Ordinary and singular points · Differential Equations

Intuition

Where does the method work, and how far does the series reach? Put the equation in standard form, with the coefficient of the second derivative equal to one. A point where both remaining coefficients are convergent power series is an ordinary point, and there the method always works; a point where one of them blows up is singular. The series about an ordinary point converges at least out to the nearest singular point — even a singular point that is a complex number, off the real line.

A lamp in a dark room lights everything up to the nearest wall. The singular points are the walls, and the series about an ordinary point reaches at least as far as the nearest one, in every direction.

plane diagram: the-nearest-singular-point

The complex plane of $x$ for $(1+x^{2})y''+y=0$ . Its leading coefficient vanishes at $\pm i$ , at distance $1$ from $0$ , so the series about $0$ is guaranteed only inside the dashed circle — on the real line, for $|x|<1$ — although the equation has no real singular point at all.

Where the method is guaranteed

Write the equation as $y''+p(x)y'+q(x)y=0$ . The point is ordinary if and are analytic there — equal to convergent power series about — and singular otherwise. For with polynomial coefficients and no common factor, the singular points are the zeros of , complex ones included. About an ordinary point there are two independent series solutions, and they converge at least out to the nearest singular point: Fuchs's theorem, used here without proof. Its algebraic half is the theorem below.

\text{radius}\ \ge\ \min\bigl\{|x_{0}-z|:\ z\ \text{singular}\bigr\}

Finding the reach

$(1-x^{2})y''-2xy'+6y=0$ has the singular points ; about its series converge for at least.

At an ordinary point the first two coefficients determine the rest

\begin{aligned}&p,\ q\ \text{power series at}\ 0\\&\Longrightarrow\ a_{n+2}\ \text{is fixed by}\ a_{0},\ldots,a_{n+1}\end{aligned}

Substitute the series into the equation in standard form. The second derivative, shifted, puts n + 2 times n + 1 times the coefficient of index n + 2 at the power xⁿ. The terms with p and q, products of power series, put there only coefficients of lower index. Setting the total to zero gives each coefficient from the earlier ones, because the factor in front of it is never zero, and only the first two are left free.

Proof steps

y''\ \longmapsto\ \sum_{n\ge0}(n+2)(n+1)\,a_{n+2}\,x^{n}

The second derivative contributes the coefficient of highest index.

py'+qy\ \longmapsto\ \text{only}\ a_{0},\ldots,a_{n+1}\ \text{at}\ x^{n}

Products of power series put only lower coefficients at the power xⁿ.

(n+2)(n+1)\,a_{n+2}=-\bigl[\,py'+qy\,\bigr]_{n}

Setting the coefficient of xⁿ to zero, and (n + 2)(n + 1) is never zero.

a_{0},\ a_{1}\ \text{free}

Only the first two coefficients are left free: two independent solutions.

Applications

Practice

Standard Form First

Divide by the coefficient of y″. A point where the remaining coefficients blow up is singular.

P(x)\,y''+Q(x)\,y'+R(x)\,y=0

P(x_{0})=0\ \Longrightarrow\ x_{0}\ \text{singular}

Try it

What are the singular points of $x(x-3)\,y''+y'+y=0$ ?