Series solution with regular singular points?

[Destroxia]

1. Homework Statement

##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.

 

[vela]

Homework Statement

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

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.

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

then I composed an equation to solve for the roots:

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

This is called the indicial equation.

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

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

\begin{align*}
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}
\end{align*} 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.

Looks good so far.

 

[Destroxia]

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.

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?

 

[Destroxia]

Looks good so far.

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?

 

[vela]

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?

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##.

 

[Destroxia]

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##.

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?

 

[vela]

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##.

 

[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),

 

[Destroxia]

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##.

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##

 

[vela]

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

The lefthand side is 0 regardless of the value of ##a_0##.

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

 

[Destroxia]

The lefthand side is 0 regardless of the value of ##a_0##.

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

 

[vela]

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.

 

[Destroxia]

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.

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!