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The Abel equation

  1. Feb 18, 2005 #1

    saltydog

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    I wish to study the Abel equation and think it's solution involves several aspects of Calculus that others here may find interesting. I'll post the derivation of the solution in steps; that helps me learn it and I hope will provide incentive to others to comment about the process. I'm not really good with integral equations and welcome any comments; my understanding is quite lacking. I just like them and see the potential of them (and IDEs) being used in artificial intelligence.

    Here's the equation:

    [tex]f(x)=\int_0^x{\frac{\phi(y)}{\sqrt{x-y}}}[/tex]

    With [itex] f(x)[/itex] given and [itex]\phi(x)[/itex] to be determined.

    The first step in solving the equation is to define an integral operator, that is, an operator which takes a function and processes it with an integral. For this problem, Abel used the following integral operator:

    [tex]I[f(x)]=\int_0^\xi{\frac{f(x)dx}{\sqrt{\xi-x}}}[/tex]

    Thus, applying this operator to both sides of the equation yields:

    [tex]\int_0^\xi{\frac{f(x)dx}{\sqrt{\xi-x}}}=\int_0^\xi{\frac{dx}{\sqrt{\xi-x}}}\int_0^x{\frac{\phi(y)}{\sqrt{x-y}}}dy[/tex]

    The order of integration on the RHS needs to be changed next.
     
  2. jcsd
  3. Feb 19, 2005 #2

    saltydog

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    Switching the order of integration

    It's easy for me to switch the order of integration when I consider the integral in the following form:

    [tex]\int_0^\xi\int_0^x f(x,y;\xi)dydx[/tex]

    The area of integration is then the triangular region under the graph [itex]y=x[/itex] in the range [itex][0,\xi][/itex]. In this case, the integral becomes:

    [tex]\int_0^\xi\int_y^\xi f(x,y;\xi)dxdy[/tex]

    And thus:

    [tex]\int_0^\xi{\frac{dx}{\sqrt{\xi-x}}}\int_0^x{\frac{\phi(y)}{\sqrt{x-y}}}dy=\int_0^\xi\int_y^\xi \frac{1}{\sqrt{\xi-x}}\frac{\phi(y)}{\sqrt{x-y}}dxdy[/tex]

    The transformed integral equation becomes:

    [tex]\int_0^\xi\frac{f(x)}{\sqrt{\xi-x}}dx=\int_0^\xi\phi(y)\int_y^\xi\frac{1}{\sqrt{\xi-x}\sqrt{x-y}}dxdy[/tex]

    The next step is to find a suitable change of variable that will facilitate evaluating the first iterated integral.
     
  4. Feb 20, 2005 #3

    saltydog

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    Using the change of variable [itex]x=y+(\xi-y)u[/itex], the integral on the RHS becomes:

    [tex]\int_0^\xi \phi(y)\int_0^1\frac{(\xi-y)du}{\sqrt{(\xi-y-\xi u+yu)(\xi u-yu)}}}dy[/tex]

    Simplifying, we get:

    [tex]\int_0^\xi \phi(y)\int_0^1\frac{du}{\sqrt{u(1-u)}}dy [/tex]

    Well, how convenient. This seems to be the critical part of the solution process. In this case, everything cancels. I'd like to come back to this part with a kernel that does not cancel so nicely but for now, I stick with the easy problem.

    The equation is now:

    [tex]\int_0^\xi \frac{f(x) dx}{\sqrt{\xi-x}}=\int_0^\xi \phi(y)\int_0^1\frac{du}{\sqrt{u(1-u)}}dy [/tex]

    The first iterated integral is easily solved.
     
  5. Feb 20, 2005 #4

    saltydog

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    Incorrect usage in first post

    I now realize it's incorrect to call the first step in the solution process an "integral transform" as a transform would involve a "dummy" variable for the integral. Rather, I'm just multiplying both sides of the original integral equation by [itex]\frac{dx}{\sqrt{\xi-x}}[/itex] and then integrating from 0 to [itex]\xi[/itex].
     
  6. Feb 21, 2005 #5

    saltydog

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    Continuing:

    Completing the square in the radical, the antiderivative is in the form of ArcSin. The integral is equal to [itex]\pi[/itex]

    So that:

    [tex]\int_0^\xi \frac{f(x)dx}{\sqrt{\xi-x}}=\pi \int_0^\xi \phi(y)dy[/tex]

    In order to isolate [itex]\phi[/itex], we differentiate both sides of the equation with respect to [itex]\xi[/itex] so that:

    [tex]\phi(\xi)=\frac{1}{\pi}\frac{d}{d\xi}\int_0^\xi\frac{f(x)dx}{\sqrt{\xi-x}}[/tex]

    The interesting point here is that Leibnitz's rule cannot be used here as the integrand is discontinuous at [itex]x=\xi[/itex]. However, the integral can be simplified by parts leaving:

    [tex]\phi(\xi)=\frac{1}{\pi}\frac{d}{d\xi}\{2f(0)\sqrt{x}+2\int_0^\xi \sqrt{\xi-x}f^{'}(x)dx\}[/tex]

    Or using Leibnitz rule:

    [tex]\phi(\xi)=\frac{f(0)}{\pi \sqrt{\xi}}+\frac{1}{\pi}\int_0^\xi \frac{f^{'}(x)}{\sqrt{\xi-x}}dx [/tex]

    Wow, that's beautiful. Now it's time to begin experimenting with various functions [itex]f(x)[/itex] in Mathematica in an effort to better understand the behavior of this integral equation from the perspective of "I don't care about the application, only the math".
     
    Last edited: Feb 21, 2005
  7. Jul 8, 2008 #6
    For a numerical treatment (using FORTTRAN) visit wwwdotgeocitiessdotcomslashserienumerica
     
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