Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 8

A finite point $x_0$ is an ordinary point when the normal-form coefficients $p,q$ are analytic at $x_0$. Then arbitrary values $y(x_0),y^{\prime }(x_0)$ determine a convergent Taylor series, and two independent choices give two linearly independent local analytic solutions.

A singular point is a regular singular point when $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$. The Frobenius method then provides at least one solutionwith a locally convergent analytic factor and an exponent fixed by the indicial equation. If its two indicial roots differ by a noninteger, there are two independent Frobenius solutions. For integer-separated roots a second solution may include a logarithm; a repeated root gives a Logarithmic Frobenius solution accompanying the single pure series. Such solutions need not be finite or ordinary Taylor series at the singular point.

Expanding the derivative and dividing by $1-x$ givesThe only finite singular point is $\boxed{x=1}$. There $(x-1)p(x)=1$ and $(x-1{)}^{2}q(x)=c^2(x-1{)}^2-\lambda (x-1)$ are analytic, so it is a regular singular point. At infinity the coefficient of $y$ tends to $c^2\ne 0$, giving an irregular singular point after reciprocal substitution; the finite classification is independent of $\lambda$.

Put $t=1-x$ and $Y(t)=y(x)$. The differential equation becomesSubstitute the Frobenius solution $Y=t^{\sigma }{\sum }_{n\ge 0}{a}_n{t}^n$. The lowest power has coefficient ${\sigma }^2{a}_0$, soEquating successive coefficients gives, with $a_{-1}=0$,Thus $a_0\ne 0$ is the only free coefficient: $a_1=-\lambda a_0$, $a_2=({\lambda }^2-c^2)a_0/4$, and so on. Since every denominator is nonzero, there is exactly one independent pure Frobenius solution of this form. The repeated root supplies no second free coefficient; the other solution is logarithmic, as parts (iii) and (iv) demonstrate.

For the normal-form coefficient $p(x)=-1/(1-x)$, Abel's identity gives the WronskianUse reduction of order: if $y_2=y_{1}v$, then $W=y_1^2{v}^{\prime }$. Absorbing a nonzero Wronskian constant into the second integration constant, the general solution isThis local representation applies on intervals where $y_1\ne 0$, in particular sufficiently near $x=1$ because $y_1(1)=a_0\ne 0$.

Write $t=1-x$. The analytic solution has $y_1=a_0+O(t)$ with $a_0\ne 0$, so the integrand in the reduction of order formula satisfiesIntegrating with respect to $x$ gives $-a_0^{-2}\log (1-x)+O(1)$. Multiplication by $y_1$ therefore givesAny solution independent of $y_1$ has a nonzero coefficient of this second solution. HenceAn added multiple of $y_1$ stays bounded and cannot cancel this logarithm. On the real side $x>1$, the corresponding statement uses $\log |1-x|$.