Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on deriving Bessel's equation

  1. May 17, 2013 #1
    [tex]\hbox { Bessel's equation of order p: }\;x^2y''+xy'+(x^2-p^2)y=0\;\hbox { and let}\;y=\sum_0^{\infty}C_nx^{n+r}[/tex]
    [tex]\Rightarrow\; \left[\sum_0^{\infty}C_k(k+r)(k+r-1)x^k+\sum_0^{\infty}C_k(k+r)x^k+\sum_2^{\infty} C_{k-2}x^k-p^2\sum_0^{\infty}C_kx^k\right]x^r\;=\;0[/tex]
    [tex]\Rightarrow\; x^r\left[(r^2-p^2)C_0+[(r+1)^2-p^2]C_1 x+\sum_2^{\infty}\left([(k+r)(k+r)-p^2]C_k+C_{k-2}\right)x^k\right]\;=\;0[/tex]

    According to the book, all the terms have to be zero for the equation to be zero.
    [tex](r^2-p^2)C_0=0\;\Rightarrow r=^+_-p[/tex]
    The book also say
    [tex][(r+1)^2-p^2]C_1=0\;\Rightarrow C_1=0\;\hbox { because from above, }\;r=^+_-p[/tex]


    My problem with this assumption ##C_1=0## is I can just as easy say ##(r+1)^2-p^2=0## and claim ##C_0=0##!!! This will change the whole equation of the Bessel's equation!!!

    Why do I have to follow the book to let ##C_1=0## which result in all ##C_{2n+1}=0##? Is it just because it is simpler and more convenient this way?

    Also, the book just said let ##C_0=\frac {1}{2^p\Gamma(1+p)}## without explaining why. Why?

    I know it is for fitting the formula of:

    [tex]J_p=\sum_0^{\infty} \frac{(-1)^k}{k! \Gamma(k+p+1)}\left(\frac {x}{2}\right)^{2k+p} [/tex]

    But can you just let ##C_0## to be anything?

    Thanks
     
    Last edited: May 17, 2013
  2. jcsd
  3. May 17, 2013 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can make either choice. You will still find that the leading term in ##x^{k+r}## in the solution has degree ##p##. Choosing ##C_0\neq 0## makes ##r=p## and the sum over even powers ##k##, while choosing ##C_1\neq 0## makes ##r=p-1## and the sum over odd powers. In either case the series starts with ##x^p## and multiplies this by a polynomial in even powers. So the choice corresponds to shifting the definitions of ##k## and ##r## by ##\pm 1## in such a way that they cancel out in the complete solution.

    Yes, since the equation is linear, any multiple of a solution is still a solution. So we can choose any convenient normalization that we want when we define the Bessel functions.
     
  4. May 17, 2013 #3
    Thanks for your answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question on deriving Bessel's equation
  1. Bessel equation (Replies: 2)

  2. Bessel's equation (Replies: 1)

Loading...