Power series solution of differential equations

In mathematics, the power series method is used to seek a power series solution to certain differential equations. In general, such a solution assumes a power series with unknown coefficients, then substitutes that solution into the differential equation to find a recurrence relation for the coefficients.

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 calls for the construction of 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 to 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

{\begin{aligned}&{}\quad \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}\end{aligned}}

Making a shift on the first sum

{\begin{aligned}&=\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}\end{aligned}}

If this series is a solution, then 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, i.e. if we have an initial value problem.

So we have

{\begin{aligned}A_{4}&={1 \over 4}A_{2}=\left({1 \over 4}\right)\left({-1 \over 2}\right)A_{0}={-1 \over 8}A_{0}\\[8pt]A_{5}&={1 \over 4}A_{3}=\left({1 \over 4}\right)\left({1 \over 6}\right)A_{1}={1 \over 24}A_{1}\\[8pt]A_{6}&={7 \over 30}A_{4}=\left({7 \over 30}\right)\left({-1 \over 8}\right)A_{0}={-7 \over 240}A_{0}\\[8pt]A_{7}&={3 \over 14}A_{5}=\left({3 \over 14}\right)\left({1 \over 24}\right)A_{1}={1 \over 112}A_{1}\end{aligned}}

and the series solution is

{\begin{aligned}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 \\[8pt]&=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 \\[8pt]&=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 \end{aligned}}

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

f=A_{0}\left(1+{-1 \over 2}x^{2}+{-1 \over 8}x^{4}+{-7 \over 240}x^{6}+\cdots \right)+A_{1}\left(x+{1 \over 6}x^{3}+{1 \over 24}x^{5}+{1 \over 112}x^{7}+\cdots \right)

which can be further simplified by the use of hypergeometric series.

Nonlinear equations

The power series method can be applied to certain nonlinear differential equations, though with much less flexibility. Generally, the power series method will give solutions only to initial value problems (opposed to boundary value problems), this is not an issue when dealing with linear equations since the solution may turn up multiple linearly independent solutions which may be combined (by superposition) to solve boundary value problems as well. A further restriction is that the series coefficients will be specified by a nonlinear recurrence (the nonlinearities are inherited from the differential equation).

In order for the solution method to work, as in linear equations, it is necessary to express every term in the nonlinear equation as a power series so that all of the terms may be combined into one power series.

As an example, consider the initial value problem

FF''+2F'^{2}+\eta F'=0\quad ;\quad F(1)=0\ ,\ F'(1)=-{\frac {1}{2}}

which describes a solution to capillary-driven flow in a groove. Note the two nonlinearities: the first and second terms involve products. Note also that the initial values are given at $\eta =1$ , which hints that the power series must be set up as:

F(\eta )=\sum _{i=0}^{\infty }c_{i}(\eta -1)^{i}

since in this way

{\frac {d^{n}F}{d\eta ^{n}}}{\Bigg |}_{\eta =1}=n!\ c_{i}

which makes the initial values very easy to evaluate. It is necessary to rewrite the equation slightly in light of the definition of the power series,

FF''+2F'^{2}+(\eta -1)F'+F'=0\quad ;\quad F(1)=0\ ,\ F'(1)=-{\frac {1}{2}}

so that the third term contains the same form $\eta -1$ that shows in the power series.

The last consideration is what to do with the products; substituting the power series in would result in products of power series when it's necessary that each term be its own power series. This is where the identity

\left(\sum _{i=0}^{\infty }a_{i}x^{i}\right)\left(\sum _{i=0}^{\infty }b_{i}x^{i}\right)=\sum _{i=0}^{\infty }x^{i}\sum _{j=0}^{i}a_{i-j}b_{j}

is useful; substituting the power series into the differential equation and applying this identity leads to an equation where every term is a power series. After much rearrangement, the recurrence

\sum _{j=0}^{i}\left((j+1)(j+2)c_{i-j}c_{j+2}+2(i-j+1)(j+1)c_{i-j+1}c_{j+1}\right)+ic_{i}+(i+1)c_{i+1}=0

is obtained, specifying exact values of the series coefficients. From the initial values, $c_{0}=0$ and $c_{1}=-1/2$ , thereafter the above recurrence is used. For example, the next few coefficients:

c_{2}=-{\frac {1}{6}}\quad ;\quad c_{3}=-{\frac {1}{108}}\quad ;\quad c_{4}={\frac {7}{3240}}\quad ;\quad c_{5}=-{\frac {19}{48600}}\ \dots

A limitation of the power series solution shows itself in this example. A numeric solution of the problem shows that the function is smooth and always decreasing to the left of $\eta =1$ , and zero to the right. At $\eta =1$ , a slope discontinuity exists, a feature which the power series is incapable of rendering, for this reason the series solution continues decreasing to the right of $\eta =1$ instead of suddenly becoming zero.

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