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Series solution

  1. Dec 19, 2005 #1
    Can someone please explain some steps of a worked example.

    Q. Find the power series solutions about z = 0 of 4zy'' + 2y' + y = 0.

    (note: y = y(z))

    Writing the equation in standard form:

    [tex]
    y'' + \frac{1}{{2z}}y' + \frac{1}{{4z}}y = 0
    [/tex]

    Let [tex]y = z^\sigma \sum\limits_{n = 0}^\infty {a_n z^n }[/tex]

    The indicial equation has roots zero and (1/2).

    Using the substitution for y above the DE becomes:

    [tex]
    \sum\limits_{n = 0}^\infty {\left[ {\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right) + \frac{1}{2}\left( {n + \sigma } \right) + \frac{1}{4}z} \right]} a_n z^n = 0
    [/tex]...(1)

    Demanding that the coefficients of z^n vanish separately in the above equation we obtain the recurrence relation:

    [tex]
    \left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0
    [/tex]

    What happened to the factor of z which was multiplied by (1/4) in (1)?

    If we choose the larger root, 1/2, then the recurrence relation becomes [tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}[/tex]

    Setting a_0 = 1 we find...etc

    I am told that the larger of the two roots of the indicial equation always leads to a solution. However, I don't understand why a_0 is set equal to one? Is that an arbitrary selection or is it convenient?

    Any help would be good thanks.
     
    Last edited: Dec 19, 2005
  2. jcsd
  3. Dec 19, 2005 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    It's there, it's the [itex]\frac{1}{4}a_{n-1}[/itex]. Do you see why?
    I've never done problems like this before, but if you are getting this by substituting [itex]\sigma = \frac{1}{2}[/itex] into:

    [tex]\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0[/tex]

    then you should get:

    [tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {\mathbf{2}n + 1} \right)}}[/tex]
    Different choices of a0 will give different solutions, and you should indeed expect a family of solutions, not just one solution. Are there any given boundary conditions that might explain the choice of a0 = 1? If not, then perhaps it is just arbitrary/convenient/conventional, I don't know myself.
     
  4. Dec 19, 2005 #3
    There aren't any ICs or BCs in the example so a_0 = 1 might have been chosen simply to illustrate the solution method. Thansk for your help.
     
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