Asymptotic Solutions of Linear ODEs. 4

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$

Irregular Singular Points of Differential Equation

  1. Toy-model: 1st order equation
  2. Toy-model: 2nd order equation
  3. Remarks

Toy-model: 1st order equation

See Definition 3.1.3 for irregular singular points.

Again we can assume that $z_0=0$. Consider first order equation \begin{equation} u'+p(z)u=0. \label{eq-3.4.1} \end{equation} As known in the vicinity of isolated singular point function could be expanded into Laurent series \begin{equation} p(z)=\sum_{-N\le n <\infty} p_n (z-z_0)^n \label{eq-3.4.2} \end{equation} (with $N\ge 0$ is taken the smallest possible) where $N=0$ iff $p(z)$ is analytic in $z_0$, $0< N < \infty$ iff $z_0$ is a pole (then $N$ is an order of the pole) and $N=\infty$ iff $z_0$ is an essential singularity of $p(z)$.

According to our classification $z_0$ is an ordinary point iff $N=0$, a regular singular point iff $N=1$ and an irregular singular point iff $N \ge 2$.

Consider solution \begin{equation} u(z)= Ce^{-\int p(z)\,dz}, \qquad C=\const. \label{eq-3.4.3} \end{equation} From (\ref{eq-3.4.2}) \begin{equation} \int p(z)\,dz = \sum_{-N\le n <\infty, n\ne -1 } \frac{p_n}{n+1} (z-z_0)^{n+1} + p_{-1} \log (z-z_0) \label{eq-3.4.4} \end{equation} and therefore \begin{equation} u(z)= Ce^{-\sum_{-N\le n <\infty, n\ne -1 } \frac{p_n}{n+1} (z-z_0)^{n+1}} (z-z_0)^{p_{-1}} \label{eq-3.4.5} \end{equation} Clearly, as $N=0$ this solution is analytic at $z_0$ (as it follows from Section 3.2); as $N=1$ this solution has at $z_0$ either pole or a branching point (as it follows from Section 3.3) or could be even analytic there (find out when).

On the other hand, as $N\ge 2$ $\exp\bigl(-\sum_{-N\le n <\infty, n\ne -1 } \frac{p_n}{n+1} (z-z_0)^{n+1}\bigr)$ has at $z_0$ essential singularity for sure and factor $(z-z_0)^{p_{-1}}$ can add branching to this singularity.

Toy-model: 2nd order equation

Consider second-order equation \begin{equation} u''+p(z)u'+q(z)u=0. \label{eq-3.4.6} \end{equation} Let us look for solution in the form $u(z)=e^{S(z)}$; plugging into equation we get $u'=S'u$, $u''= ({S'}^2 +S'')u$ and \begin{equation} {S'}^2 +S'' +p(z)S' + q(z)=0 \label{eq-3.4.7} \end{equation} This is a first-order equation (with respect to $S'$) albeit non-linear (and trying to solve equation we gained nothing). However under certain assumption we can derive asymptotical properties of $S(z)$ and $u(z)$. Namely, let us assume that \begin{equation} S''= o({S'}^2)\qquad \text{as }\ z\to z_0. \label{eq-3.4.8} \end{equation} Why it could be true? If $S(z)\sim \kappa (z-z_0)^{-m}$ then it is true as $m>1$. But we need to check this condition a posteriori.

Then we replace (\ref{eq-3.4.7}) by \begin{equation} {S'}^2 + p(z)S' + q(z)=0 \label{eq-3.4.9} \end{equation} which is an algebraic equation with respect to $S'$ and we can find \begin{equation} S'(z)=\frac{1}{2}\bigl( - p(z) \pm \sqrt{p^2(z)-4q(z)}\bigr). \label{eq-3.4.10} \end{equation}

One can see easily that if both $p(z)$ and $q(z)$ have only poles (not essential singularities) and if $z_0$ is an irregular singular point (thus either $p(z)$ has a pole or order $\ge 2$ or $q(z)$ has a pole of order $\ge 3$ then at least one of $S'$ satisfies (\ref{eq-3.4.8}).

Then we can find a complete expansion of $S'$ satisfying (\ref{eq-3.4.7}) by the methods which are rather perturbational.

Example 1. Consider equation \begin{equation} z^3u''-u=0. \label{eq-3.4.11} \end{equation} Then (\ref{eq-3.4.7}) becomes \begin{equation} {S'}^2 +S'' -z^{-3}=0 \label{eq-3.4.12} \end{equation} Then (\ref{eq-3.4.10}) gives $S'(z)= \sigma z^{-\frac{3}{2}}$ with $\sigma=\pm 1$. So, let \begin{equation} S'_0(z):=\sigma z^{-\frac{3}{2}}. \label{eq-3.4.13} \end{equation} If we plug $S'_0(z)$ into (\ref{eq-3.4.7}) we will get an error $S''_0=-\frac{3}{2}\sigma z^{-\frac{5}{2}}$. Let us set $S'(z)=S'_0(z)+ s(z)$ where $s(z)$ is a perturbation. Then the main term in (\ref{eq-3.4.12}) will be $2S'_0 s' +-\frac{3}{2}\sigma z^{-\frac{5}{2}} $ and to make it $0$ we need to take $s(z)= -\frac{S''_0}{2S'_0}= \frac{3}{4}z^{-1}$. So, let \begin{equation} S'_1(z):=\sigma z^{-\frac{3}{2}}+\frac{3}{4}z^{-1}. \label{eq-3.4.14} \end{equation} If we plug $S'_1(z)$ into (\ref{eq-3.4.7}) we will get an error $s^2 + s'=-\frac{9}{16}z^{-2}$.

Let us set $S'(z)=S'_1(z)+ s(z)$ (we use the same notation $s(z)$ for a new function as we do not need an old function anymore) the main term in (\ref{eq-3.4.12}) will be $2S'_0 s' -\frac{9}{16}z^{-2} $ and to make it $0$ we need to take $s(z)= -\frac{9}{32}\sigma z^{-\frac{1}{2}}$.

Continuing this process we arrive to the asymptotic expansion \begin{equation*} S'(z)\sim \sigma z^{-\frac{3}{2}}+\frac{3}{4}z^{-1}+ \sum_{n\ge 0} b_n z^{(n-1)/2} \end{equation*} and therefore \begin{equation*} S(z)\sim -2\sigma z^{-\frac{1}{2}}+\frac{3}{4}\log z + \ln C \sum_{n\ge 0} c_n z^{(n+1)/2} \end{equation*} and therefore \begin{equation} u(z)\sim C z^{3/4}e^{-2\sigma z^{-\frac{1}{2}}}. \label{eq-3.4.15} \end{equation}

One can investigate behaviour of $u(x)$ for real $x\to \pm 0$.


We claim that in the general case at least one solution is "bad": not of the type derived in Section 3.3. Indeed, if $u_1(z), u_2(z),\ldots,u_n(z)$ is a fundamental system of solutions then equation is \begin{equation} \left| \begin{matrix} u &u' &\dots &u^{(n)} &u^{(n)}\\ u_1 &u'_1 &\dots &u^{(n-1)}_1 &u^{(n)}_1\\ \vdots &\vdots &\ddots &\vdots &\vdots\\ u_n &u'_n &\dots &u^{(n)}_n &u^{(n)}_n \end{matrix}\right|=0 \label{eq-3.4.16} \end{equation} and if all solutions were of the type derived in Section 3.3 then as one can prove easily equation would have either an ordinary or regular singular point.

$\Leftarrow$  $\Uparrow$  $\Rightarrow$