# Frobenius method questions

1. Sep 1, 2010

### TheFerruccio

I am having an issue understanding the Frobenius method. I plan on posting multiple questions in this thread as I go. I searched these forums and saw some other threads pointing to it, but the questions were slightly different. I am able to get through the problem to a certain extent, but after a point, I don't know where to go on.

Example 1:

$$x^{2}y''-3xy'+4y = 0$$

This can be put into the format:

$$y''-\frac{3x}{x^{2}}y'+\frac{4}{x^{2}}y$$

Which is consistent with the form that can be solved using the Frobenius method.

I assume the solution $$y(x)$$ and its derivatives to be in this form, multiplying by the power $$x^{r}$$:

$$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}$$
$$y'(x)=\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r-1}$$
$$y''(x)=\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r-2}$$

I plug these into the original equation, getting:

$$\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r}-3\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r}+4\sum_{n=0}^{\infty}a_{n}x^{n+r} = 0$$

I find that the lowest power in this equation is $$x^{r}$$. Plugging $$n=0$$ into the equation, and pulling out $$x^{r}$$, I get:

$$(r-1)ra_{0} - 3ra_{0} + 4a_{0} = 0$$
$$a_{0}(r^{2}-r-3r+4) = 0$$
$$a_{0}(r^{2}-4r+4) = 0$$

Assuming $$a_{0} \neq 0$$, making the solution non-trivial, I get the roots for r, which are $$r_{1},r_{2} = 2$$

I use one of the roots, r=2, and plug in (here's my attempt to get the TeX to work again)

$$\sum_{n=0}^{\infty}(n+1)(n+2)a_{n}x^{n+2}-3\sum_{n=0}^{\infty}(n+2)a_{n}x^{n+2}+4\sum_{n=0}^{\infty}a_{n}x^{n+2}$$

I do not know where to go from this point. I cannot get $$a_{n+1}$$ or $$a_{n+2}$$ in terms of $$a_{n+1}$$ or $$a_{0}$$. Every single result results in some constant, multiplied by $$a_{n}$$ equals 0. So, at this point. I am lost.

NOTE: I don't want the answer. I just want some clues on how I should continue.

Last edited: Sep 1, 2010
2. Sep 1, 2010

### ross_tang

In fact, what you have show is that, $$x^2$$ is one of the solution. Since for all other terms other than $$a_0$$, they have to be zero in order for the expression to equal to zero.

3. Sep 2, 2010

### TheFerruccio

I'm a tad confused, because what I'm ending up with is a kind of useless term.

The $$x^{2}$$ terms in the last equation reduce to $$0a_{0} = 0$$ which doesn't tell me anything!

For the higher order terms, I do end up with some coefficient * $$a_{n}$$, which shows that the higher order terms have to equal 0.

I'm still left with not enough information, though. I'm confused as to how $$x^{2}$$ term can be a solution.

4. Sep 2, 2010

### ross_tang

Please substitute $$y = x^2$$ to your differential equation, and you will find that it is indeed one of the solution.

$$y = x^2$$
$$y' = 2x$$
$$y'' = 2$$

Therefore:

$$x^2 y'' -3x y' + 4y = 2 x^2 - 6 x^2 + 4x^2 = 0$$

5. Sep 2, 2010

### HallsofIvy

This can be written as a single sum:
$$\sum_{n=0}^\infty [(n+1)(n+2)- 3(n+2)+ 4)a_n= 0$$
so that you must have either $a_n= 0$ or $(n+1)(n+2)- 3(n+2)+ 4$$= n^2+ 3n+ 2-3n- 6+ 4= n^2= 0$
That is, $a_n= 0$ for all n> 0. Since your solution is of the form "$Cx^{n+ 2}$" that means that your solution is $Cx^2$.

And ross_tang has verified that, indeed, $y= Cx^2$ is a solution.

6. Sep 2, 2010

### TheFerruccio

Yes, I can see that it is verified that $$x^{2}$$ is a solution, just by plugging in, I'm still having trouble seeing how that answer was arrived at. Let me think a bit.

The ODE reduces to the form of $$\sum_{n=0}^{\infty}n^{2}a_{n}x^{n+2}$$ So either $$a_{n} = 0$$ or $$n^{2} = 0$$ for all n. It's been assumed that $$a_{n}\neq0$$ so $$n^{2}=0$$ So $$n=0$$ for the statement to be true. So, one solution becomes the form:

$$\sum_{n=0}^{\infty}a_{n}x^{n+2}$$ Thus, the first solution is $$a_{0}x^{2}$$?

I think I realized where my confusion came from. Of course the equation in the form of summations would reduce to 0, because that's not one of the solutions! That's the ODE. I was mixing up the solution with the ODE. I just put it in summation form to solve for $$n$$, then plugged that n back into $$y(x)$$.

Thanks for the help. If what I typed is correct, then I understand this problem so far. I will work on finding the second solution.

7. Sep 2, 2010

### LawlQuals

I am not sure I understand what you are saying with the whole $$n^2 = 0$$ business. Maybe a closer inspection of all posts on this thread would help me see. Put another way (or, perhaps the same way),

You found a value for $$r = 2$$ under the proviso that $$a_0 \neq 0$$. Your "recursive" relation (which relates the coefficients of one term to the coefficients of others), reduces to (as you have shown, with $$r = 2$$):

$$(const)*a_n = 0$$

so that all $$a_n$$ terms are indeed zero, except the one that you enforced to not be so ($$a_0$$ as you had stated). You can piece together the Frobenius solution then by noting you were seeking a solution of the form:

$$y(x) = \sum_{n = 0}^{\infty} a_n x^{n+r}$$

but $$r = 2$$,

$$y(x) = \sum_{n = 0}^{\infty} a_n x^{n+2} = y(x) = x^2 \sum_{n = 0}^{\infty} a_n x^{n} = x^2 \left( a_0 x^0 + a_1 x^1 + a_2 x^2 + \ldots\right)$$

you have just shown that $$a_0 \neq 0$$, while $$a_n = 0 ,\, n\neq 1$$. this means all terms vanish except $$a_0$$, so we have the above reducing to:

$$y(x) = a_0x^2$$

$$a_0$$ may have been the first coefficient, but in a Frobenius solution, the first term $$n = 0$$ corresponds to a power of $$x^{n+r} = x^{r} = x^2$$.

8. Sep 4, 2010

### TheFerruccio

Thanks for the help. I understand the process for doing this DE with its indicial equation having repeated roots of $$r$$ now. This time, it is a very specific question.

I have another question, regarding an ODE with two solutions. I understand what a basis of solutions is, and why you need at least two coefficients in a second order ODE to form the solution space. However, the following confuses me.

Given an ODE:

$$x^2y''-4xy'+(x^2+6)y = 0$$
$$y_1(x) = a_0x^2sin(x)$$
$$y_2(x) = A_0x^2cos(x)+A_1x^2sin(x)$$

Why does the $$A_1x^2sin(x)$$ term get removed from $$y_2$$? It's been said that it is $$y_1$$, so removing it simplifies the solution. However, how is $$A_1x^2sin(x)$$ the same as $$a_0x^2sin(x)$$ given that $$a_0$$ is not $$A_1$$? The book doesn't explain this part, nor do the solutions.

I understand that, if $$A_1 = a_0$$, then removing $$y_1$$ makes sense.

Last edited: Sep 5, 2010
9. Sep 5, 2010

### LawlQuals

Just to be clear, the fact that this differential equation happened to have a one-term solution has nothing to do the fact that equation held repeated roots in the parameter $$r$$. This was a coincidence. We are not even usually so fortunate to be able to express a Frobenius solution in closed form, which does not have anything to do with whether or not the roots are repeated. Did you end up finding the second solution to the first differential equation by the way? Seeking solutions by way of Frobenius' method is a delicate process in that how you go about finding those solutions strongly depends on how the two roots (if a second order ODE) $$r$$ are related (repeated roots, distinct, differ by an integer, perhaps I am preaching to the choir).

For your followup question, where and when does $$A_1x^2\sin x$$ get removed? It sounds like your confusion possibly is coming from the phrasing used in the textbook (which may have translated into the question you have posed, as I cannot follow exactly what the question really is.). What I suspect though is that your text provides the ODE. It then lists two representative solutions, $$y_1(x)$$ and $$y_2(x)$$:

$$y_1(x) = a_0 x^2\sin x$$
$$y_2(x) = A_0x^2\cos x + A_1 x^2\sin x$$

Put any of these into the ODE, and you may verify they are indeed solutions. It sounds like your text was trying to construct the most general form of a solution to an ODE from two known solutions $$y_1$$ and $$y_2$$. One characteristic of a linear differential equation is that a linear combination of any solutions is also a solution. So, that you may write a general solution as:

$$y(x) = y_1(x) + y_2(x) = a_0x^2\sin x + A_0x^2\cos x + A_1x^2\sin x = (a_0 + A_1)x^2\sin x + A_0\cos x = A_2 x^2\sin x + A_0x^2 \cos x$$

where $$A_2 = a_0 + A_1 = const$$, it really does not matter what the constant actually is, we are seeking a general solution. The most general form is of the above form, there is no need to represent

$$A_2x^2 \sin x$$ as $$\alpha x^2\sin x + \beta x^2\sin x$$ any more than there is reason to write $$\gamma x$$ as $$\kappa x + \epsilon x + \delta x + \ldots$$. These are redundant, you may consolidate them all into one constant.

I imagine that is what your textbook is getting at, but I could be off. In which case, a more exact excerpt from your text would help clarify the context.

Last edited: Sep 5, 2010
10. Sep 5, 2010

### TheFerruccio

Yes, I managed to answer the remainder of the previous problem in the first part of the thread.

I guess I am still having trouble understanding what, exactly, these solutions mean for a differential equation. It looks to me like you combined $$y_1(x)$$ and $$y_2(x)$$ to equal something that looks exactly like another $$y_2(x)$$, especially given that it doesn't matter what the constants are. Why is $$y_1(x)$$ even necessary to list as a solution, anymore? Adding it to $$y_2(x)$$ seems to just result in $$y_2(x)$$.

What do you mean by linear combination?

I'm sorry, but I'm left with even more questions than before. Now, I'm completely lost as to the meaning of these solutions, when I previously thought I had some grasp. All I know is that it can be verified that they are solutions, and I have verified them. I just don't understand what constitutes a basis of solutions.

I was told that a basis of solutions is much like having basis vectors in space, sufficient to describe every point in that space by multiplying those vectors by some constant. I'm having trouble seeing how this carries over to a basis of solutions.

11. Sep 5, 2010

### LawlQuals

No problem about being confused, I should be clearer.

I do not have your text in front of me so I cannot gauge the actual context the author is presenting without speculation, but what I was suggesting is that the text's motive was to list two candidate solutions $$y_1$$ and $$y_2$$, it then proceeds to establish a basis in solution space (exactly as you understand it, both solutions constitute a basis to describe every solution in solution space constrained by any boundary conditions) by showing that the most general form of the solution is:

$$y(x) = b_0x^2\sin x + b_1 x^2\cos x$$ (This is a linear combination, a sum of two solutions multiplied each by arbitrary constants).

where $$b_0$$ and $$b_1$$ are arbitrary constants that may be uniquely determined given boundary conditions. I suspect what the author wants to do is to label the two solutions as $$y_1$$ and $$y_2$$, then we could write the most general solution to the differential equation as

$$y(x) = y_1(x) + y_2(x)$$

Proceeding, the author suggests two solutions that do happen to solve the differential equation, but the question is: are they distinct solutions, and further what is the general solution? Inputting the suggestions by the author into the above form:

$$y(x) = \underbrace{a_0x^2\sin x}_{y_1} + \underbrace{A_0x^2\cos x + A_1x^2\sin x}_{y_2}$$
$$= \underbrace{(a_0 + A_1)}_{A_2}x^2\sin x + A_0x^2\cos x$$
$$\Rightarrow y(x) = A_2x^2\sin x + A_0 x^2\cos x$$

is the most general representation of the above, since $$a_0$$ and $$A_1$$ can be combined. Nothing is being removed, the solution is preserved. All that was found was that $$y_1$$ + $$y_2$$ as written, was not the most general form of the solution since we had a repeated term ($$x^2\sin x$$). Now, we may relabel the solutions to be compact, where we write one independent solution in $$y_1$$ and the other as $$y_2$$,

$$y_1(x) = a_0 x^2\sin x$$ (does not have to be $$a_0$$, it can be anything, it is only a constant, I am just sticking to the notation given)
$$y_2(x) = A_0 x^2\cos x$$

Such that,

$$y(x) = y_1(x) + y_2(x)$$ is the most general form (no repeated terms, each solution is independent). That was all they were doing I believe. Alternatively, you could just relabel $$y_2$$ as having two solutions that are independent and discard $$y_1$$, same thing. The bottom line is just that you need two independent solutions, the extra term in $$y_2$$ had a little "extra" on it that was unneeded, since just the other term itself also satisfies the differential equation. Two solutions, independent of each other, need only be written as just above, there is no reason to have an extra term in there for $$y_2$$

I think the confusion you have is just that the author seems to be doing something that is possible backwards to your thinking. The thought experiment was to seek the general solution a posteriori, rather than working from the beginning towards the end. Given two solutions, construct the general solution (rather than, given an ODE, find the general solution. In which case, you would have found those two solutions anyway). In doing it the way the author did, given these two solutions, you can combine two of the terms as described above because they are, for all intents and purposes, the same.

Last edited: Sep 5, 2010
12. Sep 5, 2010

### TheFerruccio

Thanks for the information. I think this is beginning to make more sense. I will need to do some more practice to really nail down the concept of independent solutions, and what terms can be eliminated, but your help has given me lots of direction in understanding this.

13. Sep 5, 2010

### LawlQuals

Glad to hear it and happy to help.