Power series method

In mathematics, the power series method is used to seek a power series solution to certain differential equations.

Method

Consider the second-order linear differential equation: $$a_2(z)f"(z)+a_1(z)f'(z)+a_0(z)f(z)=0;!Suppose "a"₂ is nonzero for all "z". Then we can divide throughout to obtain: $$f"+{a_1(z)over a_2(z)}f'+{a_0(z)over a_2(z)}f=0Suppose further that "a"₁/"a"₂ and "a"₀/"a"₂ are analytic functions.

The power series method calls for the construction of a power series solution:$$f=sum_{k=0}^infty A_kz^k

If "a"₂ is zero for some "z", then the Frobenius method, a variation on this method, is suited to deal with so called "singular points".

Example usage

Let us look at the Hermite differential equation,: $$f"-2zf'+lambda f=0;;lambda=1

We can try and construct a series solution: $$f=sum_{k=0}^infty A_kz^k: $$f'=sum_{k=0}^infty kA_kz^{k-1}: $$f"=sum_{k=0}^infty k(k-1)A_kz^{k-2}

Substituting these in the differential equation: $$sum_{k=0}^infty k(k-1)A_kz^{k-2}-2zsum_{k=0}^infty kA_kz^{k-1}+sum_{k=0}^infty A_kz^k=0: $$=sum_{k=0}^infty k(k-1)A_kz^{k-2}-sum_{k=0}^infty 2kA_kz^k+sum_{k=0}^infty A_kz^kMaking a shift on the first sum: $$=sum_{k+2=0}^infty (k+2)((k+2)-1)A_{k+2}z^{(k+2)-2}-sum_{k=0}^infty 2kA_kz^k+sum_{k=0}^infty A_kz^k: $$=sum_{k=-2}^infty (k+2)(k+1)A_{k+2}z^k-sum_{k=0}^infty 2kA_kz^k+sum_{k=0}^infty A_kz^k: $$=(0)(-1)A_{0}z^{-2} + (-1)(0)A_{1}z^{-1}+sum_{k=0}^infty (k+2)(k+1)A_{k+2}z^k-sum_{k=0}^infty 2kA_kz^k+sum_{k=0}^infty A_kz^k: $$=sum_{k=0}^infty (k+2)(k+1)A_{k+2}z^k-sum_{k=0}^infty 2kA_kz^k+sum_{k=0}^infty A_kz^k: $$=sum_{k=0}^infty left((k+2)(k+1)A_{k+2}+(-2k+1)A_k ight)z^kNow, if this series is a solution, all these coefficients must be zero, so:: $$(k+2)(k+1)A_{k+2}+(-2k+1)A_k=0;!We can rearrange this to get a recurrence relation for "A"_"k"+2.: $$(k+2)(k+1)A_{k+2}=-(-2k+1)A_k;!: $$A_{k+2}={(2k-1)over (k+2)(k+1)}A_k;!

Now, we have: $$A_2 = {-1 over (2)(1)}A_0={-1over 2}A_0,, A_3 = {1 over (3)(2)} A_1={1over 6}A_1We can determine "A"₀ and "A"₁ if there are initial conditions, ie., if we have an initial value problem.

So, we have: $$A_4={1over 4}A_2 = left({1over 4} ight)left({-1 over 2} ight)A_0 = {-1 over 8}A_0

: $$A_5={1over 4}A_3 = left({1over 4} ight)left({1 over 6} ight)A_1 = {1 over 24}A_1

: $$A_6={7over 30}A_4 = left({7over 30} ight)left({-1 over 8} ight)A_0 = {-7 over 240}A_0

: $$A_7={3over 14}A_5 = left({3over 14} ight)left({1 over 24} ight)A_1 = {1 over 112}A_1

and the series solution is

: $$f=A_0x^0+A_1x^1+A_2x^2+A_3x^3+A_4x^4+A_5x^5+A_6x^6+A_7x^7+cdots

: $$=A_0x^0+A_1x^1+{-1over 2}A_0x^2+{1over 6}A_1x^3+{-1 over 8}A_0x^4+{1 over 24}A_1x^5+{-7 over 240}A_0x^6+{1 over 112}A_1x^7+cdots

: $$=A_0x^0+{-1over 2}A_0x^2+{-1 over 8}A_0x^4+{-7 over 240}A_0x^6+A_1x+{1over 6}A_1x^3+{1 over 24}A_1x^5+{1 over 112}A_1x^7+cdots

which we can break up into the sum of two linearly independent series solutions:

: $$f=A_0(1+{-1over 2}x^2+{-1 over 8}x^4+{-7 over 240}x^6+cdots)+A_1(x+{1over 6}x^3+{1 over 24}x^5+{1 over 112}x^7+cdots)

which can be further simplified by the use of hypergeometric series (which goes beyond the scope of this article).

External links

*
* [http://math.fullerton.edu/mathews/n2003/FrobeniusSeriesMod.html Module for Frobenius Series Solution]
* [http://www.ntu.edu.sg/home/mwtang/odesite.htm A Concise Introductory Course in Ordinary Differential Equations (with a chapter on series solutions)]