Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 7A (Differential Equations)

Classification and characteristic exponents. For a normalized second-order linear differential equation, an ordinary point at zero means $p$ and $q$ are real analytic functions there. A regular singular point is a singular point where $xp$ and $x^{2}q$ nevertheless extend as real analytic functions to zero. Write $p_0={\lim }_{x\to 0}xp(x)$ and $q_0={\lim }_{x\to 0}{x}^{2}q(x)$. The lowest-power coefficient in the Frobenius method gives the indicial equationIts roots are the characteristic exponents at a regular singular point. If their difference is not an integer, there are two linearly independent solutions $x^{{\sigma }_j}{A}_j(x)$, with $A_j$ real analytic and $A_j(0)\ne 0$. For distinct real roots ${\sigma }_1>{\sigma }_2$ with integer difference, a solution for the larger root exists, and a second has the form $x^{{\sigma }_2}B(x)+c{y}_1\log x$; the logarithmic coefficient can vanish. For a repeated root, the second solution has the form $y_1\log x+x^{{\sigma }_1}B(x)$ after normalization. This is a Logarithmic Frobenius solution. Complex conjugate roots $a\pm ib$ give real solutions whose leading behaviour is $x^a\cos (b\log x)$ and $x^a\sin (b\log x)$, with analytic-series corrections. All these descriptions are on $x>0$, or on a fixed complex branch.

For the present equation, division by $x$ gives $p=1+(1-m)/x$ and $q=(1-m)/x$. Therefore $xp=x+1-m$ and $x^{2}q=(1-m)x$ are real analytic. Zero is regular singular when $m\ne 1$; when $m=1$, it is ordinary after removing the common factor $x$. In that exceptional case the normalized equation is $y^{\prime \prime }+y^{\prime }=0$.

For the Frobenius method substitution, the lowest coefficient and the general linear recurrence relation areThese equations come respectively from the coefficients of $x^{\sigma -1}$ and $x^{n+\sigma -1}$. At $m=1$, the same formal exponents $0,1$ describe the two possible leading powers at an ordinary point, rather than singular behaviour.

For the exponent $\sigma =m$, the linear recurrence relation reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:The ratio test gives an infinite radius of convergence for the power series factor. Hence a solution isFor nonintegral $m$, choose instead the exponent $\sigma =0$. Since $n-m\ne 0$ for every positive integer $n$, its linear recurrence relation gives $n{a}_n=-a_{n-1}$. ThusThe distinct nonintegral leading exponents establish linear independence. More explicitly, $W(e^{-x},y_m)=m{e}^{-x}{x}^{m-1}$ by the Abel identity and the leading coefficient at zero, so the Wronskian is nonzero. The factor $x^m$ uses the real branch on $x>0$.

For positive integral $m$, the exponent-zero linear recurrence relation fixes $a_n=(-1{)}^n{a}_0/n!$ for $1\le n<m$, but at $n=m$ its factor $n-m$ vanishes and leaves $a_m$ free. Choose $a_0=1$ and $a_m=0$. All later coefficients then vanish by the same linear recurrence relation.

A polynomial solution of degree $m-1$ isFor $m=1$ this is the constant solution $1$, in agreement with the ordinary point exception. The terminating Frobenius series arises because the coefficient equation at the resonant index becomes $0=0$, permitting truncation rather than forcing a logarithm.

Here the indicial equation has the repeated root $0$. The first Frobenius method solution is $y_1=e^{-x}$. Since $p=1+1/x$, the Abel identity gives $W=C{e}^{-x}/x$. Reduction of order with $C=1$ produceswhere the additive constant is absorbed into $y_1$. In particular the logarithmic coefficient is nonzero: an additional independent real analytic solution does not exist.

The general local form isThis is the repeated-root Logarithmic Frobenius solution. On the negative real side one may use $\log |x|$, or use a fixed logarithm branch for a complex solution.