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 by z2 to obtain a differential equation of the form
- <math>u+{p(z) \over z}u'+{q(z) \over z^2}u=0</math>
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).
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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(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}</math>
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 zk+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 should 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 Ak 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 Ur(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-{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</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}+\left({1\over z^2}-{1\over z}\right)\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 zk+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)A_k-A_{k-1}=0\,</math>
- <math> A_k = {A_{k-1}\over k^2} </math>
Given some initial conditions, we can either solve the recurrence entirely or obtain a solution in power series form. Since the ratio of coefficients <math>A_k/A_{k-1}</math> is a rational function, the power series can be written as a hypergeometric series.
