Frobenius Method - Example

Example

Let us solve

Divide throughout by z2 to give

which has the requisite singularity at z = 0.

Use the series solution

Now, substituting

\begin{align} & \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} \end{align}

We need to shift the final sum.

\begin{align} & = \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} \end{align}

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

\begin{align} & = ((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} \end{align}

We obtain one 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

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

Since the ratio of coefficients is a rational function, the power series can be written as a generalized hypergeometric series.

Read more about this topic:  Frobenius Method

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