# The Abel equation

1. Feb 18, 2005

### 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.

2. Feb 19, 2005

### 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.

3. Feb 20, 2005

### 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.

4. Feb 20, 2005

### 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 $\xi$.

5. Feb 21, 2005

### 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".

Last edited: Feb 21, 2005
6. Jul 8, 2008

### reiabretti

For a numerical treatment (using FORTTRAN) visit wwwdotgeocitiessdotcomslashserienumerica