solving ODE using power series

Discussion in 'Physics & Math' started by sandysq, Apr 1, 2008.

  1. sandysq Registered Member

    Messages:
    4
    I'm trying to solve an ordinary differential equation, with non-constant coefficients, using a

    power series solution. The question is as follows:

    ----------------------------------------------------------------

    solve the initial value problem:

    \( x(2-x)y'' - 6(x-1)y' - 4y = 0\)
    \( y(1)=1 \)
    \(y'(1) = 0 \)

    hint: since the initial condition is given at \(x_0 = 1 \), it is best to write the

    solution as a series centered at \(x_0 = 1 \).

    ---------------------------------------------------------------

    I have attempted the question, but I got stuck in one of the steps. Here's what I have so far:

    assume the solution can be written as a power series centered about x = 1. Then the solution and

    its derivatives are:


    \( y(t) = \displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} \)

    \( y'(t) = \displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}} \)

    \( y''(t) = \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}} \)

    where the \( a_n \) are constant coefficients.

    Substitute y, y', and y'' into the ODE in order to determine the \( a_n \):

    \(x(2-x)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n= 1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0\)

    \((1-(x-1)^2)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1) \displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1) ^n} = 0\)

    \(\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n ((x-1)^{n-2}- (x-1)^n)}-6\displaystyle\sum_{n= 1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0\)

    \(\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}- \displaystyle\sum_{n=2}^{\infty} {n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0} ^{\infty}{a_n (x-1)^n} = 0\)

    now I need to shift indices to make all the exponents of (x-1) equal to n, and also to try to

    make all the summations start at n=0:

    \(\displaystyle\sum_{n=0}^{\infty}{(n+2)(n+1) a_{n+2} (x-1)^n}- \displaystyle\sum_{n=2}^ {\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4 \displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0\)


    I don't know how to proceed from here. I have 4 summations. Two of them start at n=0 and has

    the \((x-1)^n\) factor. The other two summations start at n=2 or n=1. I can't get them

    to start at n=0 because if I try to do that, then I get:

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

    and

    \(6\displaystyle\sum_{n=0}^{\infty}{(n+1) a_{n+1} (x-1)^{n+1}}\)

    As you can see, these have the factors \( (x-1)^{n+2} \) and \( (x-1)^{n+1} \)

    instead of \( (x-1)^n \). So I don't know how to combine all this, since the summations

    do not start at the same number.

    --------------------------------------------------------------
    I'm told that in the end, the answer to the coefficients should be:

    \(a_{2n} = (n+1)a_0 \) and \(a_{2n+1} = \frac{2n+3}{3}a_1 \)

    So I just don't know how to get from the point I'm stuck to this answer. I'm also quite confused

    as to how to solve the initial value problem, after obtaining the expected answer. The initial

    conditions are: \( y(1)=1 \), \(y'(1) = 0 \). Using the initial condition y(1)

    = 1, I get:

    \( y(1)= \displaystyle\sum_{n=0}^{\infty}{a_n ((1)-1)^n} \)
    \( 1 = \displaystyle\sum_{n=0}^{\infty}{a_n (0) \)
    \( 1 = 0 \)

    This does not make sense...what am I doing wrong?
     
  2. temur man of no words Registered Senior Member

    Messages:
    1,330
    You have to consider the first few terms separately. If one sum starts at 0 and the other starts at 1, it just means that the first coefficient in the latter sum is zero. So if you only consider the coefficient in front of the zero order term in the whole expression, that will consist of only the term coming from (the first term of) the first sum.
     
  3. temur man of no words Registered Senior Member

    Messages:
    1,330
    On the second thought, you don't even need to write out the first terms separately, because the expressions in this case are so that if you just extend the summation limit down to n=0, exactly the right terms become zero. For example,

    \(\sum_{n=2}^{\infty}n(n-1)a_n(x-1)^n=\sum_{n=0}^{\infty}n(n-1)a_n(x-1)^n\)

    because the factor n(n-1) is zero when n=0 or n=1.

    Your attempt to make the sums start at 0 is not exactly how it should be done in this case, because that makes the power of x inconsistent with the other sums.

    I would also suggest first make the variable substitution z=x-1 to make computation simpler.
     
  4. sandysq Registered Member

    Messages:
    4
    Thanks for your response. So it seems that I can just rewrite the two summations that are giving me trouble as starting at zero, without shifting the indices.

    I'm still not sure how to solve the initial value problem. Since the initial value is x=1, when I try to substitute x=1 into the expression for y, the whole sum becomes 0
     
  5. iceaura Valued Senior Member

    Messages:
    16,951
    edit: oh, that's what we're doing. Never mind - --
     
    Last edited: Apr 1, 2008
  6. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

    Messages:
    8,966
    Except for the n = 0 term, of course!!!
     
  7. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

    Messages:
    8,966
    Also, if you plug in n=0 in this expression, I get \(2 a_0 = a_2\) so you should have SOME confidence in this answer.

    And who says that all of your terms have to have the same power?
     
  8. D H Some other guy Valued Senior Member

    Messages:
    2,257
    The whole sum does not become zero when x=1. When you write power series in the form \(f(x) = \sum_{n=0}^{\infty}a_n(x-1)^n\) you are implicitly assuming \((x-1)^0=1\) for all x including \(x=1\). In short, \(0^0\equiv 1\) when it comes to power series. Without this shorthand one must write the power series in the form \(f(x) = a_0 + \sum_{n=1}^{\infty}a_n(x-1)^n\).

    To solve this particular problem, the first thing to do is to make the substitution \(z=x-1\). This will simplify things a lot; hence the hint. The derivatives with respect to z are easy to compute via the chain rule:
    \(\frac{dy}{dz} = \frac{dx}{dz}\, \frac{dy}{dx} = 1\cdot \frac{dy}{dx} = y'\). Thus the same differential equation applies with the substitution, but now you have

    \((z+1)(2-(z+1))\,\frac{d^2y}{dz^2} - 6z\,\frac{dy}{dz} - 4y = 0\)

    This simplifies even more (I'll let you work that out). Next write y as a power series in z about z=0 (this is the same power series as the series in x about x=1) and compute the derivatives. Next collect terms with equal powers so you have something that looks like

    \(\sum_{n=0}^{\infty} b_n z^n = 0\)

    This can be true for all z if and only if each b[sub]n[/sub] is identically zero (Jess: Justify this step.) Setting the terms to zero should give you a recursive relationship of the form

    \(a_{\{n+\text{offset}\}} + f(n)a_n = 0\)

    I'll let you figure out what that offset is. With this, you should be able to show that

    \(a_{2n} = (n+1)a_0 \) and \(a_{2n+1} = \frac{2n+3}{3}a_1 \)


    Finally, solve for a[sub]0[/sub] and a[sub]1[/sub] using the given initial conditions.
     

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