# 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

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