# Homework Help: Series solution with regular singular points?

1. Jul 18, 2015

### Destroxia

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

$x^{2}y'' + (x^{2} + 1/4)y=0$

3. The attempt at a solution

First I found the limits of a and b, which came out to be values of a = 0, and b = 1/4

then I composed an equation to solve for the roots:

$r^{2} - r + 1/4 = 0$ $r=1/2$

The roots didn't differ by an integer so the equation must take the form of

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

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

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

Now I plugged the corresponding derivatives into the differential equation:

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

Then, I distributed the x terms through the series

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

Now I pull out 2 terms from the \sum_{n=0}^\infty a_{n}x^{n+(1/2)} term:

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

Then of course I would find the recursive formula, but I just wanted to make sure this is the proper way to set everything up before I proceed with that part of the problem, as I always have issues with these series solutions.

2. Jul 18, 2015

### vela

Staff Emeritus
What are $a$ and $b$ supposed to be? You didn't define them anywhere.

This is called the indicial equation.

You have a double root, so the roots differ by 0, which is an integer.

Looks good so far.

3. Jul 19, 2015

### Destroxia

Sorry, I didn't define the a and b because I figured I knew limits well enough, but I could definitely be wrong about that, just so you can make sure I'm not doing anything incredibly stupid, I will post them (I also didn't mean to write "take the limits of a and b", I meant to write take the limit of the second term divided by the first, multiplied by x, as it goes to zero, and define that as a , and the limit of the 3rd term divided by the 1st term, multiplied by $x^{2}$, as it goes to zero, and define that as b)

$a = \lim_{x\to 0} x \frac {0} {x^2} = 0$

$b = \lim_{x\to 0} {x^2} \frac {x^{2}+\frac {1} {4}} {x^{2}} = \lim_{x\to 0} x^{2}+\frac {1} {4} = \frac {1} {4}$

$a=0 , b=\frac {1} {4}$

Then to define how I created the indical equation, I defined it as:

$r^{2} + (a-1)r + b = 0$

$r^{2} - r + \frac {1} {4} =0$

So my only question now is, even though I made the mistake of thinking the double root wasn't separated by an integer, my work still holds up right? Just to get my $y_{2}$ value now I must use the reduction of order formula?

4. Jul 19, 2015

### Destroxia

Also, I forgot to mention...

When finding the recurrence relation what do I do with the $\frac 1 4 a_{0} x^{\frac 1 2}$ and $\frac 1 4 a_{1} x^{\frac 3 2}$ terms? Since they don't have the same x value, they can't be set in the same equation to 0, so does that mean that both $a_0$ and $a_1$ are 0???

5. Jul 19, 2015

### vela

Staff Emeritus
The derivatives you calculated are incorrect. I missed that the first time. You can't drop the first and second terms in the series because $r \ne 0$. The derivative of the lowest-order term, $x^{1/2}$, doesn't vanish. That should clear up the confusion you have with $a_0$ and $a_1$.

6. Jul 19, 2015

### Destroxia

I'm confused... I did the derivatives again and I get the same thing, do you just mean that the index doesn't change as I take the derivatives? So would the derivatives be the same but n=0 always?

7. Jul 19, 2015

### vela

Staff Emeritus
Right. Try writing out the first few terms of the series to see why. Contrast it to what happens to when you have $r=0$ so that $y = a_0 + a_1 x + a_2 x^2+\cdots$.

8. Jul 19, 2015

### Noctisdark

I can give some approximisation, y'' = -(1 + 1/(4x2)y, if x is being to big then y'' ≈ - y this suggest that y ≈ Acos(x) + Bsin(x), now if x2 is small enough, then 4x2y'' ≈ -y, which suggest that y≈A√x + B√x*ln(x),

Last edited: Jul 19, 2015
9. Jul 19, 2015

### Destroxia

I understand now, but am still having the same issue with $a_0$ and $a_1$ ending up equaling 0

After changing the derivatives to all be n=0, I start out like so, after plugging into the differential equation:

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

Then I distribute:

$\sum_{n=0}^\infty (n - \frac 1 2)(n+\frac 1 2)a_{n} x^{n+\frac 1 2} + \sum_{n=0}^\infty a_n x^{n+\frac 5 2} + \sum_{n=0}^\infty \frac {1} {4} a_n x^{n+\frac 1 2}$

Then I changed the 2nd series index to n=2 to make its $x^{n+\frac 5 2}$ into $x^{n+\frac 1 2}$ and following that I took 2 terms out of both the 1st and 3rd series to make all the indexs n=2 so I can set the coefficients to 0:

$-\frac {1} {4} a_0 x^{\frac 1 2} +\frac 3 4 a_1 x^{\frac 3 2}+\sum_{n=2}^\infty (n - \frac 1 2)(n+\frac 1 2)a_{n} x^{n+\frac 1 2} + \sum_{n=2}^\infty a_{n-2} x^{n+\frac 1 2} + \frac 1 4 a_0 x^{\frac 1 2} + \frac 1 4 a_1 x^{\frac 3 2} + \sum_{n=2}^\infty \frac {1} {4} a_n x^{n+\frac 1 2}$

when I put the outside terms together $a_0$ and $a_1$ still equal 0

$-\frac {1} {4} a_0+ \frac 1 4 a_0 =0$ ends up being $a_0 = 0$

$\frac 3 4 a_1 + \frac 1 4 a_1 = 0$ ends up being $a_1 = 0$

10. Jul 19, 2015

### vela

Staff Emeritus
The lefthand side is 0 regardless of the value of $a_0$.

11. Jul 19, 2015

### Destroxia

So then what of $a_1$ do I just have to make the equation in terms of $a_2$ or something?

12. Jul 19, 2015

### vela

Staff Emeritus
In terms of $a_0$ (as there is no restriction on $a_0$). $a_1$, on the other hand, is 0, so it will turn out all of the odd-n coefficients are 0.

Note if you ever think you find $a_0=0$, you did something wrong. By assumption, $r$ is the exponent of the lowest-order non-vanishing term. If $a_0=0$, that would be a contradiction.

13. Jul 19, 2015

### Destroxia

OH!

I understand what you were trying to say now, I didn't even notice the $a_0$ terms cancelling out because I was too busy trying to make it equal something other than zero!

I think I can solve it now, Thank you for your help!