Power series solution of differential equations

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 z. Then we can divide throughout to obtain

$f''+{a_{1}(z) \over a_{2}(z)}f'+{a_{0}(z) \over a_{2}(z)}f=0$

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

$f=\sum _{k=0}^{\infty }A_{k}z^{k}$

If a₂ 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

$f=\sum _{k=0}^{\infty }A_{k}z^{k}$

$f'=\sum _{k=0}^{\infty }kA_{k}z^{k-1}$

$f''=\sum _{k=0}^{\infty }k(k-1)A_{k}z^{k-2}$

Substituting these in the differential equation

$\sum _{k=0}^{\infty }k(k-1)A_{k}z^{k-2}-2z\sum _{k=0}^{\infty }kA_{k}z^{k-1}+\sum _{k=0}^{\infty }A_{k}z^{k}=0$

$=\sum _{k=0}^{\infty }k(k-1)A_{k}z^{k-2}-\sum _{k=0}^{\infty }2kA_{k}z^{k}+\sum _{k=0}^{\infty }A_{k}z^{k}$

Making a shift on the first sum

$=\sum _{k+2=0}^{\infty }(k+2)((k+2)-1)A_{k+2}z^{(k+2)-2}-\sum _{k=0}^{\infty }2kA_{k}z^{k}+\sum _{k=0}^{\infty }A_{k}z^{k}$

$=\sum _{k=-2}^{\infty }(k+2)(k+1)A_{k+2}z^{k}-\sum _{k=0}^{\infty }2kA_{k}z^{k}+\sum _{k=0}^{\infty }A_{k}z^{k}$

$=(0)(-1)A_{0}z^{-2}+(-1)(0)A_{1}z^{-1}+\sum _{k=0}^{\infty }(k+2)(k+1)A_{k+2}z^{k}-\sum _{k=0}^{\infty }2kA_{k}z^{k}+\sum _{k=0}^{\infty }A_{k}z^{k}$

$=\sum _{k=0}^{\infty }(k+2)(k+1)A_{k+2}z^{k}-\sum _{k=0}^{\infty }2kA_{k}z^{k}+\sum _{k=0}^{\infty }A_{k}z^{k}$

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

Now, 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

$A_{2}={-1 \over (2)(1)}A_{0}={-1 \over 2}A_{0},\,A_{3}={1 \over (3)(2)}A_{1}={1 \over 6}A_{1}$

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

$f=A_{0}x^{0}+A_{1}x^{1}+A_{2}x^{2}+A_{3}x^{3}+A_{4}x^{4}+A_{5}x^{5}+A_{6}x^{6}+A_{7}x^{7}+\cdots$

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

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

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).