Notes on Whittaker & Watson, Chapter X

Whittaker & Watson, A Course of Modern Analysis. Chapter X, Linear Differential Equations.

10.1 Set-up the discussion of the homogeneous second order ODE $$ \frac{d^2u}{dz^2}+p(z)\frac{du}{dz}+q(z)u=0. $$ 10.2 Series solution in the neighbourhood $S_b$ of an ordinary point $b$. With the substitution $u=v\int_b^zp(\zeta)d\zeta$, the equation becomes $$ \frac{d^2v}{dz^2}+J(z)u=0. $$ where $J(z)=q(z)-\frac{1}{2}\frac{dp}{dz}-\frac{p^2}{4}$. Series solution $v(z)=\sum_{n-0}^{\infty}v_n(z)$ is uniform convergent and analytic in $S_b$ with $$ v_0(z)=a_0+a_1(z-b),\newline v_n(z)=\int_b^z(\zeta-z)J(\zeta)v_{n-1}(\zeta)d\zeta, $$ and $a_0=v(b)$, $a_1=v'(b)$ determined by initial conditions at $z=b$.

[This solution is also followed in Gerhard Kristensson, Second Order Differential Equations, Special Functions and Their Classification (Springer, 2010), 2.3 Solution at a regular point, pp. 6–11.]

10.3 The second order ODE has regular singularity at $z=c$ if $p$ has a pole up to order 1 and $q$ has a pole up to order 2 at $c$. Rewriting the equation as $$ (z-c)^2\frac{d^2u}{dz^2} + (z-c)P(z-c)\frac{du}{dz}+Q(z-c)u=0 $$ with $P=\sum_{n=0}^\infty p_n(z-c)^n$ and $Q=\sum_{n=0}^\infty q_n(z-c)^n$ both analytic at $c$, and assuming the power series solution $$ u=(z-c)^\alpha\left[1+\sum_{n=1}^\infty a_n(z-c)^n\right], $$ we can find the coefficients $\alpha$ and $a_n$ by solving indicial equations $$ \begin{align*} F(\alpha)\equiv\alpha^2+(p_0-1)\alpha+q_0&=0\newline a_1F(\alpha+1) + \alpha p_1+q_1&=0\newline \cdots & \quad \newline a_nF(\alpha+n) + \sum_{m=1}^{n-1}[(\alpha+n-m)p_m+q_m] + \alpha p_n+q_n&=0. \end{align*} $$ The two roots \(\rho_1\), \(\rho_2\) of \(F(\alpha)\) are called the *exponents* of the differential equation at $c$. When the difference $s=\rho_1-\rho_2$ is not zero or an integer, such that $F(\alpha+n)$ does not vanish, two distinct series solutions can be determined with $\alpha=\rho_1$ and $\alpha=\rho_2$ respectively.

10.31 The convergence of the series solution in 10.3.

10.32 When $s=\rho_1-\rho_2$ is zero or an integer, with $u=w_1(z)=(z-c)^{\rho_1}\left[1+\sum_{n=1}^\infty a_n(z-c)^n\right]$ being one solution, we assume a formal second solution $u=w_1(z)\zeta$, such that $$ (z-c)^2\frac{d^2\zeta}{dz^2}+\left[2(z-c)^2\frac{w_1'(z)}{w_1(z)}+(z-c)P(z-c)\right]\frac{d\zeta}{dz}=0 $$ which as the general solution by integrating factor $$ \begin{align*} \zeta&=A+B\int^z\frac{1}{w_1(z)^2}\exp\left\{-\int^z\frac{P(\xi-c)}{\xi-c}d\xi\right\}dz \newline &=A+B\int^z(z-c)^{-p_0-2\rho_1}g(z)dz \end{align*} $$ with $g(z)$ being analytic within the radius of convergence for $P(z-c)$’s Taylor series. Hence the second solution $$ Aw_1(z)+B\left\{g_sw_1(z)\log(z-c) + (z-c)^{\rho_2}\left[-\frac{1}{s}+\sum_{n=1}^\infty h_n(z-c)^n\right]\right\} $$ can be found when $s\neq 0$ with $g_n$ being the coefficients of $g$’s Taylor series $g(z)=1+\sum_{n=1}^\infty g_n(z-c)^n$ and $h_n=\frac{g_n}{n-s}$ for $n\neq s$.

When $s=0$, the second solution becomes $$ Aw_1(z)+B\left\{w_1(z)\log(z-c) + (z-c)^{\rho_2}\sum_{n=1}^\infty h_n(z-c)^n\right\}. $$

These solution are called regular integrals.

[Some technical details supplied in Kristensson.]

10.4 Solutions in the neighbourhood of $\infty$. Under the mapping $z\mapsto z_1=\frac{1}{z}$, the equation in the neighbourhood of $\infty$ becomes $$ z_1^4\frac{d^2u}{dz_1^2}+\left[2z_1^3-z_1^2p\left(\frac{1}{z_1}\right)\right]\frac{du}{dz_1} + q\left(\frac{1}{z_1}\right)u=0 $$ with $z_1$ in a neighourhood of 0. Thus, $\infty$ is ordinary point when $2z-z^2p$ and $z^4q$ are analytic at $\infty$, and a regular singular point when $zp$ and $z^2q$ are analytic.

10.5 Irregular singularities and confluence. The equation cannot be solved by regular integrals in the neighbourhood of points that are not a regulr singularities. However, such singularities may be derived from two or more singularities in a limiting process of confluence.

10.6 The differential equation $$ \frac{d^2u}{dz^2}+\left[\sum_{r=1}^4\frac{1-\alpha_r-\beta_r}{z-a_r}\right]\frac{du}{dz}+\left[\sum_{r=1}^4\frac{\alpha_r\beta_r}{(z-a_r)^2}+\frac{Az^2+2Bz+C}{\prod_{r=1}^4(z-a_r)}\right]u=0 $$ has 4 regular singularities at $a_1, a_2, a_3, a_4$ with exponents $\alpha_r$, $\beta_r$ at $a_r$ ($r=1,2,3,4$) and an ordinary point at $\infty$ with exponents $\mu_1$, $\mu_2$ which are the roots of $$ \mu^2+\mu\left(\sum_{r=1}^4(\alpha_r+\beta_r)-3\right) + \sum_{r=1}^4\alpha_r\beta_r + A=0. $$ When $\mu_2-\mu_1=\frac{1}{2}$, this equation is called the generalized Lamé equation. Klein and Bôcher show that a range of differential equations in mathematical physics can be regarded as confluent forms of the generalized Lamé equation.

Depending on the values of $B$ and $C$, the confluence of two singularities in the confluence form may not have the exponent-difference $\frac{1}{2}$. Irregular singularities may arise from the confluence of three or more singularities.

Six types of equations can be obtained from the generalized Lamé equation with five singularities (including $\infty$) by confluence:

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10.7 Let linear differential equations with three singularities at $a,b,c$ have exponents $\{\alpha, \alpha'\}$, $\{\beta, \beta'\}$, $\{\gamma, \gamma'\}$, with $\infty$ being an ordinary point. The exponents satisfy $\alpha+ \alpha'+ \beta+ \beta'+ \gamma+ \gamma'=1$, and $$ p(z)=\frac{1-\alpha-\alpha'}{z-a}+\frac{1-\beta-\beta'}{z-b}+\frac{1-\gamma-\gamma'}{z-c}\newline q(z)=\left.\left(\frac{\alpha\alpha'(a-b)(a-c)}{z-a}+ \frac{\beta\beta'(b-c)(b-a)}{z-b} + \frac{\gamma\gamma'(c-a)(c-b)}{z-c}\right)\right/(z-a)(z-b)(z-c). $$ The equation $\frac{d^2u}{dz^2}+p(z)\frac{du}{dz} + q(z)u=0$ becomes Riemann’s $P$-equation, first given by Papperitz. The solution is written in Riemann’s $P$ symbol $$ u=P\begin{Bmatrix}a & b & c & \newline \alpha & \beta & \gamma & z \newline \alpha' & \beta' & \gamma' & \end{Bmatrix}. $$ 10.71 Riemann’s $P$ is invariant under the Möbius transformation $f(\cdot)$ that maps $z\mapsto f(z)=z_1$, $a\mapsto f(a)=a_1$, $b\mapsto f(b)=b_1$, $c\mapsto f(c)= c_1$, in the sense that $$ P\begin{Bmatrix}a & b & c & \newline \alpha & \beta & \gamma & z \newline \alpha' & \beta' & \gamma' & \end{Bmatrix} = P\begin{Bmatrix}a_1 & b_1 & c_1 & \newline \alpha & \beta & \gamma & z_1 \newline \alpha' & \beta' & \gamma' & \end{Bmatrix}. $$ The transform $u\mapsto u_1=\left(\frac{z-a}{z-b}\right)^k\left(\frac{z-c}{z-b}\right)^lu$ maps the solution of a Riemann’s $P$ equation to that of an equation with singularities at the same points but exponents $\{\alpha+k, \alpha'+k\}$, $\{\beta-k-l, \beta'-k-l\}$, $\{\gamma+l, \gamma'+l\}$: $$ \left(\frac{z-a}{z-b}\right)^k\left(\frac{z-c}{z-b}\right)^l P\begin{Bmatrix}a & b & c & \newline \alpha & \beta & \gamma & z \newline \alpha' & \beta' & \gamma' & \end{Bmatrix} = P\begin{Bmatrix}a & b & c & \newline \alpha +k & \beta -k-l & \gamma+l & z_1 \newline \alpha'+k & \beta' -k-l & \gamma'+l & \end{Bmatrix}. $$ 10.72 Using these transforms, we obtain $$ P\begin{Bmatrix}a & b & c & \newline \alpha & \beta & \gamma & z \newline \alpha' & \beta' & \gamma' & \end{Bmatrix} = \left(\frac{z-a}{z-b}\right)^\alpha\left(\frac{z-c}{z-b}\right)^\gamma P\begin{Bmatrix}0 & \infty & 1 & \newline 0 & \beta +\alpha+\gamma & 0 & x \newline \alpha'-\alpha & \beta'+\alpha+\gamma & \gamma'-\gamma & \end{Bmatrix} $$ with $x=\frac{(z-a)(c-b)}{(z-b)(c-a)}$. This shows that any Riemann’s $P$ can be obtained from the solution to a corresponding hypergeometric equation on the RHS.

[The hypergeometric differential equation $$ z⁢(1-z)\frac{⁢d^2u}{dz^2}+(c-(a+b+1)⁢z)\frac{⁢du}{dz}-a⁢bu=0. $$ has regular singularities at $0,\infty,1$ with corresponding exponents $\{0,1-c\}$, $\{a,b\}$, $\{0,c-a-b\}$. See Chapter XIV.]

10.8 Linear differential equations with two singularities are derived from 10.7 by making $p$ and $q$ analytic at $z=c$, resulting in $$ \frac{⁢d^2u}{dz^2}+\left(\frac{1-\alpha-\alpha'}{z-a}+ \frac{1+\alpha+\alpha'}{z-b}\right)\frac{⁢du}{dz}-\frac{\alpha\alpha'(a-b)^2u}{(z-a)^2(z-b)^2}=0 $$ which has solutions in elementary form that $$ u=A\left(\frac{z-a}{z-b}\right)^\alpha+B\left(\frac{z-a}{z-b}\right)^{\alpha'} $$ when $\alpha\neq\alpha'$ and $$ u=A\left(\frac{z-a}{z-b}\right)^\alpha+B\left(\frac{z-a}{z-b}\right)^{\alpha}\log\left(\frac{z-a}{z-b}\right) $$ when $\alpha=\alpha'$.