Definition and Motivation of Regular Singular Points
In this tutorial, we talked about ordinary points of an ODE, the opposite of which are singular points. Near these singular points, the solutions to the ODE will often be more “pathological”, and series solutions are now generally unavailable. However, if the singular points are mild enough, we may have a chance to salvage them and arrive at some other satisfactory solutions of a similar form. We will now work on the specific case of regular singular points: \( x_0 \) is a regular singular point of the second-order linear ODE
\begin{align}
\frac{d^2y}{dx^2} + P(x) \frac{dy}{dx} + Q(x) y = 0 \tag{1}
\end{align}
if \( (x-x_0)P(x) \) and \( (x-x_0)^2 Q(x) \) are analytic (compared to \( P(x) \) and \( Q(x) \) alone are analytic for an ordinary point). In other words, the singularity of \( P(x) \) and \( Q(x) \) is not worse than \( 1/(x-x_0) \) and \( 1/(x-x_0)^2 \) respectively. Or, they can be written as
\begin{align}
\begin{aligned}
P(x) &= \frac{p_{-1}}{x-x_0} + p_0 + p_1(x-x_0) + p_2(x-x_0)^2 + \cdots \\
&= \sum_{k=-1}^{\infty} p_k(x-x_0)^k \\
Q(x) &= \frac{q_{-2}}{(x-x_0)^2} + \frac{q_{-1}}{x-x_0} + q_0 \\
&\quad + q_1(x-x_0) + q_2(x-x_0)^2 + \cdots = \sum_{k=-2}^{\infty} q_k(x-x_0)^k
\end{aligned} \tag{2}
\end{align}
Frobenius Solution
The reason why we consider this kind of regular singular point is motivated by the possibility of deriving the exact solution to an Euler form ODE (divided by \(x^2\)):
\begin{equation}
\frac{d^2y}{dx^2} + \frac{\alpha}{x} \frac{dy}{dx} + \frac{\beta}{x^2} y = 0 \tag{3}
\end{equation}
\( \alpha/x \) and \( \beta/x^2 \) are singularities of the order of \( 1/x \) and \( 1/x^2 \) at \( x = 0 \). Therefore, we believe that similarly, there may be a way to obtain a solution to (1) at regular singular points where \( P(x) \) and \( Q(x) \) are in the more general form of (2). ((3) is just a special case of (1) with \( p_{-1} = \alpha \) and \( p_k = 0 \) where \(k \geq 0\), and \( q_{-2} = \beta\), \(q_k = 0 \) when \(k \geq -1\).) As we know an Euler form ODE can be solved by the substitution \( z = \ln x \) and the new auxiliary equation is \( r^2 + (\alpha -1)r + \beta = 0\), if the roots are denoted as \( m_1, m_2 \), the general solution will then be formed by \( x^{m_1}, x^{m_2} \). Hence, we are tempted to try a variant of the series solution as
\begin{align}
y = x^{m}(a_0 + a_1x + a_2x^2 + \cdots) = \sum_{n=0}^{\infty} a_nx^{m+n} \tag{4}
\end{align}
(assumed the regular singular point \(x_0 = 0 \) is at the origin, replaced by \( x-x_0 \) for the general case.) where \( m = m_1, m_2 \). To see why it makes sense, we write down the derivatives of (4):
\begin{align}
y’ &= ma_0x^{m-1} + (m+1)a_1x^m + \cdots = \sum_{n=0}^{\infty} (m+n)a_nx^{m+n-1} \tag{5} \\
y^{\prime\prime} &= m(m-1)a_0x^{m-2} + (m+1)ma_1x^{m-1} + \cdots \\
&= \sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n-2} \tag{6}
\end{align}
(The justification of term-by-term differentiation is very similar to that in this and that tutorial.) Plugging all of these into (1), where the coefficients are written out as in (2), gives
\begin{align}
(m(m-1)a_0x^{m-2} + (m+1)ma_1x^{m-1} + \cdots) & \\
{}+ (\frac{p_{-1}}{x} + p_0 + \cdots) (ma_0x^{m-1} + (m+1)a_1x^m + \cdots) & \\
{}+ (\frac{q_{-2}}{x^2} + \frac{q_{-1}}{x} + \cdots) (a_0x^{m} + a_1x^{m+1} + \cdots ) &= 0 \\
m(m-1)a_0x^{m-2} + p_{-1}ma_0x^{m-2} + q_{-2} a_0x^{m-2} + \cdots &= 0 \tag{7}
\end{align}
where the lowest-degree terms from all three contributions are \( x^{m-2} \) and hence compatible. (Otherwise, if the singularity is more severe and there are terms like \(p_{-2}\) or \(q_{-3}\), the lowest power from the first contribution will be unable to match the latter two.) Hence, we may compare the coefficients and obtain the so-called indicial equation:
\begin{align}
m(m-1)a_0 + p_{-1}ma_0 + q_{-2} a_0 &= 0 \\
m^2 + (p_{-1}-1) m + q_{-2} &= 0 \tag{8}
\end{align}
Notice the resemblance to the auxiliary equation. If we can find two (real) roots \(m_1, m_2\) to (8), then substituting them into (4) may lead us to the desired solutions. If it is indeed the case, (4) will be subsequently referred to as the Frobenius solution. To generate the coefficients \(a_n\) in the Frobenius solution is very similar to what we have done for a series solution, by considering the higher-degree terms in (7) and extracting a recurrence relation. In general,
\begin{align}
\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n-2} & \\
{}+ (\sum_{k=-1}^{\infty} p_{k}x^k) \sum_{n=0}^{\infty} (m+n)a_nx^{m+n-1} + (\sum_{k=-2}^{\infty} q_{k}x^k)\sum_{n=0}^{\infty} a_nx^{m+n} &= 0 \\
\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{n-2} & \\
{}+ \sum_{k=-1}^{\infty} \sum_{n=0}^{\infty} p_{k}(m+n)a_nx^{n+k-1} + \sum_{k=-2}^{\infty} \sum_{n=0}^{\infty} q_{k}a_nx^{n+k} &= 0 \\
\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{n-2} & \\
{}+ \sum_{n=0}^{\infty}\sum_{k=-1}^{n-1} p_{k}(m+n-k-1)a_{n-k-1}x^{n-2} + \sum_{n=0}^{\infty} \sum_{k=-2}^{n-2}q_{k}a_{n-k-2}x^{n-2} &= 0 \tag{9}
\end{align}
where in the last step we switch the order of the double summations and replace \( n \) by \( n-k -1\) and \(n-k-2\) respectively. The recurrence relation then reads
\begin{align}
(m+n)(m+n-1)a_n & \\
{}+ \sum_{k=0}^{n} p_{k-1}(m+n-k)a_{n-k} + \sum_{k=0}^{n}q_{k-2}a_{n-k} &= 0 \tag{10}
\end{align}
where we further shifted the indices for \( k \) by \(1\) and \(2\).
Example
If the two roots \(m_1, m_2\) for the indicial equation are neither the same nor differ by an integer, then we can generate two Frobenius solutions in the form of (4) without any problem using (10), as the respective powers will be “interleaved”. Now we will try to solve the following ODE
\begin{equation}
2x y^{\prime\prime} + (4x + 1) y’ + (2x + 1) y = 0 \tag{11}
\end{equation}
by the Frobenius method as an example. It is not hard to see that \( x = 0 \) is a regular singular point of (11) if we divide it by \(2x\):
\begin{align}
y^{\prime\prime} + (2 + \frac{1}{2x}) y’ + (1 + \frac{1}{2x}) y = 0 \tag{12}
\end{align}
The indicial equation is then
\begin{align}
m^2 + (\frac{1}{2} -1)m &= 0 \\
m^2 -\frac{1}{2} m &= 0 \\
m(m-\frac{1}{2}) &= 0 \tag{13}
\end{align}
so the roots are \(m_1 = 0\), \(m_2 = 1/2\). For the sake of demonstration, we will deal with the part of \(m_2 = 1/2\) first. Substituting (4)-(6) into (11) yields
\begin{align}
2x \sum_{n=0}^{\infty} (n+\frac{1}{2})(n-\frac{1}{2})a_nx^{n-\frac{3}{2}} & \\
{}+ (4x + 1) \sum_{n=0}^{\infty} (n+\frac{1}{2})a_nx^{n-\frac{1}{2}} & \\
{}+ (2x + 1) \sum_{n=0}^{\infty} a_nx^{n+\frac{1}{2}} &= 0 \\
\sum_{n=0}^{\infty} 2(n+\frac{1}{2})(n-\frac{1}{2})a_nx^{n-\frac{1}{2}} & \\
{}+ \sum_{n=0}^{\infty} (n+\frac{1}{2})a_nx^{n-\frac{1}{2}} + \sum_{n=0}^{\infty} 4(n+\frac{1}{2})a_nx^{n+\frac{1}{2}} & \\
{}+ \sum_{n=0}^{\infty} a_nx^{n+\frac{1}{2}} + \sum_{n=0}^{\infty} 2a_nx^{n+\frac{3}{2}} &= 0 \\
\sum_{n=0}^{\infty} 2(n+\frac{1}{2})(n-\frac{1}{2})a_nx^{n-\frac{1}{2}} & \\
{}+ \sum_{n=0}^{\infty} (n+\frac{1}{2})a_nx^{n-\frac{1}{2}} + \sum_{n=1}^{\infty} 4(n-\frac{1}{2})a_{n-1}x^{n-\frac{1}{2}} & \\
{}+ \sum_{n=1}^{\infty} a_{n-1}x^{n-\frac{1}{2}} + \sum_{n=2}^{\infty} 2a_{n-2}x^{n-\frac{1}{2}} &= 0 \tag{14}
\end{align}
We then compare the coefficients. When \(n = 0\), the sum only contains the first two terms and readily reduces to zero as the lowest-degree terms always have to satisfy the indicial equation and vanish. When \(n = 1\), it is
\begin{align}
2(\frac{3}{2})(\frac{1}{2})a_1 + \frac{3}{2}a_1 + 4(\frac{1}{2})a_0 + a_0 &= 0 \\
3a_1 + 3a_0 &= 0 \\
a_1 &= -a_0 \tag{15}
\end{align}
and generally for \( n \geq 2 \):
\begin{align}
2(n+\frac{1}{2})(n-\frac{1}{2})a_n + (n+\frac{1}{2})a_n & \\
{}+ 4(n-\frac{1}{2})a_{n-1} + a_{n-1} + 2a_{n-2} &= 0 \\
n(2n+1) a_n + (4n-1) a_{n-1} + 2a_{n-2} &= 0 \tag{16}
\end{align}
Now we try to compute the first few subsequent terms:
\begin{align}
\begin{aligned}
2(2(2)+1) a_2 -(4(2)-1) a_0 + 2a_0 &= 0 \\
10 a_2 -5a_0 = 0 &\Rightarrow a_2 = \frac{1}{2}a_0 \\
3(2(3)+1) a_3 + (4(3)-1) (\frac{1}{2}a_0) -2a_0 &= 0 \\
21 a_3 + \frac{7}{2} a_0 = 0 &\Rightarrow a_3 = -\frac{1}{6}a_0 \\
4(2(4)+1) a_4 + (4(4)-1) (-\frac{1}{6}a_0) + 2(\frac{1}{2}a_0) &= 0 \\
36 a_4 -\frac{3}{2} a_0 = 0 &\Rightarrow a_4 = \frac{1}{24}a_0 \\
\vdots &
\end{aligned} \tag{17}
\end{align}
If we keep going on, we will observe that \( a_n = ((-1)^n/n!) a_0 \), and so the corresponding Frobenius solution will be
\begin{align}
y &= x^{\frac{1}{2}} \sum_{n=0}^{\infty} a_n x^n \\
&= a_0 x^{\frac{1}{2}} \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{n!} \\
&= a_0 \sqrt{x} e^{-x} \tag{18}
\end{align}
In a similar vein, the other Frobenius solution for \(m_1 = 0\) can be found as just \( y = e^{-x} \), and so the overall general solution is \( y = c_1e^{-x} + c_2\sqrt{x}e^{-x} \).
Exercise
Find the (first few terms of the) two Frobenius solutions for
\begin{align}
2x^2 y^{\prime\prime} + 3xy’ -(x+1)y = 0 \tag{19}
\end{align}
Answer
The standardized equation is \( y^{\prime\prime} + \frac{3}{2x}y’ -(\frac{1}{2x}+\frac{1}{2x^2})y = 0 \) and the corresponding indicial equation for this regular singular point of \(x_0 = 0\) is \( m^2 + (\frac{3}{2}-1)m -\frac{1}{2} = m^2 + \frac{1}{2}m -\frac{1}{2} = 0 \) which has roots \( m_1 = -1, m_2 = 1/2 \). Substituting (4)-(6) into (19) leads to
\begin{align}
2x^2 \sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n-2} & \\
{}+ 3x\sum_{n=0}^{\infty} (m+n)a_nx^{m+n-1} &\\
{}-(x+1)\sum_{n=0}^{\infty} a_nx^{m+n} &= 0 \\
\sum_{n=0}^{\infty} 2(m+n)(m+n-1)a_nx^{m+n} &\\
{}+ \sum_{n=0}^{\infty} 3(m+n)a_nx^{m+n} &\\
{}-\sum_{n=0}^{\infty} a_nx^{m+n} -\sum_{n=0}^{\infty} a_nx^{m+n+1} &= 0 \\
\sum_{n=0}^{\infty} 2(m+n)(m+n-1)a_nx^{m+n} &\\
{}+ \sum_{n=0}^{\infty} 3(m+n)a_nx^{m+n} &\\
{}-\sum_{n=0}^{\infty} a_nx^{m+n} -\sum_{n=1}^{\infty} a_{n-1}x^{m+n} &= 0
\end{align}
where we have shifted the index for the last summation by \(1\). Comparing the coefficients then leads to the recurrence relation (for \(n \geq 1\), while \(n = 0\) represents the indicial equation) of
\begin{align}
2(m+n)(m+n-1)a_n + 3(m+n)a_n -a_n -a_{n-1} = 0 \\
\Rightarrow a_n = \frac{a_{n-1}}{2(m+n)(m+n-1) + 3(m+n) -1}
\end{align}
For the case of \(m_1 = -1\), the first few coefficients are generated as
\begin{align}
a_1 &= \frac{a_0}{2(0)(-1) + 3(0) -1} \\
&= -a_0 \\
a_2 &= \frac{a_1}{2(1)(0) + 3(1) -1} \\
&= \frac{a_1}{2} = -\frac{a_0}{2} \\
a_3 &= \frac{a_2}{2(2)(1) + 3(2) -1} \\
&= \frac{a_2}{9} = -\frac{a_0}{18} \\
&\vdots
\end{align}
and similar for the case of \(m_2 = 1/2\):
\begin{align}
a_1 &= \frac{a_0}{2(3/2)(1/2) + 3(3/2) -1} \\
&= \frac{a_0}{5} \\
a_2 &= \frac{a_1}{2(5/2)(3/2) + 3(5/2) -1} \\
&= \frac{a_1}{14} = \frac{a_0}{70} \\
a_3 &= \frac{a_2}{2(7/2)(5/2) + 3(7/2) -1} \\
&= \frac{a_2}{27} = \frac{a_0}{1890} \\
&\vdots
\end{align}
and hence the two Frobenius solutions are \( y_1 = \frac{1}{x} (1 -x -\frac{x^2}{2} -\frac{x^3}{18} -\cdots) \) and \( y_2 = \sqrt{x} (1 + \frac{x}{5} + \frac{x^2}{70} + \frac{x^3}{1890} + \cdots) \).
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