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Thread: The Method of Frobenius

  1. #1
    Super Member Aryth's Avatar
    Feb 2007

    The Method of Frobenius

    Hermite's Equation is an equation of the type:

    $\displaystyle y''(x) - 2xy'(x) + 2ny(x) = 0, n \in \mathbb{N}_0$

    Construct series solutions (about the origin) for n=1 and n=2.

    I tried to solve this generally so I could just plug in n later for any other n's that I may need, but somewhere I may have messed up because my coefficients ended up all being zero for the n=1 series... So, all I need is someone to see if I messed up...

    I started off by noting that the equation was in general form and that p(x) and q(x) are both analytical everywhere so x = 0 is a valid point for a convergent series solution.

    Next (and this part is where I had my doubts because it is an ordinary point, not a singular one), I took the limits of xp(x) and $\displaystyle x^2q(x)$ and got zero for both. This lead to the roots:

    $\displaystyle r_1 = 0$ and $\displaystyle r_2 = 1$

    Now, I prepared a generalized solution:

    $\displaystyle y(x) = x^r\sum_{i=0}^{\infty}a_ix^i=\sum_{i=0}^{\infty}a_ ix^{i+r}$

    Plugging this into the DE yields:

    $\displaystyle \sum_{i=0}^{\infty}(i+r-1)(i+r)a_ix^{i+r-2} - 2x\sum_{i=0}^{\infty}(i+r)a_ix^{i+r-1} + 2n\sum_{i=0}^{\infty}a_ix^{i+r}=0$

    When the x in the second term is distributed, the second and third sum can be added together:

    $\displaystyle \sum_{i=0}^{\infty}(i+r-1)(i+r)a_ix^{i+r-2} - 2(1 - n)\sum_{i=0}^{\infty}(i + r + 1)a_ix^{i+r} = 0$

    Next I multiplied through by $\displaystyle x^2$ to eliminate negative indices:

    $\displaystyle \sum_{i=0}^{\infty}(i+r-1)(i+r)a_ix^{i+r} - 2(1-n)\sum_{i=0}^{\infty}(i+r+1)a_ix^{i+r + 2} = 0$

    Then I shifted the indices, the first sum gets the replacement k=i the second one gets k=i+2:

    $\displaystyle \sum_{k=0}^{\infty}(k+r-1)(k+r)a_kx^{k+r} - 2(1-n)\sum_{k=2}^{\infty}(k+r-1)a_{k-2}x^{k+r} = 0$

    Then I took out the first two terms of the first sum:

    $\displaystyle a_0[r(r-1)]x^r + a_1[r(1+r)]x^{1 + r} + \sum_{k=2}^{\infty}(k+r-1)(k+r)a_kx^{k+r} - $ $\displaystyle 2(1-n)\sum_{k=2}^{\infty}(k+r-1)a_{k-2}x^{k+r} = 0$

    Then I combined the two sums:

    $\displaystyle a_0[r(r-1)]x^r + a_1[r(1+r)]x^{1+r} + \sum_{k=2}^{\infty}(k+r-1)x^{k+r}[(k+r)a_k - 2(1-n)a_{k-2}] = 0$

    Then I set the coefficients equal to 0 and used r = 1:

    $\displaystyle a_0[1(0)] = 0$ -> $\displaystyle a_0$ is arbitrary.

    $\displaystyle a_1[1(2)] = 0$ -> $\displaystyle a_1 = 0$

    $\displaystyle (k+1)a_k - 2(1-n)a_{k-2} = 0$

    $\displaystyle a_k = \frac{2(1-n)a_{k-2}}{(k+1)}, k \geq 2$

    Now, when n=1, every coefficient is zero except $\displaystyle a_0$ so the solution for n=1 is:

    $\displaystyle y(x) = a_0x$

    When n=2, All of the odd numbered coefficients are zero so that the odd number powers remain, and I get:

    $\displaystyle y(x) = a_0[x - \frac{2}{3}x^3 + \frac{4}{15}x^5 - \frac{8}{105}x^7 + ...]$

    Thanks in advance for any help.
    Last edited by Aryth; Feb 18th 2010 at 04:45 PM.
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  2. #2
    MHF Contributor

    Apr 2005
    That equation is NOT singular at any point and you do not need to use
    Frobenius' method. Just use a solution of the form $\displaystyle \sum_{n=0}^\infty a_nx^n$.
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