Past exam of the differential equations 2 course of the University of Oxford / 2023 / 2 / a

Put $N=n(n+1)$. After division by $1-x^2$, the Legendre differential equation has coefficientsBoth are analytic functions at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm 1$, $p$ has a genuine simple pole; hence these are singular points even when $n=0$ makes $q$ vanish. However,and both products extend analytically across $x_0$. By the regular singular point criterion for a second-order equation,None of the three specified points is an irregular singular point.

Use $t=x+1$. The Legendre differential equation becomesSubstituting a Frobenius series $y=t^r{\sum }_{k\ge 0}{a}_k{t}^k$, $a_0\ne 0$, gives the indicial equation $2r^2=0$: the indicial root zero is repeated. For $r=0$, the coefficient of $t^k$ yieldsorFor $0\le k<n$ the multiplying factor is nonzero. At $k=n$ it vanishes, so $a_{n+1}$ and all later coefficients are zero. Thus one solution is a polynomial of exact polynomial degree $n$, including a nonzero constant when $n=0$. Denote a chosen normalization by $P_n$.

The linear recurrence relation determines every analytic coefficient from $a_0$, so there cannot be a second independent analytic solution at $t=0$. To see the required logarithm explicitly, reduction of order gives the second solution locally asNear $t=0$, the integrand is $1/(2a_0^{2}t)$ plus an analytic function. After rescaling the second solution, its form is thereforeThe analytic remainder converges locally; its constant term may be set to zero by adding a multiple of $P_n$. The nonzero logarithmic coefficient makes $y_2$ linearly independent of $P_n$. This Frobenius expansion of the Legendre equation at an endpoint gives a Logarithmic Frobenius solution; the same endpoint calculation at $x=1$ explains why the independent Legendre function of the second kind is excluded when boundedness at both ends is required.