E, here is the link : https://physics.nist.gov/PhysRefData/XrayMassCoef/ElemTab/z82.html
    - E1 is the exponential integral function

    My goal is to find f(r).

    This can be rewritten in the following form :

    $$ y_{E} = \int_{0}^{\infty} K(E;r) * f(r) dr $$

    which (thx wikipedia) is called a Fredholm integral equation of the first kind. And which is very ugly to solve from what i've read. :'(

    I would like to introduce a parametrization of f(r) as the following :
    $$f(r) = C*e^{-\lambda_{1}*(r+\lambda_{3})} * [1-e^{-\lambda_{2}*(r+\lambda_{3})}]$$

    The justification for this form is that i know f(r) usually has a profile where it increases then exponentially descreases. So I think such an expression is okay. Of course if there was a way to solve this without postulating such a hard thing, that would be better.

    Anyway, so basically an expression of 4 parameters : lambdas 1, 2, 3 and C. If I can find a decent approximation of these 4 parameters, I will be a happy man.

    From there, if you have any suggestion how to solve this, my ears are wide open.

    I have tried the following :

    First I introduced a quadrature form. I choosed a Gauss-Laguerre one, because of the limits on the integral from 0 to infinity, but I have no idea if that is a good choice. I've read Gauss-Laguerre works great to estimate an integral with a polynomial function multiplied by an exponential, but here I have no polynomial function but rather 2 ugly exponential integral E1 which probably behaves very differently, particulary close to 0+. So is it a wise idea?

    Anyway, by doing so, I can get some kind of linear system :
    $$\hat{K} \hat{f} = \hat{g}$$

    where :
    Kij = K'(Ei;rj)
    fj=f(rj) with j going from 1 to n
    gi=y(Ei) with i going from 1 to m

    n is something like 15 or more (to get enough points in the G-L quadrature for a decent fit)
    m is, as previously stated, around 5 or so.

    So K is not a square matrix, and the system is clearly undertermined because n>m.
    But since all the fj only depend on 4 parameters only, this looks like some kind of 5 equations with 4 unknown, but non-linear (because of the expression of f). Am I correct on this ?

    And if so, how to solve this to find my 4 parameters ?

    As already said, any idea would be greatly appreciated, and thank you for reading this !
    Last edited: Nov 3, 2017
  2. jcsd
  3. Nov 7, 2017 #2


    User Avatar
    2016 Award

    Staff: Mentor

    Don’t use your values of f as unknowns, use the lambdas and C. Then your system is overdetermined and you can do a least square fit or something similar. It also means you can use much more than 15 points to evaluate the integral.
    The downside: The system is not linear any more. But it should behave well enough for a fit.
  4. Nov 9, 2017 #3
    One alternative could be to try to approximate ##y(E_i)## by polynomial interpolation, or as a polynomial times an asymptotic function. Whether or not this is possible depends of course on how the measured ##y(E_i)## looks like. In that they to can approximately compute y(E_i) at all n points, and you get a system of $$n\times n$$ equations to be solved.
  5. Nov 13, 2017 #4
    Thank you for these answers !

    Yes, this is what I was saying in the last part of my post. About using more than 15 points, yes I can do that but this doesnt give me more data. It just approximates the integral better, so the error is on the 4 unknown parameters should be smaller. The problem is that each of these 15 points costs me some measurement time so I think 15 is a good number.

    Interesting alternative indeed. I don't think I can reasonnably find a good interpolation when I see how y(Ei) looks like. But that's definitively an idea.

    Now, by using the method suggested by mfb, I can indeed get some values. The next question I am dealing with now is :

    How can I estimate the uncertainty on these 4 parameters value that i find using the method described above ?

    So I have like 10 equations, non-linear in my 4 parameters. Some of the coefficients in these equations have a value with an associated standard deviation. I use a least-square method to kind the best value of my 4 parameters but i'd like to know their standard deviation.

    I'm kinda blocked here :-/
  6. Nov 15, 2017 #5


    User Avatar
    2016 Award

    Staff: Mentor

    Every fitting package will be able to do both minimization and an error estimate- typically both together as they come from the same procedure.

    Using more points for the integral shouldn’t be related to the number of measurements you make - that doesn’t enter there.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted