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
  • $\begingroup$ I'm not quite sure why there's the insistence of multiplying the second derivative term by $x^2$ before applying Frobenius, but you don't really need that. All you have to do is assume that your function has an appropriate power series ansatz, which you then substitute into your DE. Then, you match up coefficients. $\endgroup$ – J. M. is a poor mathematician Jul 13 '12 at 8:10
  • $\begingroup$ That's what I thought. $\endgroup$ – Ziggy Jul 13 '12 at 18:16
  • $\begingroup$ For question (1), now somebody remains us for a better approach math.stackexchange.com/questions/362089 because of the effect of the cauchy product. $\endgroup$ – doraemonpaul Apr 16 '13 at 0:42
  • $\begingroup$ 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. $\endgroup$ – SasQ Oct 16 '13 at 1:52

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

  • $\begingroup$ Wow! Instant love from me for sure. $\endgroup$ – Ziggy Oct 30 '13 at 16:00

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.

  • $\begingroup$ 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. $\endgroup$ – Ziggy Jan 4 '13 at 17:43

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