Definition of Ordinary Points
Before entering the subject of this section, we need to generalize the notions of analyticity and series solution a bit back in this tutorial. Just like a general Taylor series, a series solution can be expanded at some other location \( x_0 \neq 0 \), instead of \( x_0 = 0 \) (Maclaurin series) previously, i.e.
\begin{align}
y(x) &= a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + a_3 (x-x_0)^3 + \cdots \\
&= \sum_{n=0}^{\infty} a_n (x-x_0)^n \tag{1}
\end{align}
and
\begin{align}
y'(x) &= a_1 + 2a_2 (x-x_0) + 3a_3 (x-x_0)^2 + \cdots \\
&= \sum_{n=1}^{\infty} n a_n(x-x_0)^{n-1} \tag{2} \\
y^{\prime\prime}(x) &= 2a_2 + 6a_3(x-x_0) + \cdots \\
&= \sum_{n=2}^{\infty} n(n-1) a_n(x-x_0)^{n-2} \tag{3}
\end{align}
and so on similarly for higher-order derivatives. Analyticity for a function \( f(x) \) at \( x = x_0 \) then particularly means that \( f(x) \) can be expressed as a power series now centered at \( x = x_0 \) in the form of (1) that is convergent near \( x = x_0 \).
Ordinary points of a given second-order linear homogeneous ODE
\begin{align}
\frac{d^2y}{dx^2} + P(x) \frac{dy}{dx} + Q(x) y = 0 \tag{4}
\end{align}
then refer to the set of \( x_0 \) where \( P(x) \) and \( Q(x) \) are analytic at \( x = x_0 \). Points that are otherwise not ordinary are called singular points. It can be shown that at such an ordinary point, the corresponding series solution (1) will be valid with a radius of convergence same as (the smaller of) \( P(x) \) and \( Q(x) \) (see bottom). For examples,
- \( y^{\prime\prime} + 3x^2y’ + (\sin x)y = 0 \): All \(x\) are ordinary points;
- \( y^{\prime\prime} + (1/x) y’ + (1/x^2)y = 0 \): \( x = 0 \) is a singular point and all other points are ordinary;
- \( y^{\prime\prime} + \frac{1}{x-2}y’ + \frac{1}{x^2-1}y = 0\): \( x = -1, 1, 2 \) are the three singular points;
- \( y^{\prime\prime} + \frac{\sin x}{x}y’ + y = 0 \): All \(x\), including \(x = 0\), are ordinary points.
Deriving Series Solution to Second-order ODEs at Ordinary Points
Although we just talked about ordinary points at any \( x \), for simplicity we will mostly consider the case of \( x = 0 \). We will start with a very simple example of
\begin{align}
\frac{d^2y}{dx^2} + 4y = 0 \tag{5}
\end{align}
By the essence of this tutorial, we know that its general solution is made up of \( y_1 = \cos(2x) \) and \( y_2 = \sin(2x) \). Now we will acquire a series solution and see if they are consistent. Specifically, \( x = 0 \) is an ordinary point of (5). Assuming (1) and also plugging (3) into (5):
\begin{align}
\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + 4\sum_{n=0}^{\infty} a_n x^n &= 0 \\
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + 4\sum_{n=0}^{\infty} a_n x^n &= 0 \tag{6}
\end{align}
where as usual we have replaced \( n \) by \( n+2 \) in the first summation. The recurrence relation is then, by comparing the coefficients:
\begin{align}
(n+2)(n+1) a_{n+2} + 4a_n &= 0 \\
a_{n+2} &= -\frac{4}{(n+2)(n+1)} a_n \tag{7}
\end{align}
This form suggests that the coefficients \(a_n\) can be generated with a stride of \(2\) (\(n \rightarrow n+2 \rightarrow n+4 \rightarrow \cdots\)) and the solution can be cleanly separated into two series, one for the odd indices and another for the even indices. For the even indices, we have
\begin{align}
a_{2m} &= -\frac{4}{(2m)(2m-1)}a_{2m-2} \\
&= \frac{(-4)^2}{(2m)(2m-1)(2m-2)(2m-3)}a_{2m-4} \\
&= \frac{(-1)^m 2^{2m}}{(2m)!} a_0 \tag{8}
\end{align}
inductively. And similarly for the odd indices, we can deduce that
\begin{align}
a_{2m+1} &= \frac{(-1)^m 2^{2m}}{(2m+1)!} a_1 \tag{9}
\end{align}
So the series solution will be
\begin{align}
y &= \sum_{n=0}^{\infty} a_n x^n \\
&= a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots \\
&= (a_0 + a_2 x^2 + \cdots) + (a_1 x + a_3 x^3 + \cdots ) \\
&= a_0 (1 -2x^2 + \cdots) + a_1 (x -\frac{2}{3}x^3 + \cdots) \\
&= a_0 \sum_{m=0}^{\infty} \frac{(-1)^m (2x)^{2m}}{(2m)!} + \frac{a_1}{2} \sum_{m=0}^{\infty} \frac{(-1)^m (2x)^{2m+1}}{(2m+1)!} \\
&= a_0 \cos(2x) + \frac{a_1}{2} \sin(2x) \tag{10}
\end{align}
So this coincides with the usual solution except differed by some multiplicative constants.
Proof for Series Solution Convergence
This part is largely based on the reference book Differential Equations with Applications and Historical Notes by Simmons. If the series expressions of \( P(x) \) and \( Q(x) \) in (4) converge over \( |x| < R \) for some \( R > 0 \), then
\begin{align}
P(x) &= \sum_{k=0}^{\infty} p_kx^k && Q(x) = \sum_{k=0}^{\infty} q_kx^k \tag{11}
\end{align}
where given a positive number \( r < R \) within their radius of convergence, we have a positive quantity \( M > 0 \) such that for all \(k\)
\begin{align}
|p_k|r^k &\leq M && |q_k|r^k \leq M \tag{12}
\end{align}
as the terms of a convergent series must tend to zero. Plugging the series solution (1)-(3) into (4) produces
\begin{align}
\begin{aligned}\sum_{n=2}^{\infty} n(n-1) a_nx^{n-2} + \sum_{k=0}^{\infty} p_kx^k \sum_{n=1}^{\infty} n a_nx^{n-1} \\
+ \sum_{k=0}^{\infty} q_kx^k \sum_{n=0}^{\infty} a_n x^n\end{aligned} &= 0 \\
\begin{aligned}\sum_{n=2}^{\infty} n(n-1) a_nx^{n-2} + \sum_{k=0}^{\infty} \sum_{n=1}^{\infty} n p_ka_nx^{n+k-1} \\
+ \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} q_k a_n x^{n+k}\end{aligned} &= 0 \\
\begin{aligned}\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2}x^{n} + \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} (n+1) p_ka_{n+1}x^{n+k} \\
+ \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} q_k a_n x^{n+k}\end{aligned} &= 0 \\
\begin{aligned}\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2}x^{n} + \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1) p_ka_{n-k+1}x^{n} \\
+ \sum_{n=0}^{\infty} \sum_{k=0}^{n} q_k a_{n-k} x^{n}\end{aligned} &= 0 \tag{13}
\end{align}
where we adjust the indices in the first two summations by \(2\) and \(1\), then interchange the order of the double summations while replacing \( n \) by \( n-k \). Then by comparing the coefficients, the recurrence relation reads
\begin{align}
(n+2)(n+1) a_{n+2} + \sum_{k=0}^{n} (n-k+1) p_ka_{n-k+1} + \sum_{k=0}^{n} q_k a_{n-k} &= 0 \\
(n+2)(n+1) a_{n+2} + \sum_{k=0}^{n} (k+1) p_{n-k}a_{k+1} + \sum_{k=0}^{n} q_{n-k} a_{k} &= 0 \\
(n+2)(n+1) a_{n+2} = -\sum_{k=0}^{n} ((k+1) p_{n-k}a_{k+1} + q_{n-k} a_{k}) \tag{14}
\end{align}
where we further swap \( k \) by \( n -k \). Using (12) in (14), we have
\begin{align}
(n+2)(n+1) |a_{n+2}| &\leq \sum_{k=0}^{n} ((k+1) \frac{M}{r^{n-k}}|a_{k+1}| + \frac{M}{r^{n-k}} |a_{k}|) \\
&= \frac{M}{r^n}\sum_{k=0}^{n} r^k((k+1) |a_{k+1}| + |a_{k}|) \\
&\leq \frac{M}{r^n}\sum_{k=0}^{n} r^k((k+1) |a_{k+1}| + |a_{k}|) + \color{red}{M|a_{n+1}| r} \tag{15}
\end{align}
where the last term \( M|a_{n+1}| r \) is added (highlighted in red) to facilitate the later steps to construct a desired ratio. Now we define a sequence \( b_0 = |a_0|, b_1 = |a_1| \) and
\begin{align}
(n+2)(n+1) b_{n+2} = \frac{M}{r^n}\sum_{k=0}^{n} r^k((k+1) b_{k+1} + b_{k}) + \color{red}{Mb_{n+1} r} \tag{16}
\end{align}
that looks like (15), recursively for \( n \geq 0 \). It is not hard to see that \( 0 \leq |a_n| \leq b_n \) for any \(n\) (possible by induction). Now we are going to show that \( \sum_{n=0}^{\infty} b_n x^n \) converges by computing the ratio \( b_{n+1}/b_n \). Replacing \( n \) by \( n-1\) and \(n-2\) in (16) gives
\begin{align}
\begin{aligned}
(n+1)n b_{n+1} &= \frac{M}{r^{n-1}}\sum_{k=0}^{n-1} r^k((k+1) b_{k+1} + b_{k}) + Mb_{n} r \\
n(n-1) b_{n} &= \frac{M}{r^{n-2}}\sum_{k=0}^{n-2} r^k((k+1) b_{k+1} + b_{k}) + \color{red}{Mb_{n-1} r}
\end{aligned} \tag{17}
\end{align}
Multiplying the first equation in (17) by \(r\) and utilizing the second, we get
\begin{align}
r(n+1)n b_{n+1} &= \frac{M}{r^{n-2}}\sum_{k=0}^{n-1} r^k((k+1) b_{k+1} + b_{k}) + Mb_{n} r^2 \\
&= \frac{M}{r^{n-2}}\sum_{k=0}^{n-2} r^k((k+1) b_{k+1} + b_{k}) \\
&\quad + Mr(n b_n + \color{red}{b_{n-1}}) + Mb_{n} r^2 \\
&= n(n-1) b_{n} -\color{red}{Mb_{n-1} r} \\
&\quad + Mr(n b_n + \color{red}{b_{n-1}}) + Mb_{n} r^2 \\
&= n(n-1) b_{n} + Mrn b_n + Mb_{n} r^2 \\
&= b_n(n (n-1) +Mnr + Mr^2) \tag{18}
\end{align}
Therefore
\begin{align}
\frac{b_{n+1}}{b_n} = \frac{n(n-1) +Mnr + Mr^2}{rn(n+1)} \tag{19}
\end{align}
so as \( n \to \infty \), only the first term will remain (\((n-1)/(n+1) \to 1 \)), and \( b_{n+1}/b_n \to 1/r \), thus
\begin{align}
\left|\frac{b_{n+1}x^{n+1}}{b_nx^n}\right| = \frac{b_{n+1}}{b_n} |x| \to \frac{|x|}{r} \tag{20}
\end{align}
Hence the series \( \sum_{n=0}^{\infty} b_n x^n \) converges if \( |x| < r \) by the Ratio test, and since \( |a_n| < b_n \), by the comparison test the series solution (1) converges for \( |x| < r \) too. As \( r \) can be any positive number smaller than \( R \), (1) will also converge within the radius of convergence \( |x| < R \) and we are done.
Exercise
Use the series solution method to solve
\begin{align}
\frac{d^2y}{dx^2} -y = 0 \tag{21}
\end{align}
and compare to the answer that would be obtained by the usual method.
Answer
As before, substitute (1) and (3) with \( x_0 = 0 \) into (21). This yields
\begin{align}
\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} -\sum_{n=0}^{\infty} a_n x^n &= 0 \\
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n -\sum_{n=0}^{\infty} a_n x^n &= 0
\end{align}
The necessary arguments will be the same as before. The recurrence relation is
\begin{align}
a_{n+2} = \frac{a_n}{(n+2)(n+1)}
\end{align}
For even/odd indices, we simply have \( a_{2m} = (1/(2m)!) a_0 \) and \( a_{2m+1} = (1/(2m+1)!) a_1 \). So the series solution is
\begin{align}
y &= \sum_{m=0}^{\infty} (a_{2m} x^{2m} + a_{2m+1}x^{2m+1}) \\
&= a_0 \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m)!} + a_1 \sum_{m=0}^{\infty} \frac{x^{2m+1}}{(2m+1)!} \\
&= a_0 \cosh(x) + a_1 \sinh(x)
\end{align}
where we have identified the Taylor expansion of \(\cosh(x) = (e^x+e^{-x})/2 = \sum_{m=0}^{\infty} x^{2m}/(2m)!\) and \( \sinh(x) = (e^x -e^{-x})/2 =\sum_{m=0}^{\infty} x^{2m+1}/(2m+1)! \). This is consistent with the usual solution of \( y = c_1 e^x + c_2 e^{-x} \), when \( c_1 = (a_0+a_1)/2 \) and \( c_2 = (a_0-a_1)/2 \).
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