Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 8

Set $y={\sum }_{j\ge 0}{a}_j{x}^j$ and equate coefficients in the power series. The constant coefficient gives $a_2=0$. For $m\ge 1$, the coefficient of $x^m$ givesThe square bracket is $(m+1)(m+2)$, so the recurrence relation is $a_{j+3}=a_j$ for $j\ge 0$. The initial conditions give $a_0=1$, $a_1=0$, $a_2=0$. Hence

Summing the geometric series givesThere is a particularly short verification using the product rule:Also $y(0)=1$ and $y^{\prime }(0)=0$. The radius of convergence of the series is one, set by the three complex cube roots of unity; the closed expression continues the solution beyond that disk wherever its denominator is nonzero. The real initial-value solution extends on the interval $(-\infty ,1)$.

Substitute $y=v{P}_n$ into the Legendre differential equation and expand using the product rule. The terms proportional to $v$ cancel because $P_n$ is itself a solution. What remains isMultiplying by $P_n$ identifies an exact derivative:The latter form is convenient for reduction of order on intervals where $P_n\ne 0$ and $x\ne \pm 1$.

Two integrations give the general reduction of order formulaThe integral is local on intervals avoiding zeros of $P_n$. A pole of the integral at an interior polynomial zero need not be a singularity of $y$: the product extends by reduction of order across a zero of the known solution. After a standard normalization the independent second solution is the Legendre function of the second kind $Q_n$.

For normalized Legendre polynomials, $P_n(1)=1$ and $P_n(-1)=(-1{)}^n$. Near the endpoints the integral factor has the asymptoticsandThus an independent solution, for which $C_2\ne 0$, has the logarithmic endpoint singularitiesThe constants are understood for a solution continued consistently across the ordinary interior points. Adding any multiple of $P_n$ cannot cancel these logarithms. This explains why boundedness at an endpoint selects the Legendre polynomial and excludes the independent Legendre function of the second kind.

Write $P_n(x)\sim c_n{x}^n$ with $c_n\ne 0$. For $n\ge 1$, this branch grows. Choosing the other reduction of order solution by an integral from infinity gives, on $x>1$,Thus for $n\ge 1$, every bounded exterior solution decays as $|x{|}^{-n-1}$, up to a constant factor, on each exterior interval. Equivalently the dominant-balance exponents in the Legendre differential equation are $n$ and $-n-1$.

There is an important exception at degree zero. Since $P_0=1$,These solutions are bounded even when $C_1\ne 0$; boundedness alone does not then imply decay to zero. The constants on $x>1$ and $x<-1$ are independent unless continuation across the singular endpoints is specified.