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Frobenius Method

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Use the method of Frobenius, constructing a power series about x = 0,
    to find the general solution of this equation (retain terms up to and including those in
    square brackets):

    4xy'' + 2y' + y = 0 [x7/2]

    Note: the solution can be written in closed form, can you see how?


    2. Relevant equations



    3. The attempt at a solution

    set y = [itex]\sum[/itex] anxλ+n to obtain:

    4x∑(between n=0 and ∞) an(λ+n)(λ+n-1)xλ+n-2 + 2∑(between n=0 and ∞) an(λ+n)xλ+n-1 + ∑(between n=0 and ∞) anxλ+n = 0

    Now relabelling the indices using m = n-1 for the first term, m=n-1 for the second term and m=n for the third term here, in order to get a common coefficient of xλ+m for all terms, will give:

    4∑(between m=-1 and ∞) am+1(λ+m+1)(λ+m)xλ+m + 2∑(between m =-1 and ∞) am+1(λ+m+1)xλ+m + ∑(between m=0 and ∞) amxλ+m = 0

    Now look at terms when m=-1 (coefficient xλ-1):

    4[a02-λ)] + 2[a0λ] = 0

    so 4a0λ2 - 4a0λ +2a0λ = 4a0λ2 - 2a0λ = 2a0(λ(λ-1)) = 0

    with a0 [itex]\neq[/itex] 0, we have our indicial equation:

    λ(λ-1) = 0 so λ1 = 1 and λ2 = 0

    The remaining terms give:

    ∑(between m=0 and ∞) { am+1[(λ+m+1)(λ+m) + (λ+m+1)] + am } xλ+m = 0

    so this means ∑(between m=0 and ∞) { am+1[(λ+m+1)2] + am } xλ+m = 0

    By taking the am term to the other side and diving by (λ+m+1)2 gives me the recurrence relation:

    am+1 = -am / (λ+m+1)2, m= 0, 1, 2, ...

    Now looking at λ=1:

    am+1 = -am / (m+2)2

    and by substituting in values of m=0,1,2,3,4,5,6,7,8 will give me a0, ... , a8 which are the following:

    a1 = -a0 / 4, a2 = -a1 / 9, a3 = -a2 / 16 and so on with the denominator being a squared value of (m+2) each time up until a8 = -a7 / 81

    Here is the part I'm not sure of, I tried to use these to obtain the first part of the general solution (y1)... by substituting into the general form y = a0 + a1x + a2x2 + ....

    I got, y1 = -4a1 - 9a2x - 16a3x2 - 25a4x3 - 36a5x4 - 49a6x5 - 64a7x6 - 81a8x7

    and then taking square root of this, i.e. (y1)1/2, gave me the first part of the general solution in a complex form.

    Is what I have done so far correct to here? I am confident it is up to the recurrence relation where I'm struggling to form a general solution after this.

    I know that you need to use the other value of λ=0 to find the other part of the general solution y2 and your final general solution will be y(x) = y1 + y2.

    Again here I am not sure how to find y2 because I've used the recurrence relation to get a0, ... , a8 again but by the Frobenius method if you have two values of λ that differ by an integer (which is the case here), you examine the recurrence relation for λ2 which I have done, but because the recurrence relation is singular at m=1 when λ=0, you seek a second solution y2 by using:

    y2 = y1 ln|x| + ∑bnxλ+n

    Would this be the correct way to find the second solution and are my working for the first solution correct?

    Can anyone help me please?
     
  2. jcsd
  3. Feb 24, 2014 #2

    vela

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    I got a different indicial equation: ##4\lambda^2-2\lambda=0##.
     
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