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Frobenius method (Topic)

Let us consider the linear homogeneous differential equation

$\displaystyle \sum_{\nu=0}^n k_\nu(x) y^{(n-\nu)}(x) = 0$
of order $n$.  If the coefficient functions $k_\nu(x)$ are continuous and the coefficient $k_0(x)$ of the highest order derivative does not vanish on a certain interval (resp. a domain in $\mathbb{C}$), then all solutions $y(x)$ are continuous on this interval (resp. domain).  If all coefficients have the continuous derivatives up to a certain order, the same concerns the solutions.

If, instead, $k_0(x)$ vanishes in a point $x_0$, this point is in general a singular point.  After dividing the differential equation by $k_0(x)$ and then getting the form

$\displaystyle y^{(n)}(x)+\sum_{\nu=1}^n c_\nu(x)y^{(n-\nu)}(x) = 0,$
some new coefficients $c_\nu(x)$ are discontinuous in the singular point.  However, if the discontinuity is restricted so, that the products

$\displaystyle (x-x_0)c_1(x),\quad (x-x_0)^2c_2(x),\quad \ldots,\quad (x-x_0)^nc_n(x)$
are continuous, and even analytic in $x_0$, the point $x_0$ is a regular singular point of the differential equation.

We introduce the so-called  Frobenius method  for finding solution functions in a neighbourhood of the regular singular point $x_0$, confining us to the case of a second order differential equation.  When we use the quotient forms

$\displaystyle (x-x_0)c_1(x) := \frac{p(x)}{r(x)},\quad (x-x_0)^2c_2(x) := \frac{q(x)}{r(x)},$
where $r(x)$, $p(x)$ and $q(x)$ are analytic in a neighbourhood of $x_0$ and  $r(x) \neq 0$,  our differential equation reads
$\displaystyle (x-x_0)^2r(x)y''(x)+(x-x_0)p(x)y'(x)+q(x)y(x) = 0.$ (1)
Since a simple change  $x\!-\!x_0\mapsto x$  of variable brings to the case that the singular point is the origin, we may suppose such a starting situation.  Thus we can study the equation
$\displaystyle x^2r(x)y''(x)+xp(x)y'(x)+q(x)y(x) = 0,$ (2)
where the coefficients have the converging power series expansions
$\displaystyle r(x) = \sum_{n=0}^\infty r_nx^n,\quad p(x) = \sum_{n=0}^\infty p_nx^n,\quad q(x) = \sum_{n=0}^\infty q_nx^n$ (3)
and

$\displaystyle r_0 \neq 0.$
In the Frobenius method one examines whether the equation (2) allows a series solution of the form
$\displaystyle y(x) = x^s\sum_{n=0}^\infty a_nx^n = a_0x^s+a_1x^{s+1}+a_2x^{s+2}+\ldots,$ (4)
where $s$ is a constant and  $a_0 \neq 0$.

Substituting (3) and (4) to the differential equation (2) converts the left hand side to

  $\displaystyle [r_0s(s\!-\!1)\!+\!p_0s\!+\!q_0]a_0x^s+$    
  $\displaystyle [[r_0(s\!+\!1)s\!+\!p_0(s\!+\!1)\!+\!q_0]a_1\!+\![r_1s(s\!-\!1)\!+\!p_1s\!+\!q_1]a_0]x^{s+1}+$    
  $\displaystyle [[r_0(s\!+\!2)(s\!+\!1)\!+\!p_0(s\!+\!2)\!+\!q_0]a_2\!+\![r_1(s\!... ...1(s\!+\!1)\!+\!q_1]a_1\!+\![r_2s(s\!-\!1)\!+\!p_2s\!+\!q_2]a_0]x^{s+2}\!+\ldots$    
Our equation seems clearer when using the notations  $f_\nu(s) := r_\nu{s}(s\!-\!1)+p_\nu{s}+q_nu$:
$\displaystyle f_0(s)a_0x^s+[f_0(s\!+\!1)a_1+f_1(s)a_0]x^{s+1}+[f_0(s\!+\!2)a_2+f_1(s\!+\!1)a_1+f_2(s)a_0]x^{s+2}+\ldots = 0$ (5)
Thus the condition of satisfying the differential equation by (4) is the infinite system of equations
\begin{align*}\begin{cases} f_0(s)a_0 = 0\ f_0(s\!+\!1)a_1+f_1(s)a_0 = 0\ f_... ..._1+f_2(s)a_0 = 0\ \qquad\cdots\qquad\cdots\qquad\cdots \end{cases}\end{align*} (6)
In the first place, since  $a_0 \neq 0$,  the indicial equation
$\displaystyle f_0(s) \equiv r_0s^2+(p_0-r_0)s+q_0 = 0$ (7)
must be satisfied.  Because  $r_0 \neq 0$,  this quadratic equation determines for $s$ two values, which in special case may coincide.

The first of the equations (6) leaves $a_0\,(\neq 0)$ arbitrary.  The next linear equations in $a_n$ allow to solve successively the constants $a_1,\,a_2,\,\ldots$ provided that the first coefficients $f_0(s\!+\!1)$,  $f_0(s\!+\!2),$ $\ldots$ do not vanish; this is evidently the case when the roots of the indicial equation don't differ by an integer (e.g. when the roots are complex conjugates or when $s$ is the root having greater real part).  In any case, one obtains at least for one of the roots of the indicial equation the definite values of the coefficients $a_n$ in the series (4).  It is not hard to show that then this series converges in a neighbourhood of the origin.

For obtaining the complete solution of the differential equation (2) it suffices to have only one solution $y_1(x)$ of the form (4), because another solution $y_2(x)$, linearly independent on $y_1(x)$, is gotten via mere integrations; then it is possible in the cases  $s_1\!-\!s_2 \in\mathbb{Z}$  that $y_2(x)$ has no expansion of the form (4).

Bibliography

1
PENTTI LAASONEN: Matemaattisia erikoisfunktioita.  Handout No. 261. Teknillisen Korkeakoulun Ylioppilaskunta; Otaniemi, Finland (1969).



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Cross-references: quadratic equation, system, power, regular, functions, differential equation
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Physics Classification: 02.30.Hq (Ordinary differential equations)
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