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Thread: 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 $\displaystyle \dots$

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \dots$

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

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

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

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

    Kind regards

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

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

    $\displaystyle \dots$ the general solution of which is $\displaystyle \dots$

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

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

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

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

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

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

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

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

    $\displaystyle \dots$ and setting $\displaystyle \dots$

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

    $\displaystyle \dots$ the (4) becomes $\displaystyle \dots$

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

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

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


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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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