Perturbation Theory for Linear ODEs. 1

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

Chapter 4. Perturbation Theory for Linear ODEs

Regular perturbations

  1. Introduction
  2. Small parameter in the interval
  3. Perturbation of eigenvalues


In this Chapter we consider initial value problems (IVPs) and boundary value problems (BVPs) for linear ODEs: \begin{align} &L_\varepsilon u:=(L +\varepsilon M)u =f(x) && 0\le x \le a, \label{eq-4.1.1}\\ &B_{1k\varepsilon}u:=(B_{1k}+\varepsilon C_{1k})u|_{x=0}=g_{1k} && k=1,\ldots, K_1, \label{eq-4.1.2}\\ &B_{2k\varepsilon}u:=(B_{2k}+\varepsilon C_{2k})u|_{x=a}=g_{2k} && k=1,\ldots, K_2 \label{eq-4.1.3} \end{align} where $L$, $M$ are scalar differential operators of order $\le m$, $B_{jk},C_{jk}$ are operators of order $\le m_{jk}$, $m_{j1}<\ldots < m_{jK_j}< m$, $K_1+K_2=m$, $\varepsilon\ll 1$ is a small parameter. The basic (regular) theory means that $L$, $B_1$ and $C_1$ are "proper" operator of order $m$, $m_1$ and $m_2$ respectively and the unperturbed problem \begin{align} &Lu =f(x) && 0\le x \le a, \label{eq-4.1.4}\\ B_{1k}u|_{x=0}=g_{1k} && k=1,\ldots, K_1, \label{eq-4.1.5}\\ & B_{2k}u|_{x=a}=g_{2k} && k=1,\ldots, K_2 \label{eq-4.1.6} \end{align} is well-posed (so has a unique solution for any $f,g_1$ and $g_2$).

Then we look for a solution to the perturbed problem (\ref{eq-4.1.1})-(\ref{eq-4.1.3}) in the standard form of the asymptotic series \begin{equation} u\sim \sum_{n\ge 0} u_n \varepsilon^n. \label{eq-4.1.7} \end{equation} Plugging into (\ref{eq-4.1.1})-(\ref{eq-4.1.3}) we conclude that $u_0$ must be a solution of the unperturbed problem (\ref{eq-4.1.4})-(\ref{eq-4.1.6}) while $u_n$ with $n>0$ must satisfy \begin{align} &Lu_n =- Mu_{n-1} && 0\le x \le a, \label{eq-4.1.8}\\ &B_{1k}u_n|_{x=0}=-C_{1k}u_{n-1}|_{x=0}&& k=1,\ldots, K_1, \label{eq-4.1.9}\\ & B_{2k}u_n|_{x=a}=-C_{2k}u_{n-1}|_{x=a}&& k=1,\ldots, K_2. \label{eq-4.1.10} \end{align}

Surely, dependence of operators from $\varepsilon$ could be more complicate, f.e. $L_\varepsilon$, $B_{j\varepsilon}$ could be also in the form of the asymptotic series like (\ref{eq-4.1.7}).

Small parameter in the interval

Also $\varepsilon$ could be not only in the operators but in the interval. F.e. we consider BVP with $a_varepsilon=a+\varepsilon$. While it could be reduced to the previous by introducing a new variable $y=x a/(a+\varepsilon)$ we consider a better approach. Then we still look for a solution in the form (\ref{eq-4.1.7}), and obviously (\ref{eq-4.1.8}) and (\ref{eq-4.1.9}) remain, but what about (\ref{eq-4.1.10})?

Since \begin{equation*} u (a+\varepsilon)\sim \sum_{l\le 0} \frac{1}{l!} u^{(l)} \varepsilon^l \end{equation*} we conclude that \begin{equation} u (a+\varepsilon)\sim \sum_{l\le 0, n\ge 0} \frac{1}{l!} u_n^{(l)} \varepsilon^{l+n} \label{eq-4.1.11} \end{equation} and therefore we get instead of (\ref{eq-4.1.10}) \begin{multline} B_{2k}u_n|_{x=a}=-B_{2k}\sum_{l\ge 1 }\frac{1}{l!} u^{(l)}_{n-l}|_{x=a}- C_{2k}\sum_{l\ge 0 }\frac{1}{l!} u^{(l)}_{n-l-1}|_{x=a} \\ k=1,\ldots, K_2. \label{eq-4.1.12} \end{multline}

Perturbation of eigenvalues

Consider now eigenvalue problem

\begin{align} &L_\varepsilon u:=(L+\varepsilon M)u =\lambda u && 0\le x \le a, \label{eq-4.1.13}\\ &B_1u|_{x=0}=0 && k=1,\ldots, K_1, \label{eq-4.1.14}\\ &B_{2k}u|_{x=a}=0 && k=1,\ldots, K_2 \label{eq-4.1.15} \end{align} assuming that as $\varepsilon=0$ we perturb a simple eigenvalue $\lambda_0$. In this case not only $u=u_\varepsilon$ should be decomposed into (\ref{eq-4.1.7}), but $\lambda_\varepsilon$ as well: \begin{equation} \lambda\sim \sum_{n\ge 0} \lambda_n \varepsilon^n. \label{eq-4.1.16} \end{equation} Then while (\ref{eq-4.1.9}) and (\ref{eq-4.1.10}) (or (\ref{eq-4.1.12})) with $g_{1k}=g_{2k}=0$ remain we need to modify (\ref{eq-4.1.8}) to \begin{equation} (L-\lambda_0)u_n =- Mu_{n-1} +\sum _{l\ge 1}\lambda_l u_{n-l} \qquad 0\le x \le a. \label{eq-4.1.17} \end{equation} Now we have two problems: equation $(L-\lambda_0)v=f$ (with the corresponding boundary conditions) does not have solutions for arbitrary $f$ and this solution if exists is not unique but defined modulo kernel of $(L-\lambda_0)$.

We need to explain what we mean by a simple eigenvalue. In the matrix theory we consider either Hermitian (or similar) matrices or general ones. In the former case simple means that there is just one (up to a factor) corresponding eigenvector. In the latter case we need to be more subtle and assume that the corresponding Jordan cell is also 1-dimensional which means exactly that there exists $v$ such that $\langle v,u_0\rangle=1$ and $(L-\lambda_0)w = g$ has a solution $w$ if and only if $\langle v,g\rangle=0$. In fact $v$ is an eigenfunction of the adjoint matrix $(L^*-\bar{\lambda})v=0$.

For operators we need exactly the same; I do not discuss here what means adjoint operator (with boundary conditions). Here $\langle u,v\rangle =\int_0^a u\bar{v}\,dx$ in $L^2((0,a))$. To ensure uniqueness we just request \begin{equation} \langle u_s, v\rangle =0\qquad s=1,2,\ldots. \label{eq-4.1.18} \end{equation} Then (\ref{eq-4.1.7}) is solvable if and only if \begin{equation*} 0=\langle -Mu_{n-1} +\sum _{l\ge 1}\lambda_l u_{n-l}, v \rangle= -\langle M u_{n-1} , v \rangle + \lambda_n \end{equation*} where we used (\ref{eq-4.1.8}). So $\lambda_n$ and $u_n$ are defined in the following recurrent way:

Let $\lambda_s, u_s$ be defined for $s< n$. Then \begin{equation} \lambda_n = \langle M u_{n-1} , v \rangle \label{eq-4.1.19} \end{equation} and $u_n$ is defined from (\ref{eq-4.1.17}) with the corresponding $0$ boundary conditions and (\ref{eq-4.1.18}).

We can develop the theory for perturbations also in the boundary conditions and in the interval itself but we need then to explain more precisely about adjoint BVP and how to describe solvability with inhomogeneous boundary conditions .