Orbit equation

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

$u'\equiv {{du} \over {dz}}$ Template:Pad and Template:Pad $u''\equiv {{d^{2}u} \over {dz^{2}}}$

in the vicinity of the regular singular point $z=0$. We 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 us 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 tells us that we can seek a power series solution of the form

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

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:

${\begin{aligned}z^{2}\sum _{k=0}^{\infty }&(k+r-1)(k+r)A_{k}z^{k+r-2}-zp(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)(k+r)A_{k}z^{k+r}+q(z)A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\\&=\left[r(r-1)+p(z)r+q(z)\right]A_{0}z^{r}+\sum _{k=1}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\end{aligned}}$

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 because one can end up with complicated expressions 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. 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

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

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.

Double roots

The previous example involved an indicial polynomial with a repeated root, which only have one solution to the given differential equation. In general, the Frobenius method gives two independent solutions provided that the indicial equation's roots are unique.

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

$y_{2}=y_{1}\ln x+\sum _{k=1}^{\infty }a_{k}x^{k+r}$

Where $y_{1}(x)$ is the first solution and the coefficients $a_{k}$ are to be determined.

See also

• Regular singular point
• Complex differential equation
• Laurent series

External links

• 22 year-old Systems Analyst Rave from Merrickville-Wolford, has lots of hobbies and interests including quick cars, property developers in singapore and baking. Always loves visiting spots like Historic Monuments Zone of Querétaro.
  
  Here is my web site - cottagehillchurch.com
• John H. Mathews, Module for Frobenius Series Solution
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 Chapter 4 contains the full method including proofs.