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Not able to solve this integral problem

  1. Jul 6, 2011 #1
    Hi all,

    Currently I am puzzling on a real-world problem, involving some maths which I cannot solve using my limited calculus knowledge. The problem ultimately boils down to finding an expression for the function [itex]\phi(t)[/itex] which satisfies the following equation for any x within limits [itex]0 < x < 1 - x_s[/itex]:

    [itex]1 - x - x_{s} = \int_{x_s/\phi_{max}}^{1/\phi_{2} - 1/\phi_{1}\cdot x} \phi(t) dt[/itex]

    Constants: [itex]x_{s}, \phi_{1}, \phi_{2}, \phi_{max}[/itex] with [itex]0 < x_s < 1[/itex] and [itex]0 < \phi_{1} < \phi_{2} < \phi_{max}[/itex].

    Has anybody got a clue on how to do this? An general approach would be much appreciated! Just gessing functions and trying them out somehow doesn't feel very intelligent...

    .edit: I'll put you as co-author ;)
    Last edited: Jul 7, 2011
  2. jcsd
  3. Jul 6, 2011 #2


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    There seems to be something missing from your expression. There is no x in the integral expression on the right, so it won't depend on x, while the left side does.
  4. Jul 6, 2011 #3
    Aah, I put it wrongly. Should be correct now!
  5. Jul 7, 2011 #4
    Suppose that a function Phi exists as a solution of the integral equation.
    Then differentiate (relatively to x) the left and rigth terms of the equation. You obtain :
    -1 = (-1/phi1)*Phi(t) where t=(1/phi2)-(x/phi1)
    The left term is constant while the rigth term depends on x. This is impossible exept if the function Phi(t) is constant.
    As we suppose Phi(t) is not constant, there is no function Phi(t) solution of the problem.
  6. Jul 7, 2011 #5
    Use the rule for parametric differentiation with varying limits:

    \frac{d}{d x} \int_{\varphi_{1}(x)}^{\varphi_{2}(x)}{F(x, t) \, dt} = \int_{\varphi_{1}(x)}^{\varphi_{2}(x)}{\frac{\partial F(x, t)}{\partial x} \, dt} + F(x, \varphi_{2}(x)) \varphi'_{2}(x) - F(x, \varphi_{1}(x)) \varphi'_{1}(x)

    Hmmm, testing...
    \frac{\partial F(x, t)}{\partial x}
  7. Jul 7, 2011 #6
    Is it that easy? Shame on me.. I will look into this tonight (I'm in the Netherlands) and come back to it!

    Thank you! More to study in the evening..
  8. Jul 7, 2011 #7


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    What you are trying to do is handled by a branch of analysis known as the calculus of variations.
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