# Solve Using Frobenius Method

1. Oct 3, 2013

### cjc0117

1. The problem statement, all variables and given/known data

Solve $x(1-x)\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+2y=0$ using the Frobenius Method.

2. Relevant equations

$R(x)\frac{d^{2}y}{dx^{2}}+\frac{1}{x}P(x)\frac{dy}{dx}+\frac{1}{x^{2}}V(x)y=0$

$R_{0}s(s-1)+P_{0}s+V_{0}=0$

$y=\sum^{∞}_{m=0}a_{m}x^{m+s}$

$y'=\sum^{∞}_{m=0}a_{m}(m+s)x^{m+s-1}$

$y''=\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-2}$

3. The attempt at a solution

First, I divided everything by x. The diff. eq. becomes:

$(1-x)\frac{d^{2}y}{dx^{2}}-\frac{2}{x}\frac{dy}{dx}+\frac{2}{x}y=0$

It follows that $R(x)=1-x$, $P(x)=-2$, and $V(x)=2x$. Thus, $R_{0}=1$, $P_{0}=-2$, and $V_{0}=0$. The indicial roots are then $s=0$ and $s=3$.

I then plug $y=\sum^{∞}_{m=0}a_{m}x^{m+s}$ and its derivatives into the original diff. eq. and find the recurrence relation:

$(1-x)\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-2}-\frac{2}{x}\sum^{∞}_{m=0}a_{m}(m+s)x^{m+s-1}+\frac{2}{x}\sum^{∞}_{m=0}a_{m}x^{m+s}=0$

$\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-2}-\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-1}-2\sum^{∞}_{m=0}a_{m}(m+s)x^{m+s-2}+2\sum^{∞}_{m=0}a_{m}x^{m+s-1}=0$

$\sum^{∞}_{m=0}a_{m+1}(m+s+1)(m+s)x^{m+s-1}-\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-1}-2\sum^{∞}_{m=0}a_{m+1}(m+s+1)x^{m+s-1}+2\sum^{∞}_{m=0}a_{m}x^{m+s-1}=0$

$a_{m+1}(m+s+1)(m+s-2)-a_{m}[(m+s)(m+s-1)-2]=0$

$a_{m+1}=a_{m}$

I don't understand what this result means (if I even did it right) and how I would continue from here.

2. Oct 3, 2013

### SteamKing

Staff Emeritus
It suggests that there is a single coefficient 'a' for all the terms in the series, or

y = a * SUM (x^(m+s)) for m = 0 to INF

3. Oct 4, 2013

### vanhees71

It's nearly correct (as far as I can infer from the analytic solution, I got with Mathematica, and which you will easily find when you have completed your work with the Frobenius method). You simply forgot the first thing you always should figure out first, namely to get $s$!

In the most simple case you should get two solutions for $s$ and your method of comparison of coefficients gives the solution as a (generalized power series) for each of these solutions (there are exceptions, where in this way you get only one solution, but this is not the case here) and you found then two linearly independent solutions of your 2nd-order ODE and thus already the general solution.

In your case you can even resum the very simple series to closed analystical expressions.

4. Oct 5, 2013

### cjc0117

Thanks for the replies. I had found the two s values to be $s_{1}=3$ and $s_{2}=0$. $s_{1}-s_{2}$ is equal to a positive integer. I thought in this case, there are two independent solutions:

$y_{1}=\sum^{∞}_{m=0}a_{m}x^{m+s_{1}}$
(Eqn. 1)

$y_{2}=ky_{1}ln(x)+\sum^{∞}_{m=0}b_{m}x^{m+s_{2}}$
(Eqn. 2)

I thought this was because when $s_{1}-s_{2}$ $(s_{1}>s_{2})$ is equal to a positive integer, you end up getting a trivial solution when trying to find an independent solution in the form of Eqn. 1 for $s_{2}$.

However, the recurrence relation $a_{m}=a_{m+1}$ does not result in a trivial solution for either $s_{1}$ or $s_{2}$ when using independent solutions in the form of Eqn. 1. The general solution for this particular problem seems to be:

$y=a_{0}\sum^{∞}_{m=0}x^{m+3}+b_{0}\sum^{∞}_{m=0}x^{m}$

This turns out to be true if you just let both independent solutions take the form of Eqn. 1 right from the get go, or if you go the long way and only let $s_{1}$ take the form of Eqn. 1, and then let $s_{2}$ take the form of Eqn. 2 (because it turns out that k ends up being equal to zero and the recurrence relation ends up still being $b_{m+1}=b_{m}$)

I just don't get how you can know whether you'll have to let one of your independent solutions take the form of Eqn. 2. Do you just have to see through trial and error whether you'll get a trivial solution or not?

5. Oct 6, 2013

### vanhees71

Yes, that's all correct, and you have to figure out your $k$ from the equation. Since you've found the solution now, I think it's ok to redo the full problem in a systematic way to get the last uncertainties solved. I hope I don't violate the rules of the homework forum with this, but I'm convinced that at this point of understanding of the OP it helps most to give a systematic application of the Frobenius-Fuchs theorem for this nice example.

The equation to solve was (written in standard form)
$$(1-x)y''(x)-\frac{2}{x} y'(x)+\frac{2}{x}y(x)=0.$$
This indeed fulfills the conditions of the theorem for the singular point $x=0$ (note that there is another singular point at $x=1$ here!). So here we have the simplifying case that we don't need series expansions for the coefficient functions as stated already in the OP.

So the Frobenius ansatz must lead to at least one solution of the form
$$y_1(x)=x^s \sum_{k=0}^{\infty} a_k x^k.$$
To find $s$ and the recursion for the $a_k$ we plug in the ansatz into the ODE, finding after rearrangement of the summation index and dividing by $x^{s-2}$
$$s(s-3)+\sum_{k=1}^{\infty} (s+k)(s+k-3)[a_k - a_{k-1}] x^{k}=0.$$
It's clear that this implies the two solutions $s=0$ and $s=3$ for the leading behavior of the solutions. So both solutions are analytic in $x=0$, i.e., the weak singularity is integrable here.

Further we are lucky in this case and can circumvent the somewhat cumbersome case that we need $k \neq 0$, because in this case for both values for $s$ the coefficients can be set $a_k \equiv 1$ to solve the recursion equation, which gives the two solutions
$$y_1(x)=x^3 \sum_{k=0}^{\infty} x^k=\frac{x^3}{1-x}, \quad y_2(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}.$$
Since these are obviously linearly independent functions, because $y_1(x)/y_2(x) = x^3 \neq \text{const}$ the general solution of the ODE is
$$y(x)=(a_1+a_2 x^3) \frac{1}{1-x}.$$

In the general case for $m=s_1-s_2 \in \mathbb{N}_0$ only the first solution with the larger $s_1$ is given by the Frobenius ansatz, while the second solution must be found by the ansatz given in your last posting. For the case $s_1=s_2$ you can set $k=1$, and then the $b_k$ are uniquely determined with $b_0=0$, which is always a possible choice for the initial condition for the recursion for the $b_k$ in this case. For $m \in \mathbb{N}$ the $k$ must be determined from the ODE, and it is always uniquely defined (given the choice of the solution $y_1$), and you can always set $b_m=0$, and then all coefficients $b_k$ are uniquely determined by further setting $b_0=1$.