Frobenius method

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

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

We can divide through by z² to obtain a differential equation of the form

<math>u''+{p(z) \over z}u'+{q(z) \over z}u=0<math>

which we can solve with regular power series methods if p(z)/z or q(z)/z are analytic at z = 0, but of course these functions are not. The Frobenius method enables us to create a power series solution to such a differential equation.

Explanation

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

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

Differentiating:

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

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

Substituting:

<math>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}<math>

<math>=\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}<math>

<math>=\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}<math>

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

<math>=(r(r-1)+p(0)r+q(0))A_0z^r+\sum_{k=1}^{\infty} ((k+r-1)(k+r)+p(z)(k+r)+q(z))A_kz^{k+r}<math>

The expression r(r-1)+p(0)r+q(0)=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

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

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

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

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

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

The series solution with A_k above,

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

satisfies

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

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

<math>z^2f''-zf'+(1-z)f=0\,<math>

Divide throughout by z² to give

<math>f''-{z\over z^2}f'+{1-z\over z^2}f=f''+{1\over z}f'+{1-z \over z^2}f=f''+{1\over z}f'+({1\over z^2}-{1\over z})f=0<math>

which has the requisite singularity at z=0.

Use the series solution

<math>f = \sum_{k=0}^\infty A_kz^{k+r}<math>

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

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

Now, substituting

<math> \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}-{1\over z})\sum_{k=0}^\infty A_kz^{k+r}<math>

<math> = \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}<math>

<math> = \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}<math>

We need to shift the final sum.

<math> = \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}<math>

<math> = \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}<math>

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

<math> = ((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}<math>

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

<math> = (r(r-1)-r+1)A_0z^{r-2}+\,<math>

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

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

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

<math> A_k = {-A_{k-1}\over k^2+2k+2} <math>

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

External links

The Frobenius method can be generalized to orders of ordinary differential equation greater than two, see

• A Generalization of the Frobenius Method for Ordinary Differential Equations with Regular Singular Points (pdf)