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Suppose I have an equation

$y" + P(x)y' + Q(x)y = 0$

Now we apply power series when $P$ and $Q$ are analytic at $x=0$ and apply Frobenius method when $P$ and $Q$ are not analytic at $x=0$.

Now, I want to know why do we apply the Frobenius method? We could equally have taken a power series in terms of $(x-a)$, where $P$ and $Q$ would analytic at $a$ and $a$ could be anything i.e $(2, 3 ...100)$. But we don't do that. We always use the Frobenius Method at $x=0$.

So why do we not do that? Is that wrong? Why do we always look for a series centered at $x=0$?

Is it necessary that we find the series centered at $0$? And how does the multiplication of $x^r$ in the Frobenius method with our normal power series correct everything ?

Can we use power series at points other than $0$ or Frobenius Method is the only way out?

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  • $\begingroup$ Substitute in the power series and its derivatives into to ODE, factor out and equate coefficients. You can then solve for a general solution. $\endgroup$
    – McTaffy
    Feb 11, 2017 at 22:40
  • $\begingroup$ @Sam That is ok. What I want to know is that we can use power series centered at some other point rather than 0 where P , Q are analytic instead of using Frobenius Method and having a series centered at 0. We could use a series in powers of ( x -a) , where P(x) and Q(x) get analytic . Why don't we do this ? $\endgroup$
    – Shashaank
    Feb 12, 2017 at 6:12
  • $\begingroup$ We do in some cases. Taylor series are used to centralize the harmonic oscillator in a square well. Usually the spatial analytics are done before the system's. $\endgroup$
    – McTaffy
    Feb 12, 2017 at 10:23
  • $\begingroup$ @Sam So if I have an equation in which P(x) and Q(x) are not analytic about x=0 , can I form a power series in , say terms of (x-1) or (x-2) or (x-100)... Where P and Q are analytic at 1 , 2 ...or 100. Would that be correct to use the power series there instead of Frobenius Method at x=0 $\endgroup$
    – Shashaank
    Feb 12, 2017 at 10:30
  • $\begingroup$ @Sam Sorry , I don't know much about darboux region much. The only thing I wanted to know is that if we are only given with a differential equation and P and Q are not analytic at 0 , can we form the power series in terms of (x- a) where P and Q are analytic at a. Or is it absolutely necessary to use the Frobenius Method at x=0. Could you please , if you wish write an answer explaining more on this because this has been a question for past 2 weeks. Thanks $\endgroup$
    – Shashaank
    Feb 12, 2017 at 10:39

2 Answers 2

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Now we apply power series when $P$ and $Q$ are analytic at $x=0$ and apply Frobenius method when $P$ and $Q$ are not analytic at $x=0$.

Note that the power series is just a special case of Frobenius series. You don't have to decide which of these methods to use — you just can make a shortcut if you see that $P$ and $Q$ are analytic at the point of expansion.

Now, I want to know why do we apply the Frobenius method? We could equally have taken a power series in terms of $(x−a)$, where $P$ and $Q$ would analytic at $a$...

Yes we can expand in powers of $(x-a)$ at any $a$ where $P$ and $Q$ are analytic. But there are some reasons why we usually try to find expansion at $x=0$ (or some other singular point) and not at some other, nonsingular, point:

  • We're often given initial/boundary conditions at $x=0$, and it is natural to choose the expansion which a priori satisfies the condition.

  • We may want to know analytic structure of the solutions at $x=0$. For example, our aim may be to calculate a numeric approximation to the solution. The solution often has a singularity at $x=0$, so that expansion in series at a nonsingular point of the equation will give a slowly convergent series with limited radius of convergence — due to proximity of the singularity. If we instead use Frobenius method, then e.g. for $P$ and $Q$ analytic everywhere except $x=0$ we'll get an everywhere-convergent series. In any case, Frobenius method lets you know asymptotic behavior of your solution, which you can use to speedup numeric calculations.

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  • $\begingroup$ Thanks ! I understand now. So I am dealing with this in my mathematical questions and there are no boundary conditions there. So it is all the right to think that we can use power series at some other other than 0 instead of using frobenius method at zero. At least that would be right mathematically because we haven't been given any boundary points or boundary conditions nor do we have to check the convergence. So in these cases it is alright to use power series at points other than zero instead of frobenius method. Is that right? $\endgroup$
    – Shashaank
    Mar 21, 2017 at 13:35
  • $\begingroup$ @Shashaank yes, if it is easier for you to avoid Frobenius method for a particular problem, in principle it's not wrong to go this way. $\endgroup$
    – Ruslan
    Mar 21, 2017 at 14:10
  • $\begingroup$ OK thanks I understand. Only one thing. Since the answers of the question are given using Frobenius Solution at 0 & if I use power series then I shall get another answer. So practically both are correct since both will satisfy the equation ! To enquire that the answer using power series at another x will be equally correct ? $\endgroup$
    – Shashaank
    Mar 21, 2017 at 14:35
  • $\begingroup$ @Shashaank since your equation is second order, there are two arbitrary constants, which you must choose to get a particular solution. When solving using Frobenius method around $x=0$ and when using series expansion around $x=a$ you'll in general not get the same solutions. In both cases it's possible to find such values of the arbitrary constants which will let you represent any solution of the equation. $\endgroup$
    – Ruslan
    Mar 21, 2017 at 18:32
  • $\begingroup$ Thanks a lot. All of my doubts are cleared. So that is why they are giving the exercise on Frobenius at x=0 even though we can use power series at other points where P , Q get analytic so that we can cope up with the usage of Frobenius when the boundary conditions are such other reasons. Is that the reason for forcing Frobenius at x=0 so that we are comfortable with it later , I guess. $\endgroup$
    – Shashaank
    Mar 21, 2017 at 18:39
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For an equation having singularity at $x=0$ (say), you can always obtain a power series solution at $x=a$ ($a\ne 0$). But on doing so, the interval of convergence will be $|x-a|<a$ (as $0$ is the nearest singularity and $a$ is center) or $0<x<2a$ which will not explicitly display the singular behavior of its solution $y(x)$ near $x=0$.

Moreover, for $y''+P(x).y'+Q(x).y=0$; the analyticity of $P(x)$ and $Q(x)$ at the point $x_0$ guarantees the analyticity of its solution $y(x)$ in the neighborhood of $x=x_0$, but to comment correctly about the behavior of solution $y(x)$ in the neighborhood of a point which is a singularity of $P(x)$ or $Q(x)$ we have to rely on Frobenius series.

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