Roots of Indicial Equation Differed by an Integer
In the last tutorial, we walked through the scenario where the roots of the indicial equation \(m_1, m_2\) do not differ by an integer or are the same. However, the Frobenius method may become problematic if it is the other case. To see this, we rewrite the recurrence relation (10) in the last tutorial, by defining \( f(m) = m(m-1) + mp_{-1} + q_{-2} \) which is the polynomial standing for the indicial equation, then
\begin{align}
\begin{aligned}
(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}\end{aligned} &= 0 \\
\begin{aligned}
(m+n)(m+n-1)a_n + p_{-1}(m+n)a_n + q_{-2}a_n \\
+ \sum_{k=1}^{n} p_{k-1}(m+n-k)a_{n-k} + \sum_{k=1}^{n}q_{k-2}a_{n-k}\end{aligned} &= 0 \\
f(m+n)a_n + \sum_{k=1}^{n} p_{k-1}(m+n-k)a_{n-k} + \sum_{k=1}^{n}q_{k-2}a_{n-k} &= 0 \tag{1}
\end{align}
by taking out the zeroth term in the two summations. (1) shows how \( a_n \) can be computed from \( a_0, a_1, \ldots, a_{n-1} \). Consider the extreme case where \(m_1 = m_2\), then since at each iteration we uniquely determine \(a_0 \to a_1 \to \cdots \to a_n \to \cdots \) one by one with (1) as the only way in our hands, if \( m \) takes the same starting value, we will only be able to obtain the same one Frobenius solution, and there will not exist a second Frobenius solution.
Similarly, if \( m_1 -m_2 = k \) where \( k \) is some positive integer, then for \(m = m_2\) and when \( n = k \), the first term involving \(a_n\) in (1) will vanish because \( f(m_2 + k) = f(m_1) = 0 \) too, and we cannot proceed the iteration of the coefficients, unless the latter two summations are zeros as well (then we may continue the computation for the second Frobenius solution). In the above two cases, \(m_2\) is “intercepted” by \(m_1\), and to find a second linearly independent solution, we can build up from the sole Frobenius solution obtained from \(m_1\) by the method in this post. First, notice that if \( m_1 \) and \( m_2 \) are the roots of the indicial equation, then it can be written as
\begin{align}
(m-m_1)(m-m_2) &= 0 \\
m^2 -(m_1+m_2)m + m_1m_2 &= 0 \tag{2}
\end{align}
Equating the coefficients in (2) with the original indicial equation shows that \( p_{-1} -1 = -(m_1 + m_2) \), then their integer difference can be expressed as \( m_1 -m_2 = 2m_1 -(m_1 + m_2) = 2m_1 + p_{-1} -1 \), and we shall assign an integer parameter \( k’ = 2m_1 + p_{-1} = m_1 -m_2 + 1 \). Now, we propose the second solution \( y_2 = vy_1 \), where \(y_1\) is the Frobenius solution, then substitute this into the ODE:
\begin{align}
y_2^{\prime\prime} + P(x)y_2^\prime + Q(x)y_2 &= 0 \\
(v^{\prime\prime}y_1 + 2v’y_1^\prime + vy_1^{\prime\prime}) + P(x)(v’y_1 + vy_1^\prime) + Q(x)vy_1 &= 0 \\
v^{\prime\prime}y_1 + v'(2y_1^\prime + P(x)y_1) + v(y_1^{\prime\prime} + P(x)y_1^\prime + Q(x)y_1) &= 0 \tag{3}
\end{align}
The last term is zero because \(y_1\) is precisely a solution of the ODE. So what remains is
\begin{align}
v^{\prime\prime}y_1 + v'(2y_1^\prime + Py_1) &= 0 \\
\frac{v^{\prime\prime}}{v^\prime} &= -2\frac{y_1^\prime}{y_1} -P \\
\int \frac{1}{v’} dv’ &= -2 \int \frac{1}{y_1} dy_1 -\int P dx \\
\ln v’ &= -2\ln y_1 -\int P dx \\
v’ &= \frac{1}{y_1^2} e^{-\int Pdx} \tag{4}
\end{align}
by integrating both sides. Then using the original Frobenius solution, we have
\begin{align}
v’ &= \frac{1}{y_1^2} e^{-\int Pdx} \\
&= \frac{1}{(x^{m_1}(a_0 + a_1x + \cdots))^2} e^{-\int ((p_{-1}/x) + p_0 + \cdots) dx} \\
&= \frac{1}{x^{2m_1}(a_0 + a_1x + \cdots)^2} e^{-p_{-1} \ln x -p_0x -\cdots} \\
&= \frac{1}{x^{2m_1+p_{-1}}(a_0 + a_1x + \cdots)^2} e^{-p_0x -\cdots} \\
&= \frac{1}{x^{k’}(a_0 + a_1x + \cdots)^2} e^{-p_0x -\cdots} \tag{5}
\end{align}
We condense the above expression into \( (1/x^{k’}) g(x) \) where \( g(x) \) is analytic at \( x = 0 \) (Binomial expansion for the polynomial in the denominator and note that the exponential of a polynomial function is also analytic.) This means that we can write \( g(x) = b_0 + b_1x + \cdots + b_{k’-1}x^{k’-1} + \cdots \) for some coefficients \(b_n\), and hence
\begin{align}
v’ &= b_0x^{-k’} + b_1x^{-k’+1} + \cdots + b_{k’-1}x^{-1} + b_{k’} + \cdots \\
v &= \frac{b_0}{-k’+1}x^{-k’+1} + \cdots + b_{k’-1}\ln x + b_{k’}x + \cdots \tag{6}
\end{align}
Therefore,
\begin{align}
y_2 &= vy_1 \\
&= (\frac{b_0}{-k’+1}x^{-k’+1} + \cdots + b_{k’-1}\ln x + b_{k’}x + \cdots) y_1 \\
&= b_{k’-1}y_1\ln x + (\frac{b_0}{-k’+1}x^{-k’+1} + \cdots + b_{k’}x + \cdots)y_1 \\
&= b_{k’-1}y_1\ln x \\
&\quad + (\frac{b_0}{-k’+1}x^{-k’+1} + \cdots + b_{k’}x + \cdots)(x^{m_1}(a_0 + a_1x + \cdots)) \\
&= b_{k’-1}y_1\ln x + (\frac{a_0b_0}{-k’+1}x^{m_1-k’+1} + \cdots) \\
&= b_{k’-1}y_1\ln x + x^{m_2}\sum_{n=0}^{\infty} c_n x^n \tag{7}
\end{align}
by recognizing that \( m_2 = m_1-k’+1 \) and the product of the two series will lead to another power series when \( x^{m_2} \) is taken out, with the coefficients denoted by some \(c_n\). When \( m_1 = m_2 \), \(k’-1 = 0\), and since \(b_0\) cannot be zero, the logarithm term in (7) will always exist. If instead \( m_1 \) and \(m_2\) differ by an integer, then \( k’-1 \) is a positive integer and \(b_{k’-1}\) may or may not be zero, and hence the logarithm term will or will not be present. In practice, we can simply attempt to compute the second Frobenius solution forcefully. In the situation where the logarithm is really absent, then (7) is reduced to a second Frobenius solution, and it will just work. Otherwise, if the iteration of the second Frobenius solution is hindered as suggested in the beginning, then it means that we need to try
\begin{align}
y_2 = y_1\ln x + x^{m_2}\sum_{n=0}^{\infty} c_n x^n \tag{8}
\end{align}
instead (the \( b_{k’-1} \) coefficient has been absorbed and there is no need to find it explicitly). Then we can plug (8) into the ODE and recursively compute the terms (a bit harder than) as usual.
Example
Find the general solution to the ODE
\begin{align}
x^2y^{\prime\prime} -(x+x^2)y’ + y = 0 \tag{9}
\end{align}
with the Frobenius method or if necessary, (8).
The first step is to derive the indicial equation, which is
\begin{align}
m^2 + (-1-1)m + 1 &= 0 \\
m^2 -2m + 1 = (m-1)^2 &= 0 \Rightarrow m = 1 \tag{10}
\end{align}
It is a repeated root and hence apart from the first Frobenius solution, the second solution must be in the form of (8). For the former, we substitute (4)-(6) of the last tutorial into (9) to get
\begin{align}
\begin{aligned}
x^2\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n-2} \\
-(x+x^2)\sum_{n=0}^{\infty} (m+n)a_nx^{m+n-1} \\
+ \sum_{n=0}^{\infty} a_nx^{m+n}\end{aligned} &= 0 \\
\begin{aligned}
\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n} \\
-\sum_{n=0}^{\infty} (m+n)a_nx^{m+n} -\sum_{n=0}^{\infty} (m+n)a_nx^{m+n+1} \\
+ \sum_{n=0}^{\infty} a_nx^{m+n}\end{aligned} &= 0 \\
\begin{aligned}
\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n} \\
-\sum_{n=0}^{\infty} (m+n)a_nx^{m+n} -\sum_{n=1}^{\infty} (m+n-1)a_{n-1}x^{m+n} \\
+ \sum_{n=0}^{\infty} a_nx^{m+n}\end{aligned} &= 0 \tag{11}
\end{align}
where we have adjusted the index in the third summation by \(1\). Equating the coefficients then leads to the recurrence relation for \(n \geq 1\):
\begin{align}
\begin{aligned}
(m+n)(m+n-1)a_n -(m+n)a_n \\
-(m+n-1)a_{n-1} + a_n\end{aligned} &= 0 \\
(n+1)n a_n -(n+1)a_n -na_{n-1} + a_n &= 0 \\
n^2a_n -na_{n-1} &= 0 \\
a_n &= \frac{a_{n-1}}{n} \tag{12}
\end{align}
From this, it is clear that the general form for the terms is \(a_n = 1/n!\), and hence the Frobenius solution is
\begin{align}
y_1 &= x\sum_{n=0}^{\infty} \frac{x^n}{n!} = xe^x \tag{13}
\end{align}
The second solution has a supposed form of (8):
\begin{align}
y_2 &= y_1\ln x + x^{m_2}\sum_{n=0}^{\infty} c_n x^n \\
&= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \ln x + x\sum_{n=0}^{\infty} c_n x^n \\
&= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \ln x + \sum_{n=0}^{\infty} c_n x^{n+1} \tag{14}
\end{align}
Its derivatives are
\begin{align}
\begin{aligned}
y_2^{\prime} &= \sum_{n=0}^{\infty} \frac{x^{n}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n}}{n!} \ln x + \sum_{n=0}^{\infty} (n+1) c_n x^{n} \\
y_2^{\prime\prime} &= \sum_{n=0}^{\infty} \frac{nx^{n-1}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n-1}}{n!} + \sum_{n=0}^{\infty} (n+1)n\frac{x^{n-1}}{n!} \ln x \\
&\quad + \sum_{n=0}^{\infty} (n+1)n c_n x^{n-1}
\end{aligned} \tag{15}
\end{align}
Substituting (14) and (15) into (9) and after some (messy) organizing:
\begin{align}
\begin{aligned}x^2(\sum_{n=0}^{\infty} \frac{nx^{n-1}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n-1}}{n!} \\
+ \sum_{n=0}^{\infty} (n+1)n\frac{x^{n-1}}{n!} \ln x + \sum_{n=0}^{\infty} (n+1)n c_n x^{n-1}) \\
-(x+x^2)(\sum_{n=0}^{\infty} \frac{x^{n}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n}}{n!} \ln x + \sum_{n=0}^{\infty} (n+1) c_n x^{n}) \\
+ (\sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \ln x + \sum_{n=0}^{\infty} c_n x^{n+1}) \end{aligned} &= 0 \\
\begin{aligned}(\sum_{n=0}^{\infty} \frac{nx^{n+1}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n+1}}{n!} \\
+ \sum_{n=0}^{\infty} (n+1)n\frac{x^{n+1}}{n!} \ln x + \sum_{n=0}^{\infty} (n+1)n c_n x^{n+1}) \\
-(\sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n+1}}{n!} \ln x + \sum_{n=0}^{\infty} (n+1) c_n x^{n+1}) \\
-(\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n+2}}{n!} \ln x + \sum_{n=0}^{\infty} (n+1) c_n x^{n+2}) \\
+ (\sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \ln x + \sum_{n=0}^{\infty} c_n x^{n+1}) \end{aligned} &= 0 \tag{16}
\end{align}
Notice that all terms involving the logarithm
\begin{align}
&\quad \begin{aligned} &\sum_{n=0}^{\infty} (n+1)n\frac{x^{n+1}}{n!} \ln x -\sum_{n=0}^{\infty} (n+1)\frac{x^{n+1}}{n!} \ln x \\
&- \sum_{n=0}^{\infty} (n+1)\frac{x^{n+2}}{n!} \ln x + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \ln x \end{aligned} \\
&= \begin{aligned} &\ln x (\sum_{n=0}^{\infty} (n+1)n\frac{x^{n+1}}{n!} -\sum_{n=0}^{\infty} (n+1)\frac{x^{n+1}}{n!} \\
&- \sum_{n=0}^{\infty} n\frac{x^{n+1}}{(n-1)!} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}) \end{aligned} \\
&= \begin{aligned} &\ln x (\sum_{n=0}^{\infty} (n+1)n\frac{x^{n+1}}{n!} -\sum_{n=0}^{\infty} (n+1)\frac{x^{n+1}}{n!} \\
&- \sum_{n=0}^{\infty} n^2\frac{x^{n+1}}{n!} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}) \end{aligned} \\
&= \ln x (0) = 0 \tag{17}
\end{align}
cancel out as they satisfy the original ODE (replaced \( n + 1 \) by \(n\) in the third summation). The remaining terms are
\begin{align}
\begin{aligned}(\sum_{n=0}^{\infty} \frac{nx^{n+1}}{n!} + \sum_{n=0}^{\infty} (n+1)\frac{x^{n+1}}{n!} + \sum_{n=0}^{\infty} (n+1)n c_n x^{n+1}) \\
-(\sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} + \sum_{n=0}^{\infty} (n+1) c_n x^{n+1}) \\
-(\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!} + \sum_{n=0}^{\infty} (n+1) c_n x^{n+2}) + \sum_{n=0}^{\infty} c_n x^{n+1} \end{aligned} &= 0 \\
\begin{aligned}\sum_{n=0}^{\infty} \frac{2nx^{n+1}}{n!} + \sum_{n=0}^{\infty} n^2 c_n x^{n+1} \\
-(\sum_{n=1}^{\infty} \frac{x^{n+1}}{(n-1)!} + \sum_{n=1}^{\infty} n c_{n-1} x^{n+1}) \end{aligned} &= 0 \tag{18}
\end{align}
where we again have replaced \(n+1\) by \(n\) in appropriate summations. Comparing the coefficients, when \( n = 0 \) there are only the first two contributions but they are also just zeros, so we can just assign \( c_0 = 0 \). For \( n \geq 1 \), the recurrence relation is
\begin{align}
\frac{2n}{n!} + n^2 c_n -\frac{1}{(n-1)!} -n c_{n-1} &= 0 \\
n^2 c_n + \frac{1}{(n-1)!} &= n c_{n-1} \\
c_n &= -\frac{1}{n(n!)} + \frac{1}{n}c_{n-1} \tag{19}
\end{align}
The first few values are \( c_1 = -1, c_2 = -3/4, c_3 = -11/36, \ldots\), and hence the second linearly independent solution is \(y_2 = xe^x\ln x -x(x +(3/4)x^2 +(11/36)x^3 + \cdots) \).
Exercise
Show that for all second-order constant-coefficient ODEs in the form of
\begin{align}
\frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = 0 \tag{20}
\end{align}
the usual ordinary point \(x = 0\) can be viewed as a special case of a regular singular point. Hence find the corresponding indicial equation, and argue that in this case, the roots differ by an integer but we can still find two Frobenius solutions.
Answer
Any ordinary point is kind of a regular singular point because if \( P(x) \) and \( Q(x) \) are analytic then \( (x-x_0)P(x) \) and \( (x-x_0)^2 Q(x) \) will also be clearly analytic at \( x = x_0\), and this will also be the case for this question. Note that \( p_{-1} = 0 = q_{-2} \), and hence the indicial equation is \( m^2 -m = m(m-1) = 0 \) with the roots \( m_1 = 1, m_2 = 0 \), differed by an integer. However, by previous knowledge, we know that such a constant-coefficient ODE always has two series solutions, which are a limiting case of Frobenius solutions. For example, if we are given \( y^{\prime\prime} + y = 0 \), the two complementary solutions are clearly \( y_1 = \sin x = x -x^3/3! + \cdots \) and \( y_2 = \cos x = 1 -x^2/2! + \cdots \), which match the exponents \(1,0\) respectively from the indicial equation.
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