Results 1 to 2 of 2

Math Help - Method of Frobenius

  1. #1
    Junior Member
    Joined
    Aug 2006
    Posts
    34

    Method of Frobenius

    The question asks to use the method of frobenius to find two linearly independent solution. i am very lost on how to go about this

    x^2y"-(x+2)y=0 near the point x=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,897
    Thanks
    327
    Awards
    1
    Quote Originally Posted by action259 View Post
    The question asks to use the method of frobenius to find two linearly independent solution. i am very lost on how to go about this

    x^2y"-(x+2)y=0 near the point x=0
    This is going to be tough without LaTeX...

    I'm going to define a symbol to try to make things easier. Let S(f(i),i,a,b) be the sum of the function f(i) as i goes from a to b.

    The method of Frobenius, for expanding a solution about a non-singular point is that
    y(x) = S(a_n x^(n+k),n,0,infinity)
    where a_n is a "subscript" n are a series of constants, and k is a constant (I believe it is required to be an integer, but I'm not certain of that.) So basically:
    y(x) = a_0 x^k + a_1 x^(k+1) + a_2 x^(k+2) + ...

    (Note: Our expansion is in powers of x, and not powers of (x-c) because we are expanding about x = 0.)

    Plug this into your differential equation, and reduce the sums so that you can write the equation in terms of a single summation, then set each coefficient of x to zero.

    So
    y(x) = S(a_n x^(n+k),n,0,infinity)
    y'(x) = S(a_n (n+k)x^(n+k-1),n,1,infinity)
    y''(x) = S(a_n (n+k)(n+k-1)x^(n+k-2),n,2,infinity)
    (Write these out explicitly as a power series if you can't follow the notation. And for the sake of your sanity, rewrite these on paper in terms of summation symbols!)

    Thus:
    x^2y"-(x+2)y=0 becomes
    x^2 S(a_n (n+k)(n+k-1)x^(n+k-2),n,2,infinity) - (x+2) S(a_n x^(n+k),n,0,infinity) = 0

    S(a_n (n+k)(n+k-1)x^(n+k),n,2,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2 S(a_n x^(n+k),n,0,infinity) = 0

    Now we want to get these three summations to be all one sum. The first step is to note that the first and third summations already have the same power of x listed, but they have a different lower sum limit. (n = 2 for the first, and n = 0 for the last.) So I'm going to rewrite the third summation as:
    S(a_n x^(n+k),n,0,infinity) = a_0 x^k + a_1 x^(k+1) + S(a_n x^(n+k),n,2,infinity)
    by listing out the first two terms explicity. Thus the equation becomes:

    S([a_n (n+k)(n+k-1) - 2a_n]x^(n+k),n,2,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2a_0 x^k - 2a_1 x^(k+1) = 0

    Now we need to match the coefficients of x in the two summations. (The summation variable is a dummy index, so we can change it to just about whatever we like.) I am going to find a new summation variable n' such that
    n + k = n' + k + 1
    Thus n = n' + 1 and n' = n - 1.

    So replace n with n' + 1 in the function in the summation, and we have a lower limit on the summation of n = 2, so this becomes n' = 2 - 1 = 1. The upper limit, being infinity, stays the same.

    Thus:
    S([a_(n'+1) (n'+1+k)(n'+1+k-1) - 2a_(n'+1)]x^(n'+1+k),n',1,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2a_0 x^k - 2a_1 x^(k+1) = 0

    Or, changing the n' back to n for clarity:
    S([a_(n+1) (n+k+1)(n+k) - 2a_(n+1)]x^(n+k+1),n,1,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2a_0 x^k - 2a_1 x^(k+1) = 0

    Now we have matching exponents, we need matching summation limits. So write the second summation as:
    S(a_n x^(n+k+1),n,0,infinity) = a_0 x^(k+1) + S(a_n x^(n+k+1),n,1,infinity)

    And we get:
    S([a_(n+1) (n+k+1)(n+k) - 2a_(n+1)]x^(n+k+1),n,1,infinity) - S(a_n x^(n+k+1),n,1,infinity) - 2a_0 x^k - 2a_1 x^(k+1) - a_0 x^(k+1) = 0

    Or, collapsing the two summations into one:
    S([a_(n+1) (n+k+1)(n+k) - 2a_(n+1) - a_n]x^(n+k+1),n,1,infinity) - 2a_0 x^k - 2a_1 x^(k+1) - a_0 x^(k+1) = 0

    Now we start setting powers of x equal to zero.

    The lowest power of x in the expansion is x^k. It has a coefficient of:
    -2a_0 = 0 Thus a_0 = 0.

    The x^(k+1) coefficient gives:
    -2a_1 - a_0 = 0. Since a_0 = 0, so a_1 = 0.

    The x^(k+2) coefficient gives:
    a_2 (1+k+1)(1+k) - 2a_2 - a_1 = 0 ==> a_2[(k+2)(k+1) - 2] = 0 since a_1 = 0.
    I am going to assume that a_2 is NOT zero (else we will wind up with a_n = 0 for all n which gives the trivial solution y(x) = 0), because we can set (k+2)(k+1) - 2 = 0. This will give two values of k, representing two linearly independent solutions. So find these values and use each one individually to get a recursion relation relating the rest of the a_n s.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Frobenius method
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 10th 2011, 07:12 PM
  2. Frobenius Method
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 26th 2010, 02:51 PM
  3. The Method of Frobenius
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 20th 2010, 08:47 AM
  4. Help needed (Frobenius method)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: July 1st 2009, 12:24 PM
  5. power series method and frobenius method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 26th 2006, 12:25 PM

Search Tags


/mathhelpforum @mathhelpforum