Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 8

For the normalized second-order equation, $x_0$ is an ordinary point if both coefficient functions $p$ and $q$ are holomorphic near $x_0$. If it is singular but $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are holomorphic there, it is a regular singular point. Thus $p$ may have at most a simple pole and $q$ at most a double pole.

The distinction tells us which local series method is justified. At an ordinary point, two independent solutions are represented by ordinary Taylor series. At a regular singular point, the Frobenius method uses $(x-x_0{)}^r{\sum }_{n\ge 0}{a}_n(x-x_0{)}^n$; the indicial equation determines the exponents, and logarithmic terms may appear in a second solution. Singularities worse than these are irregular and need not have this Frobenius form.

The Wronskian isSuppose constants $A,B$ satisfy $Af+Bg=0$ throughout the interval. Differentiating gives $A{f}^{\prime }+B{g}^{\prime }=0$. At $x_0$, these form the homogeneous linear systemIts determinant is nonzero by hypothesis, so $A=B=0$. This is exactly linear independence of the two functions on the interval. No differential equation is needed for this implication; the converse for arbitrary differentiable functions would require additional hypotheses.

The origin is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{x}^n$ into the equation and collect the coefficient of $x^n$:Hence the recurrence relation isThere are two free initial coefficients. The independent even and odd power series beginThe general solution is $\boxed{y=a_0{y}_{\mathrm{e}}+a_1{y}_{\mathrm{o}}}$. As a useful independent check, the substitution $x=\sin t$ reduces the equation to $d^{2}y/d{t}^2+k^{2}y=0$. ThusFor a nonterminating parity series, $|a_{n+2}/a_n|\to 1$, so the ratio test in $x^2$ gives $\boxed{R=1}$. The regular singular points at $\pm 1$ explain the natural obstruction.

There is an important exception to an unconditional radius-one answer. If $k^2=m^2$ for an integer $m\ge 0$, the series of parity $m$ terminates at degree $m$ and is a multiple of a Chebyshev polynomial. Its radius of convergence is $\boxed{R=\infty }$; the other independent parity solution still has radius one. For $m=0$ the even solution is constant. Any nonzero combination containing the nonterminating solution has radius one. These are the full power series of the Chebyshev differential equation, including the terminating cases.