Frobenius method

In mathematics, the method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a second-order ordinary differential equation of the form

z^{2}u''+p(z)zu'+q(z)u=0

with

u'\equiv {{du} \over {dz}}

and

u''\equiv {{d^{2}u} \over {dz^{2}}}

in the vicinity of the regular singular point $z=0$ . One can divide by $z^{2}$ to obtain a differential equation of the form

u''+{p(z) \over z}u'+{q(z) \over z^{2}}u=0

which will not be solvable with regular power series methods if either p(z)/z or q(z)/z² are not analytic at z = 0. The Frobenius method enables one to create a power series solution to such a differential equation, provided that p(z) and q(z) are themselves analytic at 0 or, being analytic elsewhere, both their limits at 0 exist (and are finite).

Explanation

The method of Frobenius is to seek a power series solution of the form

u(z)=z^{r}\sum _{k=0}^{\infty }A_{k}z^{k},\qquad (A_{0}\neq 0)

Differentiating:

u'(z)=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}

u''(z)=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}

Substituting the above differentiation into our original ODE:

z^{2}\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}+zp(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}

=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}

=\sum _{k=0}^{\infty }[(k+r-1)(k+r)A_{k}z^{k+r}+p(z)(k+r)A_{k}z^{k+r}+q(z)A_{k}z^{k+r}]

=\sum _{k=0}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}

=\left[r(r-1)+p(z)r+q(z)\right]A_{0}z^{r}+\sum _{k=1}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}

The expression

r\left(r-1\right)+p\left(0\right)r+q\left(0\right)=I(r)

is known as the indicial polynomial, which is quadratic in r. The general definition of the indicial polynomial is the coefficient of the lowest power of z in the infinite series. In this case it happens to be that this is the rth coefficient but, it is possible for the lowest possible exponent to be r − 2, r − 1 or, something else depending on the given differential equation. This detail is important to keep in mind. In the process of synchronizing all the series of the differential equation to start at the same index value (which in the above expression is k = 1), one can end up with complicated expressions. However, in solving for the indicial roots attention is focused only on the coefficient of the lowest power of z.

Using this, the general expression of the coefficient of z^k + r is

I(k+r)A_{k}+\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}

,

These coefficients must be zero, since they should be solutions of the differential equation, so

I(k+r)A_{k}+\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}=0

\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}=-I(k+r)A_{k}

{1 \over -I(k+r)}\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}=A_{k}

The series solution with A_k above,

U_{r}(z)=\sum _{k=0}^{\infty }A_{k}z^{k+r}

satisfies

z^{2}U_{r}(z)''+p(z)zU_{r}(z)'+q(z)U_{r}(z)=I(r)z^{r}

If we choose one of the roots to the indicial polynomial for r in U_r(z), we gain a solution to the differential equation. If the difference between the roots is not an integer, we get another, linearly independent solution in the other root.

Example

Let us solve

z^{2}f''-zf'+(1-z)f=0\,

Divide throughout by z² to give

f''-{1 \over z}f'+{1-z \over z^{2}}f=f''-{1 \over z}f'+\left({1 \over z^{2}}-{1 \over z}\right)f=0

which has the requisite singularity at z = 0.

Use the series solution

{\begin{aligned}f&=\sum _{k=0}^{\infty }A_{k}z^{k+r}\\f'&=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}\\f''&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}\end{aligned}}

Now, substituting

{\begin{aligned}\sum _{k=0}^{\infty }&(k+r)(k+r-1)A_{k}z^{k+r-2}-{\frac {1}{z}}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+\left({\frac {1}{z^{2}}}-{\frac {1}{z}}\right)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-{\frac {1}{z}}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+{\frac {1}{z^{2}}}\sum _{k=0}^{\infty }A_{k}z^{k+r}-{\frac {1}{z}}\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }A_{k}z^{k+r-1}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k-1=0}^{\infty }A_{k-1}z^{k-1+r-1}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=\left\{\sum _{k=0}^{\infty }\left((k+r)(k+r-1)-(k+r)+1\right)A_{k}z^{k+r-2}\right\}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=\left\{\left(r(r-1)-r+1\right)A_{0}z^{r-2}+\sum _{k=1}^{\infty }\left((k+r)(k+r-1)-(k+r)+1\right)A_{k}z^{k+r-2}\right\}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=(r-1)^{2}A_{0}z^{r-2}+\left\{\sum _{k=1}^{\infty }(k+r-1)^{2}A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\right\}\\&=(r-1)^{2}A_{0}z^{r-2}+\sum _{k=1}^{\infty }\left((k+r-1)^{2}A_{k}-A_{k-1}\right)z^{k+r-2}\end{aligned}}

From (r − 1)² = 0 we get a double root of 1. Using this root, we set the coefficient of z^{k + r − 2} to be zero (for it to be a solution), which gives us:

(k+1-1)^{2}A_{k}-A_{k-1}=k^{2}A_{k}-A_{k-1}=0

hence we have the recurrence relation:

A_{k}={\frac {A_{k-1}}{k^{2}}}

Given some initial conditions, we can either solve the recurrence entirely or obtain a solution in power series form.

Since the ratio of coefficients $A_{k}/A_{k-1}$ is a rational function, the power series can be written as a generalized hypergeometric series.

Roots separated by an integer

The previous example involved an indicial polynomial with a repeated root, which gives only one solution to the given differential equation. In general, the Frobenius method gives two independent solutions provided that the indicial equation's roots are not separated by an integer (including zero).

If the root is repeated or the roots differ by an integer, then the second solution can be found using:

y_{2}=Cy_{1}\ln x+\sum _{k=0}^{\infty }B_{k}x^{k+r_{2}}

where $y_{1}(x)$ is the first solution (based on the larger root in the case of unequal roots), $r_{2}$ is the smaller root, and the constant $C$ and the coefficients $B_{k}$ are to be determined. Once $B_{0}$ is chosen (for example by setting it to 1) then $C$ and the $B_{k}$ are determined up to but not including $B_{r_{1}-r_{2}}$ , which can be set arbitrarily. This then determines the rest of the $B_{k}.$ In some cases the constant $C$ must be zero. For example, consider the following differential equation (Kummer's equation with $a = 1$ and $b = 2$ ):

zu''+(2-z)u'-u=0

The roots of the indicial equation are −1 and 0. Two independent solutions are $1/z$ and $(e^{z})/z,$ so we see that the logarithm does not appear in any solution. The solution $(e^{z}-1)/z$ has a power series starting with the power zero. In a power series starting with $z^{-1}$ the recurrence relation places no restriction on the coefficient for the term $z^{0},$ which can be set arbitrarily. If it is set to zero then with this differential equation all the other coefficients will be zero and we obtain the solution 1/ $z$ .