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Solution of an integral equation

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Given [tex]\frac{1}{| \int\ f(\ x)\ g(\ x)\ d\ x\ |}=\int \frac{\ f(\ x)}{\ g(\ x)}\ d\ x[/tex]
    Does the above put any condition on f(x) and g(x)?

    2. Relevant equations


    3. The attempt at a solution


    The | | in the denominator reminds me of Darboux inequality...In fact it looks impossible to solve analytically...Can it be solved numerically?
     
  2. jcsd
  3. Jun 12, 2010 #2
    What if you take the derivative of both sides?

    [tex]
    \frac{-f(x) g(x) }{| \int\ f(x)\ g(x)\ dx\ |^2}= \frac{\ f(\ x)}{\ g(\ x)}
    [/tex]

    Assuming f(x) =/= 0 and simplifying with a little algebra

    [tex]
    (g(x))^2 = - (\int\ f(x)\ g(x)\ dx )^2
    [/tex]

    Which would imply that g(x) = 0, which is impossible.

    So, no such function exists, unless we consider complex functions.
     
  4. Jun 12, 2010 #3
    Remembering |x|=x if x>0 and |x|=-x if x<0,there is another option:

    [tex]

    (g(x))^2 = (\int\ f(x)\ g(x)\ dx )^2

    [/tex]
    This leads to
    [tex]\ g(\ x)=\ e^{\pm\int\ f(\ x)\ d\ x}[/tex]

    Do you agree?
     
  5. Jun 12, 2010 #4
    Yeah, I guess that makes sense. Though it should be

    g(x) = e^int f(x).

    Not f(x).
     
  6. Jun 12, 2010 #5
    Personally I expected g(x) to have some lower bound;and that looked plausible for Darboux inequality says:| integral |>= Maximum value of integrand*length of the contour...

    Can that be a way?

    Yea...that was a typo..I am fixing it
     
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