M3M6: Methods of Mathematical Physics¶

$$\def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\E{{\rm e}} \def\dx{\D x} \def\dt{\D t} \def\dz{\D z} \def\ds{\D s} \def\C{{\mathbb C}} \def\R{{\mathbb R}} \def\H{{\mathbb H}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\FF{{\cal F}} \def\I{{\rm i}} \def\Ei{{\rm Ei}\,} \def\qqqquad{\qquad\qquad} \def\qqand{\qquad\hbox{and}\qquad} \def\qqfor{\qquad\hbox{for}\qquad} \def\qqwhere{\qquad\hbox{where}\qquad} \def\Res_#1{\underset{#1}{\rm Res}}\, \def\sech{{\rm sech}\,} \def\acos{\,{\rm acos}\,} \def\erfc{\,{\rm erfc}\,} \def\vc#1{{\mathbf #1}} \def\ip<#1,#2>{\left\langle#1,#2\right\rangle} \def\br[#1]{\left[#1\right]} \def\norm#1{\left\|#1\right\|} \def\half{{1 \over 2}} \def\fL{f_{\rm L}} \def\fR{f_{\rm R}} \def\HF(#1,#2,#3,#4){{}_2F_1\!\left({#1,#2 \atop #3}; #4 \right)} \def\questionequals{= \!\!\!\!\!\!{\scriptstyle ? \atop }\,\,\,}$$

Dr Sheehan Olver

s.olver@imperial.ac.uk

Office Hours: 3-4pm Mondays, 11-12am Thursdays, Huxley 6M40

Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 27: Singular points of ordinary differential equations¶

2. Singular points of ODEs
   • regular singular point
   • irregular singular point
   • singular points at ∞

Singular points¶

We will consider the first order ODE

$$u' + a(z) u = 0$$ and the second order ODE $$u'' + a(z) u' + b(z) u = 0$$ where $a$ and $b$ are rational.

The simplest point is an ordinary point, where we know that $u$ is analytic:

Definition (ordinary point) $z_0$ is a ordinary point of the ODE if $a(z)$ and $b(z)$ are analytic at $z_0$.

Regular singular point¶

Suppose $a$ has a simple pole, and w.l.o.g. take $z_0=0$, that is:

$$a(z) = {a_{-1} \over z} + a_0 + a_1 z + a_2 z^2 + \cdots$$ where $a_0 + a_1 z + a_2 z^2 + \cdots$ is analytic. Take as an ansatz to the first order ODE $$u(z) = z^\alpha ( 1 + z v(z)) = z^\alpha + z^{\alpha+1} v$$ where $\alpha = a_{-1}$. Note that $c u(z)$ is a solution of $u'(z) = a(z) u(z)$: the choice of $1$ above makes the solution unique.

Note that

$$u'(z) = \alpha z^{\alpha-1} + z^\alpha ((\alpha+1) v + z v')$$ The ODE $u'(z) = a(z) u(z)$ becomes an inhomogenous equation: $$\begin{align*} z^\alpha ((\alpha+1) v + z v' - z a(z) v) &= z^{\alpha} (a(z) - {\alpha \over z} ) & \Leftrightarrow\\ z v'(z) + (\alpha+1 - z a(z)) v(z) &= a_0 + a_1 z + a_2 z^2 + \cdots \end{align*}$$ In operator form, expressing $$v(z) = v_0 + v_1z + v_2 z^2 = (1,z,z^2,\ldots)\begin{pmatrix}v_0\\v_1\\\vdots \end{pmatrix}$$ we have $$\begin{align*} z v' &=(1,z,z^2,\ldots) \begin{pmatrix} 0 \cr & 1 \cr && 2 \cr &&& 3 \cr &&&&\ddots \end{pmatrix} \begin{pmatrix}v_0\\v_1\\\vdots \end{pmatrix} \\ (\alpha+1 - z a(z)) v(z) &= (1,z,z^2,\ldots) \begin{pmatrix} 1 \cr -a_0 & 1 \cr -a_1 & -a_0& 1 \cr -a_2 & -a_1&-a_0& 1 \cr \vdots&\ddots&\ddots&\ddots&\ddots \end{pmatrix} \begin{pmatrix}v_0\\v_1\\\vdots \end{pmatrix} \end{align*}$$ Thus the equation for our unknown coefficients becomes: $$\begin{pmatrix} 1 \cr -a_0 & 2 \cr -a_1 & -a_0& 3 \cr -a_2 & -a_1&-a_0& 4 \cr \vdots&\ddots&\ddots&\ddots&\ddots \end{pmatrix} \begin{pmatrix}v_0\\v_1\\\vdots \end{pmatrix} = \begin{pmatrix}a_0\\a_1\\\vdots \end{pmatrix}$$ This equation induces decay in $|v_k|$ just like last lecture, hence we know its analytic with the same radius of convergence as $a$.

Here is an example: as we can see, v(z) is analytic in a disk of radius r where r is dictated by the function a.

Here is a plot of u, which shows that it has a branch cut:

We can confirm it solves the ODE:

Definition (first order regular singular point) $z_0$ is a regular singular point of the first order ODE if $(z-z_0)a(z)$ is analytic at $z_0$.

For second order equations, we will arrive at the following analogue:

Definition (second order regular singular point) $z_0$ is a regular singular point of the second order ODE if $(z-z_0) a(z)$ and $(z-z_0)^2b(z)$ are analytic at $z_0$.

Assume

$$a(z) = {a_{-1} \over z} + a_0 + a_1 z + \cdots$$ and $$b(z) = {b_{-2} \over z^2} + {b_{-1} \over z} + b_0 + b_1 z + \cdots$$ To determine the right ansatz, consider $$u'' + {a_{-1} \over z} u' + {b_{-2} \over z^2} u'' = 0$$ which has solutions $z^\alpha$ solving the indicial equation:

Definition (indicial equation)

$$\alpha(\alpha-1) + a_{-1} \alpha + b_{-2} = 0$$

Now take as an ansatz:

$$u(z) = z^\alpha (1 + z v(z)) = z^\alpha + z^{\alpha+1} v(z)$$ Differentiating we have $$\begin{align*} u'(z) &= \alpha z^{\alpha-1} + z^\alpha((\alpha+1) v + z v') \\ u''(z) &= \alpha(\alpha-1) z^{\alpha-2} + z^{\alpha-1}( \alpha(\alpha+1) v + \alpha z v' + (\alpha+1) z v' + z v' + z^2 v'' ) \\ &= \alpha(\alpha-1) z^{\alpha-2} + z^{\alpha-1}( \alpha(\alpha+1) v + 2(\alpha+1) z v' + z^2 v'' ) \end{align*}$$ Thus $v$ solves $$z^2 v''+ z (2(\alpha+1) + z a(z)) v' + ( \alpha(\alpha+1) + z a(z) (\alpha+1) + z^2 b(z)) v = - {\alpha(\alpha-1) \over z} - a(z) \alpha -z b(z) = \underbrace{- \alpha a_0 - b_{-1} - (\alpha a_1 + b_0) z - (\alpha a_2 + b_1) z^2 + \cdots}_{f(z) = f_0 + f_1 z + \cdots}$$

The question is when can forward substitution proceed. Let's write out the operators, first we have:

$$z^2 v'' \equiv \begin{pmatrix} 0 \cr & 0 \cr &&2 \cr &&& 6 \cr &&&&12 \cr &&&&&\ddots \end{pmatrix}$$ Then we have $$2 (\alpha+1) z v' \equiv \begin{pmatrix} 0\cr & 2 (\alpha+1) \cr &&&4 (\alpha+1) \cr &&&& 6(\alpha+1) \cr &&&&&8 (\alpha+1) \cr &&&&&&\ddots \end{pmatrix}$$ and $$z^2 a(z) v' = (z a(z))(z v') \equiv \begin{pmatrix} a_{-1} \\ a_0 & a_{-1} \\ a_1 & a_0 & a_{-1} \\ a_2 & a_1 & a_0 & a_{-1} \\ \vdots &\ddots&\ddots&\ddots&\ddots \end{pmatrix} \begin{pmatrix} 0 \cr & 1 \cr &&2 \cr &&& 3 \cr &&&&4 \cr &&&&&\ddots \end{pmatrix}$$ Finally since $$\begin{align*} \alpha(\alpha+1) + z (\alpha+1) a(z) + z^2 b(z) &= \alpha(\alpha-1) + \alpha a_{-1} + b_{-2} + 2 \alpha + a_{-1} + ((\alpha+1) a_0 + b_{-1}) z + ((\alpha+2) a_1 + b_0) z^2 + \cdots \\ & = 2 \alpha + a_{-1} + ((\alpha+1) a_0 + b_{-1}) z + ((\alpha+2) a_1 + b_0) z^2 + \cdots \end{align*}$$ we have $$(\alpha(\alpha+1) + z (\alpha+1) a(z) + z^2 b(z) ) v \equiv \alpha \begin{pmatrix} 2 \alpha + a_{-1} \\ (\alpha+1) a_0 + b_{-1} & 2 \alpha + a_{-1} \\ (\alpha+1) a_1 + b_0 & (\alpha+1) a_0 + b_{-1} & 2 \alpha + a_{-1} \\ (\alpha+1) a_2 + b_1 & (\alpha+1) a_1 + b_0 & (\alpha+1) a_0 + b_{-1} & 2 \alpha + a_{-1} \\ \vdots &\ddots&\ddots&\ddots&\ddots \end{pmatrix}$$

We need to only worry about the diagonal to determine if forward substitution proceeds, in this case we have

$$\begin{pmatrix} 2 \alpha + a_{-1} \\ \times & 4 \alpha + 2 + 2a_{-1} \\ \times & \times &6 \alpha + 3a_{-1} + 6 \\ \times & \times & \times & 8 \alpha + 4a_{-1} + 12 \\ \times & \times & \times & \times & 10 \alpha + 5a_{-1} + 20\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots \end{pmatrix}$$ Thus the first condition for this to give us a solution is that $2 (k+1)\alpha + (k+1)a_{-1} + k(k+1) \neq 0$ for all $k=1,2,\ldots$. This can be thought of another way: note that plugging $\alpha+k+1$ in to the indicial equation gives us: $$(\alpha+k+1) (\alpha+k) + a_{-1} (\alpha+k+1) + b_{-2} = (\alpha) (\alpha-1) + a_{-1} \alpha + b_{-2}+ 2 (k+1)\alpha + (k+1)a_{-1} + k(k+1) = 2 (k+1)\alpha + (k+1)a_{-1} + k(k+1)$$ Thus forward substitution fails if and only $\alpha+k+1$ is the second root to the indicial equation for some $k=0,1,2,\ldots$.

Remark If this condition is satisfied, then $v$ is analytic near the singular point, which follows by bounding the forward elimination as before, but we'll skip this for now.

Remark If $\alpha+k+1$ is an integer, then using the ansatz $z^{\alpha+k+1}(1 + z v(z))$ proceeds without problem. To get the second solution, we need to introduce logarithms, but we'll skip this for now.

We now check the results of solving

$$z^2 v''+ z (2(\alpha+1) + z a(z)) v' + ( \alpha(\alpha+1) + z a(z) (\alpha+1) + z^2 b(z)) v = - {\alpha(\alpha-1) \over z} - a(z) \alpha -z b(z)$$ numerically.

We first choose a solutino to the an indicial equation:

We can solve for v, which is analytic in a neighbourhood of 0:

Thus one solution is:

Let's double check it satisfies the original ODE:

The second, linearly independent solution can be found taking the other $\alpha$:

Any general solution to the ODE is a linear combination of these two.

Irregular singular points¶

Every isolated singularity that is not regular is an irregular singular point: even something as simple as ${\D u / \dz} = {u \over z^2}$ has an irregular singular point. Irregular singular points correspond to essential singularities: in this case the solution is

$$u(z) = \E^{-1/z}$$

Singular points at infinity¶

To determine the type of singularity at $\infty$ we do our usual change of variables $u(z) = v(1/z)$. For example, if

$${\D u \over \dz} = a(z) u$$ Then $${\D v \over \dz} = -z^{-2}a(z^{-1}) v$$ Definition (first order ordinary point at ∞) $\infty$ is an ordinary point of a first order ODE if $z^2a(z)$ is analytic at $\infty$.

Definition (first order regular singular point at ∞) $\infty$ is a regular singular point of a first order ODE if $za(z)$ is analytic at $\infty$.

For second order, we see that $v$ satisfies

$$\begin{align*} v'(z) &= -z^{-2} u'(z^{-1}) \\ v''(z) &=2 z^{-3} u'(z^{-1}) + z^{-4} u''(z^{-1}) =-2 z^{-1} v'(z)+ z^{-4} u''(z^{-1}) \end{align*}$$ which gives us the ODE $$u'' + a(z) u' + b(z) u \Rightarrow z^4v'' + (2 z^3 - a z^2) v' + b v$$

Definition (second order regular singular point at ∞) $\infty$ is a regular singular point of a first order ODE if $za(z)$ and $z^2 b(z)$ are analytic at $\infty$.