Frobenius method

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In mathematics, the Frobenius method describes a way to find an infinite series solution for a second-order ordinary differential equation of the form

$z^2u''+p(z)zu'+q(z)u=0\!\;$

We can divide by z² 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 noninfinite).

1 Explanation
2 Example
3 External links
4 See also

[edit] Explanation

The Frobenius method tells us that we can seek a power series solution of the form

$u(z)=\sum_{k=0}^{\infty} A_kz^{k+r}$

Differentiating:

$u'(z)=\sum_{k=0}^{\infty} (k+r)A_kz^{k+r-1}$

$u''(z)=\sum_{k=0}^{\infty} (k+r-1)(k+r)A_kz^{k+r-2}$

Substituting:

$z^2\sum_{k=0}^{\infty} (k+r-1)(k+r)A_kz^{k+r-2}+zp(z)\sum_{k=0}^{\infty} (k+r)A_kz^{k+r-1}+q(z)\sum_{k=0}^{\infty} A_kz^{k+r}$

$=\sum_{k=0}^{\infty} (k+r-1)(k+r)A_kz^{k+r}+p(z)\sum_{k=0}^{\infty} (k+r)A_kz^{k+r}+q(z)\sum_{k=0}^{\infty} A_kz^{k+r}$

$=\sum_{k=0}^{\infty} (k+r-1)(k+r)A_kz^{k+r}+p(z)(k+r)A_kz^{k+r}+q(z)A_kz^{k+r}$

$=\sum_{k=0}^{\infty} ((k+r-1)(k+r)+p(z)(k+r)+q(z))A_kz^{k+r}$

$=(r(r-1)+p(z)r+q(z))A_0z^r+\sum_{k=1}^{\infty} ((k+r-1)(k+r)+p(z)(k+r)+q(z))A_kz^{k+r}$

The expression r(r-1)+p(z)r+q(z)=I(r) is known as the indicial polynomial, which is quadratic in r.

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

$I(k+r)A_k+\sum_{j=0}^{k-1}((j+r)p(k-j)+q(k-j))A_j$

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

$I(k+r)A_k+\sum_{j=0}^{k-1}((j+r)p(k-j)+q(k-j))A_j=0$

$\sum_{j=0}^{k-1}((j+r)p(k-j)+q(k-j))A_j=-I(k+r)A_k$

${1\over-I(k+r)}\sum_{j=0}^{k-1}((j+r)p(k-j)+q(k-j))A_j=A_k$

The series solution with A_k above,

$U_{r}(z)=\sum_{k=0}^{\infty}A_kz^{k+r}$

satisfies

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

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.

[edit] Example

Let us solve

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

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

$f = \sum_{k=0}^\infty A_kz^{k+r}$

$f' = \sum_{k=0}^\infty (k+r)A_kz^{k+r-1}$

$f'' = \sum_{k=0}^\infty (k+r)(k+r-1)A_kz^{k+r-2}$

Now, substituting

$\sum_{k=0}^\infty (k+r)(k+r-1)A_kz^{k+r-2}-{1\over z}\sum_{k=0}^\infty (k+r)A_kz^{k+r-1}+\left({1\over z^2}-{1\over z}\right)\sum_{k=0}^\infty A_kz^{k+r}$

$= \sum_{k=0}^\infty (k+r)(k+r-1)A_kz^{k+r-2}-{1\over z}\sum_{k=0}^\infty (k+r)A_kz^{k+r-1}+{1\over z^2}\sum_{k=0}^\infty A_kz^{k+r}-{1\over z}\sum_{k=0}^\infty A_kz^{k+r}$

$= \sum_{k=0}^\infty (k+r)(k+r-1)A_kz^{k+r-2}-\sum_{k=0}^\infty (k+r)A_kz^{k+r-2}+\sum_{k=0}^\infty A_kz^{k+r-2}-\sum_{k=0}^\infty A_kz^{k+r-1}$

We need to shift the final sum.

$= \sum_{k=0}^\infty (k+r)(k+r-1)A_kz^{k+r-2}-\sum_{k=0}^\infty (k+r)A_kz^{k+r-2}+\sum_{k=0}^\infty A_kz^{k+r-2}-\sum_{k-1=0}^\infty A_{k-1}z^{k+r-2}$

$= \sum_{k=0}^\infty (k+r)(k+r-1)A_kz^{k+r-2}-\sum_{k=0}^\infty (k+r)A_kz^{k+r-2}+\sum_{k=0}^\infty A_kz^{k+r-2}-\sum_{k=1}^\infty A_{k-1}z^{k+r-2}$

We can take one element out of the sums that start with k=0 to obtain the sums starting at the same index.

$= ((r)(r-1)A_0z^{r-2})+\sum_{k=1}^\infty (k+r)(k+r-1)A_kz^{k+r-2}-((r)A_0z^{r-2})-\sum_{k=1}^\infty (k+r)A_kz^{k+r-2}$

$+(A_0z^{r-2})+\sum_{k=1}^\infty A_kz^{k+r-2}-\sum_{k=1}^\infty A_{k-1}z^{k+r-2}$

$= (r(r-1)-r+1)A_0z^{r-2}+\,$

$\sum_{k=1}^\infty \left( ((k+r)(k+r-1)-(k+r)+1)A_k - A_{k-1} \right)z^{k+r-2}$

We obtain one linearly independent solution by solving the indicial polynomial r(r-1)-r+1 = r²-2r+1 =0 which gives 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 the recurrence

$((k+1)(k)-(k+1)+1)A_k - A_{k-1} =(k^2)A_k-A_{k-1}=0\,$