Frobenius method with imaginary powers

1. Jan 30, 2005

StatusX

I need to solve a linear, second order, homogeneous ODE, and I'm using the Frobenius method. This amounts to setting:

$$y = \sum_{n=0}^{\infty} c_n x^{n+k}$$

then getting y' and y'', plugging in, combining like terms, and setting the coefficient of each term to 0 to solve for the cn's. This will give solutions in the form of infinite power series. Specifically, setting the first coefficient to 0 will result in a quadratic equation in k, which is then solved to get the starting powers for the two linearly independent solutions. However, in a problem I'm doing, which I can post if anyone wants, I get imaginary values for k. Is this valid? It means the series will have the form:

$$y = c_0 x^i + c_1 x^{i+1} + c_2 x^{i+2} + ...$$

With the cn complex. Will this converge to a real, finite number for all x when I plug in real boundary conditions?

2. Jan 30, 2005

dextercioby

Of course,not.That is if your initial problem asked to solve the ODE in a certain domain of R^{2} and gave initial conditions that didn't have complex numbers,then everything in the solution should be "real"...Especially with this method...Of REAL POLYNOMIALS/POWER SERIES...

Daniel.

P.S.If u do get complex coeff.& powers for an ODE in the reals,then something is rotten in your approach.

3. Jan 30, 2005

saltydog

Would you kindly post the ODE so I can look at the problem?
Thanks,
SD

4. Jan 30, 2005

StatusX

Ok, here's the equation:

$$x^2 \frac{d^2 y}{d x^2} + (x^2 + x) \frac{d y}{d x} + y = 0$$

And after doing what I described above, up to solving for k and cn, I got:

$$c_0 (k^2 + 1) x^k + \sum_{n=1}^{\infty} ( c_n ((n+k)^2 + 1) + c_{n-1} (n + k -1) ) x^{n+k} = 0$$

If I got this right, I need to set the coefficient of each power of x to 0. This gives me two values of k (which are usually real integers, but in this case come out to be +/- i), which I then plug in the other terms to get a recursive formula for cn in terms of cn-1. But, like I said, in this case I get complex powers and coefficients. It could conceivably reduce to a real function (maybe it's a fourier series?), but I think it's more likely I made a mistake somewhere.

Last edited: Jan 30, 2005
5. Jan 30, 2005

Hurkyl

Staff Emeritus
I guess I'm missing something -- you can choose any value for k and c0, and be able to solve for all of the ci.

6. Jan 30, 2005

StatusX

I'm not sure if by that you mean I should set c0 to 0 instead of k2+1, but I tried that, and it just means I need to solve for k using the c1 term. It comes out to be 1 less, and everything else shifts by 1 as well. Effectively, all it does is rename c0 to c1.

Last edited: Jan 30, 2005
7. Jan 30, 2005

saltydog

Thanks,
Sorry if I gave you the impression I could solve it quickly. These I've worked on before but they're tedious you know. I'll spend some time with it because I enjoy the math but probably not in time for you if you need the solution soon. Post the results if you get to it before me (with a plot if you can).

SD

8. Jan 30, 2005

StatusX

I'm sorry, I don't expect you to do the whole problem for nothing, it's just that my work is alot to type. But here it is:

as before:

$$y = \sum_{n=0}^{\infty} c_n x^{n+k}$$

and the equation is:

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

so from there:

$$x^2 \sum_{n=0}^{\infty} c_n (n+k)(n+k-1) x^{n+k-2} + (x^2+x)\sum_{n=0}^{\infty} c_n (n+k) x^{n+k-1} + \sum_{n=0}^{\infty} c_n x^{n+k}$$

$$\sum_{n=0}^{\infty} c_n (n+k)(n+k-1) x^{n+k} + \sum_{n=0}^{\infty} c_n (n+k) (x^{n+k+1} + x^{n+k}) + \sum_{n=0}^{\infty} c_n x^{n+k}$$

$$\sum_{n=0}^{\infty} c_n ( ( (n+k)(n+k-1) + (n+k) + 1) x^{n+k} + (n+k) x^{n+k+1})$$

$$\sum_{n=0}^{\infty} c_n ( (n+k)^2 + 1) x^{n+k} + \sum_{n=1}^{\infty} c_{n-1} (n+k-1) x^{n+k}$$

$$c_0 ( k^2 + 1) x^{k} + \sum_{n=1}^{\infty} c_n ( (n+k)^2 + 1) x^{n+k} + c_{n-1} (n+k-1) x^{n+k}$$

9. Jan 30, 2005

Hurkyl

Staff Emeritus
Hrm, now that I look at this problem... it's icky!

First off, it should not be surprising that a Taylor series would not work: the DE is singular at x=0. I find it somewhat more disturbing that a Laurent series doesn't work: that is,

$$y := \sum_{n=-\infty}^{n=\infty} c_n x^n$$

If I've done the calculation correctly, the resulting series never converges: the coefficients increase without bound as n decreases towards negative infinity.

Using an exponent of n+k doesn't seem to help.

So, I strongly suspect that you're dealing with some ugly transcendental function that has a branch cut.

Incidentally, it should be easy enough to expand around, say (x+1) instead of x: y should be nice and analytic on the positive reals (and on the negative reals)

10. Jan 31, 2005

saltydog

StatusX, sorry. I didn't mean all the steps. That was a lot of work. I meant only the solution once you were sure it was correct. You know, do a NDSolve in Mathematica then plot the results. Then plot the actual summation function and compare the two. To me that's proof (not rigorous) that the answer is correct. Although I think there may be some problem relating the arbitrary constants c1 and c2 in the summation to the initial conditions selected for NDSOLVE

SD

Last edited: Jan 31, 2005
11. Jan 31, 2005

saltydog

Alright I'm stuck. I ain't proud: The indicial equation (c^2+1)=0 has imaginary roots +- i. Now I see what you're talking about. I'm not sure how to handle this case but I'm optimistic it can be done. I think we need to use Re(c) in some way, i.e., 0. I'll check more ODE reference books to see about it.

SD

12. Jan 31, 2005

StatusX

I thought about it some more, and the complex series should satisfy the DE if it converges. It turns out that the same derivative rule holds for complex powers of x (that is, d/dx(x^(a+bi)) = (a+bi)x^(a+bi-1) ) so the method should still work. What's more, the two complex solutions I got turned out to be conjugates of eachother, so adding them gets one real solution, and subtracting them gets a pure imaginary which can then be multiplied by i to get another, linearly independent real solution. The ones I got looked like sin(ln(x))*(power series) + cos(ln(x))*(different power series). All my steps seemed pretty sound, but when I plugged the first few terms into the DE on my calculator, it didn't seem to work. It might get better with more terms, I don't know, but I already turned it in. Anyway, thanks for your help. You can keep working on it if you want, and I'll post the solution when I get it.

13. Jan 31, 2005

saltydog

Thanks StatusX. Think I'll spend some time with them just for fun (been a while for me). Please do post the solution as I would and perhaps the group too would like some closure in the matter. I can also check my results, which may take some time, against yours if you do.

Salty

14. Jan 31, 2005

Hurkyl

Staff Emeritus
Complex exponentials don't satisfy the laws of exponents. E.G. $a^b a^c = a^{b+c}$ can fail.

If you think you have a form for the solution, you could try directly plugging it in...

15. Jan 31, 2005

StatusX

First of all, could you explain how that rule doesn't always hold? I thought it did (not that I think it matters, but the base is always real in this case). And as for plugging it in, the solution I got involves infinite power series, and I can't solve for the nth coefficient, I only have a recurrence relationship.

Also, they just posted the solutions, and they left in the x^i terms:

$$y_1 = a_0 x^i ( 1 + \frac{1}{1+2i} x + \frac{i-1}{12i-4} x^2 + ...)$$

$$y_2 = b_0 x^{-i} ( 1 + \frac{1}{1-2i} x + \frac{-i-1}{-12i-4} x^2 + ...)$$

with the generating equation:

$$a_n = \frac{n + i -1}{n^2 + 2ni} a_{n-1}$$

and the conjugate of that for the bn equation.

My solutions were (y1+y2) and -i (y1-y2), with a0=b0, which came out to be real.

Last edited: Jan 31, 2005
16. Feb 1, 2005

Hurkyl

Staff Emeritus
Bleh, that one works. I can show counterexamples for other properties though:

(-1)^i = exp(i ln -1) = exp(i (pi i)) = exp(-pi)
(-1)^i * (-1)^i = exp(-2pi)
(-1 * -1)^i = 1^i = exp(i ln 1) = exp(i 0) = 1
This contradicts (a^c) (b^c) = (ab)^c

(-1)^i = exp(-pi)
((-1)^i)^i = exp(i ln(exp(-pi))) = exp(i (-pi)) = exp(-pi i)
(-1)^(i*i) = (-1)^(-1) = -1

And, incidentally, for complex exponents, (-1)^(1/3) = (1 + i &radic;3)/2, not -1.

(I've used the principal value for everything)

17. Feb 1, 2005

arildno

At least one of your independent solutions ought to have the following form, according to Fuch's theorem:
$$Y=ln^{\alpha}(x)\sum_{i=0}^{\infty}x^{i+\beta}$$
I'll get back to this one.

18. Feb 1, 2005

saltydog

Jesus, what a solution! And I though too that that laws of exponents apply to complex numbers although logs are multi-valued (I need to check on that also). It's going to take me time just to "digest" the answer and prove to myself that a real solution can come from it. Think I'll play with it in Mathematica. Very interesting. Thanks for taking time to post it.

Salty

19. Feb 1, 2005

saltydog

Alright thanks, didn't see it when I made the above post. I'll go through it carefully to convince myself.

Salty

20. Feb 1, 2005

saltydog

$$a_1 = \frac{ i}{1 + 2i}$$
$$b_1 = \frac{- i}{1 - 2i}$$