PlanetPhysics/Frobenius Method

Let us consider the linear homogeneous [[../DifferentialEquations/|differential equation]] $\sum _{\nu =0}^n{k}_{\nu }(x)y^{(n-\nu )}(x)=0$ of order $n$.\, If the coefficient [[../Bijective/|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 $ℂ$), 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 $(x-x_0)c_1(x),(x-x_0{)}^2{c}_2(x),\dots ,(x-x_0{)}^n{c}_n(x)$ are continuous, and even analytic in $x_0$, the point $x_0$ is a [[../CoIntersections/|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 $(x-x_0)c_1(x):=\frac{p(x)}{r(x)},(x-x_0{)}^2{c}_2(x):=\frac{q(x)}{r(x)},$ where $r(x)$, $p(x)$ and $q(x)$ are analytic in a neighbourhood of $x_0$ and\, $r(x)\ne 0$,\, our differential equation reads

$\begin{array}{c}(x-x_0{)}^{2}r(x)y^{″}(x)+(x-x_0)p(x)y^{\prime }(x)+q(x)y(x)=0.\end{array}$

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

$\begin{array}{c}{x}^{2}r(x)y^{″}(x)+xp(x)y^{\prime }(x)+q(x)y(x)=0,\end{array}$

where the coefficients have the converging [[../Power/|power]] series expansions

$\begin{array}{c}r(x)=\sum _{n=0}^{\mathrm{\infty }}{r}_n{x}^n,p(x)=\sum _{n=0}^{\mathrm{\infty }}{p}_n{x}^n,q(x)=\sum _{n=0}^{\mathrm{\infty }}{q}_n{x}^n\end{array}$

and $r_0\ne 0.$ In the Frobenius method one examines whether the equation (2) allows a series solution of the form

$\begin{array}{c}y(x)=x^s\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=a_0{x}^s+a_1{x}^{s+1}+a_2{x}^{s+2}+\dots ,\end{array}$

where $s$ is a constant and\, $a_0\ne 0$.

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

$\begin{array}{cc}& [r_{0}s(s-1)+p_{0}s+q_0]a_0{x}^s+\\ & [[r_0(s+1)s+p_0(s+1)+q_0]a_1+[r_{1}s(s-1)+p_{1}s+q_1]a_0]x^{s+1}+\\ & [[r_0(s+2)(s+1)+p_0(s+2)+q_0]a_2+[r_1(s+1)s+p_1(s+1)+q_1]a_1+[r_{2}s(s-1)+p_{2}s+q_2]a_0]x^{s+2}+\dots \end{array}$

Our equation seems clearer when using the notations\, $f_{\nu }(s):=r_{\nu }s(s-1)+p_{\nu }s+q_{n}u$:

$\begin{array}{c}{f}_0(s)a_0{x}^s+[f_0(s+1)a_1+f_1(s)a_0]x^{s+1}+[f_0(s+2)a_2+f_1(s+1)a_1+f_2(s)a_0]x^{s+2}+\dots =0\end{array}$

Thus the condition of satisfying the differential equation by (4) is the infinite [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] of equations

$\begin{array}{c}\{\begin{array}{c}{f}_0(s)a_0=0\\ f_0(s+1)a_1+f_1(s)a_0=0\\ f_0(s+2)a_2+f_1(s+1)a_1+f_2(s)a_0=0\\ \cdots \cdots \cdots \end{array}\end{array}$

In the first place, since\, $a_0\ne 0$,\, the indicial equation

$\begin{array}{c}{f}_0(s)\equiv r_0{s}^2+(p_0-r_0)s+q_0=0\end{array}$

must be satisfied.\, Because\, $r_0\ne 0$,\, this [[../QuadraticFormula/|quadratic equation]] determines for $s$ two values, which in special case may coincide.

The first of the equations (6) leaves $a_0(\ne 0)$ arbitrary.\, The next linear equations in $a_n$ allow to solve successively the constants $a_1,a_2,\dots$ provided that the first coefficients $f_0(s+1)$,\, $f_0(s+2),$\,$\dots$ 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).