Determing points of singularity

**Pinkk** · November 18th 2009, 01:24 PM

I do not know how to determine if $x=0$ is a regular or irregular singular point for $x^{2}y'' + 2(e^{x}-1)y' + (e^{-x}cos(x))y=0$ . How do I determine if $2\frac{e^{x}-1}{x}$ and $e^{-x}cos(x)$ are analytic?

**shawsend** · November 18th 2009, 03:59 PM

For $y''+P(x) y'+Q(x) y=0$ , we wish to know if $P(x)$ has a pole of order greater than one or $Q(x)$ has a pole of order greater than two. If so, then the singular point is irregular. We can find out by simply taking limits that would cancel a simple pole in the former case or a double pole in the later case. So in the case above with the singular point at x=0, we evaluate the following limits:

$\lim_{x\to 0} x P(x)$

$\lim_{x\to 0} x^2 Q(x)$

If both limits exists, then $P(x)$ cannot have a pole of order greater than one, and $Q(x)$ cannot have a pole greater than two. So just take the limits and see what happens.

**Pinkk** · November 18th 2009, 04:40 PM

Hmm, but my professor said that we can only take limits if all the coefficients are polynomials; if there are non-polynomial coefficients, then we must see if the functions $xp(x)$ and $x^{2}q(x)$ are analytic; according to him we cannot take the limits then.

**shawsend** · November 19th 2009, 12:44 AM

You have:

$y''+\frac{2(e^x-1)}{x^2} y'+\frac{\cos(x)}{e^x x^2} y=0$

and $x\frac{e^x-1}{x^2}$ is analytic at zero. It has a removable singularity there. The term $x^2\frac{\cos(x)}{e^x x^2}$ is also analytic at zero. Your DE has a regular singular point at the origin. We can see a dramatic difference between ordinary and irregular singular points by considering the two DEs:

$y''+\frac{1}{x^3}y=0$

$y''+\frac{1}{x^2}y=0$

The first has an irregular singular point and the second has a regular one. I've plotted the real components of the complex solutions of both close to the origin. Note the qualitative differences: that folding pattern in the first continues infinitely often in any neighborhood of the origin and is the classic morphology of an essential singularity. However, I'm not sure if all DEs with irregular singular points give rise to solutions with essential singularities. I suspect so though.

**HallsofIvy** · November 19th 2009, 04:53 AM

Originally Posted by Pinkk

I do not know how to determine if $x=0$ is a regular or irregular singular point for $x^{2}y'' + 2(e^{x}-1)y' + (e^{-x}cos(x))y=0$ . How do I determine if $2\frac{e^{x}-1}{x}$ and $e^{-x}cos(x)$ are analytic?

Well, I would recommend using the definition of "analytic" which is that a (real) function, f, is analytic at x= a if and only if the Taylor's series for f(x) about x= a exists and converges to f(x) in some neighborhod of a. You should know that both $e^{-x}$ and cos(x) have Taylor's series that converge to them for all x and so their product does also.

As for $2\frac{e^x-1}{x}$ , it is clearly analytic everywhere except, possibly, at x= 0. Now, strictly speaking it is not analytic at x= 0 because it is is not defined there.

But what you really mean is the function that is $2\frac{e^x-1}{x}$ for x not equal to 0 and is equal to 0 when x= 0. The function $e^x$ has Taylor's series, around x= 0, $1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot$ . The function $e^x- 1$ , then, has Taylor series, around x= 0, $x+ \frac{1}{2}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+\cdot\cdot\cdot$ . Finally, then, $\frac{e^x-1}{x}= 1+ \frac{1}{2}x+ \cdot\cdot\cdot+ \frac{1}{n!}x^{n-1}+ \cdot\cdot\cdot$ .

**Pinkk** · November 22nd 2009, 08:16 PM

Ah, okay. I think I understand now, thanks.

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