Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 14E (Further Complex Methods)

After division by $z$, the coefficient functions areBoth $zP(z)=1-z$ and $z^{2}Q(z)=\lambda z$ are analytic at zero, so the regular singular point criterion for a second-order equation shows that $z=0$ is regular singular. Substitution of $y=z^r$ into the leading terms gives the indicial equationThe exponent is repeated. Directly substituting $y={\sum }_{n\ge 0}{a}_n{z}^n$ givesThus $a_0$ determines the entire power series, so there is only one such solution $y_1$ up to scale. The logarithmic solution from a repeated Frobenius exponent predictsso its leading nonanalytic term is proportional to $\log z$.

For the contour ansatz, differentiation under the integral and one integration by parts giveThe integral vanishes whenwhose solution is the Laguerre contour-integral amplitudeThe contour and branches must make $f$ single-valued along the traversed path and must kill the endpoint term

Now suppose $\lambda <0$ is nonintegral. Near $t=0$, $f(t)=O(t^{-\lambda -1})$ is integrable and the endpoint factor is $O(t^{-\lambda })\to 0$. We may therefore choosea finite loop based at the branch point $0$ and avoiding $1$ except by encirclement. A second choice isalong one bank of the negative real axis, with a consistent branch. At $-\infty$, $e^{zt}$ kills the algebraic endpoint factor because $\mathrm{Re}z>0$.

The ${\gamma }_1$ integral is analytic in $z$ because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of $y_1$. This is the Finite Laguerre contour solution construction.

Let $\lambda >0$ be nonintegral. This time $f(t)=O((t-1{)}^{\lambda })$ is integrable at $t=1$, and the endpoint factor behaves as $(t-1{)}^{\lambda +1}\to 0$. Suitable contours are thereforeandwhere the second path makes a fixed detour above or below the branch point $0$ and the branch of $t^{-\lambda -1}(t-1{)}^{\lambda }$ is continued along it. The exponential again controls the endpoint at $-\infty$ for $\mathrm{Re}z>0$.

As in part (i), the finite ${\gamma }_1$ integral is analytic near zero and hence is a constant multiple of the unique power-series solution $y_1$.

Write $\lambda =-m$ with $m\in \mathbb{N}$. The amplitude is now the rational functionChoose ${\gamma }_1$ to be a small positively oriented circle around the pole $t=1$, and choose ${\gamma }_2$ from $t=0$ along the negative real axis to $-\infty$. At zero the endpoint factor behaves as $t^m$ and vanishes; at negative infinity the exponential decays.

By the residue theorem, the finite-contour solution isEvery derivative term contains $e^z$ times a power of $z$ of degree at most $m-1$, and the highest-degree term is nonzero. Henceup to a constant, where $P_{m-1}$ is a polynomial of degree $m-1$. This is the negative-integer case of the Integer-parameter Laguerre contour residues.

Write $\lambda =N$ with $N\in {\mathbb{Z}}_{\ge 0}$. ThenTake ${\gamma }_1$ to be a small positively oriented circle around $t=0$. For ${\gamma }_2$, start at $t=1$, travel to $-\infty$, and avoid the pole at zero by a fixed upper or lower detour. The endpoint factor vanishes at $t=1$, and the exponential controls $-\infty$.

The finite contour extracts the residue at zero:Only powers $z^0,\dots ,z^N$ occur, and the coefficient of $z^N$ is nonzero. Thus this solution is a polynomial of degree $N$ and, up to normalization, is the Laguerre polynomial $L_N(z)$:This also follows from the recurrence in part (i), because $a_{N+1}=0$ and every later coefficient vanishes.