1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series solution with regular singular points?

  1. Jul 18, 2015 #1
    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. jcsd
  3. Jul 18, 2015 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  4. Jul 19, 2015 #3
    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?
     
  5. Jul 19, 2015 #4
    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???
     
  6. Jul 19, 2015 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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##.
     
  7. Jul 19, 2015 #6
    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?
     
  8. Jul 19, 2015 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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##.
     
  9. Jul 19, 2015 #8
    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
  10. Jul 19, 2015 #9
    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##
     
  11. Jul 19, 2015 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

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

     
  12. Jul 19, 2015 #11
    So then what of ##a_1## do I just have to make the equation in terms of ##a_2## or something?
     
  13. Jul 19, 2015 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  14. Jul 19, 2015 #13
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted