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[ "article:topic", "method of Frobenius", "authorname:nwalet", "license:ccbyncsa", "showtoc:yes" ]
Mathematics LibreTexts

9.1: Frobenius’ Method

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    Let us look at the a very simple (ordinary) differential equation, \[y''(t) = t\,y(t),\] with initial conditions \(y(0) = a\), \(y'(0)=b\). Let us assume that there is a solution that is analytical near \(t=0\). This means that near \(t=0\) the function has a Taylor’s series

    \[y(t) = c_0 + c_1 t + \ldots = \sum_{k=0}^\infty c_k t^k.\]

    (We shall ignore questions of convergence.) Let us proceed

    \[\begin{align} y'(t) &= c_1 + 2c_2 t +\ldots &= \sum_{k=1}^\infty k c_k t^{k-1}, \nonumber\\ y''(t) &= 2c_2+3\cdot 2 t +\ldots &= \sum_{k=2}^\infty k(k-1) c_k t^{k-2}, \nonumber\\ t y(t) &= c_0t + c_1 t^2 + \ldots &= \sum_{k=0}^\infty c_k t^{k+1}.\end{align}\]

    Combining this together we have \[\begin{align} y''-ty &= [2c_2+3\cdot 2 t +\ldots] - [c_0t + c_1 t^2 + \ldots] \nonumber\\ &= 2c_2+(3\cdot2 c_3-c_0)t+\ldots\nonumber\\ &= 2c_2+\sum_{k=3}^\infty\left\{k(k-1)c_k-c_{k-3}\right\}t^{k-2}.\end{align}\]

    Here we have collected terms of equal power of \(t\). The reason is simple. We are requiring a power series to equal \(0\). The only way that can work is if each power of \(x\) in the power series has zero coefficient. (Compare a finite polynomial....) We thus find \[c_2=0,\;\;k(k-1) c_k = c_{k-3}.\] The last relation is called a recurrence of recursion relation, which we can use to bootstrap from a given value, in this case \(c_0\) and \(c_1\). Once we know these two numbers, we can determine \(c_3\),\(c_4\) and \(c_5\):

    \[c_3= \frac{1}{6}c_0,\;\;\;c_4= \frac{1}{12}c_1,\;\;\;c_5=\frac{1}{20}c_2=0.\]

    These in turn can be used to determine \(c_6,c_7,c_8\), etc. It is not too hard to find an explicit expression for the \(c\)’s

    \[\begin{align} c_{3m} &= \frac{3m-2}{(3m)(3m-1)(3m-2)} c_{3(m-1)} \nonumber\\ &= \frac{3m-2}{(3m)(3m-1)(3m-2)} \frac{3m-5}{(3m-3)(3m-4)(3m-5)} c_{3(m-1)} \nonumber\\ &= \frac{(3m-2)(3m-5)\ldots 1}{(3m)!} c_0, \nonumber\\ c_{3m+1} &= \frac{3m-1}{(3m+1)(3m)(3m-1)} c_{3(m-1)+1} \nonumber\\ &= \frac{3m-1}{(3m+1)(3m)(3m-1)} \frac{3m-4}{(3m-2)(3m-3)(3m-4)} c_{3(m-2)+1} \nonumber\\ &= \frac{(3m-2)(3m-5)\ldots 2}{(3m+1)!} c_1, \nonumber\\ c_{3m+1} &= 0.\end{align}\]

    The general solution is thus

    \[y(t) = a \left[1+\sum_{m=1}^\infty c_{3m}t^{3m}\right] + b \left[1+\sum_{m=1}^\infty c_{3m+1}t^{3m+1}\right] .\]

    The technique sketched here can be proven to work for any differential equation \[y''(t)+p(t)y'(t)+q(t)y(t)=f(t)\] provided that \(p(t)\), \(q(t)\) and \(f(t)\) are analytic at \(t=0\). Thus if \(p\), \(q\) and \(f\) have a power series expansion, so has \(y\).