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: z^2u''+p(z)zu'+q(z)u=0\!\;
We can divide through 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 we can solve with regular Power Series Method s if ''p''(''z'')/''z'' or ''q''(''z'')/''z''2 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
: 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(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}
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=0}^{k-1}((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=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.


EXAMPLE

Let us solve
: z^2f''-zf'+(1-z)f=0\,
Divide throughout by ''z''2 to give
: 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
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}+({1\over z^2}-{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}-{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} ight)z^{k+r-2}

We obtain one linearly independent solution by solving the indicial polynomial ''r''(''r''-1)-''r''+1 = ''r''2-2''r''+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+2k+2)A_k+A_{k-1}=0\,
: A_k = {-A_{k-1}\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_{k-1} is a Rational Function , the power series can be written as a Hypergeometric Series .


SEE ALSO




EXTERNAL LINKS


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