Solving Abel Equation: Step-by-Step Guide

[saltydog]

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:

f(x)=\int_0^x{\frac{\phi(y)}{\sqrt{x-y}}}

With f(x) given and \phi(x) 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:

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

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

\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

The order of integration on the RHS needs to be changed next.

 

[saltydog]

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:

\int_0^\xi\int_0^x f(x,y;\xi)dydx

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

\int_0^\xi\int_y^\xi f(x,y;\xi)dxdy

And thus:

\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

The transformed integral equation becomes:

\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

The next step is to find a suitable change of variable that will facilitate evaluating the first iterated integral.

 

[saltydog]

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

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

Simplifying, we get:

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

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:

\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

The first iterated integral is easily solved.

 

[saltydog]

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 \frac{dx}{\sqrt{\xi-x}} and then integrating from 0 to \xi.

 

[saltydog]

Continuing:

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

So that:

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

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

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

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

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

Or using Leibnitz rule:

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

Wow, that's beautiful. Now it's time to begin experimenting with various functions f(x) 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".

 

[reiabretti]

For a numerical treatment (using FORTTRAN) visit wwwdotgeocitiessdotcomslashserienumerica