as $\varepsilon =10$ (blue), $\varepsilon=50$ (red)
$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$
Consider now the same problem as before but with perturbation $M$ being of higher order than $L$. Then even for $\varepsilon \ll 1$ perturbation is not really small. In particular, for perturbed operator we need to have more boundary conditions than for unperturbed one and the different scenarios are possible. We consider only one scenario here when $L$ and $M$ are both positive operators: $Lu=\alpha(x)u$, $Mu=-(\beta (x)u')'$ with positive smooth $\alpha, \beta$: \begin{align} &L_\varepsilon u:=\alpha u -\varepsilon^2 (\beta (x)u')'=f\qquad 0\le x\le a, \label{eq-4.2.1}\\ &u(0)=g_1, \label{eq-4.2.2}\\ &u(a)=g_2. \label{eq-4.2.3} \end{align}
Remark 1.
Operators are positive if the corresponding quadratic form is positive definite: $Lu:=\int_0^a \alpha |u|^2\,dx>0$ and $Mu:= \int_0^a \beta |u'|^2\,dx>0 $ as $u(0)=u(a)=0$ and $u\ne 0$. Therefore \begin{gather} (L_\varepsilon u,u)= (Lu,u)+\varepsilon^2 (Mu,u) \label{eq-4.2.4} \end{gather} where $(.,.)$ and $\|.\|$ an inner product and norma in $L^2([0,a])$.
In particular \begin{gather} (L_\varepsilon u,u)\asymp \|u\|^2 +\varepsilon^2\|u'\|^2 . \label{eq-4.2.5} \end{gather} Therefore $f=0$, $u_\varepsilon(0)=u_\varepsilon (a)=0$ imply that $u_\varepsilon=0$.
On the other hand, if either $\alpha >0$ and $\beta$ is not non-negative or $\beta >0$ and $\alpha$ is not non-negative, there exist $\varepsilon _n\to +0$ and $u_n\ne 0$ such that $L_{\varepsilon_n}u_n=0$, $u_n(0)=u_n(a)=0$.
So we assume \begin{equation} \alpha(x) >0, \qquad \beta (x) >0. \label{eq-4.2.6} \end{equation} This assumption guarantees that problem (\ref{eq-4.2.1})--(\ref{eq-4.2.3}) is well-posed.
Then boundary conditions (\ref{eq-4.2.2}) and (\ref{eq-4.2.3}) are important for $L_\varepsilon$ with $\varepsilon\ne 0$ but should be ignored for $L_0$.
Observe that we can construct by methods of the previous Section 4.1 \begin{equation} U_\varepsilon \sim \sum_{n\ge 0} U_n \varepsilon^n \label{eq-4.2.7} \end{equation} satisfying (\ref{eq-4.2.1}): \begin{align} &U_0=\alpha^{-1}f, \label{eq-4.2.8}\\ &U_{2m+2} = (\beta (\alpha^{-1}U_{2m})')' \label{eq-4.2.9}\\ &U_{2m+1}=0. \label{eq-4.2.10} \end{align} Then plugging $u= U_\varepsilon +v$ we get the same problem with $\ f=O(\varepsilon^\infty)$. Here $g_1$ and $g_2$ will depend on $\varepsilon$ but it does not matter much.
Remark 2.
If $f=O(\varepsilon^\infty)$ then Remark 1 implies that solution $w$ of the problem $L_\varepsilon w=f$, $w(0)=w(a)=0$ is $O(\varepsilon^\infty)$. Then plugging $u_\varepsilon:=U_\varepsilon+w$ we arrive to the same problem with $\ f=0$ exactly.
If $f=0$ then $u$ can reach positive maximum or negative minimum only on the ends of the interval $[a,b]$. Indeed, if $u$ reaches positive maximum in
$c:0< c < a$, then $u(c) > 0$, $u'(c)=0$, $u''(c) \le 0$ and
$L_\varepsilon u>0$. So,
\begin{gather}
\min (g_1,g_2)\le u_\varepsilon \le \max (g_1,g_2).
\label{eq-4.2.11}
\end{gather}
Let us start from toy-model.
Example 1. Let $\alpha,\beta$ be constant. Then solution to (\ref{eq-4.2.1}) with $f=0$ is $u= C_1 e^{-\varepsilon^{-1}\sigma xn} + C_2 e^{-\varepsilon^{-1}\sigma (x-a)}$ with $\sigma= (\alpha/\beta)^{-\frac{1}{2}}$ and plugging into (\ref{eq-4.2.2}) and (\ref{eq-4.2.3}) we can find $C_i=g_i + O(e^{-\varepsilon^{-1}\sigma a})$ and \begin{equation*} u= g_1 e^{-\varepsilon^{-1}\sigma x} + g_2e^{-\varepsilon^{-1}\sigma (x-a)}. \end{equation*}
|
| plot solutions $-\varepsilon^2u'' +u =1$, $u(0)=u(1)=0$ as $\varepsilon =10$ (blue), $\varepsilon=50$ (red) |
We see that the ends should be treated separately and $u_\varepsilon $ is negligible outside of the boundary layers $\{x \ll \varepsilon^{1-\delta} \}$ and $\{a-x\ll \varepsilon^{1-\delta}\}$ with arbitrarily small $\delta>0$.
So we concentrate on the left end $x=0$ and we look at the solution in the form \begin{equation} u_\varepsilon = w e^{-\varepsilon^{-1}\phi(x)}. \label{eq-4.2.12} \end{equation} Then plugging into (\ref{eq-4.2.1}) and considering first only terms with $\varepsilon^0$ we get equation $\beta(\phi')^2 =\alpha$ and then \begin{equation} \phi(x) = \int_0^x (\alpha/\beta)^{\frac{1}{2}}\,dx. \label{eq-4.2.13} \end{equation} We take here $\phi(0)=0$ and $\phi(x)\sim \sigma x$ as $x>0$ with \begin{equation} \sigma=(\alpha(0)/\beta(0))^{\frac{1}{2}}. \label{eq-4.2.14} \end{equation} After $\phi(x)$ is fixed we consider all terms we get \begin{equation} \beta\varepsilon\Bigl( Pw+ \varepsilon Qw)\Bigr)e^{-\phi\varepsilon^{-1}}=0 \label{eq-4.2.15} \end{equation} with \begin{align} &Pw:=2\phi'w'+ (\phi''-\phi'\beta'\beta^{-1})w, \label{eq-4.2.16}\\ &Qw:=-w''+\beta'\beta^{-1}w'. \label{eq-4.2.17} \end{align} We can find asymptotic solution to this equation satisfying boundary condition \begin{equation} w|_{x=0}= g_1\sim \sum_{n\ge 0} g_{1n}\varepsilon ^n \label{eq-4.2.18} \end{equation} in the form \begin{equation} w\sim \sum_{n\ge 0} w_{n}\varepsilon ^n \label{eq-4.2.19} \end{equation} solving \begin{align} &Pw_n + Qw_{n-1}=0, \label{eq-4.2.20}\\ &w_n(0)=g_{1n}. \label{eq-4.2.21} \end{align}
Remark 3. We see that asymptotics inside of the segment should match asymptotics near its ends where this inner part is overlapping with the boundary layer. So we have a method of matching asymptotics. In this method traditionally the boundary layer is called inner zone while regular part of the domain is called outer zone, which is really counter-intuitive.
One can consider different boundary condition, f.e. \begin{equation} \bigl(A \varepsilon u'+(B+C\varepsilon) u\bigr)|_{x=0}=0 \label{eq-4.2.22} \end{equation} Then (\ref{eq-4.2.21}) should be replaced by \begin{equation} (-A\sigma +B)w_n(0)= g_{1n}- Aw'_{n-1}(0)-Cw_{n-1}(0). \label{eq-4.2.23} \end{equation} Here $(-A\sigma +B)\ne 0$ is required but if it is not the case the trouble is much deeper than in our construction.