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# 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.