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

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

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

The power series method calls for the construction of a power series solution

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

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". The method works analogously for higher order equations as well as for systems.

Example usage

Let us look at the Hermite differential equation,

$${\displaystyle f''-2zf'+\lambda f=0;\;\lambda =1}$$

We can try to construct a series solution

$${\displaystyle {\begin{aligned}f&=\sum _{k=0}^{\infty }A_{k}z^{k}\\f'&=\sum _{k=1}^{\infty }kA_{k}z^{k-1}\\f''&=\sum _{k=2}^{\infty }k(k-1)A_{k}z^{k-2}\end{aligned))}$$

Substituting these in the differential equation

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

Making a shift on the first sum

$${\displaystyle {\begin{aligned}&=\sum _{k=0}^{\infty }(k+2)(k+1)A_{k+2}z^{k}-\sum _{k=1}^{\infty }2kA_{k}z^{k}+\sum _{k=0}^{\infty }A_{k}z^{k}\\&=2A_{2}+\sum _{k=1}^{\infty }(k+2)(k+1)A_{k+2}z^{k}-\sum _{k=1}^{\infty }2kA_{k}z^{k}+A_{0}+\sum _{k=1}^{\infty }A_{k}z^{k}\\&=2A_{2}+A_{0}+\sum _{k=1}^{\infty }\left((k+2)(k+1)A_{k+2}+(-2k+1)A_{k}\right)z^{k}\end{aligned))}$$

If this series is a solution, then all these coefficients must be zero, so for both k=0 and k>0:

$${\displaystyle (k+2)(k+1)A_{k+2}+(-2k+1)A_{k}=0}$$

We can rearrange this to get a recurrence relation for A_k+2.

$${\displaystyle (k+2)(k+1)A_{k+2}=-(-2k+1)A_{k))$$

$${\displaystyle A_{k+2}={(2k-1) \over (k+2)(k+1)}A_{k))$$

Now, we have

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

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

So we have

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

$${\displaystyle {\begin{aligned}f&=A_{0}z^{0}+A_{1}z^{1}+A_{2}z^{2}+A_{3}z^{3}+A_{4}z^{4}+A_{5}z^{5}+A_{6}z^{6}+A_{7}z^{7}+\cdots \\[8pt]&=A_{0}z^{0}+A_{1}z^{1}+{-1 \over 2}A_{0}z^{2}+{1 \over 6}A_{1}z^{3}+{-1 \over 8}A_{0}z^{4}+{1 \over 24}A_{1}z^{5}+{-7 \over 240}A_{0}z^{6}+{1 \over 112}A_{1}z^{7}+\cdots \\[8pt]&=A_{0}z^{0}+{-1 \over 2}A_{0}z^{2}+{-1 \over 8}A_{0}z^{4}+{-7 \over 240}A_{0}z^{6}+A_{1}z+{1 \over 6}A_{1}z^{3}+{1 \over 24}A_{1}z^{5}+{1 \over 112}A_{1}z^{7}+\cdots \end{aligned))}$$

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

A simpler way using Taylor series

A much simpler way of solving this equation (and power series solution in general) using the Taylor series form of the expansion. Here we assume the answer is of the form

$${\displaystyle f=\sum _{k=0}^{\infty }{A_{k}z^{k} \over {k!))}$$

If we do this, the general rule for obtaining the recurrence relationship for the coefficients is

$${\displaystyle y^{[n]}\to A_{k+n))$$

$${\displaystyle x^{m}y^{[n]}\to (k)(k-1)\cdots (k-m+1)A_{k+n-m))$$

In this case we can solve the Hermite equation in fewer steps:

$${\displaystyle f''-2zf'+\lambda f=0;\;\lambda =1}$$

$${\displaystyle A_{k+2}-2kA_{k}+\lambda A_{k}=0}$$

$${\displaystyle A_{k+2}=(2k-\lambda )A_{k))$$

$${\displaystyle f=\sum _{k=0}^{\infty }{A_{k}z^{k} \over {k!))}$$

Nonlinear equations

The power series method can be applied to certain nonlinear differential equations, though with less flexibility. A very large class of nonlinear equations can be solved analytically by using the Parker–Sochacki method. Since the Parker–Sochacki method involves an expansion of the original system of ordinary differential equations through auxiliary equations, it is not simply referred to as the power series method. The Parker–Sochacki method is done before the power series method to make the power series method possible on many nonlinear problems. An ODE problem can be expanded with the auxiliary variables which make the power series method trivial for an equivalent, larger system. Expanding the ODE problem with auxiliary variables produces the same coefficients (since the power series for a function is unique) at the cost of also calculating the coefficients of auxiliary equations. Many times, without using auxiliary variables, there is no known way to get the power series for the solution to a system, hence the power series method alone is difficult to apply to most nonlinear equations.

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

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

${\displaystyle \eta =1}$

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

$${\displaystyle \left.{\frac {d^{n}F}{d\eta ^{n))}\right|_{\eta =1}=n!\ c_{n))$$

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

${\displaystyle \eta -1}$

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 Cauchy product

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

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

${\displaystyle c_{0}=0}$

${\displaystyle c_{1}=-1/2}$

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

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 ${\displaystyle \eta =1}$, and zero to the right. At ${\displaystyle \eta =1}$, 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 ${\displaystyle \eta =1}$ instead of suddenly becoming zero.