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Math Help - Series Solutions Near a Regular Singular Point

  1. #1
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    Series Solutions Near a Regular Singular Point

    I need to find the series solution at the point x=0 for the equation 2xy''+y'+xy=0. I have already found the point to be regular singular and the roots to be 0 and 1/2. I need to solve the equation using the root 1/2.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let’s write the equation in the form \dots

    y'' + \frac {y'}{2x} + \frac {y}{2}=0 (1)

    That is an linear DE and its general solution is in the form \dots

     y(x) = c_{1} \cdot \lambda (x) + c_{2}\cdot \gamma (x) (2)

    \dots where \lambda(*) and \gamma(*) are two independently linear solutions of the (1) and c_{1} ,c_{2} two constants. Now we will research a solution of (1) analytic in x=0, so that il can be written as \dots

    \lambda (x) = \sum_{n=0}^{\infty} a_{n}\cdot x^{n}(3)

    The way for finding the a_{n} is to substitute the (3) into the (1) obtaining \dots

    \sum_{n=0}^{\infty} a_{n}\cdot x^{n}= -2\cdot \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-2} - \sum_{n=1}^{\infty} n\cdot a_{n}\cdot x^{n-2} (4)

    The term a_{0} is pratically the constant c_{1} in (2). Observing the (4) it is easy to see that is a_{1}=0 and the same holds for all the a_{n} of odd index. For the a_{n} of even index we have \dots

    a_{0}= -2\cdot 3\cdot a_{2} \rightarrow a_{2}=-\frac {a_{0}}{2\cdot 3}

    a_{2}= -2\cdot 2\cdot 7\cdot a_{4} \rightarrow a_{4}=-\frac {a_{2}}{2\cdot 2\cdot 7}

    a_{4}= -2\cdot 3\cdot 11\cdot a_{6} \rightarrow a_{6}=-\frac {a_{4}}{2\cdot 3\cdot 11}

    \dots

     a_{2k}= -2\cdot (k+1)\cdot (4k+3)\cdot a_{2(k+1)} \rightarrow a_{2(k+1)}=-\frac {a_{2k}}{2\cdot (k+1)\cdot (4k+3)} (5)

    The recursive relations (5) permit us to arrive to the Taylor expansion around x=0 of the analytic solutions of (1) \dots

    \lambda (x)= 1+\sum_{k=1}^{\infty} (-1)^ {k}\cdot \frac {x^{2k}}{2^{k}\cdot k!\cdot 3\cdot 7\dots (4k-1)} (6)

    That is about the analytic solutions of the form c_{1}\cdot \lambda(x). For the non analytic solutions of the form c_{2}\cdot \gamma (x) the problem is a little more difficult…

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    In previous post it has been ‘attacked’ the DE \dots

    y'' + \frac {y'}{2x} + \frac {y}{2}=0 (1)

    \dots the general solution of which is \dots

     y(x) = c_{1} \cdot \lambda (x) + c_{2}\cdot \gamma (x) (2)

    For the ‘analytic’ solution \lambda(*) the following series expansion has been found \dots

    \lambda (x) = 1+\sum_{k=1}^{\infty} (-1)^ {k}\cdot \frac {x^{2k}}{2^{k}\cdot k!\cdot 3\cdot 7\dots (4k-1)} (3)

    … and now we will try to arrive to the ‘non analytic’ solution \gamma (*). Since \lambda (*) and \gamma (*) are both solutions of (1) it is \dots

    \lambda '' + \frac {\lambda '}{2x} + \frac {\lambda}{2}=0

    \gamma '' + \frac {\gamma '}{2x} + \frac {\gamma}{2}=0

    Multiplying the first equation by \gamma (*), the second by \lambda (*) and making the difference we obtain \dots

    \lambda '' \cdot \gamma - \gamma '' \cdot \lambda + \frac {\lambda ' \cdot \gamma - \gamma ' \cdot \lambda}{2x}=0 (4)

    \dots and setting \dots

    \phi (x)= \lambda ' \cdot \gamma - \gamma ' \cdot \lambda (5)

    \dots the (4) becomes \dots

    \phi ' + \frac {\phi}{2x}=0 (6)

    The (5) is a linear DE of first order whose solution is relatively easy to find \dots

    \phi (x)= \frac {c}{2x} (7)


    Substituting the (7) into (5) and setting c=2 without losing anything we obtain \dots

    x \cdot \lambda ' \cdot \gamma - x \cdot \lambda \cdot \gamma '=1 (8)

    In a succesive post we will perform succesive steps...

    Kind regards

    \chi \sigma
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