10. Power Series Solutions - Introduction to ODEs and Linear Algebra

Power series solutions: Basic method

In the simplest cases, we simply try to determine the coefficients of a power series expansion of the solution to a differential equation:

$y(x) = \sum_{n=0}^{\infty} a_n (x - x_0)^n = a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 \ldots$

Usually for convenience we will take the basepoint of the expansion to be zero ( $x_0 = 0$ ).

Example: Airy's Equation

We will illustrate the method with a relatively simple example, the Airy equation. This equation arises in the study of optics, as well as quantum mechanics. It is related to work of the astronmer George Biddell Airy. The ODE is:

$y'' = x y$

As a second-order ODE with smooth coefficient functions this is guaranteed to have a basis of two linearly independent solutions. We can pick initial conditions $y_1(0) = 1$ , $y_1'(0) = 0$ and $y_2(0) = 0$ , $y_2'(0) = 1$ to get two such basis functions $y_1$ and $y_2$ .

Let us calculate the series for $y_1$ at $x_0 = 0$ . To relate it to the ODE, we need to take two derivatives:

$y_1 = \sum_{n=0}^{\infty} a_n x^n$

$y_1' = \sum_{n=1}^{\infty} n a_n x^{n-1}$

$y_1'' = \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2}$

Pluggin in the series for $y_1''$ and $y_1$ into the ODE gives us:

$\sum_{n=2}^{\infty} n (n-1) a_n x^{n-2} = x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+1}$

In order to compare the right- and left-hand sides more easily we can shift the index variable for the first sum:

$\sum_{n=2}^{\infty} n (n-1) a_n x^{n-2} = 2 a_2 + 6 a_3 x + 12 a_4 x^2 + 20 a_5 x^3 + \ldots = \sum_{n=-1}^{\infty} (n+3) (n+2) a_{n+3} x^{n+1}$

and with this rewritten form it is easy to compare the terms of each side:

$\sum_{n=-1}^{\infty} (n+3) (n+2) a_{n+3} x^{n+1} = \sum_{n=0}^{\infty} a_n x^{n+1}$

Since the left-hand side starts at $n=-1$ , but right-hand side starts at $n=0$ , we must have $a_2 = 0$ . We also know from the initial conditions that $y_1(0) = 1 = a_0$ and $y_1'(0) = 0 = a_1$ . So we know the first three terms of the series for $y_1$ .

For $n \ge 0$ we have series terms on both sides which must be equal, giving us the condition:

$(n+3) (n+2) a_{n+3} = a_n$

or solving for the higher index coefficient:

$a_{n+3} = \frac{a_n}{(n+3) (n+2)}$

This is a recursive formula for each coefficient in terms of a previous one. Since $a_1 = a_2 = 0$ , this means that $a_4 = a_5 = 0$ , and $a_7 = a_8 = 0$ , and so on. The only nonzero coefficients are the $a_{3m}$ for a nonnegative integer $m$ . The first few are $a_3 = a_0/6 = 1/6$ , $a_6 = a_3/30 = 1/180$ , $a_9 = a_6/23760$ .

Additional Exercises

Find the power series solution to the ODE $y' = 3x^2y$ (expanded at $x=0$ ). You should be able to determine each coefficient as an explicit function of its index (rather than just a recurrence relation).
Show that the coefficients of the power series solution to the initial value problem $y'' - y' -y = 0$ , $y(0) = 0$ , $y'(0) = 1$ have the form $c_n = F_n/n!$ where $F_n$ is the $n$ th Fibonacci number. (The Fibonacci numbers are $1,1,2,3,5,8,13,21,34,\ldots$ , satisfying the recursion relation that each number is the sum of the previous two in the sequence.)
Determine the power series solution and radius of convergence of the ODE $y'' + x^2y = 0$ .

Notes

Notes: (Local storage in a cookie)

License

This work (text, mathematical images, and javascript applets) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Biographical images are from Wikipedia and have their own (similar) licenses.