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 through by z² to obtain a differential equation of the form

[u''+u'+u=0]

which we can solve with regular power series methods if p(z)/z or q(z)/z² 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.

Contents

1 Explanation
2 Example
3 See also
4 External links

Explanation

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

[u(z)=\sum_^ A_kz^]

Differentiating:

[u'(z)=\sum_^ (k+r)A_kz^]

[u''(z)=\sum_^ (k+r-1)(k+r)A_kz^]

Substituting:

[z^2\sum_^ (k+r-1)(k+r)A_kz^+zp(z)\sum_^ (k+r)A_kz^+q(z)\sum_^ A_kz^]

[=\sum_^ (k+r-1)(k+r)A_kz^+p(z)\sum_^ (k+r)A_kz^+q(z)\sum_^ A_kz^]

[=\sum_^ (k+r-1)(k+r)A_kz^+p(z)(k+r)A_kz^+q(z)A_kz^]

[=\sum_^ ((k+r-1)(k+r)+p(z)(k+r)+q(z))A_kz^]

[=(r(r-1)+p(0)r+q(0))A_0z^r+\sum_^ ((k+r-1)(k+r)+p(z)(k+r)+q(z))A_kz^]

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

[I(k+r)A_k+\sum_^((j+r)p(k-j)+q(k-j))A_j]

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

[I(k+r)A_k+\sum_^((j+r)p(k-j)+q(k-j))A_j=0]

[\sum_^((j+r)p(k-j)+q(k-j))A_j=-I(k+r)A_k]

[\sum_^((j+r)p(k-j)+q(k-j))A_j=A_k]

The series solution with A_k above,

[U_(z)=\sum_^A_kz^]

satisfies

[z^2U_(z)''+p(z)zU_(z)'+q(z)U_(z)=I(r)z^\!\;]

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

Divide throughout by z² to give

[f-f'+f=f-f'+f=f''-f'+\left(-\right)f=0]

which has the requisite singularity at z=0.

Use the series solution

[f = \sum_^\infty A_kz^]

[f' = \sum_^\infty (k+r)A_kz^]

[f'' = \sum_^\infty (k+r)(k+r-1)A_kz^]

Now, substituting

[ \sum_^\infty (k+r)(k+r-1)A_kz^-\sum_^\infty (k+r)A_kz^+(-)\sum_^\infty A_kz^]

[ = \sum_^\infty (k+r)(k+r-1)A_kz^-\sum_^\infty (k+r)A_kz^+\sum_^\infty A_kz^-\sum_^\infty A_kz^]

[ = \sum_^\infty (k+r)(k+r-1)A_kz^-\sum_^\infty (k+r)A_kz^+\sum_^\infty A_kz^+\sum_^\infty A_kz^]

We need to shift the final sum.

[ = \sum_^\infty (k+r)(k+r-1)A_kz^-\sum_^\infty (k+r)A_kz^+\sum_^\infty A_kz^+\sum_^\infty A_z^]

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^)+\sum_^\infty (k+r)(k+r-1)A_kz^-((r)A_0z^)+\sum_^\infty (k+r)A_kz^]

:[+(A_0z^)+\sum_^\infty A_kz^+\sum_^\infty A_z^]

[ = (r(r-1)-r+1)A_0z^+\,]

: [\sum_^\infty \left( ((k+r)(k+r-1)+(k+r)+1)A_k + A_ \right)z^]

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^2+2k+2)A_k+A_=0\,]

[ A_k = \over k^2+2k+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_] is a rational function, the power series can be written as a hypergeometric series.

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)

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Frobenius method

Explanation

Example

See also

External links