Power Series Solutions: Method/Example

Section 15.10 Power Series Solutions: Method/Example

The power series method is one of the most powerful analytic methods that physicists have for solving linear differential equations. The idea is very simple, make an ansatz that a power series solution exists, but the coefficients in the power series are unknown. Plug this ansatz into the differential equation and use an iterative strategy to solve for the unknown coefficients. The method is easiest to see using a simple example:

Activity 15.6. A simple example of power series solutions of linear ODEs.

Find a power series solution of

\begin{equation} y^{\prime}=A y\tag{15.10.1} \end{equation}

for \(A\) a constant, expanded around the point \(z=1\text{.}\)

Solution.

Ansatz: Assume the solution IS a power series of the form:

\begin{equation} y=\sum_{n=0}^{\infty} c_n\, (z-1)^n\tag{15.10.2} \end{equation}

Make sure that the powers that appear in the sum are powers of \(z-z_0\text{,}\) where \(z_0\) is the point you are expanding around. Also, make sure that you use the correct name of the independent variable (in this case, the \(y^{\prime}\) notation makes the independent variable unclear, so I have arbitrarily chosen \(z\)). And choose the correct name for the dependent variable (in this case, \(y\)).

Differentiating this expression, we obtain:

\begin{equation} y^{\prime} =\sum_{n=0}^{\infty} c_n\, n\, (z-1)^{n-1}\tag{15.10.3} \end{equation}

Now, rewrite Equation (15.10.1) so that all the terms are on the same side of the equation, then insert Equation (15.10.2) and Equation (15.10.3) to obtain:

\begin{equation} 0=\sum_{n=0}^{\infty} c_n\, n\, (z-1)^{n-1} -A\sum_{n=0}^{\infty} c_n\, (z-1)^n\tag{15.10.4} \end{equation}

Now the strategy is to manipulate the indices \(n\) in the sums (in this case, just the first sum) so that each sum contains the same power of \(z-z_0\) so that the sums can be combined into a single sum. Notice that each sum has an index that is summed over so that the index does not appear in the final expression once the sum has been completed. This index is analogous to the variable of integration in a definite integral. Such an indexis called a "dummy index" and you are free to rename it any way you want. On the first sum in Equation (15.10.4), rename \(m=n-1\rightarrow n=m+1\text{.}\) Don't forget to change every \(n\) to an equivalent expression with \(m\)s, including the lower index in the sum.

\begin{equation} 0=\sum_{m=-1}^{\infty} c_{m+1}\, (m+1)\, (z-1)^{m} -A\sum_{n=0}^{\infty} c_n\, (z-1)^n\tag{15.10.5} \end{equation}

Finally, relabel the dummy variable on the first sum, again: let \(n=m\text{,}\) so that the sums can be combined. Be careful, the limits of the sums are not the same, so you must separate out any extra terms that are in one sum and not the other.

\begin{align} 0=\amp \sum_{n=-1}^{\infty} c_{n+1}\, (n+1)\, (z-1)^{n}-A\sum_{n=0}^{\infty} c_n\, (z-1)^n\tag{15.10.6}\\ =\amp \sum_{n=-1}^{n=-1} c_{n+1}\, (n+1)\, (z-1)^{n}\tag{15.10.7}\\ \amp +\sum_{n=0}^{\infty} [c_{n+1}\, (n+1)-A\, c_n]\, (z-1)^{n}\tag{15.10.8} \end{align}

The first term is identically zero when you plug in \(n=-1\text{.}\)

A crucial observation is that Equation (15.10.8) must be true for all values of \(z\text{.}\) This fact can only be true if the coefficients of each power of \(z-1\) vanishes separately, i.e. if

\begin{equation} 0 =[c_{n+1}\, (n+1)-A\, c_n]\tag{15.10.9} \end{equation}

for every value of \(n\text{.}\) This relationship is called a recurrence relation. You can use it to find values of the coefficients for high values of \(n\) if you know the coefficients for lower values of \(n\text{.}\) Solve for \(c_{n+1}\text{:}\)

\begin{equation} c_{n+1}=\frac{A}{n+1}\, c_n\tag{15.10.10} \end{equation}

By keeping \(c_0\) as an unknown constant, we can find other coefficients \(c_n\) by an iterative process.

\begin{align} c_1\amp =\frac{A}{1}\, c_0\tag{15.10.11}\\ c_2\amp =\frac{A}{2}\, c_1=\frac{A^2}{(2)(1)}\, c_0\tag{15.10.12}\\ c_3\amp =\frac{A}{3}\, c_2=\frac{A^3}{(3)(2)(1)}\, c_0\tag{15.10.13}\\ c_4\amp =\frac{A}{4}\, c_3=\frac{A^4}{(4)(2)(1)}\, c_0\tag{15.10.14}\\ \amp \vdots\tag{15.10.15}\\ c_n\amp =\frac{A}{n}\, c_n=\frac{A^n}{n!}\, c_0\tag{15.10.16} \end{align}

In this case, we are able to see what the formula will be for the generic value of \(c_n\text{.}\) In most cases, this we will not be able to determine the general formula and will be forced to trucate the series when we have as many terms of the series as we need for the accuracy that we care about.

Our final answer is:

\begin{align*} f(x)\amp =c_0 (1 + \frac{A}{1!} (z-1) + \frac{A^2}{2!} (z-1)^2 + \dots + \frac{A^n}{n!} (z-1)^n+\dots\\ \amp = c_0\, \sum_{n=0}^{\infty} \frac{A^n}{n!} (z-1)^n\\ \amp = c_0 e^{A(z-1)} \end{align*}

Since we are solving a first order differential equation, we expect a single solution multiplied by an overall arbitrary constant. Notice that the coefficient of the first term in the series \(c_0\text{,}\) through the recurrence relation, becomes a factor in every term and therefore becomes the overall arbitrary constant.

For an \(n\)th order equation, we expect to get \(n\) linearly independent solutions using this method. In exceptional cases, the extra solutions will be a power series multiplied by a logarithm.