math lessons - 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 a₂ is nonzero for all x. Then we can divide throughout to obtain

$f''+{a_1(x)\over a_2(x)}f'+{a_0(x)\over a_2(x)}f=0$

Suppose further that a₁/a₂ and a₀/a₂ are analytic functions.

The power series method tells us we may be able to construct a power series solution

$f=\sum_{k=0}^\infty A_kz^k$

If a₂ is zero for some x, 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

$\sum_{k=0}^\infty k(k-1)A_kz^{k-2}-2z\sum_{k=0}^\infty kA_kz^{k-1}+\sum_{k=0}^\infty A_kz^k=0$

$=\sum_{k=0}^\infty k(k-1)A_kz^{k-2}-\sum_{k=0}^\infty 2kA_kz^k+\sum_{k=0}^\infty A_kz^k$

$=\sum_{k=0}^\infty k(k-1)A_kz^{k-2}-\sum_{k=0}^\infty 2kA_kz^k+A_kz^k$

Making a shift on the first sum

$=\sum_{k+2=0}^\infty (k+2)((k+2)-1)A_{k+2}z^{(k+2)-2}-\sum_{k=0}^\infty 2kA_kz^k+A_kz^k$

$=\sum_{k=-2}^\infty (k+2)(k+1)A_{k+2}z^k-\sum_{k=0}^\infty 2kA_kz^k+A_kz^k$

$=(0)(-1)A_{0}z^{-2} + (-1)(0)A_{1}z^{-1}+\sum_{k=0}^\infty (k+2)(k+1)A_{k+2}z^k-\sum_{k=0}^\infty 2kA_kz^k+A_kz^k$

$=\sum_{k=0}^\infty (k+2)(k+1)A_{k+2}z^k-\sum_{k=0}^\infty 2kA_kz^k+A_kz^k$

$=\sum_{k=0}^\infty \left((k+2)(k+1)A_{k+2}+(-2k+1)A_k\right)z^k$

Now, if this series 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

$A_2 = {-1 \over (2)(1)}A_0={-1\over 2}A_0,\, A_3 = {1 \over (3)(2)} A_1={1\over 6}A_1$

We can determine A₀ and A₁ if there are initial conditions, ie., if we have an initial value problem.

So, we have

$A_4={1\over 4}A_2 = \left({1\over 4}\right)\left({-1 \over 2}\right)A_0 = {-1 \over 8}A_0$

$A_5={1\over 4}A_3 = \left({1\over 4}\right)\left({1 \over 6}\right)A_1 = {1 \over 24}A_1$

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

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

and the series solution is

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

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

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

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

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

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

Categories: Ordinary differential equations

01-04-2007 01:18:14
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Power series method

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Power series method

Method

Example usage

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