Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 14A (Further Complex Methods)

The Papperitz symbolrepresents the two-dimensional solution space of a homogeneous linear differential equation for a second-order Fuchsian differential equation on the Riemann sphere. Its only singularities are the three regular singular points $a,b,c$. The two entries beneath each point are its characteristic exponents at a regular singular point: for example, in the nonresonant case a local basis near $a$ has leading behaviours $(z-a{)}^{\alpha }$ and $(z-a{)}^{{\alpha }^{\prime }}$. If one of the points is infinity, exponent $\rho$ conventionally means behaviour $z^{-\rho }$ there.

The six exponents obey the Fuchs relationSubject to this relation, the symbol determines the corresponding Riemann differential equation; resonant differences between exponents may produce Logarithmic Frobenius solutions without changing the exponent data recorded by the symbol.

Let $\omega$ solve the equation represented by the original symbol and define a dependent-variable rescaling of a Papperitz symbolwith a consistent local choice of branch. Near $a$, the multiplier is a nonzero analytic factor times $(z-a{)}^{\delta }$, so both exponents at $a$ increase by $\delta$. Near $b$, it is a nonzero analytic factor times $(z-b{)}^{-\delta }$, so both exponents at $b$ decrease by $\delta$. It is analytic and nonzero near $c$, so the exponents there are unchanged. Therefore the transformed equation has symbolThe opposite shifts also preserve the Fuchs relation.

A Möbius transformation of a Papperitz symbol changes the independent variable by a Möbius transformation. In particular,maps $a,b,c$ to $0,1,\infty$, respectively. Because a Möbius map is locally biholomorphic away from its pole, it carries each exponent pair with its singular point and givesNow define a dependent-variable rescaling of a Papperitz symbol by writing the old dependent variable asso that $v=u^{-\alpha }(1-u{)}^{-\beta }y$. This dependent-variable rescaling of a Papperitz symbol subtracts $\alpha$ from both exponents at zero, subtracts $\beta$ from both exponents at one, and adds $\alpha +\beta$ to both exponents at infinity. Hence $v$ has symbolDefineThe Fuchs relation gives ${\alpha }^{\prime }-\alpha =1-C$ and ${\beta }^{\prime }-\beta =C-A-B$. After renaming $u$ as $z$, the symbol is thereforewhich is the Gauss hypergeometric equation. Renaming $A,B,C$ as $a,b,c$ gives exactly the displayed symbol in the question.

In the hypergeometric symbol, select the exponent $1-c$ at zero and factor out its local power by writingThe Exponential shifting of a Papperitz symbol makes the exponents of $v$ equal to $0,c-1$ at zero, $0,c-a-b$ at one, and $a-c+1,b-c+1$ at infinity. These are the hypergeometric parametersbecause $1-C=c-1$ and $C-A-B=c-a-b$. The solution for $v$ that is analytic at zero and equals one there is therefore the Gauss hypergeometric function $F(A,B,C;z)$. Thus the second local hypergeometric solution isFor noninteger $c$ this is linearly independent from $F(a,b,c;z)$; at resonant values, the usual limiting or logarithmic modification may be required.