Power Series Solutions to Linear Differential Equations

Power Series Solutions to Linear Differential Equations

It may be helpful to review shifting the summation index before proceeding.

We’ll also use the following theorem about power series vanishing on an interval: If $\sum_{n=0}^{\infty}{a_0(x-x_0)^n} = 0$ for all $x$ in some open interval, then each coefficient $a_n$ equals zero.

I’ll proceed with an example (from Nagle, Section 8.3).

Let’s find a power series solution about x = 0 to

\[y' + 2xy = 0. \tag{a}\]

The coefficient of $y$ here is $2x$, which is analytic everywhere, so $x = 0$ is an ordinary point of equation (a). So, we should expect to find a power series solution of the form:

\[y(x) = a_0 + a_1 x + a_2 x^2 + \cdots = \sum_{n=0}^{\infty} a_n x^n. \tag{b}\]

We simply need to find the unknown coefficients $a_n$.

We can differentiate (4) to find the expansion of $y’(x)$:

\[y'(x) = 0 + a_1 + 2a_2 x + 3a_3x^2 + \cdots = \sum_{n=0}^{\infty} n a_n x^{n-1}. \tag{c}\]

Substituting in these series for $y$ and $y’$ into (a) gives:

\[\sum_{n=0}^{\infty} n a_n x^{n-1} + 2x \sum_{n=0}^{\infty} a_n x^n = 0 \tag{d}\]

which simplifies to

\[\sum_{n=0}^{\infty} n a_n x^{n-1} + \sum_{n=0}^{\infty} 2 a_n x^{n+1} = 0 \tag{e}\]

We could expand these out for a few terms and solve for the coefficients from here to get the first few terms of the power series solution, but we’ll proceed with finding a formula for the general term of the power series solution.

Using summation index shifting techniques, we can combine the series from (e) to get:

\[a_1 + \sum_{k=1}^{\infty}{ \left [ (k+1)a_{k+1} + 2a_{k-1} \right ] x^k} = 0 \tag{f}\]

Matching coefficients on the left and right, it’s clear that $a_1 = 0$, and that for all $k \geq 1$

\[(k+1)a_{k+1} + 2a_{k-1} = 0. \tag{g}\]

Here, (g) is a recurrence relation that we can use to determine $a_{k+1} in terms of $a_{k-1}$, that is

\[a_{k+1} = - \frac{2}{k+1}a_{k-1}.\]

Now, to find $a_2$ we set $k=1$ and get

\[a_2 = - \frac{2}{2}a_0 = -a_0.\]

and for $k=2$

\[a_3 = - \frac{2}{3}a_1 = 0\]

and for $k=3$

\[a_4 = -\frac{2}{4}a_2 = \frac{1}{2}a_0\]

and for $k=4$:

\[a_5 = -\frac{2}{5}a_3 = 0\]

and so on.

If we continue this and inspect the pattern that arises we see that

\[a_{2n} = \frac{(-1)^n}{n!} a_0, \quad n = 1,2,...\] \[a_{2n+1} = 0 , \quad n = 0,1,2,...\]

and when we substitute this back into (b) we get

\[a_0 \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}x^{2n}.\]

Since $a_0$ is left undetermined as an arbitrary constant, this is a general solution to equation (a).

Also note that the radius of convergence is infinite, and that it converges to

\[y(x) = a_0 e^{-x^2}.\]

Existince of Analytic Solutions

Given the equation

\[y''(x) + p(x)y'(x) +q(x)y(x) = 0, \tag{1}\]

Suppose $x_0$ is an ordinary point for equation (1). Then (1) has two linearly independent analytic solutions of the form

\[y(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^n. \tag{2}\]

Moreover, the radius of convergance of any power series solution of the form given by (2) is at least as large as the distance from $x_0$ to the nearest singular point (real or complex-valued) of equation (1).

Translation

It’s generally a lot easier to compute with series that are centered at $x_0=0$ than at other points. We can make a substitution by saying $x_0 = a, t_0 = 0, x = t + a$. Then, we can follow the procedure outlined above, and in the final series, we can replace $t$ with $t = x - a$.