# 9.2: Singular Points

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As usual there is a snag. Most equations of interest are of a form where \(p\) and/or \(q\) are *singular* at the point \(t_0\) (usually \(t_0=0\)). Any point \(t_0\) where \(p(t)\) and \(q(t)\) are singular is called a *singular point*. Of most interest are a special class of singular points called *regular singular points*, where the differential equation can be given as

\[(t-t_0)^2 y''(t) + (t-t_0) \alpha(t) y'(t) + \beta(t)y(t) = 0,\]

with \(\alpha\) and \(\beta\) analytic at \(t=t_0\). Let us assume that this point is \(t_0=0\). Frobenius’ method consists of the following technique: In the equation

\[x^2 y''(x) + x \alpha(x) y'(x) + \beta(x)y(x) = 0,\]

we assume a generalised series solution of the form

\[y(x)=x^\gamma \sum_{n=0}^\infty c_n x^k .\]

Equating powers of \(x\) we find \[\gamma(\gamma-1) c_0 x^\gamma + \alpha_0 \gamma c_0 x^\gamma + \beta_0c_0 x^\gamma = 0,\] etc. The equation for the lowest power of \(x\) can be rewritten as

\[\gamma(\gamma-1) + \alpha_0\gamma + \beta_0 = 0. \label{indicial}\]

Equation \ref{indicial} is called the** indicial equation. **It is a quadratic equation in \(\gamma\), that usually has two (complex) roots. Let me call these \(\gamma_1\), \(\gamma_2\). If \(\gamma_1-\gamma_2\) is not integer one can prove that the two series solutions for \(y\) with these two values of \(\gamma\) are independent solutions.

Let us look at an example \[t^2 y''(t) + \frac{3}{2} t y'(t) + ty = 0.\] Here \(\alpha(t)=3/2\), \(\beta(t)=t\), so \(t=0\) is indeed a regular singular point. The indicial equation is

\[\gamma(\gamma-1)+\frac{3}{2}\gamma = \gamma^2+\gamma/2 = 0.\]

which has roots \(\gamma_1=0\), \(\gamma_2=-1/2\), which gives two independent solutions

\[\begin{align} y_1(t)&= \sum_{k}c_kt^k,\nonumber\\ y_2(t)&= t^{-1/2}\sum_{k}d_kt^k.\nonumber\end{align}\]

INDEPENDENT SOLUTIONS

Independent solutions are really very similar to independent vectors: Two or more functions are independent if none of them can be written as a combination of the others. Thus \(x\) and \(1\) are independent, and \(1+x\) and \(2+x\) are dependent.