text:
\begin{document}

\begin{abstract}
We study the behavior of the limit of the spectrum of a non self-adjoint
 Sturm-Liouville operator with  analytic potential as
 the semi-classical parameter $h\to 0$. We get a good description of the spectrum and  limit spectrum near $\infty$. We also study the action of one  special perturbation of the operator (adding a Heaviside function), and prove
that the limit spectrum is very unstable.
 As an illustration we describe the limit spectrum as  $h\to 0$ for $P^h=-h^2\Delta+i x^2$ and the effect of this perturbation.
\end{abstract}
\maketitle

\section{Introduction}

This paper is devoted to  non self-adjoint Sturm-Liouville problems. We study  the spectrum  of the   1-dimensional, semiclassical Schr\"odinger operator on $L^2([-1,1])$ with Dirichlet boundary condition, given by
\begin{equation}
H^h=-h^2 \frac d {dx^2} +V(x).
\label{Hi}
\end{equation}
The potential $V$ is a complex valued function  on $[-1,1]$, which extends holomorphically to some domain in $\C$.
The boundary value at $\pm 1$ play no special rule but are fixed to avoid more notation.

The study of such operators is motivated  by  the Orr-Som\-merfeld equation
with linear profile \cite{sh} or by the non linear Zakharov-Shabat eigenvalue problem, cf.  work of Miller \cite{mi}.

As an application  we will focus on the case where
\begin{equation}
V(x)=i x^2,
\label{Vquad}
\end{equation}
and shall write $P^h$ for the corresponding operator. From this one could also
study the slightly  more general case
$-h^2 \frac d {dx^2} +e^d x^2$, $d\in\cit$ using a change of variable.
The spectrum of this operator on $\rit$  was analyzed by Davies \cite{davies},
cf. also the recent work of Hitrik \cite{hit}.

It is well known that the spectrum of a non self-adjoint operator is unstable
under perturbation of the operator. This motivates the introduction of the
pseudo-spectrum, which has now been studied by many people, particularly
Trefethen (who maintains the web archive http:web.comlab.ox.ac.uk/projects/pseudospectra)
and Davies \cite{dav},\cite{dave}; we note also the recent paper of
Denker, Sj\"ostrand and  Zworski \cite{speudo}.

We  also consider the following perturbation of $H$:

For $\beta\in (-1,1)$ and $\delta\geq 0$, let $H_{\delta,\beta}$
\begin{equation}
H^h_{\delta,\beta}=-h^2 \frac d {dx^2} +V_{\delta,\beta}(x), \quad V_{\delta,\beta}(x)=
\left\{
\begin{array}{l}
V(x)+i\delta,\quad x>\beta\\
V(x)-i\delta,\quad x<\beta
\end{array}
\right.,
\label{Hdelta}
\end{equation}
with domain
\begin{equation}
\CD_{\delta,\beta}=\{u\in L^2([-1,1]), u''\in L^2([-1,1]), u(-1)=u(1)=0, \}.
\label{domain}
\end{equation}
Notice that $H^h_{0,\beta}=H^h$.

Redparth \cite{red} has obtained similar results for arbitrary piecewise
linear complex-valued potentials, a class which includes perturbations as above
of the operator $-h^2\Delta+ix$.

We  focus on the semi classical limit, and denote by
$\lim_h\Sp(H^h_\delta)$ the set of all values of $E$ which arise as limits  as $h\to 0$ of some sequence of eigenvalues  of $ H^h_\delta$.

We give  a geometric criterion for $E$ not to  belong to  $\lim_h\Sp(H^h_\delta)$:   namely
$\lim_h\Sp(H^h_\delta)\subset \CT_\delta^c,\quad \CT_\delta^c=\cit\subset\CT$ where $\CT_{\delta}$ is the set of $E\in\C$ such that there exists
a progressive path from  $-1$ to $1$ with respect to $V_{\delta,\beta}$. ( We
define this concept later). This is done in section \ref{semi},  see Theorems \ref{inclus}.
 This criterion is not sharp, and also not so easy to use.
It can be proved using either  ellipticity  or  exact  WKB .

As a second step we give an alternate criterion for $E$ to belong to
\par\noindent $\lim_h\Sp(H^h_\delta)$, when $E$ is large enough, which is sharp and
computable, see Theorem \ref{reci} .

As an application we prove that the perturbation $\delta$ changes the spectrum
near  $\infty$ quite drastically, see Theorem \ref{curveper}.
The pseudo spectrum is defined to be the limit as $\delta\to 0$ of the union of spectrum over all pertubations of $H^h$ of size $\delta$, and we show that  this perturbation is sufficient to reach
the entire pseudo spectrum of $H^h$, see section \ref{speudospec}.
We note also the work  in progress of M. Hager that studies the effect of different pertubations on the spectrum.

We use the geometric criterion for  $P^h$ to  characterize the full set
$\CT^c$, see Theorem \ref{forme}, using special properties of the potential,
and see that $\CT^c$ forms a $Y$ shape. After this article was completed, we
learned that some of our results and methods are similar to those of Shkalikov
\cite{sh}, \cite{sha}.
The algorithm proposed by Miller in \cite{mi} to study the spectrum is
the same as the one we use here, though he applies it to a different operator.

We also obtain a description of the large spectrum of $H^h$ which is near to
\par\noindent $\lim_h\Sp(P^h)$ see Theorem \ref{quanti}.
In the  self-adjoint case; this can be found in the book of Marchenko \cite{ma} or the paper of Kappeler and M\"ohr  \cite{KM}, and in  some special  non self-adjoint cases in work of Carlson, Threadgill and Shubin \cite{ca}.
However, these results apply for  potentials $V$ with  $V'\notin L^1_{loc}$,
hence do not pertain to  $V_{\delta,\beta}$.

Even though   it is not self-adjoint, the operator $H^h_{\delta,\beta}$ has discrete spectrum. Indeed, $V_{\delta,\beta}$ is  compact relative to $\Delta$, and from Weyl's theorem it follows that  its essential  spectrum is empty. In particular $H^h_{\delta,\beta}$ has no residual spectrum.

\section{Generals tools}

In this section we are interested in the Dirichlet  eigenvalues  of the operator $H^h$ on $L^2([-1,1])$ defined by
\begin{equation}
\left\{
\begin{array}{ll}
H^hu=Eu,\\
u(-1)=u(1)=0.
\end{array}
\right .
\label{schr}
\end{equation}

\begin{definition}\sl A point $x_0$ in the complex plane is a turning point of order $k$ for  the operator $H^h-E$ if $V(x)-E$ vanishes to order $k$ at $x=x_0$.
\label{turningpoint}
\end{definition}

\begin{definition}\sl
Let $x$ be a point in the complex plane. The Stokes line starting from $x$ is the set
$$
\CL(x)=\{ y\in \C, \re\int_x^y (V(t)-E)^{1/2} \,dt=0\}.
$$
\label{defLS}
\end{definition}
This is independant of the branch of the square root.
Denote by $S_{x,y}(E)=\int_x^y (V(t)-E)^{1/2} \,dt$.

Notice that the Stokes lines are integral curves of the Stokes vector field \begin{equation}
s(x)=i\;\overline{(V(x)-E)^{1/2}}.
\label{stokesfield}
\end{equation}

The local structure of Stokes lines can be easily investigated (see
e.g. \cite{fe}, \cite{fr}).  Away from  turning points or singularities
of $V$,  the Stokes lines form a non singular analytic foliation.
 Indeed, if $\Omega\subset \C$ does not contain any singularities of $(V-E)^{1/2}$,  then $x\mapsto z(x,E)=\int^x_{x_0}(V(t)-E)^{1/2}\,dt$ is an analytic diffeomorphism.
Notice also that from a simple turning point $x_0$ emanates  three Stokes lines,  each making an angle of $2\pi/3$ with any other at $x_0$.

Using ideas from quantum  resonance theory, we shall consider as in \cite{fr}  some distorted operators associated to $H$.
Let $\gamma:[-1,1]\ni t \to \C$ be a smooth simple path in the complex plane $\gamma(-1)=-1$ and $\gamma (1)=1$.
Since $V$ is analytic, we can define an operator $H^{\gamma,h}$ on
$L^2([-1,1])$ as the restriction of  $H^h$  to $\gamma([-1,1])$. One computes that
$$H^{\gamma,h}= -h^2 ( \frac 1 {\gamma'(x)} \frac d  {dx}) ( \frac 1 {\gamma'(x)} \frac d  {dx}) +V(\gamma(x)).    $$

The correspondence between the Stokes geometry and spectral properties of $H^h$ is given in the following result  from \cite{fr}.

\begin{proposition}\sl
The operator $H^{\gamma,h} -E$ is elliptic if and only if $\gamma$ is transverse to the Stokes lines.
\label{ellipticite}
\end{proposition}

{\sl Proof: } The principal symbol of  $H^{\gamma,h} -E$ is
$$
h^\gamma(t,\tau)=\frac 1{(\gamma'(t))^2}(\tau^2+(\gamma'(t))^2(V(\gamma(t))-E)).
$$
We set  $(V(\gamma(t))-E)^{\frac 1 2}= r e^{i\theta}$ and $\gamma'(t)=e^{i\theta'}$  (one  can always suppose that $\vert\gamma'(t)\vert=1$).
Then the path $\gamma$ is transversal to the Stokes lines if and only if
$$
\det (\gamma'(t), s(\gamma(t)))\neq 0,
$$
where $s(x)$ is the Stokes field  defined in (\ref{stokesfield}). This condition is the same as
$$
\re \gamma'(t)(V(\gamma(t)-E)^{1/2}\neq 0,
$$
or finally
$$
\cos(\theta+\theta')\neq 0.
$$
Therefore, the transversality condition is equivalent to
$\theta+\theta'\not\equiv \pi/2  \mod [\pi]$.
Since $\im ({\gamma'(t)}^2 h^{\gamma})=r^2\sin  2(\theta+\theta')$  and $\re( {\gamma'(t)}^2  h^{\gamma})=\tau^2+r^2\cos 2(\theta+\theta')$,  the transversality condition  is equivalent to the invertibility of $h^\gamma$, as stated in the proposition. \cqfd

\medskip

We deduce the following proposition, but  this will be improved in Theorem
\ref{mainbis} below, where we control the dependence on $E$ and introduce the perturbation.
\begin{proposition}\sl
For $h$ small enough, $E\notin \Sp(H^h)$  if there exists  $\gamma$  form $-1$ to $1$ which is transverse to the Stokes lines.
\label{inver}
\end{proposition}
{\sl Proof: }
General properties of  distorted analytic operators imply that  $E\in \Sp(H^h)$ is equivalent to $E\in \Sp(H^{\gamma,h})$.
Let $\gamma$ be a path  transverse to the Stokes lines. Then by Proposition \ref{ellipticite},
$H^{\gamma,h} -E$ is elliptic and invertible for sufficiently small $h$ (depending on $E$).
\cqfd

\section{The semi-classical limit of the spectrum }
\label{semi}
The section is devoted to the proof of the  Theorem \ref{mainbis} and a direct application Theorem \ref{inclus}.
The next sections \ref{large} and \ref{largeper}  are devoted to the proof of
a converse
of Theorem \ref{inclus}.

We now introduce  some notation.
Denote by $\lim_h \Sp(H^h)$ the set of all $E_0$ such that
there exists a sequence $h_j\to 0$  with  $E_{h_j}\in\Sp(H^{h_j})$ and
$E_{h_j}\to E_0$. For $a\in\cit$,
we denote by $z_{0}(x)$ and $z_{l,r}(x)$ the  action integrals
\begin{equation}
\begin{array}{c}
\ds z_{0}(x)=\int_{\gamma(-1,x)} (V(t)-E)^{1/2}\,dt,\;
z_{l}(x)=\int_{\gamma_{l}(\beta,x)} (V(t)-i\delta-E)^{1/2}\,dt,
\\
\ds z_{r}(x)=\int_{\gamma_{r}(\beta,x)} (V(t)+i\delta-E)^{1/2}\,dt,
\end{array}
\label{actions}
\end{equation}
where $\gamma(-1,x)$ and $\gamma_{l,r}(\beta,x)$ are paths from $-1$ or $\beta$ to
$x$ on the Riemann surface associated to the determination of $\Sigma_{0}$ of $t\mapsto (V(t)-E)^{1/2}$ and
$\Sigma_{l,r}$ associated to the determination of $t\mapsto (V(t)\pm i\delta-E)^{1/2}$, $E\in\cit$ respectively.

We denote by $\CT_{\delta}$ the set of $E\in\C$ such that there exists $\gamma_{l}(-1,\beta)$ on which the function $t\mapsto\re(z_l(\gamma_{l}(t) ))$ is
strictly monotone, and  a path $\gamma_{r}(\beta,1)$ on which the function
$t\mapsto\re(z_r(\gamma_{r}(t) ))$ is strictly monotone.
We will call such a path 'progressive' .
For $\delta=0$ we take a different definition: $E\in\CT_0$ if and only if  there exists a path $\gamma$ from $-1$  to
$1$ on which the function $t\mapsto \re(z_0(\gamma(t))))$ is
strictly monotone.

Finally, for $E\in \CT_\delta$,  we denote by
\begin{equation}
d(E,\CT_\delta^c,\gamma)=\inf_{l,r}\inf_{x\in \gamma^{l,r}} |\re \partial_x z^{l,r}(x)| \quad \mbox{ if }  \delta\neq 0
\end{equation}
\begin{equation}
d(E,\CT_0^c,\gamma)=\inf_{x\in \gamma} |\re \partial_x z^0(x)|
\label{dist}
\end{equation}
where $\gamma$  is a progressive path associated to $E$ ($\gamma=\cup_{l,r}\gamma^{l,r}$ if $\delta\neq 0$).

\begin{remark}\sl
\label{remclose}
Notice that $\CT_\delta$ is open. More precisely, if  $c$ is small enough there exists $C>0$,  such that if $E_0$ belongs to $\CT_\delta$  then, for $E$ such that
$|E-E_0|\leq c d(E_0,\CT_\delta^c,\gamma)$, we get $E\in \CT_\delta$   and $d(E,\CT_\delta^c,\gamma)\geq C d(E_0,\CT_\delta^c,\gamma)$.
This follows from the relation
$$
\sqrt{V_{\delta,\beta}-E}=\sqrt{V_{\delta,\beta}-E_0}+O(\frac {E-E_0}{\sqrt{V_{\delta,\beta}-E_0}}),
$$
which shows that a progressive path for $E_0$  is also  a progressive path for $E$.
 \end{remark}

\begin{theorem}
\label{mainbis}\sl
Let $\delta\geq 0$ be  small enough and  $E\in \CT_{\delta}$. Then  $E\notin
  \Sp(H^h_\delta)$ as soon as $h\leq  d(E,\CT^c_{\delta},\gamma)^7$.
  (The reason for the exponent $7$ will emerge at the end of the proof.)
\end{theorem}

From Remark \ref{remclose} and Theorem \ref{mainbis} we obtain the
\begin{theorem}\label{inclus}\sl
If $E_0\in\lim_h\Sp(H^h_\delta)$, then  $E_0\in \CT_\delta^c$.
\end{theorem}

{\sl Proof:}
Suppose that $E_0\in\CT_\delta$ and $\gamma$ is a progressive path associated  to $E_0$. Let $E_h$ be a sequence in the spectrum of $H^h_\delta$ tending to $E_0$. By the remark above, $E_h$ belongs to $\CT_\delta$ as soon as $h$ is small enough, and we get $d(E_h,\CT_\delta^c,\gamma)\geq \tilde C_0$.
Theorem \ref{mainbis} shows that $E_h$ is not in the spectrum  of $H^h_\delta$ as soon as $h$ is small. This contradicts the hypothesis, so $E_0\in \CT_\delta^c$.
\cqfd

\medskip
{\sl Proof of Theorem \ref{mainbis}: }
We denote by $(u^l_+,u^l_-)$ a basis of the space of  solutions of the equation
\begin{equation}
(H^h_\delta-E)u=0,
\label{eqvp}
\end{equation}
in $L^2([-1,\beta ])$, and  by $(u^r_+, u^r_-)$  a basis  of the space of solutions of (\ref{eqvp}) in $L^2([\beta,1])$.
A function $f\in L^2([-1,1])$ is an eigenfunction of $H_\delta$ with eigenvalue $E\in \C$, if and only if
$f(1)=f(-1)=0$, and both $f$ and $\partial_x f$ are continuous at $x=\beta $. Writing
\begin{equation}
f=\alpha^l_+u^l_++\alpha^l_-u^l_-\mbox{ on } [-1,\beta] \quad f=\alpha^r_+u^r_++\alpha^r_-u^r_-\mbox{ on } [\beta,1] ,
\label{solglobale}
\end{equation}
Then $f$ is an eigenfunction of $H^h_\delta$ with eigenvalue $E\in \C$ if and only if
\begin{gather}
\begin{split}
\left(
\begin{array}{cccc}
u^l_-(-1)&u^l_+(-1)&0&0\\
 u^l_-(\beta)&u^l_+(\beta)&-u^r_-(\beta)&-u^r_+(\beta)\\
\partial_x{u^l}_-(\beta)&\partial_x{u^l}_+(\beta)&-\partial_x{u^r}_-(\beta)&-\partial_x{u^r}_+(\beta)\\
0&0&u^r_-(1)&u^r_+(1)\end{array}\right)
\left(\begin{array}{l}\alpha^l_+\\\alpha^l_-\\\alpha^r_+\\
\alpha^r_-\end{array}\right)=0.
\end{split}
\label{defI}
\end{gather}

Therefore $E$ belongs to $\Sp (H^h_\delta)$, the spectrum of $H^h_{\delta}$,  if and only if
\begin{equation}
\det\, \CI(\beta,E)=0,
\label{detI}
\end{equation}
where $\CI(\beta,E)$ is the matrix appearing here.
\begin{remark}
For $\delta=0$, the proof is slightly different, the corresponding matrix is
\begin{gather}
\begin{split}
\CI(\beta,E)=\left(
\begin{array}{cccc}
u_-(-1)&u_+(-1)\\
 u_-(1)&u_+(1)\end{array}\right),
\end{split}
\label{defIdelta}
\end{gather}
We leave details to the reader and in the following, treat only the case
$\delta\neq 0$.
\end{remark}

 In the sequel, we shall compute this determinant for two particular bases $(u^l_+,u^l_-)$ and $(u^r_+, u^r_-)$, namely for complex WKB solutions  as in  \cite{gg}, or \cite{fr}.
 Suppose $E\in T_{\delta}$. Then there is a path
 $\gamma_l=\gamma_{l}(-1,\beta)$ transversal to the Stokes lines $\CL(y)$,
 and we can suppose that $\gamma_{l}(-1,\beta)$ is of type $+$, that is
 $t\mapsto z_{l}(\gamma_{l}(t))$ is strictly increasing. Indeed if it is not
 the case, the path on the other sheet of the Riemann surface $\Sigma_{l}$
 associated to the definition of the square root with the same projection on $\C$ as
$\gamma_{l}(-1,\beta)$ is then of type $+$. (We have $z_{l}(-1)<0$.)
We define two independent complex WKB solutions $w_{\pm}$ of the equation
$(H^h_\delta-E)u=0$ on the interval $[-1,\beta]$.These have the form
\begin{gather}
\begin{split}
w^l_\pm : x\mapsto (V(x)-i\delta-E)^{-\frac 1 4} e^{\pm z_l(x)/h}W^l_\pm(z_l(x))
\\
=({\partial_x z_l (x)})^{-\frac 1 2}(x)e^{ \pm z_l(x)/h} W^l_\pm (z_l(x)).
\label{wkbsoll}
\end{split}
\end{gather}

Similarly, there exists a path $\gamma_r=\gamma_r(\beta,1)$ transverse to the
Stokes lines of type $+$ , (We have $z_r(1)>0$ and we have again  two independent complex  WKB solutions  of the equation
$H_\delta w_\pm=0$ on the interval $[\beta,1]$ :
\begin{gather}
\begin{split}
w^r_\pm : x\mapsto (V(x)+i\delta-E)^{-\frac 1 4} e^{\pm z_r(x)/h}W^l_\pm(z_r(x))\\
=({\partial_x z_r}(x))^{-\frac 1 2}e^{ \pm z_r(x)/h} W^r_\pm (z_r(x)).
\label{wkbsolr}
\end{split}
\end{gather}
The functions $W^{l,r}_\pm$ are convergent series  of the form
\begin{equation}
W^{l,r}_\pm=\sum_{n=0}^\infty W_{n,\pm}^{l,r},
\label{wlrn}
\end{equation}
where the $W_{n,\pm}^{l,r}$ are determined  by the recurrence relations:
\begin{equation}
\left\{
\begin{array}{l}
(\partial_z \pm\frac 2h)W^{l,r}_{2n+1,\pm}=-H^{l,r}W^{l,r}_{2n,\pm}\\[8pt]
\partial_z W^{l,r}_{2n,\pm}=-H^{l,r}W^{l,r}_{2n-1,\pm}
\end{array}
\right .
\end{equation}
with initial data
\begin{equation}
\left\{
\begin{array}{ll}
W^{l,r}_{0,\pm}=1,&\\[8pt]
W^{l}_{n,+}(z_l(-1))=0,\ W^{r}_{n,-}(z_r(1))=0, & n\geq 1,\\[8pt]
W^{l}_{n,-}(z_l(\beta))=0,\ W^{r}_{n,+}(z_r(\beta))=0, & n\geq 1,
\end{array}
\right .
\end{equation}
We have set here
\begin{equation}
 H^{l,r}(z_{l,r}(x))=-\frac {1} 4 \frac {\partial_xV(x)}{(\partial_x z_{l,r}(x))^3}\cdotp
\label{H}
\end{equation}
 These equations can be written in integral form as
\begin{equation}
\left\{
\begin{array}{l}
W_{2n+1,\pm}^{r,l}= I_\pm^{l,r}(W_{2n,\pm}^{r,l}),\\[8pt]
 W_{2n+2,\pm}^{r,l}= J^{l,r}(W_{2n+1,\pm}^{r,l})
 \end{array}
\right .
 \end{equation}
where
\begin{equation}
\left\{
\begin{array}{l}
\ds
I_\pm^{l,r}(v)(z)=-\int_{\tilde\gamma_{l,r}(z)}e^{\pm2(u-z)/ h} H^{l,r}(u)v(u)\,du
\\[12pt]
\ds
J^{l,r}(v)(z)=-\int_{\tilde\gamma_{l,r}(z)} H^{l,r}(u)v(u)\,du
\end{array}
\right .
\label{IetJ}
\end{equation}
and $\tilde\gamma_{l,r}(z)$ is the image by $x\mapsto z_{l,r}(x)$ of
$\gamma_{l,r}$ from the initial point  $-1$ to $z$ for $l$ and  from $\beta$ to $z$ for $r$.

Now we have the following estimates
\begin{gather}\begin{split}
\Sup{z\in\tilde\gamma_{l,r}}|I^{l,r}_+(v)(z)|
&\leq  \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
\Sup{z\in\tilde\gamma_{l,r},z=z^{l,r}(x)}
\big\{ |H^{l,r}(z)|
\frac {|\partial_x z_{l,r}(x)|}{|\re \partial_x z_{l,r}(x)| } \big\} \times
\\& |\int_{\tilde\gamma_{l,r}}e^{\re (\frac {2(u-z)} h)}\re(\,dz)|\\&
\leq  h \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
\Sup{z\in\tilde\gamma_{l,r},z=z^{l,r}(x)} |H^{l,r}(z)||\partial_x z_{l,r}(x)| \frac 1 {d(E,\CT_\delta^c,\gamma)}
\\ &\leq C(V) h \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
 \frac 1 {d(E,\CT_\delta^c,\gamma)^3}
 \end{split}
 \end{gather}

\begin{gather}\begin{split}
\Sup{z\in\tilde\gamma_{l,r}}|J^{l,r}(v)(z)| &\leq  \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
\Sup{z\in\tilde\gamma_{l,r},z=z^{l,r}(x)} |H^{l,r}(z) |\partial_x z_{l,r}(x)| \\&
\leq   \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
\Sup{z\in\tilde\gamma_{l,r},z=z^{l,r}(x)} |H^{l,r}(z)||\partial_x z_{l,r}(x)|
\\ &\leq C(V)  \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
 \frac 1 {d(E,\CT_\delta^c,\gamma)^2}.  \end{split}\end{gather}
Similarly we obtain
\begin{gather}\begin{split}
\Sup{z\in\tilde\gamma_{l,r}}|I^{l,r}_-(v)(z)|
&\leq  h \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
\Sup{z\in\tilde\gamma_{l,r},z=z^{l,r}(x)} |H^{l,r}(z)| |\partial_x z_{l,r}(x)| \frac 1 {d(E,\CT_\delta^c,\gamma)}
\\& \leq C(V) h \Sup{z\in\tilde\gamma_{l,r}} |v(z)|
 \frac 1 {d(E,\CT_\delta^c,\gamma)^3}
 \end{split}\end{gather}
 We denote by $\lfloor x\rfloor$ the integer part of $x$.
Hence on a progressive path $\gamma$ we have
$$
|W_{n,\pm}^{l,r}|_\infty \leq \frac{ h^{\Entier\frac {n+1} 2\rfloor}} {d(E,\CT_\delta^c,\gamma)^{3n}},
$$
which means that
\begin{equation}\label{vingh}
\begin{array}{l}
W^l_{+}(-1)=1,\quad W^l_{+}(\beta)=1+O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}),\\
W^l_{-}(\beta)=1,\quad W^l_{-}(-1)=1+O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}),\\
W^r_{+}(\beta)=1,\quad W^r_{+}(1)=1+O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}),\\
W^r_{-}(1)=1,\quad W^r_{-}(\beta)=1+O(\frac h
{d(E,\CT_\delta^c,\gamma)^{6}}).\end{array}
\end{equation}
The function $w^l_\pm$ form a basis of  solutions of the equation $H_\delta
w=0$, and similarly $w^r_\pm$ is a basis of  solutions of the equation
$H_\delta w=0$.
A computation gives
\begin{equation}\nonumber
\frac {dw^{l,r}_\pm}{dx}(x)=\pm {\partial_x z_{l,r}(x) }^{\frac 1 2} e^{\pm \frac {z_{l,r}(x)} h} \frac 1 h
\sum_{n=0}^\infty (-1)^n W^{l,r}_n(z_{l,r}(x)),
\end{equation}
and so
\begin{equation}\nonumber
\frac {dw^{l,r}_\pm}{dx}(x)=\pm  {\partial_x z_{l,r}(x)}^{\frac 1 2} e^{\pm
  \frac {z_{l,r}(x)} h}\frac 1 h(1+O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}
)).
\end{equation}
Now compute the determinant;  we  shall write  $\O=\O(\frac h
{d(E,\CT_\delta^c,\gamma)^{6}})$ for symplicity.
\begin{gather}\begin{split}
&\det\left(\begin{array}{llll}
w^l_-(-1)&w^l_+(-1)&0&0\\
 w^l_-(\beta)&w^l_+(\beta)&-w^r_-(\beta)&-w^r_+(\beta))\\
{w^l}'_-(\beta)&{w^l}'_+(\beta)&-{w^r}'_-(\beta)&-{w^r}'_+(\beta)\\
0&0&w^r_-(1)&w^r_+(1)\end{array}\right)=
\\&
\frac 1 h \partial_x z_r^{-\frac 1 2}(1)\partial_x z_r^{-\frac 1 2}(\beta)
\partial_x z_l^{-\frac 1 2}(\beta)\partial_x z_l^{-\frac 1 2}(-1)
e^{-\frac {z_l(-1)} h } e^{\frac {z_r(1)} h }\times
\\&\left|\begin{array}{llll}
1+\O&
 e^{2\frac {z_l(-1)} h } &0&0\\
1 & 1+\O
 &-1+\O&-1 \\
- \partial_x z_l(\beta) (1+\O )   &
 \partial_x z_l(\beta)(1+\O)&
\partial_x z_r(\beta)(1+\O)&
-\partial_x z_r(\beta)(1+\O) \\
0&0& e^{-2\frac {z_r(1)} h}&1+\O\end{array}\right|
\end{split}\end{gather}
So the sign of the  determinant is given by the sign of
\begin{gather}\begin{split}
& \partial_x z_l^{-\frac 1 2}(-1)\partial_x z_r^{-\frac 1 2}(\beta)
\partial_x z_l^{-\frac 1 2}(\beta)\partial_x z_r^{-\frac 1 2}(1)\times
\\&\quad\quad
(\det\left(\begin{array}{llll}
1& 0 &0&0\\
0 & 1 &-1&0 \\
0&\partial_x z_l(\beta)&\partial_x z_r(\beta)&0\\
0&0&0&1\end{array}\right)
+\O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}))
\\&=\partial_x z_l^{-\frac 1 2}(-1)\partial_x z_r^{-\frac 1 2}(\beta)
\partial_x z_l^{-\frac 1 2}(\beta)\partial_x z_r^{-\frac 1 2}(1)\times
\\& (\partial_x z_r(\beta)+\partial_x z_l(\beta)+\O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}) )
\\&=
{(V(-1)-i\delta-E)}^{-\frac 1 4}
{(V(\beta)-i\delta-E)}^{-\frac 1 4}\\&\times
{(V(\beta)+i\delta-E)}^{-\frac 1 4}
{(V(1)+i\delta-E)}^{-\frac 1 4}
\\&\times
(\sqrt{V(\beta)+i\delta-E}+\sqrt{V(\beta)-i\delta-E}+\O(\frac h {d(E,\CT_\delta^c,\gamma)^{6}}) )
\end{split}\end{gather}
( we have $\re(\sqrt{V(\beta)+i\delta-E}+\sqrt{V(\beta)-i\delta-E})\geq 2 d(E,\CT_\delta^c,\gamma)$.)

The determinant is non zero as soon as $\frac h {d(E,\CT_\delta^c,\gamma)^{6}}$ is small and
 also small
compared to $d(E,\CT_\delta^c,\gamma)$ (and this explain the exponent $7$).\par

\cqfd

\section{spectrum for large $E$ and $\delta=0$}
\label{large}

In this section we prove three results: the first, Theorem \ref{reci}, gives the condition for $E_0$ to belong to $\lim_h \Sp(H^h)$.
The second,  Theorem \ref{curve}, describes  $\lim_h \Sp(H^h)$  as  a curve and  gives its asymptotics near $\infty$.
The last, Theorem \ref{quanti}, describes the eigenvalues of $H^h$ which are  near to $\lim_h \Sp(H^h)$.
In the following we let denote by $Y$ a primitive of the potential $V$.
In this section, we assume that $V$ verifies the following hypothesis:
\textbf{(H1)} For any $\Omega\subset \C$, if $V^{-1}(\Omega)$ bounded  then $\Omega$ is relatively compact.

\textbf{(H2)} If  $E$ is  large enough,but  with small imaginary part, then  $1$ and $-1$ belong to the same Stokes region. We recall that this means that one can find a path going from $-1$ to $ 1$ that does not intersect the Stokes lines issuing from the turning points.
\medskip
\noindent
Notice that, under the assumption \textbf{(H2)}, $E\in \CT^c_0$ is  equivalent to $\re(\int_{-1}^1\sqrt {V-E}\,dx)=0$.
\begin{theorem}\label{reci}
For $E_0$ large  with small enough imaginary part,  then  \linebreak
$\re(\int_{-1}^1\sqrt {V-E_0}\,dx)=0$ if and only if  $E_0\in\lim_h\Sp(H^h_0)$.

Moreover there exists $C$ small enough and  $E_h$  in the spectrum of  $H^h_0$ which satisfies   $|\sqrt{ E_h}-\sqrt{ E_0}|\leq C h$.
\end{theorem}

{\sl Proof of Theorem \ref{reci}:}
The reverse implication  follows directly from Theorem  \ref{inclus}, so we prove the direct implication.

There exists two WKB solutions $w_\pm$ of $H^h_0-E=0$ of the form
\begin{equation}
w^\pm(x) ={\partial_x z_0(x)}^{-\frac 1 2}e^{\pm z_0(x)/h}W_\pm(z_0(x)),
\end{equation}
$$W_\pm=\sum_{j=0} W_{n,\pm}\quad    W_{0,\pm}(-1)=1, \quad W_{n,\pm}(-1)=0 \mbox{ for } n>0.$$
The complex number $E$ belongs to the spectrum of $H^h_0$ if and only if
$$
e^{\frac 2 h  z_0(1)}=\frac {W_+(1)}{W_-(1)},
$$
which is equivalent to the  existence of  $k\in\Z$ such that
$$
\int_{-1}^1 \sqrt{V(x)-E}\,dx-\frac h 2 \ln(W_+(1)) +\frac h 2 \ln(W_-(1)) -ihk\pi=0.
$$
Let us denote
 $$
 f(E)=\int_{-1}^1 \sqrt{V(x)-E}\,dx-\im\int_{-1}^1 \sqrt{V(x)-E_0}\,dx,
 $$
 and
  $$
  k_0=\lfloor\frac {\im\int_{-1}^1 \sqrt{V(x)-E_0}\,dx} h\rfloor
  $$
 and also
$$
g(E)=\int_{-1}^1 \sqrt{V(x)-E}\,dx-\frac h 2 \ln(W_+(1)) +\frac h 2 \ln(W_-(1)) -i hk_0\pi.
$$
We have  $f(E_0)=0$. We want to apply  Rouch\'e's theorem,
but to do so we must give  upper and lower bounds for  $W_\pm(1)$ and an estimate on
$f'(E_0)=-\frac 1 2 \int_{-1}^1\frac 1 {\sqrt {V(x)-E_0}} \,dx$.

By the hypothesis on the geometry of the Stokes lines near $-1$ and $1$ there exists a path $\gamma$  which links $-1$ to $1$  on which $\re\int_{-1}^{t}\sqrt{V(x)-E_0}\,dx=0$ for all $t\in\gamma$. Moreover $V(\gamma)$ is bounded uniformly with respect to $E_0$, and by assumption \textbf{(H1)}, $\gamma$ is also bounded uniformly with respect to $E_0$.

We have
$$
\frac 1 {\sqrt {V(x)-E_0}}=\frac 1 {i\sqrt E_0} (1+\O(\frac {sup_\gamma V} {E_0})),
$$
therefore
$$
|f(E)-f(E_0)|\geq \frac 1 {2\sqrt{| E_0|}} |(E-E_0)|,
$$
$$
\sup_{\gamma} |H(z(x))|\leq C_1 \frac {\Sup{\gamma} |V'(x)| } {|E_0|^{\frac 3 2}},
$$
and
$$
\sup_{b,b'\in\gamma} |\re\int_b^{b'} \sqrt{V(t)-E}\,\,dt|\leq C_2 \frac {E-E_0} {\sqrt {|E_0|}},
$$
 for any $E$ and $\a_0$ such that $|E-E_0|\leq \po (E_0)$  and  $|E_0|>C_0$ .
Using the expression of
$W_{n,+}(z)$ as a Volterra integral,
we obtain the estimates
$$
|W_{n,\pm}(z)|\leq \exp\big\{\frac 2 h \sup_{b,b'\in\gamma} |\re\int_b^{b'} \sqrt{V(t)-E}\,dt| (n+2) \big\} \sup_\gamma |H|^n \frac 1 {n!}  .
$$
This gives
$$
|W_{n,\pm}(z)|\leq \exp\big\{\frac   {C_2} h \frac {|E-E_0|} {\sqrt {|E_0|}} {n+2} \big\} \frac { \hat C_1^n }{|E_0|^{\frac {3n} 2} n!}
$$
for $|E-E_0|\leq h_{0}(E_0)$ and  $|E_0|>C_0$ .
So we obtain
$$
W_\pm(1)=1+  \frac { \hat C_1 }{|E_0|^{\frac 3 2} }
 \exp \big\{ \frac {2 C_2} h \frac {|E-E_0|} {\sqrt{ |E_0|}}\big\}
 \exp\big\{ \frac {\hat C_1}{|E_0|^{\frac 3 2} }e^{\frac  {2 C_2} h } \frac {|E-E_0|} {\sqrt{ |E_0|}     }\big\}  .
$$
This gives the estimate
$$|W_\pm(1)-1|\leq   \frac { \tilde C_1 }{|E_0|^{\frac 3 2} } e^{C_2 C_3} e^{ \hat C_1  {|E_0|^{-\frac 3 2}}
e^{  C_2 C_3 }}$$
for $E_0>C_0$  and $|E-E_0|\leq C_3 h\sqrt{|E_0|}$ with $h$ small enough. We also have
\begin{equation}\label{vingt}
 |\ln W_+(1)|+|\ln W_-(1)|\leq  \frac {\check C_1 } {|E_0|^{\frac 3 2}} \end{equation}
for $E_0>C(C_3)$, and $|E-E_0|\leq C_3 h\sqrt{|E_0|}$ with $h$ enough small,
and  for the functions $f$ and $g$ we get
$$
|f(E)-g(E)|\leq h +\frac h 2 (|\ln W_+(1)|+|\ln W_-(1)|).
$$
Thus if $|E-E_0|= C_3 h\sqrt{|E_0|}$, for some large constant $C_{3}>0$,  we have
\begin{equation}
\label{titus}
|f(E)-g(E)|\leq (1+C_4 |E_0|^{-\frac 3 2} )  h,\quad \quad |f(E)|\geq \frac {C_3}2h.
\end{equation}
Finally, if $E_0$ is large enough, all the assumptions of Rouch\'e's theorem are fulfilled, and  we get the result.\cqfd

\begin{theorem}\sl
\label{curve}
For any fixed $a$  large enough, the equation
$$
\re S_{-1,1}(E)_{\vert_{E=a+ib}}=\re\int_{-1}^1\sqrt{V(x)-(a+ib)}\,dx=0
$$
has a  unique solution $b(a)$. Moreover
$b(a)=  \frac 1 {2} \im(Y(1)-Y(-1))+\O( \frac 1 {a} )$.
\end{theorem}

{\sl Proof :} First we prove that if  $b$ is such that
$\re\int_{-1}^1\sqrt{V(x)-a-ib}\,dx=0$,  and
$b=\po(a)$ then
\begin{equation}
b= i \frac 1 2 \im(Y(1)-Y(-1))+\O( a^{-1} ).
\label{etape1}
\end{equation}
Indeed if we  denote by $E=a+ib$
$$
\varphi(E,\alpha,y)= \int_{\alpha}^y\sqrt{V(x)-E}\,dx,
$$
then, uniformly for $y$ in a compact set,
$$
\varphi(E,\alpha,y)=i\sqrt E (y-\alpha)-i\frac 1 {2\sqrt E} (W(y)-W(\alpha))+O(E^{-\frac 3 2}).
$$
Since $\re S_{-1,1}(E)=0$, we have
$$
\re\big(i2\sqrt E -i\frac 1 {2\sqrt E} (Y(1)-Y(-1))\big)=\O(E^{-\frac 3 2}).
$$
Writing $\sqrt E=c+i\tilde b$, then $\tilde b=\po(c)$ and
$$
-2\tilde b +\frac 1 {2c} \im(Y(1)-Y(-1))+O(\frac {\tilde b} {c^2})=\O(c^{- 3 }),
$$
or equivalently
$$
\tilde b =\frac 1 {4c} \im(Y(1)-Y(-1))+o(\frac 1 c).
$$
This gives $b= \frac 1 {2} \im(Y(1)-Y(-1))+o(1)$, and  $b= O(1)$. Therefore
$\tilde b=O(\frac 1 c)$ and recycling through this argument,
 $$\tilde b =\frac 1 {4c} \im(Y(1)-Y(-1))+O(\frac 1 {c^3}),
 $$
 which gives (\ref{etape1}).

The existence part of the theorem is straightforward:  set
$$
\varphi(c,\hat b):=\re\int_{-1}^1\sqrt{cV(x)-1-i\hat b}\,dx=c\re\int_{-1}^1\sqrt{V(x)-E}\,dx
$$
where $E=\frac 1 c (1+i\hat b)$. Since
$\partial_b\varphi(0,0)=-2$ and
$\varphi(0,0)=0$,
the implicit function theorem applies.

Finally, the uniqueness follows from the fact that the map
$\phi:b\mapsto\re\int_{-1}^1\sqrt{V(x)-a-ib}\,dx$ is injective for $a$ large enough. Indeed
$$
|\sqrt{V(x)-a-ib}-\sqrt{V(x)-a-ib'}|\leq C \sqrt{|b-b'|},
$$
for a suitable branch of the square root.\cqfd.

\begin{remark}\sl  We can compute more terms of the asymptotic expansion than in (\ref{etape1}). Indeed,
from
\begin{gather}\begin{split}
\varphi(E,-1,1)=i2 \sqrt E -i\frac 1 {2\sqrt E} (Y(1)-Y(-1))+ i E^{-\frac 3 2} \int_{-1}^1 V^2(x)\,dx
\\-iE^{-\frac 5 2} \int_{-1}^1 V^3(x)\,dx +O(E^{-\frac 7 2}).
\end{split}\end{gather}
we get
$$b(a)=   \frac 1 {2} \im(Y(1)-Y(-1)) +\frac 3 {8a^2} \im(\int_{-1}^{1} V^3 )+\O( \frac 1 {a^3} ).$$
\end{remark}

Looking more carefully at the quantization rules, we obtain the asymptotics of the eigenvalues.

\begin{theorem}
\label{quanti}
Let $E_0$ be a  solution of  $\re(\int_{-1}^1\sqrt {V-E_0}\,dx)=0$ with
$|E_0|$ large and $\im(E_0)$ small.
If $E_h$ is in the spectrum of  $H^h_0$ and satisfies  $|\sqrt{ E_h}-\sqrt{ E_0}|\leq C h$, provided $C$ is small enough then $E_{h}$
satisfies:
$$ E= (\frac {\pi hk} 2)^2 +
 \frac {(Y(1)-Y(-1))} {2} + \frac {(Y(1)-Y(-1))^2} {{(2hk\pi)}^2}   +  \O( \frac 1 {(hk)^3} ).
 $$
for some $k\in \nit$.
\end{theorem}
\begin{remark}
For a real potential $V$ bounded on $[-1,1]$ and for any $E_0\in\rit $ large we get  $\re(\int_{-1}^1\sqrt {V-E_0}\,dx)=0$.
So Theorem \ref{quanti} can be read as:
If $E_h$ is large and satisfies $E_h\in\Sp(H^h_0)$ then $E_{h}$
satisfies:
$$ E= (\frac {\pi hk} 2)^2 +
 \frac {(Y(1)-Y(-1))} {2} + \frac {(Y(1)-Y(-1))^2} {{(2hk\pi)}^2}   +  \O( \frac 1 {(hk)^3} ).
 $$
for some $k\in \nit$.
\end{remark}
{\sl Proof :}
$$\int_{-1}^1\sqrt {V-E}\,dx=i2\sqrt E-i\frac 1 {2\sqrt E} (Y(1)-Y(-1))+\O(E^{-\frac 3 2}).$$
Then $E\in\Sp(H^h_\delta)$ if and only if there exists $k\in\Z$ such that
$$\int_{-1}^1 \sqrt{V(x)-E}\,dx-\frac h 2 \ln(W_+(1)) +\frac h 2 \ln(W_-(1)) -ihk\pi=0$$
For $E_0$ big enough,  $|E-E_0|\leq C_3 h\sqrt{E_0}$, $C_3$ is as in
(\ref{titus}) and  $h$  small, we get
$$ |\ln W_+(1)|+|\ln W_-(1)|\leq  \check{ C} |E_0|^{-\frac 3 2} $$
So $E\in\Sp(H^h_\delta)$ if and only if there exist $k\in\Z$ such that
$$i2\sqrt E-i\frac 1 {2\sqrt E} (Y(1)-Y(-1)) -ihk\pi=\O(E^{-\frac 3 2})$$
In particular taking $\sqrt E=c+i\tilde b$  the real and imaginary parts give
$$2 c - hk\pi=\O(c^{-1})\quad c =\frac 1 2 hk\pi+\O((kh)^{-1})$$
$$2\tilde b=\frac 1 {2c} \im(Y(1)-Y(-1))+\O( c^{-3} )$$
Using this again gives
$$c =\frac 1 2 hk\pi+\frac 1 {2hk\pi} \re(Y(1)-Y(-1))+\O((kh)^{-3})$$
\cqfd
\medskip
\begin{remark}
One can also treat the case $P^h$ on all of $\rit$.  Then a condition for $E$ to belong to
 $\lim_h \Sp(H^h)$ is that  there exists a progressive path joining two  turning points.
 \end{remark}

\section{Large spectrum of  a perturbation of H}
\label{largeper}

In this section, we prove two theorems: Theorem \ref{reciper}  gives the condition for $E_0$ to belong to $\lim_h \Sp(H^h_\delta)$ and Theorem  \ref{curveper} gives a description  of $\lim_h \Sp(H^h_\delta )$ as
the union of two curves, the asymptotics  near $\infty$ of which are made explicit.

We assume that $V$ verifies the  hypothesis  \textbf{(H1)} and in addition~:
\par
\textbf{(H3)} If  $E$ is  large enough, with small imaginary part then  $1$,  $-1$  and $\beta$ belong to the same Stokes region.\par
\medskip
We assume $\beta\neq \pm 1$.
Recall that \textbf{(H3)}  means that one can find  paths going from $ -1$ to
$ \beta$  and from $ 1$ to $\beta$  which do not intersect the Stokes lines issuing from the turning points.
\medskip
 With the hypothesis \textbf{(H3)}, the condition $E\in \CT^c_\delta$ is then equivalent to \par\noindent
either $\re(\int_{-1}^\beta\sqrt {V-i\delta-E}\,dx)=0$
or $\re(\int_{\beta}^{1}\sqrt {V+i\delta-E}\,dx)=0$.

\begin{theorem}\label{reciper}
For $E_0$ large enough,  with small enough imaginary part,
 then $E_0$ satisfies $\re(\int_{-1}^\beta\sqrt {V-E_0}\,dx)=0$ or $\re(\int_{\beta}^1\sqrt {V-E_0}\,dx)=0$ if and only if $E_0\in\lim_h\Sp(H^h_\delta)$
 \end{theorem}

\begin{theorem}\label{curveper}
For any fixed $a$ large enough, the equation
$$\re(S^\delta_{\pm1,\beta}(E))|_{E=a+ib}:=\re\int_{\pm1}^\beta \sqrt{V(x)\pm i\delta-a-ib}\,dx=0$$
 has a unique solution $b(a)$, and this solution  satisfies
$$ b(a)= i \frac 1 {\beta-\pm 1} \im(Y(\beta)-Y(\pm 1)) \pm i \delta +O( \frac 1 a ).$$
\end{theorem}

{\sl Proof :}
The proof is the same as in Theorem \ref{curve} where the potential is $V\pm i\delta$  in each side and we just need to change the interval for the integral. \cqfd

{\sl Proof of Theorem \ref{reciper}:}
The reverse implication  is already proved in Theorem \ref{inclus}. Let us
prove the direct implication as in the proof of Theorem \ref{reci}.
There exist four WKB solutions $w^{l,r}_\pm$ of $H^h_\delta-E=0$  on the interval $[-1,\beta]$ or $[\beta,1]$
\begin{equation}\begin{array}{l}
w^l_\pm =({\partial_xz_l}(x))^{-\frac 1 2}e^{ \pm z_l(x)/h} W^l_\pm (z_l(x)),\\
w^r_\pm =({\partial_x z_r}(x))^{-\frac 1 2}e^{ \pm z_r(x)/h} W^r_\pm (z_r(x)),\end{array}
\label{wkbsolld}
\end{equation}
with initial data
\begin{equation}
\left\{
\begin{array}{ll}
W^{l,r}_{0,\pm}=1,&\\[8pt]
W^{l}_{n,+}(z_l(\beta))=0,\ W^{r}_{n,-}(z_r(\beta))=0, & n\geq 1,\\[8pt]
W^{l}_{n,-}(z_l(\beta))=0,\ W^{r}_{n,+}(z_r(\beta))=0, & n\geq 1.
\end{array}
\right .
\end{equation}

Recall that $E\in\Sp(H^h_\delta)$ if and only if
\begin{gather}\begin{split}\label{detcal}
&0=\det\left(\begin{array}{llll}
w^l_-(-1)&w^l_+(-1)&0&0\\
 w^l_-(\beta)&w^l_+(\beta)&-w^r_-(\beta)&-w^r_+(\beta)\\
{w^l}'_-(\beta)&{w^l}'_+(\beta)&-{w^r}'_-(\beta)&-{w^r}'_+(\beta)\\
0&0&w^r_-(1)&w^r_+(1)\end{array}\right)=
\\&
\frac 1 h \partial_x z_r^{-\frac 1 2}(1)\partial_x z_l^{-\frac 1 2}(-1)\times
\\& \det\left(\begin{array}{llll}
{\scriptscriptstyle e^{-\frac{ z_l(-1)} h} W^l_- (z_l(-1))}
 &{\scriptscriptstyle e^{\frac {z_l(-1)} h }W^l_+ (z_l(-1))}&0&0\\
1 & 1 &-\frac { \partial_x z_l^{\frac 1 2}(\beta)}{\partial_x z_r^{\frac 1 2}(\beta)}
&-  \frac {\partial_x z_l^{\frac 1 2}(\beta)}{\partial_x z_r^{\frac 1 2}(\beta)} \\
-1 &1& \frac {\partial_x z_r^{\frac 1 2}(\beta)} {\partial_x z_l^{\frac 1 2}(\beta)}
&-\frac {\partial_x z_r^{\frac 1 2}(\beta)}{ \partial_x z_l^{\frac 1 2}(\beta)}\\
0&0&{\scriptscriptstyle e^{-\frac {z_r(1)} h}W^r_- (z_r(1))}&
{\scriptscriptstyle e^{\frac {z_r(1)} h}  W^r_+ (z_r(1))} \end{array}\right)
\end{split}\end{gather}
To make the computation,  simplify the notation by setting
$$t=\frac { \partial_x z_l^{\frac 1 2}(\beta)}{\partial_x z_r^{\frac 1
    2}(\beta)} ,\;y=e^{\frac {z_r(1)} h}\; x= e^{\frac {z_l(-1)} h}$$
then (\ref{detcal}) becomes
 \begin{gather}\begin{split}
&\det\left(\begin{array}{llll}
x^{-1} W^l_- (z_l(-1)) & xW^l_+ (z_l(-1))&0&0\\
1 & 1 &-t &- t \\
-1 &1&  t^{-1}  &- t^{-1} \\
0&0& y W^r_- (z_r(1))& y^{-1}  W^r_+ (z_r(1)) \end{array}\right)=0,
\end{split}\end{gather}
which gives
$$ x^2 \frac {W^l_+ (z_l(-1))}{W^l_- (z_l(-1))}= \big (1+y^2  \frac {W^r_-
  (z_r(1))}{W^r_+ (z_r(1))} \frac {1-t^2}{1+t^2} \big) \big({\frac {1-t^2}{1+t^2} +y^2  \frac {W^r_- (z_r(1))}{W^r_+ (z_r(1))} }^{-1}\big).$$
We remark that
$$|\big(1-\frac {\partial_x z_l(\beta)}{\partial_x z_r(\beta)} \big) \big({1+\frac {
    \partial_x z_l(\beta)}{\partial_x z_r(\beta)}}^{-1}\big)|\leq C_8 \sqrt \delta.$$
So the condition for $E$ to be in the spectrum of $H_\delta^h$ is
\begin{gather}\begin{split}
& \big(e^{2\frac {z_l(-1)} h}  \frac {W^l_+ (z_l(-1))}{W^l_- (z_l(-1))}-
\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))} \big)\times
\\&
\big(e^{-2\frac {z_r(1)} h}  \frac {W^r_- (z_r(1))}{W^r_+ (z_r(1))}+\frac
{(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))}\big)
\\&=\big(1-\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))^2}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))^2}\big).\end{split}\end{gather}

We write this condition by taking logs as
\begin{gather}\begin{split}\label{titi}
&\ln\big( e^{2\frac {z_l(-1)} h}  \frac {W^l_+ (z_l(-1))}{W^l_- (z_l(-1))}-
\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))}\big)
\\&+\ln\big(e^{-2\frac {z_r(1)} h}  \frac {W^r_- (z_r(1))}{W^r_+
  (z_r(1))}+\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))}\big)\\
&-\ln\big (1-\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))^2}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))^2}\big)=2ik\pi,\quad k\in\zit.
\end{split}\end{gather}
\underline{Step1}
First assume that
$$\re(\int_{-1}^\beta\sqrt {V-i\delta-E_0}\,dx)=0\;\mbox{ and }\;  \re(\int_{\beta}^1\sqrt {V+i\delta-E_0}\,dx)\neq0.$$
 (The case $\re(\int_{-1}^\beta\sqrt {V-i\delta-E_0}\,dx)\neq 0$ and $\re(\int_{\beta}^1\sqrt {V+i\delta-E_0}\,dx)=0$ is treated the same way. )\par
As  (\ref{vingt}), we get
\begin{equation}\label{auauu} |\ln W^l_+(z_l(-1))|+|\ln W^l_-(z_l(-1))| \leq   C_1 |E_0|^{-\frac 3 2}
\end{equation}
 for $|E-E_0|\leq C_3 h\sqrt{|E_0|}$ and $E_0>C$ big, $h$ small.
As in (\ref{vingh}), we get
\begin{equation} \label{auau} |\ln W^r_+ (z_r(1))|+|\ln W^r_- (z_r(1))|\leq  C h \end{equation}
with $C$ uniform in $E_0$,  for $|E-E_0|\leq C_3 h\sqrt{|E_0|}$
and
\begin{equation}\label{auu}| e^{-2\frac {z_r(1)} h}|\leq e^{-\frac {C} h}.\end{equation}
Rewrite the previous condition as
\begin{gather}\begin{split}
&g_{1,k}(E)= 2 z_l(-1) +h \ln( \frac {W^l_+ (z_l(-1))}{W^l_- (z_l(-1))})
\\&
+h \ln(1-e^{-2\frac {z_l(-1)} h}  \frac{W^l_- (z_l(-1))} {W^l_+
  (z_l(-1))} \frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))})
\\&+h \ln(\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))})
\\&+ h \ln(1+
e^{-2\frac {z_r(1)} h}  \frac {W^r_- (z_r(1))}{W^r_+ (z_r(1))}\frac
{(\partial_x z_r(\beta)+\partial_x z_l(\beta))} {(\partial_x z_r(\beta)-\partial_x z_l(\beta))} )
\\&-h \ln (1-\frac {(\partial_x z_r(\beta)-\partial_x
  z_l(\beta))^2}{(\partial_x z_r(\beta)+\partial_x z_l(\beta))^2})-2ikh\pi=0.
\end{split}\end{gather}
Setting
$$f_1(E)=2 z_l(-1) +2\Im(\int_{-1}^{\beta} \sqrt{ V(x)-i\delta-E_0} \,dx)$$ and
$$k_0=\Entier  \Im(\int_{-1}^{\beta} \sqrt{ V(x)-i\delta-E_0} \,dx)h^{-1}\rfloor,$$ we have $f_1(E_0)=0$.
Using  (\ref{auauu}), (\ref{auau}) and (\ref{auu}) gives
$$|g_{1,k_0}(E)-f_1(E)|\leq h(1+C\sqrt\delta+C h\sqrt\delta^{-1} +  C_1 |E_0|^{-\frac 3 2}).$$
Now recall that
$$|f_1(E)-f_1(E_0)|\geq \frac 1 {\sqrt{| E_0|}} |(E-E_0)|.$$
We can  apply the Rouch\'e's Theorem to prove the existence of an eigenvalue
of $H^h_\delta$  for each $h$ at a distance
 $C_3(\delta) \sqrt{| E_0|} h$ of $E_0$ for $E_0$ large.
 \medskip
\underline{Step2}
Assume now  that
$$\re(\int_{-1}^\beta\sqrt {V-i\delta-E_0}\,dx)=\re(\int_{\beta}^1\sqrt {V+i\delta-E_0}\,dx) =0.$$
As in  (\ref{vingt}),
\begin{gather}\begin{split}
|\ln W^l_+ (z_l(-1))|+|\ln W^l_- (z_l(-1))|+ |\ln W^r_+ (z_r(1))|+\\
|\ln W^r_- (z_r(1))|
\leq  C_9   |E_0|^{-\frac 3 2} .\end{split}
\end{gather}
For $E_0>C$, $C$ sufficiently large, and $|E-E_0|\leq C_3 h\sqrt{|E_0|}$ with $h$ small.
We write (\ref{titi}) as
\begin{gather}\begin{split}
&g_{2,k}(E)=2 z_l(-1)-2 z_r(1)  +h \ln( \frac {W^l_+ (z_l(-1))}{W^l_- (z_l(-1))})\\&
+h \ln(1-e^{-2\frac {z_l(-1)} h}  \frac{W^l_- (z_l(-1))} {W^l_+ (z_l(-1))} \frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x z_r(\beta)+\partial_x z_l(\beta))})
\\&
h \ln( \frac {W^r_+ (z_r(1))}{W^r_- (z_r(1))})
+h \ln(1-e^{2\frac {z_r(1)} h}  \frac{W^r_- (z_r(1))} {W^r_+ (z_r(1))}
\frac {(\partial_x z_r(\beta)-\partial_x z_l(\beta))}{(\partial_x
  z_r(\beta)+\partial_x z_l(\beta))})
\\& -h \ln (1-\frac {(\partial_x z_r(\beta)-\partial_x
  z_l(\beta))^2}{(\partial_x z_r(\beta)+\partial_x z_l(\beta))^2}-2ikh\pi=0,
\end{split}\end{gather}
and write
\begin{gather}\begin{split}
f_2(E)=2 z_l(-1)-2 z_r(1) -2\Im(\int_{\beta}^{-1} \sqrt{ V(x)-i\delta-E_0} \,dx) \\+2\Im(\int_{\beta}^{1} \sqrt{ V(x)+i\delta-E_0} \,dx).\end{split}
\end{gather}
Let
$$k_0=\Entier \Im(\int_{\beta}^{-1} \sqrt{ V(x)-i\delta-E_0} \,dx-\int_{\beta}^{1} \sqrt{ V(x)+i\delta-E_0} \,dx)h^{-1}\rfloor.$$
Then
$$|f_2(E)-f_2(E_0)|\geq {\sqrt{| E_0|}}^{-1} |(E-E_0)|,$$
and for $|E-E_0|\leq C_3 h\sqrt{|E_0|}$
$$|f_2(E)-g_{2,k_0}(E)|\leq h( C_9 |E_0|^{-\frac 3 2} +C_8\sqrt\delta +1).$$
On the set where $|E-E_0|= C_3 h\sqrt{|E_0|}$, we have for $C_3=C_3(\delta)$ and  $E_0$ large
$$|(f_2-g_{2,k_0})(E)|\leq |f_2(E)| .$$
All the hypotheses of Rouch\'e's theorem are now satisfied and we conclude as before.\cqfd

\section{Application to the harmonic oscillator}
\label{harmonic}

In this section we are interested in  the Dirichlet  eigenvalues  of the operator $P^h$ on $L^2([-1,1])$,
\begin{equation}
 P^h=-h^2 \frac d {\,dx^2} +ix^2,
\label{Px^2}
\quad
\left\{
\begin{array}{ll}
P^hu=Eu,\\
u(-1)=u(1)=0.
\end{array}
\right .
\end{equation}

We obtain here only two results. Theorem \ref{forme}  describes  the shape of
the set $\lim_h\Sp(P^h)$, cf.  figure \ref{spectre}.
Theorem \ref{infiniper}  shows how the spectrum changes near $\infty$
when $P^h$ is change by a specific perturbation of size $\delta$, as illustrated in figure
 \ref{spectreper} .

For any $E\in \C^*$ there are two turning points, $\alpha_\pm (E)=\pm(-iE)^{1/2}$, which are both simple i.e. of order 1.

Denote by $S_{x,y}(E)=\int_x^y (it^2-E)^{1/2} \,dt$.
The function $S_{0,x}$ is even, so to simplify the computation one can use that
$S_{-x,x}=2S_{0,x}$.
Writing
$$z(x,y,E)=\int_x^y (it^2-E)^{1/2} \,dt, $$
$$z^{har}(x,y,E)=\int_x^y (u^2-E)^{1/2} \,du,$$
and changing  coordinate $t=e^{-\frac{i\pi} 8}u$, we get
 $$z(x,y,E)=\pm z^{har}(x e^{\frac{i\pi} 8} ,y e^{\frac{i\pi} 8} ,Ee^{-\frac{i\pi} 4}).$$
 So the Stokes lines of $-h^2\Delta+ix^2$ can be deduce from those  of the harmonic oscillator
 $-h^2\Delta+x^2$ by a rotation by  $-\frac {\pi} 8$ about the origin,  \cite{fe}.
Using the geometry of the Stokes lines we find for some $E$ a progressive path from $-1$ to $1$. Combine with previous theorems, we obtain a partial description   of  $\CT^c$~:
\begin{theorem} We have\par\noindent
 \begin{itemize}
 \item Suppose $E\in \C^*$ is such that
 $
 \re S_{ \alpha_-,\alpha_+}(E)\neq 0,$  $\re S_{-1,1}(E)\neq 0,$   $\re S_{\alpha_+,1}(E)\neq 0, $  $\re S_{\alpha_-,1}(E)\neq 0,$
 then $E\in\CT$
\item  If $E\in \CT$, then for any $h$ small enough, $E$ is not an eigenvalue for $P^h$.
\item  If $E\in\lim_h\Sp(P^h)$ then $E\in\CT^c$
\end{itemize}
\label{main}
\end{theorem}
{\sl Proof:}
Using Theorem \ref{mainbis} and Theorem \ref{inclus}, it is enough to find a path  $\gamma$ tranversal to the Stokes lines from $-1$ to $1$.
We have $\re S_{\alpha_-,\alpha_+}(E)\neq 0$,  so the Stokes line issuing from $\a_+$ do not intersect the Stokes lines issuing form $\a_-$. Therefore the complex plane is divided into exactly five region delimited by the Stokes lines issuing from the turning points.
In  Figure \ref{gen1},  we have drawn the configuration of the Stokes lines up to an analytic diffeomorphism.\par
 \begin{figure}[ht]
\begin{center}
\epsfysize=5cm
\epsfbox{gen1.eps}
\end{center}
\caption{Non degenerate Stokes lines}
\label{gen1}
\end{figure}
We have $\re S_{1,\alpha_\pm}(E)\neq 0$, so that in particular $1$, (and by
symmetry $-1$) are not on any boundary of the Stokes regions. \par
 \begin{figure}[ht]
\begin{center}
\epsfysize=5cm
\epsfbox{che.eps}
\end{center}
\caption{Progressive path issuing from $1$}
\label{che}
\end{figure}

If $1$ and $-1$ belong to different regions, then there always exists a path $\gamma$ tranversal to the Stokes lines (see the figure \ref{che});
if on the other hand  $1$ and $-1$ belong to the same region, but we also have
$\re S_{-1,1}(E)\neq 0$, then they do not belong to the same Stokes lines and  there still exists  such a path $\gamma$.  \cqfd
\par
\medskip
If $\re S_{\alpha_-,\alpha_+}(E)= 0$,  the Stokes configuration is
as in Figure \ref{gen2}.\par
 \begin{figure}[ht]
\begin{center}
\epsfysize=5cm
\epsfbox{gen2.eps}
\end{center}
\caption{Degenerate Stokes lines}
\label{gen2}
\end{figure}
We now picture the evolution of the Stokes lines as $E$ moves in the complex plane.\par
\begin{figure}[ht]
\begin{center}
\epsfysize=8cm
\epsfbox{fullstoke.eps}
\end{center}
\caption{Stokes line for different $E$}
\label{fullstoke}
\end{figure}
In fact  Theorem \ref{main} can  be improved.  Choose  the segment $[\a_-,\a_+]$ as a cut for  $x\mapsto (V(x)-E)^{1/2}$.
The pictures \ref{det2} and \ref{secondet} show the branch of  the square root  for different values of $E$. \par
\begin{figure}[ht]
\begin{center}
\epsfysize=4cm
\epsfbox{det2.eps}
\end{center}
\caption{branch for $E=e^{i\frac \pi 4}$, the cut is the dashed
  line and also a Stokes line}
\label{det2}
\end{figure}
\par

\begin{figure}[ht]
\begin{center}
\epsfysize=4cm
\epsfbox{secondet.eps}
\end{center}
\caption{branch  for $E=e^{i\frac \pi 4}e^{i\beta}$,$\beta>0$ small. The
  cut is the dashed line }
\label{secondet}
\end{figure}
 Denote by
 $$\Gamma_{-1,1}=\{E\in\cit;\re S_{-1,1}(E)= 0,\quad \re S_{\alpha_-,\alpha_+}(E)\neq 0\}$$
The next theorem is proved by listing all the situations where there do not exist  progressive paths from $-1$ to $1$ using the geometry of the Stokes lines.
\pagebreak
\begin{theorem}\sl\label{main2}We have\par\noindent
 \begin{itemize}
 \item
 Suppose $E\in \C^*$  belongs to
$$\Gamma_{\alpha_-,\alpha_+}=\{E\in\cit;  \re S_{\alpha_-,\alpha_+}(E)= 0, \re S_{\alpha_+,1}(E)\leq 0\},$$ then $E\in\CT$.
\item
Suppose $E\in \C^*$  belong to
 $$\Gamma_{\alpha_+,1}=\{E\in\cit; \re S_{\alpha_+,1}(E)= 0; \im S_{\alpha_+,1}(E)\leq 0\},$$ then $E\in\CT$.
\item
Suppose $E\in \C^*$  belong to
$$\Gamma_{\alpha_-,1}=\{E\in\cit; \re S_{\alpha_-,1}(E)= 0; \im S_{\alpha_-,1}(E)\geq 0\},$$ then $E\in\CT$.
 \item $E\notin \Gamma_{\alpha_+,1}\cup\Gamma_{\alpha_-,1}\cup\Gamma_{\alpha_-,\alpha_+}\cup\Gamma_{-1,1}$ then there is no progressive path form $-1$ to $1$.
  \end{itemize}
\end{theorem}
\begin{remark}\label{stupid}
If a progressive path enters a  Stokes region crossing a Stokes line issuing from a turning point $\alpha$, then it cannot leave this region by crossing any others    Stokes lines issuing from the same turning point $\alpha$.
\end{remark}
The proof consists in listing the cases where progressive paths do not exist.
Recall first that if  $1$ and $-1$ belong to the same region then there exists a progressive path if and only if
$\re S_{-1,1}(E)\neq 0$.
We get   $\Gamma_{-1,1}=\{E\in\cit; 1\mbox{ and }-1 \mbox{ belong to the same region }\re S_{-1,1}(E)= 0\}$.
Now  assuming  $1$ and $-1$ belong to different regions, we get two different figures for the Stokes lines either
  $\re S_{\alpha_-,\alpha_+}(E)= 0$ (Figure\ref{gen2}) or  $\re S_{\alpha_-,\alpha_+}(E)\neq 0$ (Figure \ref{gen1}).
In the first case, using Remark \ref{stupid}, we see that we cannot find a path if and only if  $1$ belongs to the hatched region ( and $-1$ by symmetry to  the opposite one)(Figure \ref{sit1}).\par
\begin{figure}[ht]
\begin{center}
\epsfysize=3cm
\epsfbox{sit1.eps}
\end{center}
\caption{The position of $1$ is in one of hatched regions}
\label{sit1}
\end{figure}
The condition $\re S_{\alpha_-,\alpha_+}(E)= 0,\quad \re S_{\alpha_+,1}(E)>0$ corresponds to this situation, i.e.
 Figure \ref{deter} shows  the sign of the quantity $\re S_{\alpha_+,x}(e^{i\frac \pi 4}\lambda)$ depends on the position of $x$ for  $\lambda\in\rit$. \par
\begin{figure}[ht]
\begin{center}
\epsfysize=3cm
\epsfbox{deter.eps}
\end{center}
\caption{ Sign of the quantity $x\to\re S_{\alpha_+,x}(e^{i\frac \pi 4})$}
\label{deter}
\end{figure}
Now  assuming  $1$ and $-1$ belong to different regions and   $\re S_{\alpha_-,\alpha_+}(E)\neq 0$ (Figure \ref{secondet}).
Using Remark \ref{stupid}, we see that we cannot find a path if and only if $1$ belong the dotted curves ( and $-1$ by symetry to the opposite one )(Figure \ref{sit2}).
The condition  $\re S_{\alpha_+,1}(E)= 0,\quad \im S_{\alpha_+,1}(E)>0$ correspond to Figure \ref{sit2}.\par
\begin{figure}[ht]
\begin{center}
\epsfysize=5cm
\epsfbox{sit2.eps}
\end{center}
\caption{The position of $1$ is in one of the dotted line, that are Stokes lines}
\label{sit2}
\end{figure}
The condition  $\re S_{\alpha_+,1}(E)= 0,\quad \im S_{\alpha_+,1}(E)>0$ corresponds to this case.
i.e. Figure \ref{seco} shows the sign of the quantity
$\im S_{\alpha_\pm,x}(e^{i\frac \pi 4}e^{i\beta}\lambda)$ with $\beta>0$ and small depending on $x$ near to $\alpha_\pm$.\par
\begin{figure}[ht]
\begin{center}
\epsfysize=8cm
\epsfbox{seco.eps}
\end{center}
\caption{ Sign of the quantity
$\im S_{\alpha_\pm,x}(e^{i\frac \pi 4}\lambda e^{i\beta})$}
\label{seco}
\end{figure}
Now we want to describe the set $\CT^c$.
\pagebreak
 \begin{theorem}\sl\label{forme}We have\par\noindent
 \begin{enumerate}
 \item The set $\CT^c$ is the union of three curves
$\Gamma_{\alpha_+,1}$, $\Gamma_{\alpha_-,\alpha_+}$, $\Gamma_{-1,1}$ with only one of infinite length $\Gamma_{-1,1}$.
\item The three curves meet at a common point $\lambda_0e^{\frac {i\pi} 4}$ with $\lambda_0\in\rit^+$.
\item The curve $\Gamma_{-1,1}$ goes from $\lambda_0e^{\frac {i\pi} 4}$ to $\rit+\frac i 3$.
\item $\Gamma_{\alpha_-,\alpha_+}=\{\lambda e^{\frac {i\pi} 4}; \lambda\in\rit, 0\leq \lambda<\lambda_0\} $.
\item The curve $\Gamma_{\alpha_+,1}$ goes from  $i$ to  $\lambda_0e^{\frac {i\pi} 4}$.
\end{enumerate}
\end{theorem}

{\sl Proof:}
(1)~:\par
Let $E=\lambda^2 e^{i\beta}$ with $\beta\in\rit$ and $\lambda\in\rit$. We fix $\beta$ and increase $\lambda$.
We get $z(x,y,\lambda^2 e^{i\beta})=z(\frac x \lambda,\frac y \lambda,
e^{i\beta})$. The shape of the Stokes lines remains invariant up to
dilation by $\lambda$.
Let $\beta$ be such that $\re S_{\alpha_-.\alpha_+}(e^{i\beta})\neq 0$ then under symmetry we get
 $\re S_{0,\alpha_+}(e^{i\beta})\neq 0$ , and $\re S_{\alpha_-.0}(e^{i\beta})\neq 0$.
So if $\beta$ is such that there is no Stokes line issuing from the turning
points and going through  $0$, then there exists a neighborhood  of $0$ with no point of any  Stokes lines issuing from the turning points.
For $\lambda$ big enough, we  get that $-1$ and $1$  belong to this neighborhood of $0$.
Then $E$ will belong to $\CT^c$ if and only if $\re S_{-1,1}(E)\neq 0$.

If $\beta$ is such that $\re S_{\alpha_-,\alpha_+}(e^{i\beta})= 0$   then
$\beta=\pm \frac \pi 4$ and Figure \ref{stokesaxe} and the fact that the
spectrum is included in the set $\{E\in\cit; E=\rit^++ix^2, x\in[-1,1]\}$
(i.e. the pseudo spectrum or the values of the principal symbol) shows that
$E\in\CT^c$, $E=\lambda e^{i\frac \pi 4}$, $\lambda\in\rit$,   implies $E$ bounded.
So we have that if $E$ is big enough and belongs to $\CT^c$, then  $E\in\Gamma_{-1,1}$.

We  choose  the determination of $\alpha_\pm$ so that the set
$$\Gamma_{\alpha_-,1}=\{E\in\cit;\, \re S_{\alpha_-,1}(E)= 0;\, \im
 S_{\alpha_-,1}(E)\geq 0\}$$
 is empty see Figure \ref{fullstoke}.
\par
The second statement is proved by  the relation
$$ S_{0,\alpha_+}(E) +  S_{\alpha_+,1}(E) + S_{1,0}(E)=0,$$\par
and illustrated in Figure \ref{evo}.\par
 \begin{figure}[ht]
\begin{center}
\epsfysize=4cm
\epsfbox{evo.eps}
\end{center}
\caption{Stokes line for $E$ near  $\lambda_0 e^{\frac {i\pi} 4}$, the arrows locate the positions of $-1$ and $1$ }
\label{evo}
\end{figure}

Point 3 is proved in Theorem \ref{alinfini} below.\par
Point 4 is deduced from
 $$z(x,y,E)=\pm \int_{xe^{\frac{i\pi} 8}} ^{y e^{\frac{i\pi} 8}} (u^2-Ee^{-\frac{i\pi} 4} )^{1/2} \,du$$
 So for $E= \{\lambda e^{\frac {i\pi} 4},\lambda\in\rit\}$ we obtain
  $$\re S_{0,\alpha_+}(E)=\re \pm \int_{0} ^{\pm \sqrt\lambda} (u^2-\lambda )^{1/2} \,du=0.$$
The existence of $\lambda_0$ is obvious in Figure \ref{extre}.\par
 \begin{figure}[ht]
\begin{center}
\epsfysize=7cm
\epsfbox{stokesaxe.eps}
\end{center}
\caption{Stokes line for $E=\lambda e^{\frac {i\pi} 4}$}
\label{stokesaxe}
\end{figure}

Point 5 is obvious from Figure \ref{extre}.\par
 \begin{figure}[ht]
\begin{center}
\epsfysize=4cm
\epsfbox{extre.eps}
\end{center}
\caption{Evolution of the Stokes line for $E\in \Gamma_{\alpha_+,1}$, the arrows locate the position of $-1$ and $1$}
\label{extre}
\end{figure}

\cqfd\par

The hypothesis  \textbf{(H1)} is obviously satisfied by the potential $V(x)=ix^2$.
The hypothesis \textbf{(H2) (H3)} are satisfied by the potential $ix^2$ ( and by any  potential $e^d x^2$  if $e^{\frac d 2}\notin \rit $ ) but not by the potential $x^2$.

For large energies, we summarize the result of Proposition \ref{inver} , Theorem \ref{reci}   and Theorem \ref{curve} in
\begin{theorem}\sl We have\par\noindent
\begin{enumerate}
\item Large values of $\lim_h\Sp P^h$  are  close to the curve $\Gamma_{-1,1}$.
\item The curve  $\Gamma_{-1,1}=\{E\in \C, \re S_{-1,1}(E)=0\}$ tends  to infinity, and  is asymptotic to the line $\R+i/3$.
\end{enumerate}
\label{alinfini}
\end{theorem}
 One could compute the set $\CT^c$ with Matlab or Mathematica. In Figure \ref{spectre}  we have  drawn the set $\CT^c$ using Theorems \ref{forme} and \ref{alinfini}. \par
\begin{figure}[ht]
\begin{center}
\epsfysize=6cm
\epsfbox{spectre.eps}
\end{center}
\caption{Limit spectrum of $P^h$ }
\label{spectre}
\end{figure}\par\sl
For large energies and perturbation , we summarize the results of Theorem \ref{reciper} and Theorem \ref{curveper} in
\begin{theorem}\sl We have\par\noindent
\begin{enumerate}
\item Large  $E\in\lim_h\Sp P_\delta^h$  are  near to   the union of the two  curves
$\Gamma_{-1,\beta}$, $\Gamma_{\beta,1}$.
\item The curve  $\Gamma_{-1,\beta}=\{E\in \C, \re S_{-1,\beta}(E)=0\}$ tends to infinity, and is asymptotic to  $\R+i \frac 1 3 (\beta^2-\beta+1) -i\delta$.
\item The curve  $\Gamma_{\beta,1}=\{E\in \C, \re S_{\beta,1}(E)=0\}$ goes to infinity, and  is asymptotic to $\R+i \frac 1 3 (\beta^2+\beta+1) +i\delta$.
\end{enumerate}
\label{infiniper}
\end{theorem}
We remark that the two curves $\Gamma_{-1,\beta}$ $\Gamma_{\beta,1}$ are distinct for $\beta\neq0$.

Figure \ref{spectreper} represents the two Theorems,   \ref{alinfini} and \ref{infiniper}.\par

\begin{figure}[ht]
\begin{center}
\epsfysize=6cm
\epsfbox{spectreper.eps}
\end{center}
\caption{Spectrum of $P^h$ in plain  lines and spectrum of $P^h_\delta$  for $\delta$ small near to $\infty$ as dotted lines.}
\label{spectreper}
\end{figure}\par\sl

\section{pseudo-spectrum}
\label{speudospec}
We have defined here the pseudo-spectrum of $H_0^h$ as the set
$${\overline{ \{\xi^2+V(x), x\in(-1,1), \xi\in\rit, \im(\xi V'(x))\neq 0\}}}.$$
We remark that for any $z$ is this set, we could contruct a function $u^h$ in the domain of $H_0^h$  such that
$$\|(H^h_0-z)u^h\|=\O(h^\infty) \|u^h\|.$$

Let suppose now that we put two jumps in the potential, one at  $\beta$ and one at $\beta'$.
i.e $V_{\beta\beta'}=V+i\delta_1 H_\beta+i\delta_2 H_{\beta'}.$

Assume that $V$ verifies the  hypothesis  \textbf{(H1)} and~:

\textbf{(H4)} If  $E$ is  large enough, with small enough imaginary part then  $1$,  $-1$, $\beta$ and $\beta'$  belong to the same Stokes region.

Then we obtain just as in Theorems \ref{reciper} and \ref{curveper}  the next two theorems
\begin{theorem}
For $E_0$ large, the following two conditions are equivalent
\begin{enumerate}
\item $\re(\int_{-1}^\beta\sqrt {V_{\beta\beta'}-E_0}\,dx=0)$ or $\re(\int_{\beta'}^1\sqrt {V_{\beta\beta'}-E_0}\,dx=0)$ or \par\noindent
$\re(\int_{\beta}^{\beta'}\sqrt {V_{\beta\beta'}-E_0}\,dx=0)$ .
\item $E_0\in\lim_h\Sp H^h_\delta$.
\end{enumerate}
\end{theorem}

\begin{theorem}
For  $a$ large enough, the equation
$$
\re S_{\beta,\beta'}(E)|_{E=a+ib}= \re \int_{\beta}^{\beta'} \sqrt{V(x)+i\delta_1-a-ib}\,dx=0
$$
has  a unique solution $b(a)$, the solution  satisfies
$$ b(a)=i \frac 1 {\beta'-\beta} \im(Y(\beta')-Y(\beta))-i \delta_1 +O( \frac 1 a ).$$
\end{theorem}
We remark that
$$\lim_{\beta'\to \beta}\frac 1 {\beta'-\beta} \im(Y(\beta')-Y(\beta))=\im(V(\beta))$$
This means, letting $\beta'\to\beta$, and $\delta\to 0$ any values of the form
$\{\xi^2+i\im( V(x)); \xi \in\rit, |\xi|>>1, x\in[-1,1] \}$ is  in the
spectrum of this kind of  perturbation of $H^h$.
 Thus we can reach all the large values of the  pseudo-spectrum  with this special type of perturbation.

config: arxiv


TITLE: Prove that the union of countably many countable sets is countable.
QUESTION [74 upvotes]: I am doing some homework exercises and stumbled upon this question. I don't know where to start.

Prove that the union of countably many countable sets is countable.

Just reading it confuses me.
Any hints or help is greatly appreciated! Cheers!

REPLY [1 votes]: Using
The wiki definition for countable sets,
The axiom of countable choice,
The make-it-a-disjoint-union technique found here,
The padding-if-necessary the index out to $\omega$ technique,
The prime integers form an infinite set,
The prime factorization theorem.
the problem is (eventually) reduced to observing that
$\quad U =\;\bigsqcup_{p \text{ is a prime}} \{p^n \mid n \ge 1\}$
is a subset of $\Bbb N$.
Note: When there are duplicates in the original sets you'll need to get the smallest index where the element occurs to define the injective mapping into $U$.


TITLE: Given a p.d.f. of $X$, find the p.d.f. of $Y=4-3X$
QUESTION [0 upvotes]: The probability density function of a random variable $X$ is given by
$$f_X (x) =\begin{cases}
\frac{5}{x^2} & x > 5\\
0 & \text{otherwise}\\
\end{cases}
$$
Find the probability density function of $Y = 4-3X $. Don't forget to state the support of the p.d.f (set of values $y$ for which $y > 0$).

Going to solve this using this formula from my textbook
$$f_Y(y)\frac{f_X(h^{-1}(y))}{|h'(h^{-1}(y))|}$$
Let $h^{-1}(y) = 4 - 3y$.
Then $h(y) = \dfrac{4-y}{3}$ and $h'(h^{-1}(y)) = 0-3y^{1-1} = -3$
$$\dfrac{\dfrac{5}{(4-3y)^2}}{|-3|} = \dfrac{5/3}{(4-3y)^2}$$
with support
$$\frac{4-y}{3} > 5 \iff4 - y > 15 \iff y < -11$$
Answer is
$$\dfrac{15}{(4-y)^2}$$
which is different from mine. Why?
If I let $h^{-1}(y) = \frac{4-y}{3}$,
then
$$\frac{5 \cdot 9}{(4-y)^2} \cdot \frac{1}{\left|h'\left(\frac{4-y}{3}\right)\right|} = \frac{45}{(4-y)^2} \cdot \frac{1}{\frac{1}{3}} = \frac{45 \cdot 3}{(4-y)^3}$$
which is again, not the same as the solution.

I understand it for the most part except I don't get the answer.
I still don't get how $$h'(h^{-1}(y)) = h'\left(\frac{4-y}{3}\right) = -3$$

REPLY [0 votes]: You're given $y = 4-3x$. Let this $y=h(x)=4-3x$. Then  $x=h^{-1}(y)=\dfrac{4-y}{3}$. So by the given formula,$$f_Y(y)=\frac{f_X(h^{-1}(y))}{|h'(h^{-1}(y))|}=\frac{\dfrac{5}{\left(\frac{4-y}{3}\right)^2}}{|-3|}=\frac{15}{(4-y)^2}.$$
Note that $h'$ is constant, since $Y$ is linear: $$h'(\color{blue}{x})=-3\,\,\forall x\implies h'(\color{blue}{h^{-1}(y)})=-3.$$ On the other hand, if you've noticed, a shorthand for this when $Y=\alpha X+\beta$, $$f_Y(y)=\dfrac{f_X(\dfrac{y-\beta}{\alpha})}{|\alpha|}$$ for $\alpha\ne0.$


text:
\section{Numerical Examples}\label{sec:Num}

In this section, we present some numerical experiments using the constructed schemes. We focus on the influence
of the different numerical fluxes on the long time behaviour of the error.
From \cite{kopriva2017error, oeffner2018error}, we know that in case of constant coefficients the choice of the
numerical flux has an essential influence on the error behaviour, especially in the Gauß-Lobatto-Legendre case.

We consider our model problem, the linear advection equation
\begin{equation*}\tag{\ref{eq:Model_problem}}
\begin{aligned}
   \partial_t u(t,x)+\partial_x (a(x)u(t,x))&=0, &t>0,\; x\in (x_L,\;x_R),\\
   u(t,x_L)&=g_L(t), & t\geq 0, \\
   u(0,x)&=u_0(x), & x\in (x_L,\;x_R),
\end{aligned}
\end{equation*}
with smooth speed $a(x)>0$, initial condition $u_0$ and boundary condition $g_L$.
The solution $u$ of the corresponding Cauchy problem can be calculated by the method of characteristics,
see e.g. \cite[Chapter 3]{bressan2000hyperbolic}.
As time integrator, we use the fourth order, ten stage, strong stability
preserving Runge-Kutta method of \cite{ketcheson2008highly} and the time step is
chosen such that the time integration error is negligible. Although the term
``strong-stability preserving'' means the preservation of stability properties of the
explicit Euler method and the explicit Euler method is not stable for our numerical
experiments, this fourth order Runge-Kutta method is strongly stable for linear
equations \cite{ranocha2018L2stability}. All elements are of uniform size.

\subsection{Coefficient \texorpdfstring{$a(x)=x$}{a(x)=x}}\label{subsec:a(x)=x}

In our first experiment, we choose $a(x)=x$ with initial condition $u_0(x) =\sin (12(x-0.1))$.
The interval is $[x_L,x_R] = [0,2\pi]$ and we choose the inflow boundary condition such that
we get the solution
\begin{equation*}
 u(t,x)= \exp(-t) u_0\bigl( x \exp(-t) \bigr).
\end{equation*}
For the coefficient $a(x)=x$, the first derivative of $a$ is strictly positive,
implying $Int_d > 0$.

In our first simulation, we use $K=40$ elements and calculate the solutions up to $t=20$ with
\num{200000} time steps. In \autoref{fig:Splitted_unsplitt_gauss_Lobatto}, we plot the long
time error behaviour using polynomial degrees three and four. One recognizes that in all
cases the error remains bounded in time.

In the first row of \autoref{fig:Splitted_unsplitt_gauss_Lobatto}, all terms (surface, flux and volume) are split whereas in the second row
they are not. We see that the error for the split version behaves like in the case of
constant coefficients \cite{kopriva2017error, oeffner2018error}. We mean that the errors using
the upwind fluxes are always lower than the ones using central fluxes and one may recognize
that we have some noisy behaviour using the central fluxes. Using upwind fluxes, the error
reaches its asymptotic value faster than for the central fluxes.

In the second row, the unsplit discretisation is used. We recognize that we lose the predictions
from \cite{kopriva2017error, oeffner2018error} that applying the upwind flux yields a more
accurate solution. The absolute value is also bigger applying the unsplit versions and we have
again the noisy behaviour by applying the central fluxes.

\begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.495\textwidth}
  \centering
    \includegraphics[width=\textwidth]{Spl_u_fu_p_3_N_40_Nu_200000_k_to_20.pdf}
    \caption{$N=3$, $K=40$, $t=20$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.495\textwidth}
  \centering
    \includegraphics[width=\textwidth]{Spl_u_fu_p_4_N_40_Nu_200000_k_to_20.pdf}
    \caption{$N=4$, $K=40$, $t=20$.}
  \end{subfigure}\\%

 \begin{subfigure}[b]{0.495\textwidth}
  \centering
    \includegraphics[width=\textwidth]{Nonspl_u_fu_p_3_N_40_Nu_200000_k_to_20.pdf}
    \caption{$N=3$, $K=40$, $t=20$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.495\textwidth}
  \centering
    \includegraphics[width=\textwidth]{Nonspl_u_fu_p_4_N_40_Nu_200000_k_to_20.pdf}
    \caption{$N=4$, $K=40$, $t=20$.}
  \end{subfigure}
  \caption{Errors of numerical solutions using split and unsplit discretisations, $a(x) = x$.}
  \label{fig:Splitted_unsplitt_gauss_Lobatto}
\end{figure}

Comparing all four plots, we recognize that the best results are obtained by using Gauß-Legendre nodes and
the split discretisation. Therefore, we have a closer look on this.
In \autoref{fig:flux_split_unsplit_gauss}, we consider only Gauß nodes and compare the split
numerical fluxes and the unsplit numerical fluxes (with split surface and volume terms).
\emph{True} in the legend indicates the split numerical fluxes and
\emph{false} the unsplit ones.
The experiment on the left-hand side demonstrates clearly that the noisy behavior for the central flux
transfers also to the application of Gauß-Legendre nodes if all terms are split. Furthermore, we can
hardly indicate some difference between the usage of split and unsplit upwind fluxes here, whereas
we have a slight different behaviour in the usage of the central fluxes.
The test indicates that the split discretisation (volume/surface and numerical fluxes)
should be preferred, matching our stability analysis.

 \begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.495\textwidth}
    \includegraphics[width=\textwidth]{Fluxspl_Gauss_spl_Spl_u_fu_p_3_N_40_Nu_200000_k_to_20.pdf}
    \caption{$N=3$, $K=40$, $t=20$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.495\textwidth}
    \includegraphics[width=\textwidth]{Fluxspl_Gauss_u_fu_p_4_N_40_Nu_200000_k_to_20.pdf}
    \caption{$N=4$, $K=40$, $t=20$.}
  \end{subfigure}
  \caption{Errors of numerical solutions using the split form and both split (true) and
           unsplit (false) numerical fluxes.}
  \label{fig:flux_split_unsplit_gauss}
\end{figure}

\subsection{Coefficient \texorpdfstring{$a(x)=x^2$}{a(x)=x²}}\label{subsec:a(x)=x2}

In our second experiment, we choose $a(x)=x^2$ with initial condition
$u_0(x) = \cos\left( \frac{\pi x}{2} \right)$. The interval is $[x_L,x_R] = [0.1,1]$
and we choose the inflow boundary condition according to the solution
\begin{equation}\label{sol_a_x_x}
  u(t,x)= \frac{u_0\bigl(x/(1+tx)\bigr)}{(1+tx)^2}.
\end{equation}

In our simulation shown in \autoref{fig:flux_splitted_unsplit_gauss2},
we apply different numbers of time steps up to $t = 200$.
\begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.495\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu2_p_3_N_40_Nu_500000_k_to_50.pdf}
    \caption{$N=3$, $K=40$, final time $t = 50$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.495\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu2_p_3_N_40_Nu_2000000_k_to_200.pdf}
    \caption{$N=3$, $K=40$, final time $t = 200$.}
  \end{subfigure}
  \caption{Errors of discretisations for $a(x) = x^2$, $[x_L,x_R] = [0.1,1]$,
           and $u_0(x) = \cos\left( \frac{\pi x}{2} \right)$.}
  \label{fig:flux_splitted_unsplit_gauss2}
\end{figure}
First, we recognize that all errors are bounded in time,
but different from the first case we do not have any noisy behavior
of the central fluxes, at least we can not identify some.
Simultaneously, the unsplit central flux error  with Lobatto nodes
increases at first rapidly before it finally tends to its
asymptotic value. In all cases, the errors are small
but we get always the \emph{best} results by applying Gauß-Legendre nodes.
Nevertheless, it takes a lot of time for the errors to reach the asymptotic values.
Even at time $t=800$, the asymptotic is still not reached, cf. \autoref{fig:flux_splitted_unsplit_gauss2_log}.
\begin{figure}
\centering
  \includegraphics[width=0.5\textwidth]{Log_u2_800_sec}
  \caption{Error behaviour on a logarithmic scale for $a(x) =  x^2$, $t=800$.}
  \label{fig:flux_splitted_unsplit_gauss2_log}
\end{figure}

In the first simulation, the interval has been chosen as $[x_L,x_R] = [0.1,1]$ to guarantee the
positivity of the derivative of $a$ and also of its interpolation. Now, we change the interval
to $[x_L,x_R] = [-0.1,1]$, resulting in two major issues. First, the first derivative of $a$
is not strictly
positive anymore and the solution develops a pole at time $t=10$. Here, the solution is also
not uniformly bounded in its Sobolev norm and our error bounds \eqref{eq:Errorbound3}
and \eqref{eq:Errorbound4} do not hold. Nevertheless, the error behaviour can be investigated.
Using only the split discretisation for different times, we see in
\autoref{fig:flux_splitted_unsplit_gauss3}
that the errors increase and will increase further. They are unbounded.
Simultaneously, we also recognize
that the errors using Gauß-Legendre nodes still increase slower
due to the fact that the methods using
these nodes are more accurate.
 \begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.324\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu2_p_3_N_40_Nu_50000_k_to_5.pdf}
    \caption{$N=3$, $K=40$, $t=5$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.324\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu2_p_3_N_40_Nu_90000_k_to_9.pdf}
    \caption{$N=3$, $K=40$, $t=9$.}
  \end{subfigure}
    ~
  \begin{subfigure}[b]{0.324\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu2_p_3_N_40_Nu_99900_k_to_10.pdf}
    \caption{$N=3$, $K=40$, $t=9.9$.}
  \end{subfigure}
  \caption{Errors of discretisations for $a(x) = x^2$, $[x_L,x_R] = [-0.1,1]$,
           and $u_0(x) = \cos\left( \frac{\pi x}{2} \right)$.}
  \label{fig:flux_splitted_unsplit_gauss3}
\end{figure}

Furthermore, by changing the initial condition to $u_0(x)=\exp(-x^4)$ instead of
$u_0(x) =\cos\left( \frac{\pi x}{2} \right)$, we are able to avoid the pole in the solution
\eqref{sol_a_x_x} since the exponential function will tend fast enough to zero compared to
$(1+tx)^2$ and we can extend the solution. Nevertheless we get further problems here.
If we have a look on the error behaviour in \autoref{fig:flux_splitted_unsplit_gauss3_3},
we see that we get a similar increase of the errors like in \autoref{fig:flux_splitted_unsplit_gauss3},
but they are much smaller. Nevertheless they are still unbounded, but why do we have this behaviour?
The analytical solution is for fixed times bounded, nevertheless we demand
as one assumption right at the beginning at equation \eqref{eq:Model_problem} the solution to be
\textbf{uniformly} bounded in time. However, this is not the case anymore. This demonstrates
again how essential this assumption is.

The same issue arises if we are investigate $a(x)=\cosh(x)+1$ as in \cite{ranocha2018generalised}.
Therefore, we skip this case here.

\begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.324\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu4_p_3_N_40_Nu_50000_k_to_5.pdf}
    \caption{$N=3$, $K=40$, $t=5$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.324\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu4_p_3_N_40_Nu_90000_k_to_9.pdf}
    \caption{$N=3$, $K=40$, $t=9$.}
  \end{subfigure}
    ~
  \begin{subfigure}[b]{0.324\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu4_p_3_N_40_Nu_99900_k_to_10.pdf}
    \caption{$N=3$, $K=40$, $t=9.9$.}
  \end{subfigure}
  \caption{Errors of discretisations for $a(x) = x^2$, $[x_L,x_R] = [-0.1,1]$, and $u_0(x) = \exp(-x^4)$.}
  \label{fig:flux_splitted_unsplit_gauss3_3}
\end{figure}

\subsection{Coefficient \texorpdfstring{$a(x) = 1-x^2$}{a(x) = 1-x²}}\label{subsec:test_1_x_x}

Here, we choose the coefficient $a(x) = 1 - x^2$. The solution of the Cauchy problem is
\begin{equation}\label{solutions_ax_1_x}
  u(t,x)
  =
  \frac{u_0\bigl( (-x \cosh(t) + \sinh(t)) / (x \sinh(t) - \cosh(t)) \bigr)}
       {(\cosh(t) - x \sinh(t))^2}.
\end{equation}
Using the domain $[x_L,x_R] = [-1, 0.9]$ and the initial condition $u_0(x) = \sin(\pi x)$,
the solution remains bounded but $a'(x) < 0$ for $x > 0$.
\begin{figure}
\centering
  \includegraphics[width=0.5\textwidth]{Spl_u_fu5_p_4_N_40_Nu_200000_k_to_20}
  \caption{Error behaviour for $a(x) = 1 - x^2$, $t=20$.}
  \label{fig:flux_splitted_no_convergece}
\end{figure}
If we investigate now the long time error behaviour, we get a huge  increase of the errors
if we apply the central fluxes, cf. \autoref{fig:flux_splitted_no_convergece}. This matches
perfectly our theoretical investigations in \autoref{sec:Error_Lobatto} and
\autoref{sec:Error_Gauss}, cf. Remark~\ref{remark_int}.
We explain the reasons again in detail in the next test case and a physical interpretation
and illustration is given afterwards.

\subsection{Coefficient \texorpdfstring{$a(x) = \cos(x)$}{a(x) = cos(x)}}\label{subsec:a(x)=cos(x)}

Here, we choose $a(x) = \cos(x)$ and $u_0(x) = \sin(5 x)$. The solution of the Cauchy problem is
\begin{equation}\label{eq:Test_a_x_cos_x}
\begin{aligned}
  u(t,x)& = u_0\bigl( x_0(t,x) \bigr) \frac{\cos \bigl( x_0(t,x) \bigr) }{\cos (x)},
  \\
  x_0(t,x)
  &=
  - 2 \arctan\bigl( \tanh\bigl( t/2 - \artanh( \tan(x/2) ) \bigr) \bigr).
\end{aligned}
\end{equation}
We can find an interval for our solution \eqref{eq:Test_a_x_cos_x} so that $a'(x)\leq 0$
and $u(t,x)$ does not blow up, e.g. $[x_L, x_R] = [0.1, \pi/3]$. The solution remains
bounded but $a'(x) < 0$.

 \begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.332\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu7_p_3_N_30_Nu_50000_k_to_5.pdf}
    \caption{$N=3$, $K=30$, $t=5$.}
  \end{subfigure}
   ~
  \begin{subfigure}[b]{0.332\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu7_p_3_N_30_Nu_400000_k_to_40.pdf}
    \caption{$N=3$, $K=30$, $t=40$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.332\textwidth}
    \includegraphics[width=\textwidth]{Spl_u_fu7_p_3_N_30_Nu_1000000_k_to_100.pdf}
    \caption{$N=3$, $K=30$, $t=100$.}
  \end{subfigure}
  \caption{For $a(x) = \cos(x)$, the slope of $a$ is negative. The errors of numerical solutions
           using the central flux increase, whereas the upwind flux results in bounded errors.}
  \label{fig:flux_splitted_unsplit_gauss4}
\end{figure}

In \autoref{fig:flux_splitted_unsplit_gauss4}, we see the behaviour of the error for different times.
First, one may suppose that the error remains bounded in time, but this is not the case as can be seen
stepping further in time.
Using the central fluxes ($\sigma=0$), the $BTs$ terms are zero and we do not find an $\eta$
which is bounded with a positive constant from below away from zero.
One may recognize also that for Gauß-Legendre nodes, the error increases much slower
(second picture).
Surely, one reason for this is the smaller error in the Gauß-Legendre case. Furthermore, also
the term $\Theta$ may have a positive impact of the error behaviour.

However, this example demonstrates well that the condition $a'(x)>0$ is essential for the
boundedness of the error, also in the test case of section~\ref{subsec:test_1_x_x}. One can
rescue \eqref{eq:Errorbound3} and \eqref{eq:Errorbound4} by applying the upwind flux like it can be seen
in this test case and especially in \autoref{fig:flux_splitted_unsplit_gauss4_log}.
\begin{figure}[!htp]
\centering
  \begin{subfigure}[b]{0.49\textwidth}
    \includegraphics[width=\textwidth]{Spl_Log_u_fu7_p_3_N_30_Nu_400000_k_to_40.pdf}
    \caption{$N=3$, $K=30$, $t=40$.}
  \end{subfigure}
  ~
  \begin{subfigure}[b]{0.49\textwidth}
    \includegraphics[width=\textwidth]{Spl_Log_u_fu7_p_3_N_30_Nu_1000000_k_to_100.pdf}
    \caption{$N=3$, $K=30$, $t=100$.}
  \end{subfigure}
  \caption{Error of numerical solutions for $a(x) = \cos(x)$ in logarithmic scale.}
  \label{fig:flux_splitted_unsplit_gauss4_log}
\end{figure}

By applying an SBP-SAT finite difference scheme with one block, the internal terms $BTs$
do not exist. Using the SBP difference operator of \cite{mattsson2004summation} with interior
order of accuracy eight, the split form, and \num{100} nodes for this problem, the error
is unbounded, as can be seen in \autoref{fig:SBP_FD__a_cos__order_8__N_100__split_true}.
However, if the high-order artificial dissipation operator of \cite{mattsson2004stable} is
applied additionally, the error remains bounded.
\begin{figure}
  \centering
  \captionsetup{skip=0pt}
  \includegraphics[width=\textwidth]{SBP_FD__a_cos__order_8__N_100__split_true}
  \caption{Errors of numerical solutions using SBP finite difference schemes.}
  \label{fig:SBP_FD__a_cos__order_8__N_100__split_true}
\end{figure}

Comparing Figures~\ref{fig:flux_splitted_unsplit_gauss4_log} and
\ref{fig:SBP_FD__a_cos__order_8__N_100__split_true} demonstrates that stabilisation
induced by upwind fluxes or artificial dissipation operators is crucial and comparable.
Furthermore, Gauss-Legendre nodes not including boundary points provide some stabilisation.

\subsection{A first analytical study}\label{subsec:analytical}
\revsec{
As can be seen in Figures \ref{fig:flux_splitted_unsplit_gauss4_log}
and \ref{fig:SBP_FD__a_cos__order_8__N_100__split_true}, if $a'(x)$ is not positive
the long time errors show different behaviours depending on the dissipation which is
added to the scheme by numerical fluxes or artificial dissipation terms.
Here, we give a short rough analysis on this topic under what conditions we can guarantee boundedness.
A more detail analysis should follow in future research with more validations. \\
We are starting considering  $\eta_G(t)$ from \eqref{eq:eta_def}.
It is
\begin{equation}\label{eq:eta_analysis}
 \eta(t):=\frac{BTs +Int_d+\Theta_2}{||\epsilon_1||^2_N}
\end{equation}
with $Int_d:= \frac{1}{2} \sum_{k=1}^K \frac{\Delta x_k}{2} \l(
\partial_x \vec{a}^k, \mat{\epsilon}_1^k
  \vec{\epsilon}_1^k \r)_N$.
A sufficient condition for the mean of $\eta(t)$  to be positive is that every value of $\eta(t)$  is positive.
Therefore, we require
\begin{equation*}
 \frac{BTs +Int_d+\Theta_2}{||\epsilon_1||^2_N} >0.
\end{equation*}
If the derivative of $a$ is negative, we can reformulate the inequality above as
\begin{equation*}
 \left( BTs +\Theta_2 \right)\frac{1}{||\epsilon_1||^2_N} >
 \frac{1}{2||\epsilon_1||^2_N} \left( \sum_{k=1}^K \frac{\Delta x_k}{2}
 \l(   \left|\partial_x \vec{a}^k \right|,   \mat{\epsilon}_1^k
  \vec{\epsilon}_1^k  \r)_N\right),
\end{equation*}
and even strengthen our assumptions by requiring
\begin{equation}\label{eq:condtion_on_a}
\begin{split}
  &\left( BTs +\Theta_2 \right)\frac{1}{||\epsilon_1||^2_N} >
 \frac{\max_{x\in (x_0,x_K)} \left|\partial_x \vec{a} \right| }{2||\epsilon_1||^2_N}
 \left( \sum_{k=1}^K \frac{\Delta x_k}{2} \l(    \vec{1}, \mat{\epsilon}_1^k
  \vec{\epsilon}_1^k \r)_N \right)\\
 \text{ or }&\left( BTs +\Theta_2 \right)\frac{1}{||\epsilon_1||^2_N} >
 \frac{\max_{x\in (x_0,x_K)} \left|a'(x) \right| }{2} .
\end{split}
\end{equation}
From \eqref{eq:condtion_on_a} we realize the $BTs$-terms are responsible to guarantee
that this sufficient condition is fulfilled. In case of a central numerical flux, $BTs\equiv0$
and we have to add additional dissipation to the scheme as it is done in the SBP-SAT schemes in
Figure \ref{fig:SBP_FD__a_cos__order_8__N_100__split_true}. However, also the dependence of the error is
important and we may also realize that in case of using Gau\ss-Legendre we rather get the condition
\eqref{eq:condtion_on_a} fulfilled. However, this estimation is rough and should be improved in
further research.
}
\subsection{Physical Interpretation and Illustration}
\label{sec:phys-interpretation}

In order to understand some results better, a physical interpretation of the
advection equation can be used. This serves also as illustration and explains
the rational behind some of the choices regarding for example the numerical
experiments.

The advection equation $\partial_t u + \partial_x (a u) = 0$ with non-negative
velocity $a(x)$ is a conservation law with varying coefficients. Thus, the total
mass $\int u$ is conserved and $u$ is transported from left to right due to
$a(x) \geq 0$. In order to compute analytical solutions of the Cauchy problem,
the method of characteristics can be used, cf. \cite[Chapter 3]{bressan2000hyperbolic}.
\begin{itemize}
  \item
  Solve the ODE $x'(t) = a\bigl( x(t) \bigr)$, $x(0) = x_0$, for $x(t) = x(t; x_0)$.
  Compute also the inverse function $x_0 = x_0(t; x)$.

  \item
  Solve the ODE $z'(t) = - a'\bigl( x(t;x_0) \bigr) z(t)$, $z(0) = z_0$, for
  $z(t) = z(t; z_0, x_0)$.

  \item
  Set $z_0 = z_0(x_0) = u_0(x_0) = u_0\bigl( x_0(t;x) \bigr)$ and obtain the analytical
  solution $u(t,x) = z(t; z_0, x_0) = z(t; z_0\bigl(x_0(t;x)), x_0(t;x)\bigr)$.
\end{itemize}
In the second step, if $a' > 0$, the absolute value of $z(t)$ decreases. Contrary,
if $a' < 0$, the absolute value of $z(t)$ increases. This corresponds directly
to the physical interpretation as transport problem. Since $u$ is conserved and
transported with velocity $a(x)$, there is a loss of $u$ if $a' > 0$, since
there is less new mass coming from the left than going to the right. Similarly,
$a' < 0$ yields an increase of $u$, since more mass is coming from the left than
transported to the right. This explains also the critical role of $a'(x)$.
If $a' < 0$, there can be blow-up phenomena in the solution $u$, resulting in
possibly finite life spans and increasing energies and errors of numerical
solutions. If $a' > 0$, this cannot happen.

If one wants to investigate a situation with $a'(x) > 0$ in some parts and
$a'(x) < 0$ in other parts of the domain, there are basically two possibilities.
Firstly, there can be a local minimum of $a(x)$, e.g. for $a(x) = x^2$.
In this case, there can be a blow-up of the solution $u$,
since more mass is coming from the left than transported to the right at this
minimum. However, this blow-up phenomenon caused by the varying transport velocity
$a(x)$ can be balanced by the initial condition $u_0$. If there is simply not
enough mass on the left, than the higher transport speed there can not cause a
blow-up of the solution $u$. This explains our choice of the intervals and the
initial conditions for these cases.

Secondly, there can be a local maximum of $a(x)$, e.g. for $a(x) = 1 - x^2$ or
$a(x) = \cos(x)$. Now, there is no blow-up at the critical point, since more
mass is transported to the right. However, both examples have stagnation points
with $a(x) = 0$. At such points, there will be a  blow-up of the solution, since
mass is coming from the left but not transported to the right. In order to
avoid this phenomenon of the Cauchy problem, the interval can be chosen adequately,
i.e. bounded away at the right from the point with $a(x) = 0$. Then, the blow-up
of the solution of the Cauchy problem does not cause any problems for the corresponding
solution of the initial value problem. This explains our choices of the domains
for these cases.


TITLE: Explanation how it can be that $f'(x) = g(x)$ but wolfram alpha says $\int g(x) \neq f(x)$?
QUESTION [3 upvotes]: I've just started learning about antiderivatives/primitive functions/indefinite integrals, and I have the functions
$f(x) = 3 \ln((\frac{x+2}{3})^2 + 1)$
$g(x) = \dfrac{2\frac{x+2}{3}}{(\frac{x+2}{3})^2 + 1}$
I came to the conclusion that $f'(x) = g(x)$ so that $\int g(x) = f(x)$ and I wanted to check with wolfram alpha, but wolfram says that $\int g(x) \neq f(x)$ even though it says $f'(x) = g(x)$.
It seems to me that this violates the definition of antiderivative?

REPLY [1 votes]: The function $f(x)$ is one of the infinite number of antiderivatives of the function $g(x)$ which differ only by a constant. In the case of $f(x)$, that constant, typically denoted as $C$, is 0. When you integrate $g(x)$, you get a whole set of functions. Not just one function. That set is usually denoted $f(x)+C$ where $C\in\mathbb{R}$. Therefore, this means that there are going to be as many functions in that set as there as real numbers—that is, an infinite number. That's why, technically speaking, $\int g(x)\,dx\neq f(x)$. The result of the integration process is not a function, but a set of functions. Those are slightly different concepts.


text:
\section{Direct proof of identity \eqref{eq:channelSymmetry} for symmetric distributions}
\label{proof:equi_channel_symmetry}
If $\sfx(-dh) = e^{-2h} \sfx(dh)$ holds, then we have
\begin{align*}
&\int_{-\infty}^{\infty} (\tanh h)^{2k-1} \, \sfx(dh)
	 = \int_{0}^{\infty} (\tanh h)^{2k-1} \, \sfx(dh) - \int_{0}^{\infty} (\tanh h)^{2k-1} \, \sfx(-dh)  \\
	 = \,&\int_{0}^{\infty} (\tanh h)^{2k-1} (1-e^{-2h}) \, \sfx(dh)   = \int_{0}^{\infty} (\tanh h)^{2k} (1+e^{-2h}) \, \sfx(dh)  \\
	 = \,&\int_{0}^{\infty} (\tanh h)^{2k} \, \sfx(dh) + \int_{0}^{\infty} (\tanh h)^{2k} \, \sfx(-dh)  = \int_{-\infty}^{\infty} (\tanh h)^{2k} \, \sfx(dh) .
\end{align*}


TITLE: rank deficient least squares with minimum $\ell_1$ norm
QUESTION [2 upvotes]: When a least square solution of a rank deficient least square problem is sought, there are multitude of solutions that give the same minimal residual vector $r=\mathbf{A}x-b$. I am familiar with Moore-Penrose pseudo-inverse that picks minimum $\ell_2$-norm least square solution among these solutions  i.e. $$ x_{LS} = \mathbf{A}^{\dagger}b$$
Computing this usually involves getting the SVD of A and screening all singular vectors associated with zero singular values. Is there a similar expression that gives the minimum $\ell_1$-norm? Even if no expression can be given, is there a algorithm to compute the minimum $\ell_1$-norm among all solutions that give the minimum residual vector? This seems to be a convex optimization problem but I could be mistaken.
If it helps I am interested in a problem with column-rank deficient $\mathbf{A}$ and it has more rows than columns.
Update: A related problem where the residual norm is known can be solved via LASSO (using the homotopy approach to find the regularization parameter that matches the residual norm). LASSO indeed seems to give the minimum $\ell_1$ norm solution for the given residual norm. In this problem the residual vector is given and a minimum $\ell_1$ norm solution is sought.
Edit: After the comment below by @littleO I realized the mistake I was making that motivated the original question. I was planning to project the $b$ vector on to null space of $\mathbf{A}$ and call it the noise or residual vector but there could be a component of the noise vector in the range of $\mathbf{A}$ as well. So if I may change the question, how does one find out or estimate the norm of the noise?

REPLY [1 votes]: First solve the unconstrained least squares problem, for example by using the Moore-Penrose pseudo-inverse (which picks the minimum $\ell_2$-norm least square solution), or any other method. Evaluating the two-norm of the residual at that optimal value of $x$ provides the $\epsilon$.  Then solve the $\ell_1$-norm minimization problem subject to the  $\ell_2$-norm constraint.
Using the Moore-Penrose pseudo-inverse least squares solution, that would be
Minimize $\|x\|_1$ subject to $\|\mathbf{A}x - b\|_2 \le \|\mathbf{A}\mathbf{A}^{\dagger}b - b\|_2$
The CVX code for this convex optimization problem is:
cvx_begin
variable x(n)
minimize(norm(x,1))
norm(A*x-b) <= norm(A*pinv(A)*b-b)
cvx_end