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I find the Frobenius Method quite beautiful, and I would like to be able to apply it. In particular there are three questions in my text book that I have attempted. In each question my limited understanding has stopped me. Only one of these questions (the last) is assigned homework. The rest are examples I found interesting*.

1) $ L[y] = xy'' + 2xy' +6e^xy = 0 $ (1)

The wikipedia article begins by saying that the Frobenius method is a way to find solutions for ODEs of the form

$ x^2y'' + xP(x) + Q(x)y = 0 $

To put (1) into that form I might multiply across by x, giving me

$ x^2y'' + x[2x]y' + [6xe^x]y = 0 $ (2)

But is that OK? The first step in the method seems to be dividing by $ x^2 $, so can't I just leave the equation in its original form? I'll assume I can.

Now we let $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $

then, $ y_1' = \sum _{n=0}^{\infty} (r+n)a_nx^{r+n-1} $

and, $ y_1'' = \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-2} $

substituting into (2) we get,

$ x\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 2x\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 6e^x\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $

But now what? I am aware that $ 6e^x = 6\sum _{n=0}^{\infty}x^n/n! $, but my powers stop there. Can I multiply the two series together? I would have to multiply each term in one series by every term in the other, and I don't know how to deal with that. The text provides no worked examples in which P(x) or Q(x) are not polynomials... so for now my work stops here.

2) $ L[y] = x(x-1)y'' + 6x^2y' + 3y = 0 $

Again, I will leave the question in its original form, rather than try to get that x^2 in front (I realise I am not checking that the singular point is a regular singular point, but checking the answer in the back of the book, x = 1 and x = 0 are indeed regular points). With two regular singular points, I expect I will get 2 sets of answers: one near x = 1 and the other near x = 0. Is it enough to just proceed with one case and then the next? I will assume so, and begin with the case close to x = 0.

Again, letting $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $, and taking the appropriate derivatives, we find by substitution,

$ x(x-1)\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 6x^2\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 3\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $

$ x^2\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} - x\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 6x^2\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 3\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $

$ \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n} - \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=0}^{\infty}6(r+n)a_nx^{r+n+1} + \sum _{n=0}^{\infty}3a_nx^{r+n} = 0 $

we shift the indexes of the above sums, so that everything will be in terms of the same power of x.

$ \sum _{n=1}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + \sum _{n=1}^{\infty}3a_{n-1}x^{r+n-1} = 0 $

we synchronise the indexes in order to group like terms, by extracting early terms from each series,

$ r(r-1)a_0x^r + \sum _{n=2}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r - \sum _{n=2}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + 3a_0x^{r-1} + \sum _{n=2}^{\infty}3a_{n-1}x^{r+n-1} = 0 $

rearranging, we get

$ r(r-1)a_0x^r - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r + 3a_0x^{r-1} + \sum _{n=2}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - \sum _{n=2}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + \sum _{n=2}^{\infty}3a_{n-1}x^{r+n-1} = 0 $

At this point I expect the indicial equation to emerge, and I expect it to be similar to an Euler Equation. That is, I expect a polynomial that I can solve to get two 'exponents at the singularity'. Unfortunately, I cannot see an indicial equation and am at a loss to know precisely why.

3) $ L[y] = xy'' + y = 0 $

Finally we come to the assigned question, which I have been able to manipulate into an almost final form.

Again, letting $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $, taking derivatives, and substituting into L, we get

$ x\sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-2} + \sum _{n=0}^{\infty} a_n x^{r+n} = 0 $

$ \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=0}^{\infty} a_n x^{r+n} = 0 $

Now shifting indexes,

$ \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=1}^{\infty} a_{n-1} x^{r+n-1} = 0 $

and extracting the $ 0^{th} $ term of the first sum,

$ r(r-1)a_0x^{r-1} + \sum _{n=1}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=1}^{\infty} a_{n-1} x^{r+n-1} = 0 $

$ r(r-1)a_0x^{r-1} + \sum _{n=1}^{\infty}[(r+n)(r+n-1)a_n + a_{n-1}]x^{r+n-1} = 0 $

And voila! We have an indicial term with solutions $r_1 = 1$ and $r_2 = 0$, and a recurrence relation. From my text I expect that

$y_1 = |x|^{r_1}[1+\sum _{n=0}^{\infty}a_nx^n]$

and since $ r_1 - r_2 \in \mathbb{Z} $,

$y_2 = ay_1ln|x| + |x|^{r_2}[1 + \sum _{n=0}^{\infty}b_nx^n]$

to find $a_n$ we observe the recurrence relation with $ r = r_1 = 1 $,

$ (r+n)(r+n-1)a_n + a_{n-1} $

$ a_n = -a_{n-1}/n(n+1) $

so, $ a_1 = -a_0/2*1 $

$ a_2 = -a_1/2*3 = a_0/3*2*1*2*1 = a_0/3!2! $

$ a_3 = -a_2/3*4 = -a_0/4!3! $

and in general, $ a_n = (-1)^na_0/n!(n+1)! $

so we have $ y_1 = |x| + \sum _{n=0}^{\infty} (-1)^na_0x^{n+1}/n!(n+1)! $

Not so easily done with r = r_2 = 0, I'm afraid...

since the relation becomes $ a_n = -a_{n-1}/n(n-1) $, which means we can't have a_1 for fear of division by zero. Never the less, starting at n = 2 we get,

$ a_2 = -a_1/2*1 $

$ a_3 = -a_2/2*3 = a_1/3*2*1*2*1 = a_1/3!2! $

$ a_4 = -a_3/3*4 = -a_1/4!3! $

and in general, $ a_n = (-1)^{n-1}a_1/n!(n-1)! $

so we have $ y_2 = ay_1ln|x| + 1 + \sum _{n=0}^{\infty} (-1)^{n-1}a_1x^{n+1}/n!(n-1)! $

Which I feel may not be correct... and even if it is, how should one man solve for a in a single lifetime?

Thanks everyone for looking at this. I want to stress that I am not just a student looking for help in his homework: I would really like to understand this method because it appeals to me. I particularly like the way we extract the indicial expression from the sums, in order to synchronise them. That is so cool. And how you get 1 recurrence relation that you can use for both solutions: neat.

PS sorry if my Latex is not perfect? I'm just getting started with it.

  • questions taken from "Elementary Differential Equations and Boundary Value Problems" by William E. Boyce and Richard C. DiPrima (9th ed), from sectin 5.6 pp 290
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    For your equation (2) after synchronizing indexes you've extracted $3a_0x^{r-1}$ while for n=1 it should be: $3a_{n-1}x^{r+n-1} = 3a_{1-1}x^{r+1-1} = 3a_0x^r$ So all the terms extracted outside the sums should be: $r(r-1)a_0x^r - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r + 3a_0x^r$ Now you can collect them as: $[r(r-1)a_0 - r(r+1)a_1 + 3a_0] x^r - r(r-1)a_0x^{r-1} \\ [(r(r-1)+ 3)a_0 - r(r+1)a_1] x^r - r(r-1)a_0x^{r-1}$ and simplify a bit: $[(r^2-r+3)a_0 - (r^2+r)a_1] x^r - r(r-1)a_0x^{r-1}$ and you can get your indicial equation from this part of your equation.2013-10-16

2 Answers 2

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but my powers stop there. Can I multiply the two series together? I would have to multiply each term in one series by every term in the other, and I don't know how to deal with that.

I cracked my head about that a little while and I figured out how you can multiply two infinite series without getting lost. Here's an animation explaining my technique:

enter image description here

  • 0
    Wow! $I$nstant love from me for sure.2013-10-30
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I'd prefer give a self-contained solution, but (as it seems you are already well aware of) this requires a lot of typing.

However, the answers to at least your (2) and (3) will be available once you work through the very well explained examples found here.

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    I appreciate the link, and I will surely read the given PDF and work the examples. However, I took a class on ODE's which had a textbook, and I did all the examples there. The reason I came to stack overflow was that the answers to my questions weren't available, even after careful study.2013-01-04