Frobenius Method O.D.E

1. Sep 11, 2013

T-chef

1. The problem statement, all variables and given/known data
The task is to find an analytic solution to the O.D.E
$$4(1-x^2)y''-y=0 \hspace{20mm} y'(1)=1$$
by using an appropriate series solution about x=1.

3. The attempt at a solution
The singularity at x=1 is regular, which makes me think the Frobenius method is what's meant by appropriate series solution. But I've always done these about x=0, so I suppose the form would be
$$y=\sum_{n=0}^{\infty} a_n (x-1)^{n+r}$$

with

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

At this point I'd substitute into the O.D.E and find an indicial equation, but I'm not really sure how this works when we're expanding about x=1, or indeed if this is the right track. Any help or advice would be greatly appreciated!

2. Sep 11, 2013

Zondrina

That all looks good to me in terms of the concept, but there's a slightly easier way to do this. Whenever you have a regular point $x_0 ≠ 0$, substitute this :

$$t = x - x_0$$

This will make your life so much easier when dealing with this.

3. Sep 11, 2013

T-chef

Hey Zondrina! Cool, so making the substituion $t=x-1$ the D.E then becomes
$$-4(t^2+2t)y''(t)-y(t)=0$$
and the assumed solution is
$$y=\sum_{n=0}^{\infty} a_n t^{n+r}$$
so we end up with
$$4\sum_{n=0}^\infty a_n (n+r)(n+r-1)t^{n+r}+8\sum_{n=0}^\infty a_n (n+r)(n+r-1)t^{n+r-1}+\sum_{n=0}^\infty a_n t^{n+r}=0$$
Or perhaps more neatly,
$$\sum_{n=0}^\infty [4a_n (n+r)(n+r-1)+a_n]t^{n+r}+8\sum_{n=0}^\infty a_n (n+r)(n+r-1)t^{n+r-1} =0$$

Now, if I'm not mistaken (I might be :) ), this second sum will give me my indicial equation so I can find r?

4. Sep 11, 2013

Zondrina

You need to satisfy this equation for all $t$:

$$\sum_{n=0}^\infty [4a_n (n+r)(n+r-1)+a_n]t^{n+r}+8\sum_{n=0}^\infty a_n (n+r)(n+r-1)t^{n+r-1} =0$$

That is, the coefficient of each power of $t$ must be zero. This will yield your indicial equation and then your exponents of singularity.

5. Sep 12, 2013

T-chef

Alright, well, messing with that last sum:
$$\sum_{n=0}^{\infty}(4a_n(n+r)(n+r-1)+a_n)t^{n+r}+8\sum_{n=1}^{\infty}a_n(n+r)(n+r-1)t^{n+r-1}+8a_0(r)(r-1)t^{r-1}=0$$
So the lowest power is this $t^{r-1}$ fella, in which case $r(r-1)=0$. But from the wording of later parts in this question it seems there's only one solution to be found here. Would I be right in assuming this is from r=1, since r=0 would give a $t^{-1}$ term in the above equation, which is not the form of a power series?

Last edited: Sep 12, 2013
6. Sep 12, 2013

Zondrina

You need to be a bit more careful about the indices of your summations. They really do matter.

$-4(t^2+2t)y'' - y = 0$
$-4t^2y''- 8ty'' - y = 0$

$y = \sum_{n=0}^{\infty} a_n t^{n+r}$
$y' = \sum_{n=1}^{\infty} (n+r)a_n t^{n+r-1}$
$y'' = \sum_{n=2}^{\infty} (n+r)(n+r-1)a_n t^{n+r-2}$

Therefore the DE becomes:

$-4t^2 \sum_{n=2}^{\infty} (n+r)(n+r-1)a_n t^{n+r-2} - 8t\sum_{n=2}^{\infty} (n+r)(n+r-1)a_n t^{n+r-2} - \sum_{n=0}^{\infty} a_n t^{n+r} = 0$

$- 4 \sum_{n=2}^{\infty} (n+r)(n+r-1)a_n t^{n+r} - 8 \sum_{n=2}^{\infty} (n+r)(n+r-1)a_n t^{n+r-1} - \sum_{n=0}^{\infty} a_n t^{n+r} = 0$

$- 4 \sum_{n=2}^{\infty} (n+r)(n+r-1)a_n t^{n+r} - 8 \sum_{n=1}^{\infty} (n+r+1)(n+r)a_n t^{n+r} - \sum_{n=0}^{\infty} a_n t^{n+r} = 0$

Okay, this should be nothing too far from what you've already done, but there's still a problem to be tackled. Notice that all the powers of $t$ are the same, which is what we want. The indices of the three summations are completely different though, which needs to be rectified to be able to find the indicial equation.

Notice you can re-write the second sum as:

$8 [\sum_{n=1}^{\infty} (n+r+1)(n+r)a_n t^{n+r}] = 8[(r+2)(r+1)a_1t^{r+1} + \sum_{n=2}^{\infty} (n+r+1)(n+r)a_n t^{n+r} ]$

Do the same for the third sum. What does your equation look like now? Can you combine the sums effectively now?

Last edited: Sep 12, 2013
7. Sep 12, 2013

T-chef

$$\sum_{n=0}^{\infty} a_n t^{n+r} = \sum_{n=2}^{\infty} a_n t^{n+r}+a_0t^r+a_1t^{r+1}$$
Putting this, and your expansion into the D.E we can group all the sums, since they start at n=2,
$$\sum_{n=2}^{\infty} [4(n+r)(n+r-1)+8(n+r)(n+r+1)+1]a_nt^{n+r}+8(r+2)(r+1)a_1t^{r+1}+a_0t^r+a_1t^{r+1}=0$$
$$\sum_{n=2}^{\infty} [4(n+r)(n+r-1)+8(n+r)(n+r+1)+1]a_nt^{n+r}+[8(r+2)(r+1)+1]a_1t^{r+1}+a_0t^r=0$$
How's this look :) ? This suggests a_0=0, and then equating the coefficient of $t^{1+r}$ to zero gives non-integer r, is that okay?

8. Sep 12, 2013

Zondrina

Lookin good.

Now you have to solve for the roots and satisfy the equation for all t. That means even if $t≠0$, the equation still has to be zero.

So $a_0 = 0$ is good.

Now $8(r+2)(r+1) + 1 = 0$ yields $r_1 = \frac{1}{4} (\sqrt{2} - 6)$ and $r_2 = \frac{1}{4} (-\sqrt{2} - 6)$.

These are the exponents of singularity.

Now, solve the equation for the coefficient of $t^{n+r}$ and simplify it. This yields your recurrence relation for $n≥1$.

Now you have to break this down into cases to find the general $a_n$ term. One for the first root and one for the second root and for each case you have to test values of $n≥1$.

9. Sep 12, 2013

vela

Staff Emeritus
I haven't read through this thread carefully, but generally, this condition isn't correct. By assumption, $a_0 \ne 0$, otherwise you'd just replace r by r+1. In other words, r is defined as the value for which $a_0 \ne 0$.

10. Sep 12, 2013

Zondrina

Yes, the assumption is usually $a_0 ≠ 0$.

I found it odd that $a_0t^r$ was grouped. The problem is if $t≠0$, then $a_0t^r ≠ 0$, which prevents the solving of the equation. It has to be valid for all $t$ I believe.

11. Sep 12, 2013

vela

Staff Emeritus
That's the right indicial equation. The two roots differ by an integer, so the larger value will yield a solution. The smaller value may or may not give you a second independent solution. It depends on how the coefficients work out.

It's okay to have a term like t-1 in your solution, so that's not a reason to discard the root r=0. After all, you're expanding about a singular point. It wouldn't be surprising if the solution diverged around there too.

12. Sep 12, 2013

vela

Staff Emeritus
You're thinking of what happens when you differentiate a Taylor series. With the method of Frobenius, you can't assume the lowest order terms vanish because of $r$.

For example, suppose r=2, the series would be of the form
$$\sum_{n=0}^\infty a_n x^{r+n} = a_0 r^2 + a_1 r^3 + \cdots.$$ If you differentiate twice, you'd get
$$\sum_{n=0}^\infty a_n (r+n)(r+n-1) x^{r+n-2} = 2 a_0 + 6 a_1 r + \cdots.$$ If you started with n=2, the sum would be missing the constant and linear terms.