# PlanetPhysics/Frobenius Method

Let us consider the linear homogeneous differential equation $\sum _{\nu =0}^{n}k_{\nu }(x)y^{(n-\nu )}(x)=0$ 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 $y^{(n)}(x)+\sum _{\nu =1}^{n}c_{\nu }(x)y^{(n-\nu )}(x)=0,$ some new coefficients $c_{\nu }(x)$ are discontinuous in the singular point.\, However, if the discontinuity is restricted so, that the products $(x-x_{0})c_{1}(x),\quad (x-x_{0})^{2}c_{2}(x),\quad \ldots ,\quad (x-x_{0})^{n}c_{n}(x)$ 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 $(x-x_{0})c_{1}(x):={\frac {p(x)}{r(x)}},\quad (x-x_{0})^{2}c_{2}(x):={\frac {q(x)}{r(x)}},$ 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
   ${\begin{matrix}(x-x_{0})^{2}r(x)y''(x)+(x-x_{0})p(x)y'(x)+q(x)y(x)=0.\end{matrix}}$
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
   ${\begin{matrix}x^{2}r(x)y''(x)+xp(x)y'(x)+q(x)y(x)=0,\end{matrix}}$
where the coefficients have the converging power series expansions
   ${\begin{matrix}r(x)=\sum _{n=0}^{\infty }r_{n}x^{n},\quad p(x)=\sum _{n=0}^{\infty }p_{n}x^{n},\quad q(x)=\sum _{n=0}^{\infty }q_{n}x^{n}\end{matrix}}$
and $r_{0}\neq 0.$ In the Frobenius method one examines whether the equation (2) allows a series solution of the form
   ${\begin{matrix}y(x)=x^{s}\sum _{n=0}^{\infty }a_{n}x^{n}=a_{0}x^{s}+a_{1}x^{s+1}+a_{2}x^{s+2}+\ldots ,\end{matrix}}$
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
   ${\begin{matrix}&[r_{0}s(s\!-\!1)\!+\!p_{0}s\!+\!q_{0}]a_{0}x^{s}+\\&[[r_{0}(s\!+\!1)s\!+\!p_{0}(s\!+\!1)\!+\!q_{0}]a_{1}\!+\![r_{1}s(s\!-\!1)\!+\!p_{1}s\!+\!q_{1}]a_{0}]x^{s+1}+\\&[[r_{0}(s\!+\!2)(s\!+\!1)\!+\!p_{0}(s\!+\!2)\!+\!q_{0}]a_{2}\!+\![r_{1}(s\!+\!1)s\!+\!p_{1}(s\!+\!1)\!+\!q_{1}]a_{1}\!+\![r_{2}s(s\!-\!1)\!+\!p_{2}s\!+\!q_{2}]a_{0}]x^{s+2}\!+\ldots \end{matrix}}$
Our equation seems clearer when using the notations\, $f_{\nu }(s):=r_{\nu }{s}(s\!-\!1)+p_{\nu }{s}+q_{n}u$:
   ${\begin{matrix}f_{0}(s)a_{0}x^{s}+[f_{0}(s\!+\!1)a_{1}+f_{1}(s)a_{0}]x^{s+1}+[f_{0}(s\!+\!2)a_{2}+f_{1}(s\!+\!1)a_{1}+f_{2}(s)a_{0}]x^{s+2}+\ldots =0\end{matrix}}$
Thus the condition of satisfying the differential equation by (4) is the infinite system of equations
   ${\begin{matrix}{\begin{cases}f_{0}(s)a_{0}=0\\f_{0}(s\!+\!1)a_{1}+f_{1}(s)a_{0}=0\\f_{0}(s\!+\!2)a_{2}+f_{1}(s\!+\!1)a_{1}+f_{2}(s)a_{0}=0\\\qquad \cdots \qquad \cdots \qquad \cdots \end{cases}}\end{matrix}}$
In the first place, since\, $a_{0}\neq 0$,\, the indicial equation
   ${\begin{matrix}f_{0}(s)\equiv r_{0}s^{2}+(p_{0}-r_{0})s+q_{0}=0\end{matrix}}$
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).