1. Not finding help here? Sign up for a free 30min 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 to ODE

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine a series solution to the following ODE about x0 = 0:

    [tex]xy'' + y' + xy = 0[/tex]

    3. The attempt at a solution

    I'll try to keep this concise.

    I first divided through by x and made the usual guesses for the form of the series. Subbing those in gave:

    [tex]\sum_{2}^{\infty}n(n-1)a_{n}x^{n-2}+ \sum_{1}^{\infty}na_{n}x^{n-2} + \sum_{0}^{\infty}a_{n}x^{n}[/tex]

    Then I shifted the first two series up 1 and the third one down 1 and multiplied through by x:

    [tex]\sum_{1}^{\infty}n(n+1)a_{n+1}x^{n}+ \sum_{0}^{\infty}(n+1)a_{n+1}x^{n} + \sum_{1}^{\infty}a_{n-1}x^{n}[/tex]

    Then to get the second series to start at 1, I moved the lower index up 1 and added the n = 0 term to make up for it:

    [tex]\sum_{1}^{\infty}n(n+1)a_{n+1}x^{n}+ \sum_{1}^{\infty}(n+1)a_{n+1}x^{n} + \sum_{1}^{\infty}a_{n-1}x^{n}+a_{1}[/tex]

    Now you can combine the series:

    [tex]\sum_{1}^{\infty}[n(n+1)a_{n+1}+(n+1)a_{n+1}+a_{n-1}]x^{n} + a_{1}[/tex]

    But then setting coefficients equal to zero gives a1 = 0, but don't you need to determine recurrence relations for a0 and a1?
     
  2. jcsd
  3. Jul 9, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, your solution will have only even terms. The coefficient of the odd ones will be 0.
     
  4. Jul 9, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Because the coefficient of y'' is 0 at x= 0, you will not be able to find two independent solutions as power series. However, once you have found a single solution, even as a series, you can use the standard "order-reducing" method to reduce top a first order equation for the second independent solution- which should be something like ln(x) times the first solution.
     
  5. Jul 9, 2011 #4
    Thanks for the replies vela and Halls! Okay, so I'm just going to disregard the odd terms then:

    [tex]n(n + 1) a_{n+1} + (n + 1) a_{n+1} + a_{n-1} = 0[/tex]
    [tex]a_{n-1} = -a_{n+1}(n+1)^2[/tex]
    [tex]a_{0} = - 4a_{2} = - 16a_{4} = - 36a_{6} = - 64a_{8} = ...[/tex]
    [tex]a_{0} = - 2^2 a_{2} = - 4^2a_{4} = - 6^2a_{6} = - 8^2a_{8} = ...[/tex]
    [tex]a_{0} = - (2k)^2 a_{2k}:k=1,2,3...[/tex]
    [tex]a_{2k} = - \frac{a_{0}}{(2k)^2} :k=1,2,3...[/tex]

    [tex]y(x) = a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+a_{5}x^5+a_{6}x^6...[/tex]
    [tex]y(x) = a_{0}+a_{2}x^2+a_{4}x^4+a_{6}x^6...[/tex]
    [tex]y(x) = a_{0}- \frac{a_{0}}{(2(1))^2}x^2-\frac{a_{0}}{(2(2))^2}x^4-\frac{a_{0}}{(2(3))^2}x^6...[/tex]

    [tex] y(x) = a_{0}[1 + \sum_{n=1}^{\infty}\frac{x^{2n}}{(2n)^2}][/tex]
    [tex] y(x) = a_{0}[1 + \frac{1}{4}\sum_{n=1}^{\infty}(\frac{x^{n}}{n})^2][/tex]

    How's that? Also,

    Does that mean that I should guess a solution of the form y2(x) = v(x) y1(x) (for some v(x) and the y1 I just found) and then solve for v(x)?
     
  6. Jul 9, 2011 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's not quite right. You found [itex]a_{n+2} = -a_n/(n+2)^2[/itex], so the first few terms are
    \begin{align*}
    a_2 & = -a_0/2^2 = -a_0/4 \\
    a_4 & = -a_2/4^2 = -(-a_0/4)/16 = a_0/64 \\
    a_6 & = -a_4/6^2 = -(a_0/64)/36 = -a_0/2304
    \end{align*}
    which isn't what you got. The solution you should get is the Bessel function J0(x).
     
    Last edited: Jul 9, 2011
  7. Jul 9, 2011 #6
    I did? Oh yeah, I guess I did. So in this part:

    [tex]a_{n-1} = -a_{n+1}(n+1)^2[/tex]

    It's cool if I just sub in n + 1 for n (?) and then get:

    [tex]a_{n} = -a_{n+2}(n+2)^2[/tex]
    [tex]a_{n+2} = -\frac{a_{n}}{(n+2)^2}[/tex]

    Then

    [tex]y(x) = a_{0}- \frac{a_{0}}{4}x^2-\frac{a_{0}}{4\cdot16}x^4-\frac{a_{0}}{4\cdot16\cdot36}x^6...-\frac{a_{0}}{4\cdot16\cdot36\cdots(2n)^2}x^{2n}[/tex]

    I'm having a hard time determining a general formula for the series other than just

    [tex]y(x) = a_{0}[1-\sum_{n=1}^{\infty}\frac{x^{2n}}{4 \cdot16 \cdot36 \cdots(2n)^2}][/tex]
     
  8. Jul 9, 2011 #7

    Char. Limit

    User Avatar
    Gold Member

    Well, let's see. Ignoring the x^(2n) for now, what I see is a product that can be written like this:

    [tex]\prod_{k=1}^n \frac{1}{(2k)^2} = \prod_{k=1}^n \frac{1}{4 k^2}[/tex]

    Now, I'm not certain if this is true, but I believe that you can separate the two, to get this:

    [tex]\left( \prod_{k=1}^n \frac{1}{4}\right) \left( \prod_{k=1}^n \frac{1}{k^2} \right)[/tex]

    The first product is rather easy to evaluate, and the second one can be manipulated like so to give another product that's easy to evaluate:

    [tex]\prod_{k=1}^n \frac{1}{k^2} = \left( \prod_{k=1}^n \frac{1}{k} \right)^2[/tex]

    Can you see what the product that's being squared evaluates to?

    Note: After a bit of testing, I now am relatively certain you can separate the one product into two.
     
  9. Jul 9, 2011 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yup!
     
  10. Jul 9, 2011 #9
    Nice! I didn't know that trick.

    Regarding Char.Limit's suggestions, I think that:

    [tex] \prod_{k=1}^n \frac{1}{4} = \frac{1}{4^n} [/tex]

    And:

    [tex]\left( \prod_{k=1}^n \frac{1}{k} \right)^2 = \left[ (1/1)(1/2)(1/3)(1/4) \cdots (1/n) \right]^2 = \left[ \frac{1}{n!} \right]^2 = \frac{1}{(n!)^2}[/tex]

    So I've got:

    [tex]y(x) = a_{0}[1-\sum_{n=1}^{\infty}\frac{x^{2n}}{(4^n)(n!)^2}] [/tex]

    Thanks for the responses everyone by the way. How's that looking so far? Apparently it's a "modified Bessel function of the first kind..."

    http://www.wolframalpha.com/input/?i=sum+%28x^%282n%29%29+%2F+%28%284^n%29%28n%21%29^2%29+from+1+to+inf
     
  11. Jul 9, 2011 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You forgot the factor of (-1)n.
     
  12. Jul 9, 2011 #11
    But wouldn't that make some terms positive?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Series Solution to ODE
  1. Series Solution of ODE (Replies: 2)

Loading...