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Integral equation

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve for u(x):

    [tex]0 = e^{2\int u(x) dx} + u(x) e^{\int u(x) dx} - a(x)[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I tried using the quadratic formula,

    [tex]e^{\int u(x) dx} = \frac{-u(x) \pm \sqrt{u^2(x) + 4a(x)}}{2}[/tex]

    , converting to log notation and differentiating, but from there I didn't know how to solve for u(x). I thought maybe I could use something on the lines of the log-definitions of the inverse trig functions. Any ideas?
  2. jcsd
  3. Mar 2, 2008 #2


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    Homework Helper

    what are the limits on the integration? if they are fixed then [itex]\int dx u(x)[/itex] is just a number (call it C) and
    u=(a-e^{2 C})/e^{C}
  4. Mar 2, 2008 #3
    You could convert it to a differential equation and try to solve this, however it turns out that this new differential equation is severely determined by the unknown function a(x). In order to do this, set:

    [tex]e^{\int u(x)dx}=f(x)[/tex]



    And putting this into the equation gives:


    This is a Riccati equation, which can be transformed into a linear one by transforming:



    [tex]\frac{df(x)}{dx}=-\frac{1}{[u(x)]^2}\left(\frac{du(x)}{dx}\right)^2 +\frac{1}{u(x)} \frac{d^2u(x)}{dx^2}[/tex]

    The equation becomes:

    [tex]\frac{d^2u(x)}{dx^2}-u(x)\cdot a(x)=0[/tex]

    And this one can be solved if a(x) is known. P.e. a(x)=-1 gives sin and cos functions, a(x)=x gives airy functions, a(x)=1 gives hyperbolic ones, etc.

    This might be the complete wrong idea, but I don't see it in another way.

    [Edit] Someone was faster....
  5. Mar 2, 2008 #4
    Actually, I got the integral equation by trying to solve

    [tex]\frac{d^2u(x)}{dx^2}-u(x)\cdot a(x)=0[/tex]
  6. Mar 3, 2008 #5
    foxjwill, to my knowledge there is no solution in terms of a(x). As I pointed out the solution depends so heavily on this function a(x) that you can't solve it without knowing it explicitly. The few examples I gave did show this, no? A sin or cos function compared to a hyperbolic one or even an Airy function (which is closely related to the functions of Bessel) are so different, even for the simple assumed functions of a(x) equal to -1, 1 and x. Maybe there is an explicit integral representation of the solution, but I think it will be closely related to the one you originally posted.
  7. Mar 4, 2008 #6


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    also, you could try using Green's functions and getting an approximation... the utility of this approach probably depends on the form of a(x). E.g., find the "homogeneous" solutions
    \frac{d^2 f}{dx^2}=0
    and the "free" green's function
    and then consider the term
    as an inhomogeneous term so that the "solution" is given by
    u(x)=f(x)+\int dx' G(x,x')a(x')u(x')

    Then, supposing u(x) is only a little different from f(x) once can develop succesive approximations for u(x) as
    u(x)\approx f(x) + \int dx' G(x,x')a(x')f(x')+\int dx' G(x,x')a(x')\int dx'' G(x',x'')a(x'')f(x'')+\ldots
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