\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \usepackage{graphicx} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 47, pp. 1--54.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/47\hfil Regularity for a clamped grid equation] {Regularity for a clamped grid equation $u_{xxxx}+u_{yyyy}=f$ on a domain with a corner} \author[T. Gerasimov, G. Sweers\hfil EJDE-2009/47\hfilneg] {Tymofiy Gerasimov, Guido Sweers} % in alphabetical order \address{Tymofiy Gerasimov \newline DIAM-EWI, Delft University of Technology, PO box 5031, 2600 GA Delft, The Netherlands} \email{t.gerasimov@tudelft.nl} \address{Guido Sweers \newline MI, Universit\"{a}t zu K\"{o}ln, D 50931 Cologne, Germany} \email{gsweers@math.uni-koeln.de} \thanks{Submitted December 10, 2008. Published April 2, 2009.} \subjclass[2000]{35J40, 46E35, 35P30} \keywords{Nonisotropic; fourth order PDE; domain with corner; \hfill\break\indent clamped grid; weighted Sobolev space; regularity} \begin{abstract} The operator $L=\frac{\partial ^{4}}{\partial x^{4}} +\frac{\partial ^{4}}{\partial y^{4}}$ appears in a model for the vertical displacement of a two-dimensional grid that consists of two perpendicular sets of elastic fibers or rods. We are interested in the behaviour of such a grid that is clamped at the boundary and more specifically near a corner of the domain. Kondratiev supplied the appropriate setting in the sense of Sobolev type spaces tailored to find the optimal regularity. Inspired by the Laplacian and the Bilaplacian models one expect, except maybe for some special angles that the optimal regularity improves when angle decreases. For the homogeneous Dirichlet problem with this special non-isotropic fourth order operator such a result does not hold true. We will show the existence of an interval $( \frac{1}{2}\pi ,\omega _{\star })$, $\omega _{\star }/\pi \approx 0.528\dots$ (in degrees $\omega _{\star }\approx 95.1\dots^{\circ}$), in which the optimal regularity improves with increasing opening angle. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{condition}[theorem]{Condition} \newtheorem{definition}[theorem]{Definition} \newtheorem{notation}[theorem]{Notation} \newtheorem{claim}[theorem]{Claim} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \tableofcontents \section{Introduction} \subsection{The model} A model for small deformations of a thin isotropic elastic plate is $% u_{xxxx}+2u_{xxyy}+u_{yyyy}=f$. Here $f$ is a force density and $u$ is the vertical displacement of a plate; the model neglects the influence of horizontal deviations. Non-isotropic elastic plates are still modeled by fourth order differential equations but the coefficients in front of the derivatives of $u$ may vary. Two interesting extreme cases are $L_1=\tfrac{% \partial ^{4}}{\partial x^{4}} +\tfrac{\partial ^{4}}{\partial y^{4}}$ and $% L_2=\frac{1}{2}\tfrac{\partial ^{4}}{\partial x^{4}}+3\tfrac{\partial ^{4} }{% \partial x^2\partial y^2}+\frac{1}{2}\tfrac{\partial ^{4}}{\partial y^{4}}$. One may think of these operators as of the operators appearing in the model of an elastic medium consisting of two sets of intertwined (not glued) perpendicular fibers: $\tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{ \partial ^{4}}{\partial y^{4}}$ for fibers running in cartesian directions (Figure \ref{figure1}, left). The differential operator is not rotation invariant. For a diagonal grid the rotation of $\frac{1}{4}\pi$ transforms $% L_1$ into $L_2$ (Figure \ref{figure1}, right). We will call such medium \emph{a grid}. We should mention that sets of fibers are connected such that the vertical positions coincide but there is no connection that forces a torsion in the fibers. Such torsion would occur if the fibers are glued or imbedded in a softer medium. For those models see \cite{NSS}. The appropriate linearized model in that last situation would contain mixed fourth order derivatives. \begin{figure}[ht] \begin{center} \includegraphics[width=0.44\textwidth]{fig1a} % Grid(alligned) \includegraphics[width=0.44\textwidth]{fig1b} % Grid(digaonal) \end{center} \caption{A fragment of a rectangular grid with aligned and diagonal fibers.} \label{figure1} \end{figure} A first place where operator $L_1$ appears is J. II. Bernoulli's paper \cite% {Ber}. He assumed that it was the appropriate model for an isotropic plate. It was soon dismissed as a model for such a plate, since indeed it failed to have rotational symmetry. \subsection{The setting} We will focus on $L_1$ supplied with homogeneous Dirichlet boundary conditions. This problem, which we call \emph{a clamped grid'}, is as follows: $$\begin{gathered} u_{xxxx}+u_{yyyy}=f \quad\text{in }\Omega , \\ u=\frac{\partial u}{\partial \nu }=0\quad\text{on }\partial \Omega . \end{gathered} \label{0}$$ Here $\Omega \subset \mathbb{R}^2$ is open and bounded, and $\nu$ is the unit outward normal vector on $\partial \Omega$. The boundary conditions in (\ref{0}) correspond to the clamped situation meaning that the vertical position and the angle are fixed to be $0$ at the boundary. One verifies directly that the operator $L_1=\tfrac{\partial ^{4}}{ \partial x^{4}}+\tfrac{\partial ^{4}}{\partial y^{4}}$ is elliptic in $\overline{% \Omega }$. One may also prove, if the normal is well-defined, that the boundary value problem (\ref{0}) is regular elliptic. Indeed, the Dirichlet problem which fixes the zero and first order derivatives at the boundary, is regular elliptic for any fourth order uniformly elliptic operator. Hence, under the assumption that $\Omega$ is bounded and $\partial \Omega \in C^{\infty }$ the full classical regularity result (see e.g. \cite{LM}) for problem (\ref{0}) can be used to find for $k\geq 0$ and $p\in 1,\infty )$: $$\text{if f\in W^{k,p}(\Omega ) then u\in W^{k+4,p}(\Omega )}. \label{2.0}$$ If $\Omega$ in (\ref{0}) has a piecewise smooth boundary $\partial \Omega$ with, say, one angular point, the result (\ref{2.0}) in general does not apply. Instead, one may use the theory developed by Kondratiev \cite{K}. This theory provides the appropriate treatment of problem (\ref{0}) by employing the weighted Sobolev space $V_{\beta }^{k,p}(\Omega )$ (see Definition \ref{Def9}), where $k\geq 0$ is the differentiability index and $% \beta \in \mathbb{R}$ characterizes the powerlike growth of the solution near the angular point. Within the framework of the Kondratiev spaces $% V_{\beta }^{k,p}(\Omega )$ the regularity result `analogous'' to (\ref{2.0}% ) will then be as follows. There is a countable set of functions $\{ u_{j}\} _{j\in \mathbb{N}}$ such that for all $k\in \mathbb{N}$: $$\text{if f\in V_{\beta }^{k,p}(\Omega ) then u=w+% \sum_{j=1}^{J_{k}}c_{j}u_{j} with w\in V_{\beta }^{k+4,p}(\Omega ).} \label{2}$$ We will restrict our formulations to $p=2$. Partial differential equations on domains with corners have obtained a lot of attention in the literature. After the seminal paper by Kondatiev \cite{K} many authors of which we would like to mention Kozlov, Maz'ya, Rossmann \cite% {KMR, KMR2}, Grisvard \cite{Gr}, Dauge \cite{Dauge}, Costabel and Dauge \cite% {Costabel}, Nazarov and Plamenevsky \cite{NPl} have contributed. For applications in elasticity theory we refer to Leguillon and Sanchez-Palencia \cite{Leg}, Blum and Rannacher \cite{Blum}. A recent paper of Kawohl and Sweers \cite{KSw} concerned the positivity question for the operators $L_1$ and $L_2$ in a rectangular domain for hinged boundary conditions. \subsection{The target} In this paper, we will focus particularly on the optimal regularity for the boundary value problem which depends on the opening angle of the corner. For the sake of a simple presentation, we will consider (\ref{0}) in a domain $% \Omega \subset \mathbb{R}^2$ which has one corner in $0\in \partial \Omega$% \ with opening angle $\omega \in ( 0,2\pi]$. A more appropriate formulation of the problem should read as: $$\begin{gathered} u_{xxxx}+u_{yyyy}=f \quad\text{in } \Omega , \\ u=0\quad\text{on }\partial \Omega , \\ \frac{\partial u}{\partial \nu }=0\quad\text{on }\partial \Omega \backslash \{0\}, \end{gathered} \label{1}$$ with prescribed growth behaviour near $0$. To be more precise in the description of a domain $\Omega$, we assume the following condition. \begin{condition} \label{Condition0} \rm The domain $\Omega$ has a smooth boundary except at $(x,y)=0$, and is such that in the vicinity of $0$ it locally coincides with a sector. In other words, \begin{enumerate} \item $\partial \Omega \backslash \{0\}$ is $C^{\infty }$, \item there exists $\varepsilon >0,\omega \in ( 0,2\pi] :\Omega \cap B_{\varepsilon }(0)=\mathcal{K}_{\omega }\cap B_{\varepsilon }(0)$, \end{enumerate} where $B_{\varepsilon }(0)=\{(x,y):| (x,y)| <\varepsilon\}$ is the open ball of radius $\varepsilon$ centered at $(x,y)=0$ and $\mathcal{K}_{\omega }$\ an infinite sector with an opening angle $\omega$: \mathcal{K}_{\omega }=\left\{ (r\cos (\theta ),r\sin (\theta )):\ 00$and$\mathcal{K}_{\omega }$\ is an infinite sector with an opening angle$\omega \in ( 0,2\pi) $. These weighted spaces are as follows: \begin{definition}\label{Def9} \rm Let$l\in \{ 0,1,2,\dots \} $and$\beta \in \mathbb{R}$. Then$V_{\beta }^{l,2}(\Omega )$is defined as a completion: \begin{gather} V_{\beta }^{l,2}(\Omega ) = \overline{C_{c}^{\infty }\big( \overline{ \Omega }\backslash \{0\}\big) }^{\| \cdot \| }\quad \text{with} \label{pre50} \\ \| u\| :=\| u\| _{V_{\beta }^{l,2}(\Omega )} =\Big( \sum_{| \alpha | =0}^{l}\int_{\Omega }( x^2+y^2) ^{\beta -l+| \alpha | }| D^{\alpha }u| ^2dx\,dy\Big) ^{1/2}, \label{50} \end{gather} where \begin{equation*} C_{c}^{\infty }\left( \overline{\Omega }\backslash \{0\}\right) :=\left\{ u\in C_{c}^{\infty }\left( \overline{\Omega }\right) :\text{\rm support} (u)\subset \overline{\Omega }\backslash B_{\varepsilon }(0)\right\} . \end{equation*} \end{definition} The space$V_{\beta }^{l,2}(\Omega )$consists of all functions$u:\Omega \to \mathbb{R}$such that for each multiindex$\alpha =( \alpha_1,\alpha _2) $with$| \alpha | \leq l$,$D^{\alpha }u=\frac{\partial ^{| \alpha | }u}{% \partial x^{\alpha _1}\partial y^{\alpha _2}}$exists in the weak sense and$% r^{\beta -l+| \alpha | }D^{\alpha }u\in L^2(\Omega )$. Here$r=( x^2+y^2) ^{1/2}$. Straightforward from the definition of the norm the following continuous imbeddings hold (see \cite[Section 6.2, lemma 6.2.1]{KMR}): $$V_{\beta _2}^{l_2,2}(\Omega )\subset V_{\beta _1}^{l_1,2}(\Omega ) \quad% \text{if }l_2\geq l_1\geq 0, \; \beta _2-l_2\leq \beta _1-l_1. \label{66}$$ To have the appropriate space for zero Dirichlet boundary conditions in problem \eqref{1} we also define the corresponding space. \begin{definition} \rm For$l\in \{ 0,1,2,\dots\} $and$\beta \in \mathbb{R}$, set $${\mathaccent"7017 V}_{\beta }^{l,2}(\Omega )=\overline{C_{c}^{\infty }\left( \Omega \right) }^{\| \cdot \| }, \label{51}$$ with$\| \cdot \| $as the norm (\ref{50}) and$C_{c}^{\infty }\left( \Omega \right) :=\left\{ u\in C_{c}^{\infty }\left( \bar{\Omega}\right) :\mathop{\rm support}(u)\subset \Omega \right\} $. \end{definition} \begin{remark} \rm For$u\in {\mathaccent"7017 V}${}$_{\beta }^{l,2}(\Omega )$one finds$D^{\alpha }u=0$on$\partial \Omega $for$| \alpha | \leq \ell -1$where$D^{\alpha }u=0$is understood in the sense of traces. \end{remark} \section{Homogeneous problem in an infinite sector, singular solutions} \label{SingSol} The first step in order to improve the regularity of a weak solution is to consider the homogeneous problem in an infinite cone: $$\begin{gathered} u_{xxxx}+u_{yyyy}=0 \quad\text{in } \mathcal{K}_{\omega }, \\ u=\frac{\partial u}{\partial \nu }=0\quad\text{on }\partial \mathcal{K}_{\omega}\backslash \{0\}. \end{gathered} \label{3}$$ Here$\mathcal{K}_{\omega }$is as in (\ref{Komega}). We will derive almost explicit formula's for power type solutions to (\ref{3}). \subsection{Reduced problem\label{Reduced problem}} The reduced problem for (\ref{3}) is obtained in the following way. By Kondratiev \cite{K} one should consider the power type solutions of (\ref{3}% ): $$u=r^{\lambda +1}\Phi (\theta ), \label{3.10}$$ with$x=r\cos (\theta )$and$y=r\sin (\theta )$. Here$\lambda \in \mathbb{C% } $and$\Phi :[ 0,\omega] \to \mathbb{R}$. We insert$ufrom (\ref{3.10}) into problem (\ref{3}) and find \begin{equation*} \left( \tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{\partial ^{4}}{\partial y^{4}}\right) r^{\lambda +1}\Phi (\theta )=r^{\lambda -3}\mathcal{L}\left( \theta ,\tfrac{d}{d\theta },\lambda \right) \Phi (\theta ), \end{equation*} with \begin{aligned} \mathcal{L}\left( \theta ,\tfrac{d}{d\theta },\lambda \right) &=\tfrac{3}{4} \left( 1+\tfrac{1}{3}\cos (4\theta )\right) \tfrac{d^{4}}{d\theta ^{4}} +\left( \lambda -2\right) \sin (4\theta )\tfrac{d^{3}}{d\theta ^{3}}+ \\ &\quad +\tfrac{3}{2}\left( \lambda ^2-1-\left( \lambda ^2-4\lambda -\tfrac{7}{ 3}\right) \cos (4\theta )\right) \tfrac{d^2}{d\theta ^2}+ \\ &\quad +\left( -\lambda ^{3}+6\lambda ^2-7\lambda -2\right) \sin (4\theta ) \tfrac{d}{d\theta }+ \\ &\quad +\tfrac{3}{4}\left( \lambda ^{4}-2\lambda ^2+1+\tfrac{1}{3}\left( \lambda ^{4}-8\lambda ^{3}+14\lambda ^2+8\lambda -15\right) \cos (4\theta )\right) . \label{5} \end{aligned} Then we obtain a\lambda $-dependent boundary value problem for$\Phi $: $$\begin{gathered} \mathcal{L}\left( \theta ,\tfrac{d}{d\theta },\lambda \right) \Phi =0 \quad\text{in }(0,\omega ), \\ \Phi =\tfrac{d\Phi }{d\theta }=0\quad\text{on }\partial (0,\omega ). \end{gathered} \label{4}$$ \begin{remark} \label{rmk4.1} \rm The nonlinear eigenvalue problem \eqref{4} appears by a Mellin transformation: \begin{equation*} \Phi (\theta )=(\mathcal{M}u)(\lambda )=\int_0^{\infty }r^{-\lambda -2}u(r,\theta )dr. \end{equation*} \end{remark} So, the reduced problem for (\ref{3}) we mentioned above is problem \eqref{4}% . Before we start analyzing it, let us fix some basic notions. \begin{definition}\label{Def5} \rm Every number$\lambda _0\in \mathbb{C}$, such that there exists a nonzero function$\Phi _0$satisfying \eqref{4}, is said to be an eigenvalue of problem \eqref{4}, while$\Phi _0\in C^{4}[ 0,\omega ] $is called its eigenfunction. Such pairs$\left( \lambda _0,\Phi _0\right) $are called solutions to problem \eqref{4}. If$( \lambda _0,\Phi _0) $solves \eqref{4} and if$\Phi _1 $is a nonzero function that solves $$\begin{gathered} \mathcal{L}(\lambda _0)\Phi _1+\mathcal{L}'(\lambda _0)\Phi _0=0\quad\text{in }(0,\omega ), \\ \Phi =\tfrac{d\Phi }{d\theta }=0\quad\text{on }\partial (0,\omega ), \end{gathered} \label{4bis}$$ then$\Phi _1$is a generalized eigenfunction (of order$1$) for \eqref{4} with eigenvalue$\lambda _0$. \end{definition} \begin{remark} Similarly, one may define generalized eigenfunctions of higher order. \end{remark} The following holds for \eqref{4}. \begin{lemma}\label{lemma2} Let$\theta \in (0,\omega )$,$\omega \leq 2\pi $. For every fixed$\lambda \notin \{ \pm 1,0\} $in \eqref{4}, let us set \begin{gather*} \varphi _1(\theta )=\left( \cos (\theta )+\tau _1\sin (\theta )\right) ^{\lambda +1}, \quad \varphi _2(\theta )=\left( \cos (\theta )+\tau _2\sin (\theta )\right) ^{\lambda +1}, \\ \varphi _{3}(\theta )=\left( \cos (\theta )-\tau _1\sin (\theta )\right) ^{\lambda +1}, \quad \varphi _{4}(\theta )=\left( \cos (\theta ) -\tau _2\sin (\theta )\right) ^{\lambda +1}, \end{gather*} where$\tau _1=\frac{\sqrt{2}}{2}(1+i) $,$\tau _2=\frac{ \sqrt{2}}{2}(1-i) $and$i=\sqrt{-1}$. The set$S_{\lambda }:=\{ \varphi _{m}\} _{m=1}^{4}$is a fundamental system of solutions to the equation$\mathcal{L}\left( \theta ,\tfrac{\partial }{\partial \theta },\lambda \right) \Phi =0$on$(0,\omega )$. \end{lemma} \begin{proof} The derivation of$\varphi _{m}$,$m=1,\dots,4$in$S_{\lambda }$is rather technical and we refer to Appendix \ref{AppendixC}. There we also compute the Wronskian: \begin{equation*} W\left( \varphi _1(\theta) ,\varphi _2(\theta) ,\varphi _{3}(\theta) ,\varphi _{4}(\theta) \right) =16\left( \lambda +1\right) ^{3}\lambda ^2\left( \lambda -1\right) \left( \cos ^{4}(\theta )+\sin ^{4}(\theta) \right) ^{\lambda -2}. \end{equation*} It is non-zero on$\theta \in (0,2\pi ]$except for$\lambda \in \left\{ \pm 1,0\right\} $. Hence, for every fixed$\lambda \notin \left\{ \pm 1,0\right\} $the set$\{ \varphi _{m}\} _{m=1}^{4}$consists of four linear independent functions on$(0,\omega )$,$\omega \leq 2\pi $. \end{proof} \begin{lemma}\label{lemma3} In the particular cases$\lambda \in \left\{ \pm 1,0\right\} $in \eqref{4}, one finds the following fundamental systems: \begin{gather*} S_{-1}=\{ 1,\arctan ( \cos (2\theta )) , \mathrm{arctanh}( \tfrac{\sqrt{2}}{2}\sin (2\theta )) , \varphi _{4}(\theta )\} , \\ S_0=\left\{ \sin (\theta ),\text{ }\cos (\theta ),\text{ }\varphi _{3}(\theta ),\text{ }\varphi _{4}(\theta )\right\} , \\ S_1=\left\{ 1,\text{ }\sin (2\theta ),\text{ }\cos (2\theta ),\text{ } \varphi _{4}(\theta )\right\} , \end{gather*} where the explicit formulas for$\varphi _{4}\in S_{-1}$,$\left\{ \varphi_{3},\varphi _{4}\right\} \in S_0$and$\varphi _{4}\in S_1$\ are given in Appendix \ref{AppendixC}. \end{lemma} \begin{proof} The fundamental systems$S_{-1},S_0,S_1$are given in Appendix \ref% {AppendixC}. By straightforward computations one finds that for every above$% S_{\lambda }$,$\lambda \in \left\{ \pm 1,0\right\} $the corresponding Wronskian$W$is proportional to$\left( \cos ^{4}(\theta )+\sin ^{4}(\theta) \right) ^{\lambda -2}$,$\lambda \in \left\{ \pm 1,0\right\} $and hence is nonzero on$\theta \in (0,2\pi ]$. \end{proof} In terms of the fundamental systems$S$we have$\Phi $that solves$% \mathcal{L}\left( \theta ,\tfrac{\partial }{\partial \theta },\lambda \right) \Phi =0$as \begin{equation*} \Phi (\theta )=\sum_{m=1}^{4}b_{m}\varphi _{m}(\theta ), \end{equation*} where$b_{m}\in \mathbb{C}$. Inserting this expression into the boundary conditions of problem \eqref{4}, we find a homogeneous system of four equations in the unknowns$\left\{ b_{m}\right\} _{m=1}^{4}$reading as \begin{equation*} Ab:=% \begin{pmatrix} \varphi _1(0)) & \varphi _2(0)) & \varphi _{3}(0)) & \varphi _{4}(0)) \\ \varphi _1^{\prime }(0)) & \varphi _2^{\prime }(0)) & \varphi _{3}^{\prime }(0)) & \varphi _{4}^{\prime }(0)) \\ \varphi _1(\omega) & \varphi _2(\omega) & \varphi _{3}(\omega) & \varphi _{4}(\omega) \\ \varphi _1^{\prime }(\omega) & \varphi _2^{\prime }(\omega) & \varphi _{3}^{\prime }(\omega) & \varphi _{4}^{\prime }(\omega)% \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \\ b_{3} \\ b_{4}% \end{pmatrix} =0, \end{equation*} where$\omega \in ( 0,2\pi] $. It admits non-trivial solutions for$\{ b_{m}\} _{m=1}^{4}$if and only if$\det (A)=0$. Hence, the eigenvalues$% \lambda $of problem \eqref{4} in sense of Definition \ref{Def5} will be completely determined by the characteristic equation$\det(A)=0$. We deduce the following four cases: $$\det (A):=% \begin{cases} P(\omega ,\lambda ) & \text{when } \lambda \notin \{ \pm 1,0\}, \\ P_{-1}(\omega ) & \text{when } \lambda =-1, \\ P_0(\omega ) & \text{when } \lambda =0, \\ P_1(\omega ) & \text{when } \lambda =1.% \end{cases} \label{6}$$ The explicit formulas for$Preads as \begin{aligned} P(\omega ,\lambda) &=\left( 1-\tfrac{\sqrt{2}}{2}\sin (2\omega )\right) ^{\lambda }+\left( 1+\tfrac{\sqrt{2}}{2}\sin (2\omega )\right) ^{\lambda } \\ &\quad +\left( \tfrac{1}{2}+\tfrac{1}{2}\cos ^2(2\omega )\right) ^{\frac{1}{2} \lambda }\Big[ 2\cos \left( \lambda \left( \arctan \left( \tfrac{\sqrt{2}}{2 }\tan (2\omega )\right) +\ell \pi \right) \right) \\ &\quad -4\cos \left( \lambda \arctan \left( \tan ^2(\omega )\right) \right) \Big] , \end{aligned} \label{6.1} where \begin{gather*} \ell =0\quad\text{if }\omega \in ( 0,\frac{1}{4}\pi ] ,\quad \ell =1\quad% \text{if }\omega \in ( \frac{1}{4}\pi ,\frac{3}{4}\pi ], \\ \ell =2\quad\text{if }\omega \in ( \frac{3}{4}\pi ,\frac{5}{4}\pi ], \quad \ell =3\quad\text{if }\omega \in ( \frac{5}{4}\pi ,\frac{7}{4}\pi ], \\ \ell =4\quad\text{if }\omega \in ( \frac{7}{4}\pi ,2\pi ] . \end{gather*} In particular, for\omega \in \{ \frac{1}{2}\pi ,\pi ,\frac{3}{2}\pi ,2\pi\} $in (\ref{6.1}) we have \begin{gather*} P( \tfrac{1}{2}\pi ,\lambda ) = 2+2\cos (\pi \lambda )-4\cos ( \tfrac{1}{2}% \pi \lambda ) , \\ P( \pi ,\lambda ) =-4+4\cos ^2(\pi \lambda ), \\ P( \tfrac{3}{2}\pi ,\lambda ) = 8\cos ^{3}(\pi \lambda )-6\cos (\pi \lambda )-4\cos ( \tfrac{1}{2}\pi \lambda ) +2, \\ P( 2\pi ,\lambda ) =16\cos ^{4}(\pi \lambda )-16\cos ^2(\pi \lambda ). \end{gather*} Formulas for$P_{-1},P_0,P_1$in (\ref{6}) are available in Appendix \ref% {AppendixC}. \subsection{Analysis of the eigenvalues$\protect\lambda $} To describe the eigenvalues$\lambda $of \eqref{4} for a fixed$\omega $and, what is more important, their behavior in dependence on$\omega $, we analyze the equation$\det (A)=0$on the interval$\omega \in( 0,2\pi] $. First, we find that the equations$P_{-1}(\omega )=0$and$P_1(\omega )=0$have identical solutions on$( 0,2\pi] $, that are denoted$\omega \in \{ \pi ,\omega _0,2\pi \} $. The approximation$\omega _0/\pi \approx 1.424\dots $(in degrees$\omega _0\approx 256.25\dots^{\circ }$) is obtained by the Maple 9.5 package.\ Equation$P_0(\omega )=0$has no solutions on$\omega \in (0,2\pi ]$. Hence,$\lambda \in \{ \pm 1\} $are the eigenvalues of \eqref{4} for the above values of$\omega $, while$% \lambda =0$is not an eigenvalue of \eqref{4}. Now we consider$P(\omega ,\lambda )=0$on$\omega \in ( 0,2\pi] $; here$P$is given by (\ref{6.1}). We note that for every$\lambda \in \mathbb{C} \backslash \{ \pm 1,0\} $it holds that \begin{equation*} P(\omega ,-\lambda )=( \tfrac{3}{4}+\tfrac{1}{4}\cos (4\omega )) ^{-\lambda }P(\omega ,\lambda ), \end{equation*} that is, the solutions$\lambda $of$P(\omega ,\lambda )=0$are symmetric with respect to the$\omega $-axis. It is immediate that if$\lambda $is an eigenvalue then so is$\overline{\lambda }$. It is convenient\ to introduce the following notation. \begin{notation} \label{Notation1} \rm For every fixed$\omega \in ( 0,2\pi] $we write$\{ \lambda _{j}\} _{j=1}^{\infty }$for the collection of the eigenvalues of problem \eqref{4} in the sense of Definition \ref{Def5}, which have positive real part$\mathop{\rm Re}(\lambda) >0$and are ordered by increasing real part. \end{notation} The complete set of eigenvalues to problem \eqref{4} will then read as$\{ -\lambda _{j},\lambda _{j}\} _{j=1}^{\infty }$. Now the following lemma can be formulated. \begin{lemma}\label{lemma4} \rm Let$\mathcal{L}$be the operator given by (\ref{5}). \begin{itemize} \item For every fixed$\omega \in (0,2\pi ]\backslash \left\{ \pi ,\omega _0,2\pi \right\} $the set$\{ \lambda _{j}\} _{j=1}^{\infty }$from Notation \ref{Notation1} is given by \begin{equation*} \{ \lambda _{j}\} _{j=1}^{\infty }=\left\{ \lambda \in \mathbb{C} :\mathop{\rm Re}(\lambda) \in \mathbb{R}^{+}\backslash \{1\},\text{ \ }P(\omega ,\lambda )=0\right\} . \end{equation*} \item For every fixed$\omega \in \left\{ \pi ,\omega _0,2\pi \right\} $the set$\{ \lambda _{j}\} _{j=1}^{\infty }$from Notation \ref {Notation1} is given by \begin{equation*} \{ \lambda _{j}\} _{j=1}^{\infty }=\left\{ \lambda \in \mathbb{C} :\mathop{\rm Re}(\lambda) \in \mathbb{R}^{+}\backslash \{1\},\text{ \ }P(\omega ,\lambda )=0\right\} \cup \{1\}. \end{equation*} \end{itemize} Here$\omega _0$is a solution of$P_1(\omega )=0$on$\omega \in (\pi,2\pi )$with the approximation$\omega _0/\pi \approx 1.424\dots$(in degrees$\omega _0\approx 256.25\dots^{\circ }$). \end{lemma} \subsection{Intermezzo: a comparison with$\Delta ^2$} Let the grid-operator$\tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{\partial ^{4}}{\partial y^{4}}$in problems \eqref{1}, (\ref{3}) be replaced by the bilaplacian$\Delta ^2=\tfrac{\partial ^{4}}{\partial x^{4}}+2\tfrac{ \partial ^{4}}{\partial x^2\partial y^2}+\tfrac{\partial ^{4}}{\partial y^{4}% }$. We recall some results for that operator, in particular, the eigenvalues$\{ \lambda _{j}\} _{j=1}^{\infty }$of the corresponding reduced problem. We will compare them to those given in Lemma \ref{lemma4}. So, for$\Delta ^2$in (\ref{3}) the reduced problem of the type \eqref{4} has an operator$\mathcal{L}$reading as (see e.g. \cite[page 88]{Gr}): $$\mathcal{L}\left( \theta ,\tfrac{d}{d\theta },\lambda \right) =\tfrac{d^{4}}{ d\theta ^{4}}+2\left( \lambda ^2+1\right) \tfrac{d^2}{d\theta ^2} +\left( \lambda ^{4}-2\lambda ^2+1\right) . \label{8.0}$$ Proceeding as above one obtains that the corresponding determinants (see \cite[page 89]{Gr} or \cite[page 561]{Blum}) are the following: $$\det (A):=% \begin{cases} \sin ^2(\lambda \omega )-\lambda ^2\sin ^2(\omega ) & \text{when } \lambda \notin \{ \pm 1,0\} , \\ \sin ^2(\omega )-\omega ^2 & \text{when } \lambda =0, \\ \sin (\omega )\left( \sin (\omega )-\omega \cos (\omega )\right) & \text{% when } \lambda \in \{ \pm 1\} .% \end{cases} \label{8.1}$$ Note that for every$\lambda \in \mathbb{C} \backslash \{ \pm 1,0\} $the function$\sin ^2(\lambda \omega )-\lambda ^2\sin ^2(\omega )$is even with respect to$\omega $and hence the Notation \ref{Notation1} is applicable here. Analysis of\$\det (A)=0$\ with$\det (A)$as in (\ref{8.1}) enables to formulate the analog of Lemma \ref{lemma4}. Namely, \begin{lemma}\label{lemma4.1} Let$\mathcal{L}$be the operator given by (\ref{8.0}). \begin{itemize} \item For every fixed$\omega \in (0,2\pi ]\backslash \left\{ \pi ,\omega _0,2\pi \right\} $the set$\{ \lambda _{j}\} _{j=1}^{\infty }$from Notation \ref{Notation1} is given by \begin{equation*} \{ \lambda _{j}\} _{j=1}^{\infty }=\left\{ \lambda \in \mathbb{C} :\mathop{\rm Re}(\lambda) \in \mathbb{R}^{+}\backslash \{1\}:\sin ^2(\lambda \omega )-\lambda ^2\sin ^2(\omega )=0\right\} . \end{equation*} \item For every fixed$\omega \in \left\{ \pi ,\omega _0,2\pi \right\} $the set$\{ \lambda _{j}\} _{j=1}^{\infty }$from Notation \ref {Notation1} is given by \begin{equation*} \{ \lambda _{j}\} _{j=1}^{\infty }=\left\{ \lambda \in \mathbb{C} :\mathop{\rm Re}(\lambda) \in \mathbb{R}^{+}\backslash \{1\}:\sin ^2(\lambda \omega )-\lambda ^2\sin ^2(\omega )=0\right\} \cup \{1\}. \end{equation*} \end{itemize} Here$\omega _0$is a solution of$\tan (\omega )=\omega $on$\omega \in (\pi ,2\pi )$with the approximation$\omega _0/\pi \approx 1.430\dots$(in degrees$\omega _0\approx 257.45\dots^{\circ }$). \end{lemma} \subsection{Analysis of the eigenvalues$\protect\lambda $(continued)} Let$(\omega ,\lambda) $be the pair that solves the equations of Lemmas \ref% {lemma4} and \ref{lemma4.1}. In Figure \ref{figure3} we plot the pairs$% \left( \omega ,\mathop{\rm Re}(\lambda) \right) $\ inside the region$\left( \omega ,\mathop{\rm Re}(\lambda) \right) \in \left( 0;2\pi \right] \times % \left[ 0,7.200\right] $. \begin{remark} \label{Remark10}\rm The numerical computations are performed with the Maple 9.5 package in the following way: at a first cycle for every$\omega _{n}=\frac{ 21}{180}\pi +\frac{1}{60}\pi n$,$n=0,\dots,113$we compute the entries of the set$\{ \lambda _{j}\} _{j=1}^{N}$. Here,$N$is determined by the condition:$\mathop{\rm Re}\left( \lambda _{N}\right) \leq 7.200$and$\mathop{\rm Re}\left( \lambda _{N+1}\right) >7.200$. The points$(\omega ,\lambda )$where$\lambda _{j}$transits from the complex plane to the real one or vice-versa are solutions to the system$P(\omega ,\lambda )=0$and$\frac{\partial P}{\partial \lambda }(\omega ,\lambda )=0$(the justification for the second condition will be discussed in Lemma \ref{lemma8}). \end{remark} \begin{figure}[ht] \begin{center} \includegraphics[width=0.44\textwidth]{fig3a} % Figure1(Clamped).pdf} Figure2(Clamped).pdf} \includegraphics[width=0.44\textwidth]{fig3b} \end{center} \caption{Some first eigenvalues$\protect\lambda _{j}$in$(\protect\omega , \mathop{\rm Re}(\protect\lambda)) \in ( 0,2\protect\pi ] \times [ 0,7.200] $of problem \eqref{4}, where$\mathcal{L}$is related respectively to$\frac{% \partial ^{4}}{\partial x^{4}}+\frac{\partial ^{4}}{\partial y^{4}}$(on the top) and$\Delta ^2$(on the bottom). Dashed lines depict the real part of those$\protect\lambda_{j}\in \mathbb{C} $, solid lines are for purely real$% \protect\lambda _{j}$; the vertical thin lines mark out values$\left\{\frac{% 1}{2}\protect\pi ,\protect\pi ,\frac{3}{2}\protect\pi ,2\protect\pi \right\} $on$\protect\omega$-axis.} \label{figure3} \end{figure} In Figure \ref{figure3} one sees the difference in the behavior of the eigenvalues in the corresponding cases. In particular, in the top plot (the case$L=\tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{\partial ^{4}}{\partial y^{4}}$) there are the loops and the ellipses in the vicinities of$\omega \in \left\{ \frac{1}{2}\pi ,\frac{3}{2}\pi \right\} $(we inclose them in the rectangles). The bottom plot (the case$L=\Delta ^2$) looks much simpler near the same region. As mentioned, the contribution of the first eigenvalue$\lambda _1$to the regularity of the solution$u$to our problem \eqref{1} is the most essential. So, it is important for us to know the dependence of the eigenvalues${\lambda }$on the opening angle$\omega $. In this sense, the region$\left( \omega ,\mathop{\rm Re}(\lambda) \right) \in V$(Figure % \ref{figure3}, top) seems to be the most interesting part and the model one. One observes that inside$V$the graph of the implicit function${P(}${$% \omega $}${{{,}\lambda )=0}}$looks like a deformed$8$-shaped curve. So, if one proves that everywhere in$V$,${P(}${$\omega $}${{{,}\lambda )=0}}$allows its local parametrization in {$\omega \mapsto \lambda =\psi ({\omega }% )$} or$\lambda \mapsto \omega =\varphi (\lambda )$, then the bottom part of this graph is${\lambda }_1$and there is a subset of the this bottom part where${\lambda }_1$as a function of$\omega $increases with increasing$% \omega $. \subsubsection{Behavior of${\protect\lambda }$in$V$\label{Behavior}} So let us fix the open rectangular domain$V=\{ (\omega ,\lambda) :[ \frac{70% }{180}\pi ,\frac{110}{180}\pi] \times[ 2.900,5.100]\}$, the function$P{\in C% }^{\infty }(V,\mathbb{R} ) $is given by (\ref{6.1}) with$\ell =1: \begin{aligned} P(\omega ,\lambda) &=\left( 1-\tfrac{\sqrt{2}}{2}\sin (2\omega )\right) ^{\lambda }+\left( 1+\tfrac{\sqrt{2}}{2}\sin (2\omega )\right) ^{\lambda }& \\ &\quad +\left( \tfrac{1}{2}+\tfrac{1}{2}\cos ^2(2\omega )\right) ^{\frac{1}{2} \lambda }\Big[ 2\cos \left( \lambda \left( \arctan \left( \tfrac{\sqrt{2}}{2 }\tan (2\omega )\right) +\pi \right) \right) \\ &\quad -4\cos \left( \lambda \arctan \left( \tan ^2(\omega )\right) \right) \Big] .& \end{aligned} \label{9} and set $$\Gamma {:=\{{({\omega ,}\lambda )\in V:P({\omega ,}\lambda )=0\},}} \label{10}$$ as a zero level set ofP$in$V$. \begin{remark}\label{Remark11}\rm To plot the set$\Gamma $we perform the computations to${{P({\omega ,}\lambda )=0}}$in$V$in the spirit of Remark \ref{Remark10}. \end{remark} In particular, for$\omega =\frac{1}{2}\pi $being set in (\ref{9}) we obtain${{P\left( {{\tfrac{1}{2}\pi {,}\lambda }}\right) =}}2+2\cos (\pi \lambda )-4\cos \left( \tfrac{1}{2}\pi \lambda \right) $. The equation${{\ P\left( {{\tfrac{1}{2}\pi {,}\lambda }}\right) =0}}$\ admits exact solutions for$\lambda $in the interval$\left( 2.900,5.100\right) $, namely,${\ \lambda }\in \left\{ {3},{{{4},{5}}}\right\} $. This yields the points \begin{equation*} \big( \frac{1}{2}\pi ,3\big) =:c_1, \quad \big( \frac{1}{2}\pi ,4\big) =:a, \quad \big( \frac{1}{2}\pi ,5\big) =:c_{4}, \end{equation*} of$\Gamma $. It also holds straightforwardly that$\frac{\partial P}{ \partial \omega }\left( c_1\right) =\frac{\partial P}{\partial \omega } \left( c_{4}\right) =0$and hence one may guess that horizontal tangents to the set$\Gamma $exist at those points (in Lemma \ref{lemma7} this situation will be discussed in details for the point$c_1$). For$a$we find directly that$\frac{\partial P}{\partial \omega }(a) = \frac{\partial P}{% \partial \lambda }(a) =0$and hence more detailed analysis is required. Additionally to$c_1,c_{4}$, we will also specify four other points of the set$\Gamma $. Denoted as$c_2,c_{3},c_{5},c_{6}$, they are defined by the system$P(\omega ,\lambda )=0$and$\frac{\partial P}{\partial \lambda }% (\omega ,\lambda )=0$. The latter condition (we will justify it in Lemma \ref% {lemma8} for the point$c_2$) gives us a hint that vertical tangents to$% \Gamma $exist at those points. The approximations for the coordinates of$% c_{i}$,$i=1,\dots,6$are listed in the table and we plot the level set$% \Gamma $in Figure \ref{figure4}. \begin{table}[ht] \renewcommand{\arraystretch}{1.3} \par \begin{center} \begin{tabular}{|c|c|c|l|} \hline Point of$\Gamma $& Coordinates$\left( \omega /\pi ,\text{ }\lambda \right) $&$\omega $in degrees & property of$\Gamma $at$c_{k}$\\ \hline\hline$c_1$&$(\frac{1}{2}, 3)$&$90^{\circ }$& horizontal tangent \\ \hline$c_2$&$(0.528\dots, 3.220\dots)$&$\approx 95.1\dots^{\circ }$& vertical tangent \\ \hline$c_{3}$&$(0.591\dots, 4.291\dots)$&$\approx 106.4\dots^{\circ }$& vertical tangent \\ \hline$c_{4}$&$(\frac{1}{2}, 5)$&$90^{\circ }$& horizontal tangent \\ \hline$c_{5}$&$(0.477\dots, 4.746\dots)$&$\approx 85.96\dots^{\circ }$& vertical tangent \\ \hline$c_{6}$&$(0.412\dots, 3.655\dots)$&$\approx 74.2\dots^{\circ }$& vertical tangent \\ \hline \end{tabular}% \end{center} \caption{Approximations for the points of the level set$\Gamma $.} \label{table2} \end{table} \begin{figure}[ht] \begin{center} \includegraphics[width=0.7\textwidth]{fig4} % {Figure3.pdf} \end{center} \caption{The level set$\Gamma $(solid line) in$V$.} \label{figure4} \end{figure} As we mention in Remark \ref{Remark11}, the set$\Gamma $as in \eqref{10} was found by means of numerical computations. In order to show that the plot of$\Gamma $is adequate, we study the implicit function${{P({\omega ,} \lambda )=0}}$in$V$analytically. It is done in several steps. The first lemma studies${{P({\omega ,}\lambda )=0}}$in the vicinity of the point $$a=(\tfrac{1}{2}\pi ,4) \in \Gamma . \label{10.1}$$ \begin{lemma}\label{lemma5} Let$U=I\times J\subset V$be the closed rectangle with$I= \left[ \tfrac{88}{180}\pi ,\tfrac{92}{180}\pi \right] $,$J=[3.940,4.060] $and let point$a\in U$be as in (\ref{10.1}). The set$\Gamma $given by \eqref{10} consists of two smooth branches passing through$a$. Their tangents at$a$are$\lambda =4$and$\lambda =-\tfrac{16\sqrt{2} }{\pi }\omega +4$. \end{lemma} \begin{proof} Let$DP$stand for the gradient vector and$D^2P$is the Hessian matrix. For the given$a$we already know that$DP(a)=0$. We also find \begin{equation*} \tfrac{\partial ^2P}{\partial \omega ^2}(a)=0, \quad \tfrac{\partial ^2P}{% \partial \omega \partial \lambda }(a)=-8\sqrt{2}\pi,\quad \tfrac{\partial ^2P% }{\partial \lambda ^2}(a)=-\pi ^2. \end{equation*} That is,$\det D^2P(a)=-128\pi ^2$and by Proposition \ref{Proposition.ap.D} and remark \ref{remark2,ap.D} (Appendix \ref{AppendixD}) it holds that $$P(\omega ,\lambda )=-\tfrac{1}{2}h_2(\omega ,\lambda) \left( 16\sqrt{2}% h_1(\omega ,\lambda) +\pi h_2(\omega ,\lambda) \right) \quad\text{on }U, \label{11.0}$$ where$h_1,h_2\in C^{\infty }\left( U,\mathbb{R}\right) $are given by almost explicit formulas in (\ref{D18}), (\ref{D19}) in the same lemma. We also have that$h_1(a)=h_2(a)=0$and \begin{gather} \tfrac{\partial h_1}{\partial \omega }(a)=1, \quad \tfrac{\partial h_1}{% \partial \lambda }(a)=0, \label{12} \\ \tfrac{\partial h_2}{\partial \omega }(a)=0, \quad \tfrac{\partial h_2}{% \partial \lambda }(a)=1. \label{13} \end{gather} Due to (\ref{11.0}) we deduce that in$U$: $$P(\omega ,\lambda )=0 \quad\text{if and only if}\quad h_2(\omega ,\lambda) =0 \text{ or }16\sqrt{2}h_1(\omega ,\lambda) +\pi h_2(\omega ,\lambda) =0. \label{11}$$ By applying the Implicit Function Theorem to the functions$h_2(\omega ,\lambda) =0$and$16\sqrt{2}h_1(\omega ,\lambda) +\pi h_2(\omega ,\lambda) =0$in$U$one finds a parametrization$\omega \mapsto \lambda =\eta ({% \omega })$for each of these implicit functions. Indeed: (1) For$h_2(\omega ,\lambda) =0$it is shown in Lemma \ref{lemma3,ap.D} (Appendix \ref{AppendixD}) that \begin{equation*} \tfrac{\partial h_2}{\partial \lambda }(\omega ,\lambda) >0 \quad\text{on }U, \end{equation*} and hence there exists$\eta _1:I\to J$,$\eta _1\in {C}^{\infty}(I)$such that \begin{equation*} h_2(\omega ,\eta _1(\omega) )=0, \end{equation*} and \begin{equation*} \eta _1^{\prime }(\omega) =-\tfrac{\partial h_2}{\partial \omega }(\omega ,\eta _1(\omega) )\big[ \tfrac{\partial h_2 }{\partial \lambda }(\omega ,\eta _1(\omega) )\big] ^{-1}, \end{equation*} for all$\omega \in I$. We have that$\eta _1( \frac{1}{2}\pi ) =4$and due to (\ref{13}) we find$\eta _1^{\prime }( \tfrac{1}{2}\pi) =0$. Hence, there is a smooth branch of$\Gamma $in$U$passing through$a$, which is given by$\lambda =\eta _1(\omega) $with the tangent$\lambda =4$. (2) For$16\sqrt{2}h_1(\omega ,\lambda) +\pi h_2(\omega ,\lambda) =0$it is shown in Lemma \ref{lemma4,ap.D} (Appendix \ref{AppendixD}) that \begin{equation*} 16\sqrt{2}\tfrac{\partial h_1}{\partial \lambda }(\omega ,\lambda) +\pi \tfrac{\partial h_2}{\partial \lambda }(\omega ,\lambda) >0 \quad\text{on }U, \end{equation*} and hence there exists$\eta _2:\tilde{I}\to J$,$\eta _2\in {C} ^{\infty }(% \tilde{I})$, where$\tilde{I}\subset I$, such that \begin{equation*} 16\sqrt{2}h_1\left( \omega ,\eta _2(\omega) \right) +\pi h_2\left( \omega ,\eta _2(\omega) \right) =0, \end{equation*} and \begin{equation*} \eta _2^{\prime }(\omega) =-\frac{16\sqrt{2}\tfrac{\partial h_1}{\partial \omega }(\omega ,\eta _2(\omega) )+\pi \tfrac{ \partial h_2}{\partial \omega }(\omega ,\eta _2(\omega) )}{ 16\sqrt{2}\tfrac{\partial h_1}{\partial \lambda }(\omega , \eta _2(\omega) )+\pi \tfrac{\partial h_2}{\partial \lambda }(\omega ,\eta _2(\omega) )}, \end{equation*} for all$\omega \in \tilde{I}$. We have that$\eta _2( \frac{1}{2}\pi ) =4$and due to (\ref{12}) and (\ref{13}) we obtain \begin{equation*} \eta _2^{\prime }( \tfrac{1}{2}\pi) =-\tfrac{16\sqrt{2}}{\pi }. \end{equation*} Hence, there is another smooth branch of$\Gamma $in$U$passing through$a$and given by$\lambda =\eta _2(\omega) $. The tangent is$\lambda =-\tfrac{16% \sqrt{2}}{\pi }\omega +4$. \end{proof} The next lemma studies${{P({\omega ,}\lambda )=0}}$locally in$V$but away from the point$a$. \begin{lemma} \label{lemma6} Let \begin{gather*} H_1=\{ (\omega ,\lambda) :[ \tfrac{84}{180}\pi , \tfrac{90}{180}\pi ] \times [ 4.030,4.970] \} ,\\ H_2=\{ (\omega ,\lambda) :[ \tfrac{87}{180}\pi , \tfrac{101}{180}\pi ] \times [ 4.750,5.100] \} ,\\ H_{3}=\{ (\omega ,\lambda) :[ \tfrac{100}{180}\pi , \tfrac{108}{180}\pi ] \times [ 4.000,4.850] \} ,\\ H_{4}=\{ (\omega ,\lambda) :[ \tfrac{91}{180}\pi , \tfrac{102}{180}\pi ] \times [ 3.950,4.100] \} ,\\ H_{5}=\{ (\omega ,\lambda) :[ \tfrac{90}{180}\pi , \tfrac{96}{180}\pi ] \times [ 3.030,3.970] \} ,\\ H_{6}=\{ (\omega ,\lambda) :[ \tfrac{79}{180}\pi , \tfrac{94}{180}\pi ] \times [ 2.900,3.230] \} , \\ H_{7}=\{ (\omega ,\lambda) :[ \tfrac{72}{180}\pi , \tfrac{80}{180}\pi ] \times [ 3.150,4.000] \} ,\\ H_{8}=\{ (\omega ,\lambda) :[ \tfrac{78}{180}\pi , \tfrac{89}{180}\pi ] \times [ 3.900,4.050] \} , \end{gather*} and$U$be as in Lemma \ref{lemma5}. Then$\cup _{j=1}^{8}H_{j}$covers the set$\Gamma $in$V$(see Figure \ref{figure5}) and in each$H_{j}$the following holds: \begin{center} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|c|} \hline \rm Rectangle & \rm Property in$H_{j}$& \rm The set$\Gamma $in$H_{j}$is given by \\ \hline\hline$H_{2k-1}$&$\frac{\partial P}{\partial \omega } (\omega ,\lambda )\neq 0$&$\omega =\phi _{2k-1}(\lambda ):\phi _{2k-1}\in C^{\infty }(J_{2k-1})$\\ \hline$H_{2k}$&$\frac{\partial P}{\partial \lambda } (\omega ,\lambda )\neq 0$&$\lambda =\psi _{2k}({\omega }):\psi _{2k}\in C^{\infty }(I_{2k})$\\ \hline \end{tabular} \renewcommand{\arraystretch}{1} \end{center} Here$k=1,\dots,4$. \end{lemma} \begin{proof} In Claims \ref{Claim5,ap.D} -- \ref{Claim12,ap.D} of Appendix \ref{AppendixD} we constructed the rectangles$H_{j}\subset V$,$j=1,\dots,8$such that the results of the second column in a table above hold. In Figure \ref{figure5} we sketched the covering of the set$\Gamma $in$V$with the rectangles$% H_{j}$,$j=1,\dots,8$. \begin{figure}[ht] \begin{center} \includegraphics[width=0.7\textwidth]{fig5} %{Figure(Cover1).pdf} \end{center} \caption{For lemma \protect\ref{lemma6}.} \label{figure5} \end{figure} Due to result of the second column we can apply the Implicit Function Theorem to the function${{P({\omega ,}\lambda )=0}}$in every$H_{j}$,$% j=1,\dots,8$in order to obtain$\omega =\phi _{2k-1}(\lambda )$or${% \lambda } ${$=\psi _{2k}({\omega })$, }$k=1,\dots,4$. By assumption$P{\in C}% ^{\infty }(V,\mathbb{R})$and hence$\phi ,\psi $are$C^{\infty }$on the corresponding intervals$J,I$. \end{proof} Based on the results of the two lemmas above, we arrive at the following result. \begin{proposition}\label{Proposition1} The set$\Gamma$given by \eqref{10} is an$8$-shaped curve. That is, there exists an open set$\tilde{V}\supset [ -1,1]^2$\ and a$C^{\infty }$-diffeomorphism$S:V\to \tilde{V}$such that \begin{equation*} S\left( \Gamma \right) =\left\{ (\sin (2t),\sin (t)),\text{ }0\leq t<2\pi \right\} . \end{equation*} \end{proposition} Henceforth, we will call the set$\Gamma $a curve (having one self-intersection point) which means that every part of the set$\Gamma $is locally parametrizable in {$\omega $} or$\lambda $. \subsubsection{Eigenvalue$\protect\lambda _1$as the bottom part of$\Gamma $} The curve$\Gamma $in a rectangle$V$combines the graphs of the first four eigenvalues$\lambda _1,\dots,\lambda _{4}$of the boundary value problem ( % \ref{4}) as functions of$\omega $as far as they are real.\ Here we focus on the eigenvalue$\lambda _1$which is a bottom part of$\Gamma $(the segment$c_{6}c_1c_2\subset \Gamma $in Figure \ref{figure4}). In particular, we prove that as a function of$\omega $the eigenvalue$\lambda _1=\lambda _1(\omega )$increases between the points$c_1,c_2$(the approximations for their coordinates are given in Table \ref{table2}). The situation is illustrated by Figure \ref{figure6}. \begin{figure}[ht] \begin{center} \includegraphics[width=0.7\textwidth]{fig6} %{Figure5.pdf} \end{center} \caption{Increase of$\protect\lambda _1$between$c_1 $and$c_2$} \label{figure6} \end{figure} To prove this result, we follow the approach used in Lemmas \ref{lemma5} and % \ref{lemma6}. To be more precise, we fix two rectangles$\{ H_0,H_{\star }\} \subset V$such that$H_0\cap H_{\star }\ne \emptyset $and$H_0\cup H_{\star }$covers the part of$\Gamma $containing the segment$c_1c_2$(see Figure \ref{figure7}). We parameterize$\Gamma $in$H_0,H_{\star }$\ as {$\omega \mapsto \lambda =\psi ({\omega })$} and$\lambda \mapsto \omega =\varphi (\lambda )$, respectively, and study the properties of these parametrizations (convexity-concavity, extremum points, the intervals of increase-decrease). This will enable to gain the information about$c_1c_2$. \begin{figure}[ht] \begin{center} \includegraphics[width=0.44\textwidth]{fig7a} \quad % {Figure(Cover2).pdf} \includegraphics[width=0.44\textwidth]{fig7b} % {Figure(Cover3).pdf} \end{center} \caption{The rectangles$H_0,H_{\star }$from lemmas \protect\ref{lemma7} and \protect\ref{lemma8}, respectively (on the left); the enlarged view (on the right).} \label{figure7} \end{figure} \begin{lemma} \label{lemma7} Let$H_0=I_0\times J_0\subset V$be the closed rectangle with$I_0=\left[ \tfrac{84}{180}\pi ,\tfrac{94}{180}\pi \right] $and$J_0=\left[ 2.960,3.060\right] $. It holds that$\Gamma $in$H_0$is given by$\lambda =\psi (\omega )$,$\psi \in C^{\infty }(\omega _{\alpha },\omega _{\beta })$,$(\omega _{\alpha },\omega _{\beta })\subset I_0$and is such that it attains its minimum on$\left( \omega _{\alpha },\omega _{\beta }\right) $at$\omega =\omega _0=\frac{1}{2}\pi $and increases monotonically on$\left( \omega _0,\omega _{\beta }\right) $. Here$\omega _{\alpha },\omega _{\beta }$are the solutions to the equation$P(\omega ,3.060)=0$on$\omega \in \left( \frac{84}{180}\pi ,\frac{1}{2}\pi \right) $and on$\omega \in \left( \frac{1}{2}\pi ,\frac{94}{180}\pi \right) $, respectively, with$P$given by (\ref{9}). \end{lemma} \begin{proof} By Lemma \ref{lemma6} we know that $$P(\omega ,\lambda )=0\quad\text{if and only if } P\left( \omega ,\psi (\omega) \right) =0 \text{ in }H_{6}, \label{16}$$ and if we take the rectangle$H_0$defined as in lemma above, then due to$% H_0\subset H_{6}$, (\ref{16}) will also hold in$H_0$. Moreover, we also set$H_0$in such a way that its top boundary intersects$\Gamma $at two points, meaning that we find two solutions of$P(\omega ,3.060)=0$with$P$as in (\ref{9}). We name these two solutions$\omega _{\alpha },\omega_{\beta }$. Hence, we deduce that$\Gamma $in$H_0$is given by$\lambda =\psi (\omega ) $,$\psi \in C^{\infty }(\omega _{\alpha },\omega _{\beta })$and satisfies$\psi (\omega _{\alpha })=\psi (\omega _{\beta })=3.060$. Due to condition \begin{equation*} \psi (\omega _{\alpha })=\psi (\omega _{\beta }), \end{equation*} by Rolle's theorem there exists$\omega _0\in ( \omega _{\alpha },\omega _{\beta }) $such that$\psi ^{\prime }(\omega _0)=0$. Since$P\left( \omega _0,\psi \left( \omega _0\right) \right) =0$and due to \begin{equation*} \psi ^{\prime }(\omega )=-\tfrac{\partial P}{\partial \omega }(\omega ,\psi (\omega) )[ \tfrac{\partial P}{\partial \lambda }(\omega ,\psi (\omega) )] ^{-1}, \end{equation*} we solve the system$P(\omega ,\lambda )=0$and$\frac{\partial P}{\partial \omega }(\omega ,\lambda )=0$in$H_0$in order to find$\omega _0$. Its solution is a point$c_1=( \frac{1}{2}\pi ,3) $and hence \begin{equation*} \omega _0=\tfrac{1}{2}\pi . \end{equation*} We deduce that$\lambda =\psi (\omega )$attains its local extremum at$% \omega =\omega _0$. Next we show that$\lambda =\psi (\omega )$has a minimum at$\omega =\omega _0$on$(\omega _{\alpha },\omega _{\beta })$. For this purpose we consider a function$G\in C^{\infty }( H_0,\mathbb{R}) $such that $$G\left( \omega ,\psi (\omega )\right) =\psi ^{\prime \prime }(\omega ). \label{14}$$ For an explicit formula for$G$see Appendix \ref{AppendixD}. In Claim \ref% {Claim13,ap.D} of this Appendix we show that $$G(\omega ,\lambda) >0\quad\text{on }H_0. \label{15}$$ This condition together with (\ref{14}) yields \begin{equation*} G\left( \omega ,\psi (\omega )\right) =\psi ^{\prime \prime }(\omega )>0\quad% \text{on }(\omega _{\alpha }, \omega _{\beta }), \end{equation*} meaning that$\lambda =\psi (\omega )$is convex on$(\omega _{\alpha},\omega _{\beta })$. The result is that$\lambda =\psi (\omega )$attains its minimum on$(\omega _{\alpha },\omega _{\beta })$at$\omega =\omega _0=\frac{1}{2}\pi $and increases monotonically on the interval$\omega \in \left( \omega _0,\omega _{\beta }\right) $. \end{proof} We also have the following result. \begin{lemma}\label{lemma8} Let$H_{\star }=I_{\star }\times J_{\star }\subset V$be the closed rectangle with$I_{\star }=\left[ \tfrac{93.5}{180}\pi ,\tfrac{95.5}{ 180}\pi \right] $and$J_{\star }=\left[ 3.030,3.600\right] $. It holds that$\Gamma $in$H_{\star }$is given by$\omega =\varphi (\lambda )$,$\varphi \in C^{\infty }(\lambda _{\gamma },\lambda _{\delta })$,$(\lambda _{\gamma },\lambda _{\delta })\subset J_{\star }$and is such that it attains its maximum on$\left( \lambda _{\gamma },\lambda _{\delta }\right) $at$\lambda =\lambda _{\star }\approx 3.220\dots$and increases monotonically on the interval$\left( \lambda _{\gamma },\lambda _{\star }\right) $. Here$\lambda _{\gamma },\lambda _{\delta }$are the solutions to the equation$P\left( \tfrac{93.5}{180}\pi ,\lambda \right) =0$on$\lambda \in \left( 3.030,3.100\right) $and on$\lambda \in \left( 3.500,3.600\right) $, respectively. Also,$\lambda _{\star }$is the solution to the system$P(\omega ,\lambda )=0$and$\frac{\partial P}{\partial \lambda }(\omega ,\lambda )=0$on$\lambda \in \left( \lambda _{\gamma },\lambda _{\delta }\right) $;$P$given by (\ref{9}). \end{lemma} \begin{proof} By Lemma \ref{lemma6} we know that $$P(\omega ,\lambda )=0\quad\text{if and only if}\quad P\left( \varphi (\lambda ),\lambda \right) =0 \text{ in }H_{5}, \label{17}$$ and if we take the rectangle$H_{\star }$defined as in lemma above, then due to$H_{\star }\subset H_{5}$, (\ref{17}) will also hold in$H_{\star }$. Moreover, we also set$H_{\star }$in such a way that its left boundary intersects$\Gamma $at two points, meaning we find two solutions of$% P\left( \tfrac{93.5}{180}\pi ,\lambda \right) =0$with$P$as in (\ref{9}). We name these two solutions$\lambda _{\gamma },\lambda _{\delta }$. Hence, we deduce that$\Gamma $in$H_{\star }$is given by$\omega =\varphi (\lambda )$,$\varphi \in C^{\infty }(\lambda _{\gamma },\lambda _{\delta })$and satisfies$\varphi (\lambda _{\gamma })=\varphi (\lambda _{\delta })= \tfrac{93.5}{180}\pi $. Due to condition \begin{equation*} \varphi (\lambda _{\gamma })=\varphi (\lambda _{\delta }), \end{equation*} by Rolle's theorem there exists$\lambda _{\star }\in \left( \lambda _{\gamma },\lambda _{\delta }\right) $such that$\varphi ^{\prime }(\lambda _{\star })=0$. Since$P\left( \varphi (\lambda _{\star }),\lambda _{\star }\right) =0$and due to \begin{equation*} \varphi ^{\prime }(\lambda )=-\tfrac{\partial P}{\partial \lambda }(\varphi (\lambda ),\lambda )[ \tfrac{\partial P}{\partial \omega }(\varphi (\lambda ),\lambda )] ^{-1}, \end{equation*} we solve the system$P(\omega ,\lambda )=0$and$\frac{\partial P}{\partial \lambda }(\omega ,\lambda )=0$in$H_{\star }$in order to find$\lambda _{\star }$. Its solution is a point$c_2=\big( \tilde{\omega},\tilde{ \lambda% }\big) $, where$\tilde{\omega}/\pi \approx 0.528\dots$and$\tilde{ \lambda}% \approx 3.220\dots$. Hence, \begin{equation*} \lambda _{\star }\approx 3.220\dots\,. \end{equation*} We deduce that$\omega =\varphi (\lambda )$attains its local extremum at$% \lambda =\lambda _{\star }$. Next we show that$\omega =\varphi (\lambda )$has a maximum at$\lambda =\lambda _{\star }$on$(\lambda _{\gamma },\lambda _{\delta })$. For this purpose we consider a function$F\in C^{\infty }\left( H_{\star },\mathbb{R} \right) $such that $$F\left( \varphi (\lambda ),\lambda \right) =\varphi ^{\prime \prime }(\lambda ). \label{18}$$ For explicit formula for$F$see Appendix \ref{AppendixD}. In Claim \ref% {Claim14,ap.D} of this Appendix we show that $$\begin{tabular}{ccc} F(\omega ,\lambda) <0\quad\text{on }H_{\star }. & & \end{tabular} \label{19}$$ This condition together with (\ref{18}) yields \begin{equation*} F\left( \varphi (\lambda ),\lambda \right) =\varphi ^{\prime \prime }(\lambda )<0 \quad\text{on }(\lambda _{\gamma },\lambda _{\delta }), \end{equation*} meaning that$\omega =\varphi (\lambda )$is concave on$(\lambda _{\gamma },\lambda _{\delta })$. The result is that$\omega =\varphi (\lambda )$attains its maximum on$\left( \lambda _{\gamma },\lambda _{\delta }\right) $at$\lambda =\lambda _{\star }\approx 3.220\dots$and increases monotonically on the interval$\lambda \in \left( \lambda _{\gamma },\lambda _{\star }\right) $. \end{proof} \begin{theorem}\label{Increase} As a function of$\omega $the first eigenvalue$\lambda_1=\lambda _1(\omega )$of the boundary value problem \eqref{4} increases on$\omega \in \left( \frac{1}{2}\pi ,\omega _{\star }\right) $. Here$\omega _{\star }/\pi \approx 0.528\dots$(in degrees$\omega _{\star }\approx 95.1\dots^{\circ }$) and$\lambda _{\star }\approx 3.220\dots$. \end{theorem} \subsection{The multiplicities of$\{ \protect\lambda _{j}\}_{j=1}^{\infty }$and the structure of a singular solution} Here we proceed with the qualitative analysis of the eigenvalues$\{ \lambda _{j}\} _{j=1}^{\infty }$of problem \eqref{4}. \begin{definition} \rm Let$\omega \in ( 0,2\pi] $\ be fixed. The eigenvalue$\lambda _{j}$,$j\in \mathbb{N} ^{+}$of problem \eqref{4} is said to have an algebraic multiplicity\$\kappa ^{(j)}\geq 1$, if the following holds: \begin{equation*} P(\omega ,\lambda _{j})=0, \quad \tfrac{dP}{d\lambda }(\omega ,\lambda _{j})=0, \quad\dots,\quad \tfrac{d^{\kappa ^{(j)}-1}P}{d\lambda ^{\kappa ^{(j)}-1}}(\omega ,\lambda _{j})=0, \quad \tfrac{d^{\kappa ^{(j)}}P}{d\lambda^{\kappa ^{(j)}}}(\omega ,\lambda _{j}) \neq 0. \end{equation*} \end{definition} Based on the numerical approximations for some first eigenvalues$\lambda _{j}$,$j\in \mathbb{N}^{+}$depicted in Figure \ref{figure3} (the top one) and partly by our derivations (namely, the existence of the solution to the system$P(\omega ,\lambda )=\frac{\partial P}{\partial \lambda }(\omega ,\lambda )=0$in Lemma \ref{lemma8}) we believe that the maximal algebraic multiplicity of a certain$\lambda _{j}$of problem \eqref{4} is at most$2$. Indeed, generically$3$curves never intersect at one point, meaning that geometrically the algebraic multiplicity will always be at most$2$. \begin{definition} \rm The eigenvalue$\lambda _{j}$,$j\in \mathbb{N}^{+}$of problem \eqref{4} is said to have a geometric multiplicity$I^{(j)}\geq 1$, if the number of linearly independent eigenfunctions$\Phi $equals$I^{(j)}$. \end{definition} For given$\lambda _{j}$,$j\in \mathbb{N}^{+}$of problem \eqref{4} the three cases occur: 1.$\kappa ^{(j)}=I^{(j)}=1$one finds a solution$(\lambda _{j},\Phi _0^{(j)})$of \eqref{4} and then the solution of (\ref{3}) reads: $$u_0^{(j)}=r^{\lambda _{j}+1}\Phi _0^{(j)}(\theta ); \label{26}$$ 2.$\kappa ^{(j)}=2$,$I^{(j)}=1$one finds a solution$(\lambda _{j},\Phi _0^{(j)})$of \eqref{4} and a generalized solution$(\lambda _{j},\Phi _1^{(j)})$, with$\Phi _1^{(j)}$found from the equation \begin{equation*} \mathcal{L}(\lambda _{j})\Phi _1^{(j)}+\mathcal{L}^{\prime }(\lambda _{j})\Phi _0^{(j)}=0, \end{equation*} where$\mathcal{L}(\lambda )$is given by (\ref{5}) and$\mathcal{L}^{\prime }(\lambda )=\frac{d}{d\lambda }\mathcal{L}(\lambda )$. Then we have two solutions of (\ref{3}): $$u_0^{(j)}=r^{\lambda _{j}+1}\Phi _0^{(j)}(\theta )\text{ \ \ and \ \ } u_1^{(j)}=r^{\lambda _{j}+1}\left( \Phi _1^{(j)}(\theta )+\log (r)\Phi _0^{(j)}(\theta )\right) ; \label{27}$$ 3.$\kappa ^{(j)}=I^{(j)}=2$one finds two solutions$(\lambda _{j},\Phi _0^{(j)})$,$(\lambda _{j},\Phi _1^{(j)})$of \eqref{4}, where$\Phi _0^{(j)} $and$\Phi _1^{(j)}$are linearly independent on$\theta \in (0,\omega )$and then we again have two solutions of (\ref{3}): $$u_0^{(j)}=r^{\lambda _{j}+1}\Phi _0^{(j)}(\theta )\text{ \ \ and \ \ } u_1^{(j)}=r^{\lambda _{j}+1}\Phi _1^{(j)}(\theta ). \label{28}$$ Let us note that for an opening angle$\omega \in \{ \frac{1}{2}\pi ,\pi ,% \frac{3}{2}\pi ,2\pi \} $of the sector$\mathcal{K}_{\omega }$one can find the eigenvalues$\{ \lambda _{j}\} _{j}^{\infty }$of the problem \eqref{4} explicitly. Moreover, if$\omega \in \left\{ \frac{1 }{2}\pi ,\pi \right\} $% , then for every given$\lambda _{j}$one can compute explicitly the corresponding eigenfunctions$\Phi _{q}^{(j)}$,$q=0,\dots,\kappa ^{(j)}-1$, whereas if$\omega \in \left\{ \frac{3}{2}\pi ,2\pi \right\} $, the eigenfunctions$\Phi _{q}^{(j)}$can be computed explicitly only for some$% \lambda _{j}$. Thus, in Appendix \ref{AppendixE} we bring the formulas of some first functions$\Phi _{q}^{(j)}$(if computable) and the respective solutions$u_{q}^{(j)}$to (\ref{3}). These functions$r^{\lambda +1}\Phi (\theta )$and$r^{\lambda +1}\log (r)\Phi (\theta )$determine the bands for the regularity in Kondratiev's theory. Details are found in the next section. \section{Regularity results\label{Reg}} In this subsection we will give the regularity result for the boundary value problem \eqref{1} under consideration. In order to do this we refer to the key theorem of the Kondratiev theory (see e.g. \cite[Theorem 6.4.1]{KMR}). The general version adapted to our problem \eqref{1} will read as: \begin{theorem}[Kondratiev] \label{Theorem1} Let$u\in V_{\beta _1}^{l_1,2}(\Omega )$with$l_1\in \mathbb{N}$,$\beta _1\in \mathbb{R}$be a solution of the elliptic boundary value problem \eqref{1}. Suppose that$f\in V_{\beta _2}^{l_2,2}(\Omega )$, where$l_2\in \mathbb{N}$,$\beta _2\in \mathbb{R}$and such that$l_1-\beta _10 $and$\Phi (\theta) ,\Psi (\theta )\in C^{\infty }[ 0,\omega] $,$0<\omega <2\pi $. \begin{lemma}\label{lemma18} Let$\Phi (\theta) \in C^{\infty }[0,\omega] $be nontrivial and$0<\omega <2\pi $. Let also$\lambda \in \mathbb{C} \backslash \mathbb{Z}$with$\mathop{\rm Re}(\lambda )>0$. Suppose that$k\in \{ 0,1,2,3,\dots\} $. Then the following are equivalent: \begin{enumerate} \item$r^{\lambda +1}\Phi (\theta) \in W^{k,2}(\Omega )$, \item$r^{\lambda +1}\log (r)\Phi (\theta) \in W^{k,2}(\Omega) $, \item$\mathop{\rm Re}(\lambda )+1>k-1$. \end{enumerate} \end{lemma} \begin{proof} If$\lambda $is not an integer, then the first item and second items are equivalent with an integrability condition for the$k^{\text{th}}$-derivative that reads as$2\mathop{\rm Re}( \lambda +1-k) +1>-1$. \end{proof} \begin{remark} \rm To restrict the already heavy technical aspects we have not considered (weighted) Sobolev spaces with non-integer coefficients$k$and H\"{o}lder spaces. A similar result will hold for$k$is noninteger. Concerning H\"{o}lder spaces: \begin{equation*} r^{\lambda +1}\Phi (\theta) \in C^{k,\gamma }(\bar{\Omega})\quad \text{for }\mathop{\rm Re}(\lambda )+1\geq k+\gamma \text{ with } k\in \mathbb{N},\gamma \in [ 0,1) . \end{equation*} For the second function it holds that \begin{equation*} r^{\lambda +1}\log (r)\Phi (\theta) \in C^{k,\gamma }(\bar{ \Omega})\quad \text{for }\mathop{\rm Re}(\lambda )+1>k+\gamma \text{ with }k\in \mathbb{N},\gamma \in [ 0,1) . \end{equation*} \end{remark} A useful consequence of the above lemma is that for every fixed$\lambda \in \mathbb{C} $with$\mathop{\rm Re}(\lambda) >0$we deduce $$\left\{ r^{\lambda +1}\Phi (\theta) , r^{\lambda +1}\log (r)\Phi (\theta) \right\} \in W^{\lceil \mathop{\rm Re}(\lambda) \rceil +1,2}(\Omega ), \label{68}$$ where$\lceil \cdot \rceil $stands for the ceiling function (defined as$% \lceil x\rceil =\min \{ n\in \mathbb{Z} :x\leq n\} $). \begin{remark} \rm In the particular cases, namely,$\omega =\frac{1}{2}\pi $and$\omega =\pi $we know that each term$u_{q}^{(j)}$of the singular part$\sum_{j=1}^{N}\sum_{q=0}^{\kappa ^{(j)}-1}c_{q}^{(j)}u_{q}^{(j)}$is a polynomial in$x,y$of order$\lambda _{j}+1$(see Appendix \ref{AppendixE}). That is, for every$\lambda _{j}$,$j\in \mathbb{N}^{+}$we have $$r^{\lambda _{j}+1}\Phi ^{(j)}(\theta) =P_{\lambda _{j}+1}(x,y)\in C^{\infty }\left( \overline{\Omega }\right) . \label{69}$$ For non-polynomials the result in Lemma \ref{lemma18} even holds for$\lambda \in \mathbb{N}$. \end{remark} Now, in order to use the result of Lemma \ref{lemma18}, we proceed with Figure \ref{figure8} where we plot the$\mathop{\rm Re}(\lambda _1) $as a function of the opening angle$\omega $on the interval$\omega \in (0,2\pi ] $. The two cases are compared: the plots of$\mathop{\rm Re} (\lambda _1) $of the boundary value problem \eqref{4}\ for$\mathcal{L}$related to$L=% \tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{ \partial ^{4}}{\partial y^{4}}$and$L=\Delta ^2$. In Figure \ref{figure9} we split the plot of Figure \ref% {figure8} into two: a plot of$\mathop{\rm Re} (\lambda _1) $on$\omega \in (0,\pi ]$and$\omega \in [\pi ,2\pi] $. \begin{figure}[ht] \begin{center} \includegraphics[width=0.7\textwidth]{fig8} % Figure(FirstEigenvalue) \end{center} \caption{The plot of eigenvalue$\protect\lambda _1$in$\left(\protect% \omega ,\mathop{\rm Re}(\protect\lambda) \right) \in ( 0,2\protect\pi] \times \left[ 0,7.200\right] $of problem ( \protect\ref{4}). For$\mathcal{L% }$related to$\frac{\partial ^{4}}{\partial x^{4}}+\frac{\partial ^{4}}{% \partial y^{4}}$,$\protect\lambda _1$is represented by the red line and for$\mathcal{L}$related to$\Delta ^2$, by the blue line. Dashed lines depict the real part of$\protect\lambda _1\in \mathbb{C} $, solid lines are for purely real$\protect\lambda _1$; the vertical lines mark out$\left\{% \frac{1}{2}\protect\pi , \protect\pi , \frac{3}{2}\protect\pi ,2\protect\pi % \right\}$on$\protect\omega $-axis.} \label{figure8} \end{figure} \begin{figure}[ht] \begin{center} \includegraphics[width=0.6\textwidth]{fig9a} \\[0pt] \includegraphics[width=0.6\textwidth]{fig9b} % Figure(FirstEigenvalue2).pdf} \end{center} \caption{The plot of Figure \protect\ref{figure8} rescaled. The curves for the grid-operator and the bilaplacian intersect four times. Three of these points (we obviously exclude$\protect\omega =\protect\pi $) seem to be special: It looks like the curves intersect at$\protect\omega =\frac{3}{4}% \protect\pi $,$\protect\omega =\frac{5}{4} \protect\pi $and$\protect% \omega =\frac{7}{4}\protect\pi $(the intersection points are marked by cross). The numerical approximation shows however that the first values$% \protect\lambda _{1,Grid}$and$\protect\lambda _{1,Bilaplace}$at those points only coincide up to three digits.} \label{figure9} \end{figure} Based on the numerical approximations to$\lambda _1$and partly on the analytical estimates for$\lambda _1$we conclude the following. \begin{claim}\label{Claim2} \rm$(0,2\pi ]\ni \omega \mapsto \mathop{\rm Re}(\lambda _1(\omega )) $is a continuous function and \begin{gather*} \text{for } \omega \in (0,\omega _1)/\{ \frac{1}{2}\pi\}: \mathop{\rm Re}(\lambda _1)>3,\quad \text{for } \omega =\frac{1}{2}\pi : \mathop{\rm Re}(\lambda _1)=3, \\ \text{for } \omega \in [ \omega _1,\omega _2): 3\geq \mathrm{Re }(\lambda _1)>2, \quad \text{for } \omega \in [ \omega _2,\pi ): 2\geq \mathop{\rm Re} (\lambda _1)>1,\\ \text{for } \omega \in [ \pi ,2\pi ]: 1\geq \mathop{\rm Re}(\lambda _1)\geq \frac{1}{2}. \end{gather*} Here$\omega _1,\omega _2$are respectively the solutions of$P(\omega ,3+i\xi )=0$on$\omega \in \left( \frac{1}{2}\pi ,\frac{120}{180}\pi \right) $and of$P(\omega ,2+i\xi )=0$on$\omega \in \left( \frac{2}{3}\pi ,\frac{3}{4}\pi \right) $, where$P$as in formula (\ref{6.1}) for$\ell =1$. The approximation are$\omega _1/\pi \approx 0.555\dots$(in degrees$\omega _1\approx 99.9\dots^{\circ }$) and$\omega _2/\pi \approx 0.720\dots$(in degrees$\omega _2\approx 129.7\dots^{\circ }$). \end{claim} \subsection{Consequences} For the numerical results from Claim \ref{Claim2}, it holds by Theorem \ref% {Theorem2} that: \begin{corollary}\label{Corollary100} Let$u\in {\mathaccent"7017 W}^{2,2}(\Omega )$be a weak solution of problem \eqref{1} with$f\in L^2(\Omega )$. Then \begin{gather*} \text{for } \omega \in ( 0,\omega _2) : u\in W^{4,2}(\Omega),\quad \text{for }\omega \in (\omega _2,\pi ): u\in W^{3,2}(\Omega ),\\ \text{for }\omega =\pi : u\in W^{4,2}(\Omega ),\quad \text{for }\omega \in (\pi ,2\pi ]: u\in W^{2,2}(\Omega ). \end{gather*} Here$\omega _2$is as in Claim \ref{Claim2}. \end{corollary} \begin{remark} \rm For the opening angle$\omega =\omega _2$we have$\mathop{\rm Re}(\lambda _1) =2$and hence Theorem \ref{Theorem2} does not apply. Nevertheless, assuming$f\in L^2(\Omega )$to be more regular, e.g. in$V_0^{1,2}(\Omega )$or${\mathaccent"7017 W}^{1,2}(\Omega )$, we may show that \begin{equation*} \text{for }\omega =\omega _2: u\in W^{3,2}(\Omega ). \end{equation*} \end{remark} \begin{proof} By Theorem \ref{Theorem2} if$f\in L^2(\Omega )$, the solution$u$of problem \eqref{1} reads as $$u=w+\chi (r)\sum_{0<\lambda _{j}<2}\sum_{q=0}^{\kappa ^{(j)}-1}c_{q}^{(j)}u_{q}^{(j)}, \label{70}$$ with$w\in W^{4,2}(\Omega )$. Due to \ref{Claim2} we see that the sum in ( % \ref{70}) has no terms when$\omega \in (0,\omega _2) $and hence for$% \omega \in (0,\omega _2) $we have$u\in W^{4,2}(\Omega )$. For$\omega \in (\omega _2,\pi )\cup (\pi ,2\pi ]$the first term of sum in (% \ref{70}), depending on the algebraic multiplicity of$\lambda _1$, reads as$u_0^{(1)}=r^{\lambda _1+1}\Phi _0^{(1)}(\theta )$, or as a linear combination of$u_0^{(1)}=r^{\lambda _1+1}\Phi _0^{(1)}(\theta ) $\ and\$% u_1^{(1)}=r^{\lambda _1+1}\left( \Phi _1^{(1)}(\theta )+\log (r)\Phi _0^{(1)}(\theta )\right) $. By (\ref{68}) we have that$\left\{ u_0^{(1)},u_1^{(1)}\right\} \in W^{\lceil \mathop{\rm Re}(\lambda) \rceil +1,2}(\Omega )$, where due to Claim \ref{Claim2} we may deduce that$\lceil % \mathop{\rm Re}(\lambda _1) \rceil +1=3$, when$\omega \in (\omega _2,\pi )$and$\lceil \mathop{\rm Re}(\lambda _1) \rceil +1=2$, when$\omega \in (\pi ,2\pi ]$. This results in$u\in W^{3,2}(\Omega )$for$\omega \in (\omega _2,\pi )$and$u\in W^{2,2}(\Omega )$for$\omega \in (\pi ,2\pi ]$. Finally, for$\omega =\pi $due to (\ref{69}) the singular part is of$% C^{\infty }( \overline{\Omega }) $and hence$u\in W^{4,2}(\Omega )$in this case. \end{proof} \section{Comparing (weighted) Sobolev spaces} \subsection{One-dimensional Hardy-type inequalities} \begin{lemma}[A higher order one-dimensional Hardy inequality] \label{lemma1,ap.F} Let$w$be a function in$C_0^{\infty }[ x_1,x_2] $. For every$k\geq 1it holds that $$\int_{x_1}^{x_2}\left( \tfrac{w(x)}{(x-x_1) ^{k}} \right) ^2dx=\tfrac{4^{k}}{\left( 2k-1\right) ^2\left( 2k-3\right) ^2\dots3^21^2}\int_{x_1}^{x_2}\left( w^{(k)}(x)\right) ^2dx. \label{F2}$$ \end{lemma} \begin{proof} It holds straightforwardly that \begin{align*} &\int_{x_1}^{x_2}\left( \tfrac{w(x)}{(x-x_1) ^{k}}\right)^2dx \\ &= \frac{1}{1-2k} \left[ \left( w(x)\right) ^2(x-x_1) ^{1-2k}\right] | _{x_1}^{x_2}+\frac{2}{2k-1} \int_{x_1}^{x_2}w(x)w^{\prime }(x)(x-x_1) ^{1-2k}dx \\ &\leq \frac{2}{2k-1}\Big( \int_{x_1}^{x_2} \left( \tfrac{w(x)}{(x-x_1) ^{k}}% \right) ^2dx\Big) ^{1/2} \Big(\int_{x_1}^{x_2}\left( \tfrac{w^{\prime }(x)}{% (x-x_1)^{k-1}}\right) ^2dx \Big) ^{1/2} \end{align*} and the first step in the proof of (\ref{F2}) follows. Repeating the argument forw^{\prime }$and$k-1$etc. will give the result. \end{proof} \begin{remark} \rm Since${\mathaccent"7017 W}^{k,2}(x_1,x_2)$,$k\geq 1$is the closure of$C_0^{\infty }[ x_1,x_2] $in the$W^{k,2}$-norm, one can use the results of Lemma \ref{lemma1,ap.F} for every$w\in {\mathaccent"7017 W}^{k,2}(x_1,x_2)$,$k\geq 1$. \end{remark} \subsection{Imbeddings} As mentioned e.g. in \cite[page 240]{K} or \cite[Chapter 7, summary]{KMR}, the family of weighted spaces$V_{\beta }^{l,2}$does not contain the ordinary Sobolev spaces without weight. More precisely:$W^{k,2}\notin \left\{ V_{\beta }^{l,2}\right\} _{l,\beta }$for$k\geq 1$. We will prove the imbedding results for bounded$\Omega $that satisfy Condition \ref% {Condition0}. \begin{lemma} \label{lemma10} Let$\beta \in \mathbb{R}$and$l\in \left\{ 0,1,2,\dots \right\} $. Then the following holds: \begin{itemize} \item[(a)]$V_{\beta }^{l,2}(\Omega )\subset W^{l,2}(\Omega )$if and only if$\beta \leq 0$, \item[(b)]$ W^{l,2}(\Omega )\subset V_{\beta }^{l,2}(\Omega )$if and only if$\beta \geq l$, \item[(c)]${\mathaccent"7017 V}_{\beta }^{l,2}(\Omega )\subset {\mathaccent"7017 W}^{l,2}(\Omega )$if and only if$ \beta \leq 0$, \item[(d)]${\mathaccent"7017 W}^{l,2}(\Omega )\subset {\mathaccent"7017 V}_{\beta }^{l,2}(\Omega )$if and only if$\beta \geq 0$. \end{itemize} \end{lemma} \begin{corollary}\label{coro-imbed} For$l\in \{0,1,2,\dots\}$one has \begin{equation*} {\mathaccent"7017 W}{}^{l,2}(\Omega ) = {\mathaccent"7017 V}_0^{l,2}(\Omega ). \end{equation*} \end{corollary} \begin{proof}[Proof of Lemma \protect\ref{lemma10}] Let$\Omega $be as in Condition \ref{Condition0} and$\Omega \subset B_{M}(0)$, where$B_{M}(0)$is an open ball of radius$M>0$. The statement in a) goes as follows: for$(x,y)\in \Omega $one has$0\leq r\leq M$\ and hence$r^{2\left( \beta -l+| \alpha | \right) }\geq M^{2\left( \beta -l+| \alpha | \right) }$if and only if$\beta -l+| \alpha | \leq 0$. Since$% 0\leq | \alpha | \leq l$, we obtain$\beta \leq 0. This enables us to have the estimate \begin{align*} \| u\| _{V_{\beta }^{l,2}(\Omega )} &=\Big( \sum_{|\alpha | =0}^{l}\int_{\Omega }r^{2( \beta -l+| \alpha |) }| D^{\alpha }u|^2dx\,dy% \Big) ^{1/2} \\ &\geq \Big( \sum_{| \alpha| =0}^{l}\int_{\Omega }M^{2( \beta -l+| \alpha |) }| D^{\alpha }u| ^2dx\,dy\Big) ^{1/2} \\ &\geq \min (1,M^{\beta -l})\Big( \sum_{| \alpha |=0}^{l}\int_{\Omega }| D^{\alpha }u| ^2dx\,dy\Big) ^{1/2} \\ &=\min (1,M^{\beta -l})\| u\|_{W^{l,2}(\Omega )}, \end{align*} which is the result in (a). To prove the statement in (b) we notice thatr^{2( \beta -l+|\alpha |) }\leq M^{2( \beta -l+| \alpha|) }$if and only if$\beta -l+| \alpha | \geq 0 $. Due to$0\leq | \alpha | \leq l$, we obtain$\beta \geq land then the estimate holds \begin{align*} \| u\| _{V_{\beta }^{l,2}(\Omega )} &=\Big( \sum_{| \alpha | =0}^{l}\int_{\Omega }r^{2\left( \beta -l+| \alpha | \right) }| D^{\alpha }u|^2dx\,dy\Big) ^{1/2} \\ &\leq \Big( \sum_{| \alpha | =0}^{l}\int_{\Omega }M^{2\left( \beta -l+| \alpha | \right) }| D^{\alpha }u| ^2dx\,dy\Big) ^{1/2} \\ &\leq \max (1,M^{\beta -l})\Big( \sum_{| \alpha | =0}^{l}\int_{\Omega }| D^{\alpha }u| ^2dx\,dy\Big) ^{1/2} \\ &=\max (1,M^{\beta -l})\| u\|_{W^{l,2}(\Omega )}. \end{align*} This is the result in (b). To prove the statements in (c) and (d) we set\theta =\frac{1}{2} \omega $where the opening angle$\omega \in (0,2\pi) $. We also use the fact that for our domain there exists$c>0$such that$r>c\rho (x,y)$, where$\rho $denotes the distance from a point$(x,y)$on the lines \begin{equation*} \ell :y=\tan (\theta) x+\tau , \end{equation*} with$\tau \in \mathbb{R}$to the point$(x_1,y_1)\in \partial \Omega $. In particular, it holds that \begin{equation*} \rho ^2=(x-x_1) ^2(1+\tan ^2(\theta)) . \end{equation*} \begin{figure}[ht] \begin{center} \includegraphics[width=0.65\textwidth]{fig10}% Figure(Hardy) \end{center} \caption{Domain$\Omega $with a concave corner$\protect\omega $intersected by$\ell $} \label{figure10} \end{figure} We may integrate along the lines$\ell $and use the one-dimensional Hardy-inequality to find that there exist$\tilde{C}_{l}\in \mathbb{R}^{+}$with $$\| u\| _{V_0^{l,2}(\Omega )}\leq \tilde{C}_{l}\| u\| _{W^{l,2}(\Omega )}\quad \text{ for all }u\in C_{c}^{\infty } (\Omega) . \label{F.3}$$ On the other hand, using the same trick as in proof of a) we find$C_{l}\in \mathbb{R}^{+}$such that $$C_{l}\| u\| _{W^{l,2}(\Omega )}\leq \| u\| _{V_0^{l,2}(\Omega )}\quad \text{% for all }u\in C_{c}^{\infty } (\Omega) . \label{F.4}$$ Estimates (\ref{F.3}), (\ref{F.4}) yield \begin{equation*} {\mathaccent"7017 W}^{l,2}(\Omega )={\mathaccent"7017 V}_0^{l,2} (\Omega). \end{equation*} Due to imbedding${\mathaccent"7017 V}_{\beta _1}^{l,2}(\Omega )\subset {% \mathaccent"7017 V}_0^{l,2}(\Omega )\subset {\mathaccent"7017 V}_{\beta _2}^{l,2}(\Omega )$when$\beta _1\leq 0\leq \beta _2$one obtains the result in (c) and (d). \end{proof} \section{A fundamental system of solutions\label{AppendixC}} \subsection{Derivation of system$S_{\protect\lambda }$} Let us find the fundamental set of solutions to equation $$\mathcal{L}( \theta ,\tfrac{d}{d\theta },\lambda ) \Phi =0, \label{C1}$$ with$\mathcal{L}( \theta ,\tfrac{d}{d\theta },\lambda) $as in formula (\ref% {5}). For this$\mathcal{L}$it seems to be hard to derive a set of functions solving (\ref{C1}) explicitly. The following approach applies in this case. For$L=\tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{\partial ^{4}}{\partial y^{4}}$we find$L\left( r^{\lambda +1}\Phi \right) =r^{\lambda -3}\mathcal{% L }\left( \theta ,\tfrac{d}{d\theta },\lambda \right) \Phi $and hence instead of$\mathcal{L}\left( \theta ,\tfrac{d}{d\theta },\lambda \right) \Phi =0$we may consider the equation $$L\left( r^{\lambda +1}\Phi \right) =0. \label{C2}$$ Operator$L$admits the decomposition \begin{equation*} L=\tfrac{\partial ^{4}}{\partial x^{4}}+\tfrac{\partial ^{4}}{\partial y^{4}} =\prod_{p=1}^2\left( \tfrac{\partial }{\partial y}-\tau _{p}\tfrac{ \partial }{\partial x}\right) \prod_{p=1}^2\left( \tfrac{\partial }{ \partial y}+\tau _{p}\tfrac{\partial }{\partial x}\right) , \end{equation*} with$\tau _1=\frac{\sqrt{2}}{2}(1+i) $,$\tau _2=\frac{\sqrt{2}}{2}(1-i) $and hence every function of the form$F(x\pm \tau _{p}y)$solves (\ref{C2}). Therefore, we have that \begin{equation*} r^{\lambda +1}\Phi (\theta )=\sum_{p=1}^2c_{p}f_{p}(x+\tau _{p}y)+c_{p+2}f_{p+2}(x-\tau _{p}y), \end{equation*} and after translation$\{ f_{p},f_{p+2}\} _{p=1}^2$into polar coordinates we set \begin{gather*} f_{p}\left( r\cos (\theta )+\tau _{p}r\sin (\theta )\right) :=\left( r\cos (\theta )+\tau _{p}r\sin (\theta )\right) ^{\lambda +1}, \\ f_{p+2}\left( r\cos (\theta )-\tau _{p}r\sin (\theta )\right) :=\left( r\cos (\theta )-\tau _{p}r\sin (\theta )\right) ^{\lambda +1}, \end{gather*} So, the set of functions $$\varphi _1(\theta )=\left( \cos (\theta )+\tau _1\sin (\theta )\right) ^{\lambda +1},\quad \varphi _2(\theta )=\left( \cos (\theta )+\tau _2\sin (\theta )\right) ^{\lambda +1}, \label{C-1}$$ $$\varphi _{3}(\theta )=\left( \cos (\theta )-\tau _1\sin (\theta )\right) ^{\lambda +1},\quad \varphi _{4}(\theta )=\left( \cos (\theta )-\tau _2\sin (\theta )\right) ^{\lambda +1}, \label{C-2}$$ is a set of solutions to (\ref{C1}). The Wronskian for$\{ \varphi _{m}\} _{m=1}^{4}$reads as $$W\left( \varphi _1(\theta ),\varphi _2(\theta ),\varphi _{3}(\theta ),\varphi _{4}(\theta )\right) =\det \begin{pmatrix} \varphi _1(\theta ) & \varphi _2(\theta ) & \varphi _{3}(\theta ) & \varphi _{4}(\theta ) \\ \varphi _1^{\prime }(\theta ) & \varphi _2^{\prime }(\theta ) & \varphi _{3}^{\prime }(\theta ) & \varphi _{4}^{\prime }(\theta ) \\ \varphi _1^{\prime \prime }(\theta ) & \varphi _2^{\prime \prime }(\theta ) & \varphi _{3}^{\prime \prime }(\theta ) & \varphi _{4}^{\prime \prime }(\theta ) \\ \varphi _1^{\prime \prime \prime }(\theta ) & \varphi _2^{\prime \prime \prime }(\theta ) & \varphi _{3}^{\prime \prime \prime }(\theta ) & \varphi _{4}^{\prime \prime \prime }(\theta )% \end{pmatrix}% , \label{C0}$$ and by straightforward computations one finds \begin{equation*} W=16\left( \lambda +1\right) ^{3}\lambda ^2\left( \lambda -1\right) \left( \cos ^{4}(\theta )+\sin ^{4}(\theta) \right) ^{\lambda -2}, \end{equation*} which is non-zero on$\theta \in (0,2\pi ]$except for$\lambda \in \{\pm 1,0\} $. Hence, except for these values$\{ \varphi _{m}\} _{m=1}^{4}$given in (\ref{C-1}), (\ref{C-2}) is a fundamental system of solutions to (\ref{C1}% ). \subsection{Derivation of systems$S_{-1}$,$S_0$,$S_1$} Here we find the fundamental systems of solutions to equation$\mathcal{L} \left( \theta ,\tfrac{\partial }{\partial \theta },\lambda \right) \Phi =0$when$\lambda \in \left\{ \pm 1,0\right\} $. We will go into details in solving the corresponding equation for every$\lambda \in \left\{ \pm 1,0\right\} $. \subsubsection{Case$\protect\lambda =-1$} For$\lambda =-1$the equation (\ref{C1}) reads as $$\tfrac{1}{4}\left( 3+\cos (4\theta )\right) \Phi ^{\prime \prime \prime \prime }-3\sin (4\theta )\Phi ^{\prime \prime \prime }+\left( 3-11\cos (4\theta )\right) \Phi ^{\prime \prime }+12\sin (4\theta )\Phi ^{\prime }=0. \label{C3}$$ First we set$\Phi (\theta )=\int F(\theta )d\theta $and obtain the equation for$F$: \begin{equation*} \tfrac{1}{4}\left( 3+\cos (4\theta )\right) F^{\prime \prime \prime }-3\sin (4\theta )F^{\prime \prime }+\left( 3-11\cos (4\theta )\right) F^{\prime }+12\sin (4\theta )F=0. \end{equation*} The first integral of the above equation reads as \begin{equation*} \tfrac{1}{4}\left( 3+\cos (4\theta )\right) F^{\prime \prime }-2\sin (4\theta )F^{\prime }+3(1-\cos (4\theta ))F=c_0. \end{equation*} We use the change of variables$F(\theta )=\left( 3+\cos (4\theta )\right) ^{-1}G(\theta )$and get \begin{equation*} G^{\prime \prime }+4G=4c_0. \end{equation*} Solution of the last equation reads as \begin{equation*} G(\theta )=c_1\sin (2\theta )+c_2\cos (2\theta )+c_0. \end{equation*} and then \begin{equation*} F(\theta )=c_1\tfrac{\sin (2\theta )}{3+\cos (4\theta )}+c_2\tfrac{\cos (2\theta )}{3+\cos (4\theta )}+c_{3}\tfrac{1}{3+\cos (4\theta )}. \end{equation*} As a result,$\Phi $that solves (\ref{C3}) will read as \begin{equation*} \Phi (\theta )=A_1+A_2\int \tfrac{\sin (2\theta )}{3+\cos (4\theta )} d\theta +A_{3}\int \tfrac{\cos (2\theta )}{3+\cos (4\theta )}d\theta +A_{4}\int \tfrac{1}{3+\cos (4\theta )}d\theta , \end{equation*} and then the candidates that may form the fundamental system of solutions to (\ref{C3}) will be the following: \begin{gather*} \varphi _1(\theta )=1, \\ \varphi _2(\theta )=-4\int \tfrac{\sin (2\theta )}{3+\cos (4\theta )} d\theta =\arctan \left( \cos (2\theta )\right) , \\ \varphi _{3}(\theta )=4\sqrt{2}\int \tfrac{\cos (2\theta )}{3+\cos (4\theta ) }d\theta =\mathrm{arctanh}\left( \tfrac{\sqrt{2}}{2}\sin (2\theta )\right) , \\ \varphi _{4}(\theta )=4\sqrt{2}\int \tfrac{1}{3+\cos (4\theta )}d\theta =% \begin{cases} \arctan \left( \frac{\sqrt{2}}{2}\tan (2\theta )\right) +\ell \pi , & \theta \in \left( \frac{2\ell -1}{4}\pi ,\frac{2\ell +1}{4}\pi \right), \\ 2\theta & \theta =\frac{2\ell +1}{4}\pi ,% \end{cases}% \end{gather*} with$\ell =0,\dots,4$. Formula for$\varphi _{4}$is given on the interval$% \theta \in \left( -\frac{1}{4}\pi ,2\pi +\frac{1}{4}\pi \right) $in order to have the concise explicit form. The Wronskian$W$of$\varphi _1,\dots,\varphi _{4}$is proportional to$% \left( \cos ^{4}(\theta )+\sin ^{4}(\theta) \right) ^{-3}$and is non-zero on$\theta \in (0,2\pi ]$. Hence,$\{ \varphi _{m}\} _{m=1}^{4}$defined as above is a fundamental system of solutions to (\ref{C3}). \subsubsection{Case$\protect\lambda =0$} For$\lambda =0the equation (\ref{C1}) reads as: \begin{aligned} &\tfrac{1}{4}\left( 3+\cos (4\theta )\right) \Phi ''''-2\sin (4\theta )\Phi '''+\tfrac{1}{2}\left( 3-7\cos (4\theta )\right) \Phi ''\\ &-2\sin (4\theta )\Phi' +\tfrac{3}{4}\left( 1-5\cos (4\theta )\right) \Phi =0, \end{aligned} \label{C8} and can be split as follows: \begin{equation*} \left( \tfrac{d^2}{d\theta ^2}+1\right) \left( \tfrac{1}{4}\left( 3+\cos (4\theta )\right) \left( \tfrac{d^2}{d\theta ^2}+1\right) \right) \Phi =0. \end{equation*} So,\Phi solves \begin{equation*} \Phi ^{\prime \prime }+\Phi =A\tfrac{\sin (\theta )}{3+\cos (4\theta )}+B \tfrac{\cos (\theta )}{3+\cos (4\theta )}, \end{equation*} and after integrating this equation we obtain \begin{align*} &\Phi (\theta ) \\ &=A_1\sin (\theta )+A_2\cos (\theta )+A_{3}\Big( \tfrac{1}{ 2}\sin (\theta )\arctan \left( \cos (2\theta )\right) +4\cos (\theta )\int_0^{\theta }% \tfrac{\sin ^2(y)}{3+\cos (4y)}dy\Big) \\ &\quad +A_{4}\Big( \tfrac{1}{2}\cos (\theta )\arctan \left( \cos (2\theta )\right) +4\sin (\theta )\int_0^{\theta }\tfrac{\cos ^2(y)}{3+\cos (4y)} dy% \Big). \end{align*} %\label{C10} The candidates that may form the fundamental system of solutions to (\ref{C8}% ) will be the following: \begin{gather*} \varphi _1(\theta )=\sin (\theta ), \quad \varphi _2(\theta )=\cos (\theta ), \\ \varphi _{3}(\theta )=\tfrac{1}{2}\sin (\theta )\arctan \left( \cos (2\theta )\right) +4\cos (\theta )\int_0^{\theta }\tfrac{\sin ^2(y)}{ 3+\cos (4y)}dy, \\ \varphi _{4}(\theta )=\tfrac{1}{2}\cos (\theta )\arctan \left( \cos (2\theta )\right) +4\sin (\theta )\int_0^{\theta }\tfrac{\cos ^2(y)}{ 3+\cos (4y)}dy. \end{gather*} The WronskianW$of$\varphi _1,\dots,\varphi _{4}$is proportional to$% \left( \cos ^{4}(\theta )+\sin ^{4}(\theta) \right) ^{-2}$and is non-zero on$\theta \in (0,2\pi ]$. Hence,$\{ \varphi _{m}\} _{m=1}^{4}$defined as above is a fundamental system of solutions to (\ref{C8}). \subsubsection{Case$\protect\lambda =1$} For$\lambda =1$the equation (\ref{C1}) reads as: $$\tfrac{1}{4}\left( 3+\cos (4\theta )\right) \Phi ^{\prime \prime \prime \prime }-\sin (4\theta )\Phi ^{\prime \prime \prime }+\left( 3+\cos (4\theta )\right) \Phi ^{\prime \prime }-4\sin (4\theta )\Phi ^{\prime }=0. \label{C11}$$ We set$\Phi (\theta )=\int F(\theta )d\theta $and obtain the equation for$% F$: \begin{equation*} \tfrac{1}{4}\left( 3+\cos (4\theta )\right) F^{\prime \prime \prime }-\sin (4\theta )F^{\prime \prime }+\left( 3+\cos (4\theta )\right) F^{\prime }-4\sin (4\theta )F=0. \end{equation*} It holds that \begin{gather*} \left( 3+\cos (4\theta )\right) F^{\prime \prime \prime }-4\sin (4\theta )F^{\prime \prime }=-2g(\theta ), \\ \left( 3+\cos (4\theta )\right) F^{\prime }-4\sin (4\theta )F=\frac{1}{2} g(\theta ). \end{gather*} and we obtain, respectively, \begin{equation*} F^{\prime \prime }(\theta )=-4\tfrac{\int g(\theta )d\theta +C_1}{3+\cos (4\theta )}, \quad F(\theta )=\tfrac{\int g(\theta )d\theta +C_2}{% 3+\cos(4\theta )}. \end{equation*} Comparing the expressions for$F^{\prime \prime }(\theta )$and$F(\theta )$we deduce that$Fsolves $$F^{\prime \prime }+4F=\dfrac{c_0}{3+\cos (4\theta )}. \label{C12}$$ The solution of (\ref{C12}) reads as \begin{align*} F(\theta )&=c_1\sin (2\theta )+c_2\cos (2\theta )+c_0\Big( \tfrac{1}{4} \cos (2\theta )\arctan \left( \cos (2\theta )\right) \\ &\quad +\tfrac{\sqrt{2}}{8} \sin (2\theta )\mathrm{arctanh}\big( \tfrac{% \sqrt{2}}{2}\sin (2\theta )\big) \Big) , \end{align*} which being integrated yields \label{C13} \begin{aligned} \Phi (\theta )&=A_1+A_2\cos (2\theta )+A_{3}\sin (2\theta)\\ &\quad +A_{4}\int_0^{\theta }\left( \cos (2y)\arctan \left( \cos (2y)\right) +\tfrac{\sqrt{2}}{2}\sin (2y)\mathrm{arctanh}\left( \tfrac{\sqrt{2}}{2 }\sin (2y)\right) \right) dy \end{aligned} The candidates that may form the fundamental system of solutions to (\ref% {C11}) will be the following: \begin{gather*} \varphi _1(\theta )=1, \quad \varphi _2(\theta )=\sin (2\theta ), \quad \varphi _{3}(\theta )=\cos (2\theta ), \\ \varphi _{4}(\theta )=\int_0^{\theta }\left( \cos (2y)\arctan \left( \cos (2y)\right) +\tfrac{\sqrt{2}}{2}\sin (2y)\mathrm{arctanh} \left(\tfrac{\sqrt{% 2}}{2}\sin (2y)\right) \right) dy. \end{gather*} The WronskianW$of$\varphi _1,\dots,\varphi _{4}$is proportional to$% \left( \cos ^{4}(\theta )+\sin ^{4}(\theta) \right) ^{-1}$and is non-zero on$\theta \in (0,2\pi ]$. Hence,$\{ \varphi _{m}\} _{m=1}^{4}$defined as above is a fundamental system of solutions to (\ref{C8}). \subsection{The explicit formulas for$P_{-1}$,$P_0$,$P_1$,} For$\lambda \in \{ \pm 1,0\} we obtain: \begin{gather*} P_{-1}(\omega) =-\tfrac{32}{\pi }\sin (2\omega )\int_0^{\omega }\tfrac{\sin ^2(\theta )}{3+\cos (4\theta )} d\theta +\left( 1-\cos (2\omega )\right) \left( 1-\tfrac{4}{\pi }\arctan \left( \cos (2\omega )\right) \right) , \\ \begin{aligned} P_0(\omega) &=\pi \arctan \left( \cos (2\omega )\right) -2\arctan ^2\left( \cos (2\omega )\right) \\ &\quad +128\int_0^{\omega } \tfrac{\sin ^2(\theta )}{3+\cos (4\theta )}d\theta \int_0^{\omega }\tfrac{\cos ^2(\theta )}{3+\cos (4\theta )}d\theta , \end{aligned} \\ P_1(\omega) =-\sin (2\omega )\left( \sin (2\omega )-\tfrac{8}{ \pi }\varphi _{4}(\omega )\right) +\left( 1-\cos (2\omega )\right) \left( \cos (2\omega )-% \tfrac{4}{\pi }\varphi _{4}^{\prime }(\omega )\right) , \end{gather*} where\varphi _{4}(\theta )$is given by corresponding formula from the case$\lambda =1$. \section{Analytical tools for the numerical computation\label{AppendixD}} \subsection{Implicit function and discretization} Consider a rectangle$U=[a,b]\times [ c,d]$. For$n,m\in \mathbb{N}^{+} $and$i=0,\dots,n$,$j=0,\dots,m$we set \begin{equation*} \begin{tabular}{cc}$x_{i}=a+i\Delta x,\quad y_{j}=c+j\Delta y$, & \end{tabular}% \end{equation*} where$\Delta x=\frac{b-a}{n}$,$\Delta y=\frac{d-c}{m}$. Let$F\in C^{1}(U,\mathbb{R})$such that$F(x_{i},y_{j})>0$for all$% i=0,\dots,n$and$j=0,\dots,m$. The question to resolve is how fine should we take the\ discretization of$U$in order to be sure that$F>0$on$U$. The following result holds. \begin{lemma} \label{lemma1,ap.D}Suppose$\min_{(x_{i},y_{j})\in U} F(x_{i},y_{j})>0$. If $$\max \left\{ \Delta x,\Delta y\right\} \leq \sqrt{2} \dfrac{\min_{(x_{i},y_{j})\in U} F(x_{i},y_{j})} {\sup_{U}|DF(x,y)| }, \label{D10}$$ then$F$is strictly positive on$U$. \end{lemma} \begin{proof} For every$(x,y)\in U$there is$(x_{i},y_{j})$with$| x-x_{i}| \leq \frac{1% }{2}\Delta x$and$| y-y_{j}| \leq \frac{1}{2}\Delta y$. By the mean value theorem there exists$(\xi _1,\xi _2)\in [ (x,y) ,(x_{i},y_{j})]such that \begin{equation*} F(x,y)=F(x_{i},y_{j})+DF(\xi _1,\xi _2)\cdot (x-x_{i},y-y_{j}). \end{equation*} The following chain of estimates then holds \begin{align*} F(x,y) & = F(x_{i},y_{j})+DF(\xi _1,\xi _2)\cdot (x-x_{i},y-y_{j}) \\ & \geq \min_{(x_{i},y_{j})\in U} F(x_{i},y_{j}) -\sup_{U} | DF(x,y)| | (x-x_{i},y-y_{j})| \\ & \geq \min_{(x_{i},y_{j})\in U} F(x_{i},y_{j}) -\frac{\sqrt{2}}{2}\sup_{U}| DF(x,y)| \max \left\{ \Delta x,\Delta y\right\} . \end{align*} This last expression is positive if (\ref{D10}) holds. \end{proof} \subsection{A version of the Morse theorem} LetV\subset \mathbb{R}^2$\ be open and bounded,$F\in C^{\infty }(V,% \mathbb{R})$. For the gradient of$F$we use${D}F$and${D}^2F$is the Hessian matrix. \begin{definition} \rm A point$a\in V$is said to be a critical point of$F$if${D}F(a)=0$. Moreover, the critical point$a\in V$is said to be non-degenerate if$\det {D}^2F(a)\neq 0$. \end{definition} To study the level set$\Gamma $defined in subsection \ref{Behavior} we need the Morse theorem. The original version of the theorem reads as (see \cite{Pal}): \begin{quotation} Let$V$be a Banach space,$O$a convex neighborhood of the origin in$V$and$f:O\to \mathbb{R}$a$C^{k+2}$function ($k\geq 1$) having the origin as a non-degenerate critical point, with$f(0)=0$. Then there is a neighborhood$U$of the origin and a$C^{k}$diffeomorphism$\phi :U\to O$with$\phi (0)=0$and$D\phi (0)=I$, the identity map of$V, $such that for$x\in U$,$f(\phi (x))=\frac{1}{2}(D^2f(0)x,x) $. \end{quotation} Below we give our formulation of the theorem. This formulation is more convenient for our purposes. We will give a constructive proof that allows us to find an explicit neighborhood of a critical point where the diffeomorphism exists. \begin{theorem}\label{th1,ap.D} Let$V\subset \mathbb{R}^2$\ be open and bounded,$F\in C^{\infty }(V,\mathbb{R})$. Suppose$a=(a_1,a_2) \in V$is a non-degenerate critical point of$F$. There exists a neighborhood$W_{a}\subset V$of$a$and a$C^{\infty }$-diffeomorphism$h:W_{a}\to U_0$, where$U_0\subset \mathbb{R}^2$\ is a neighborhood\ of$0$, such that$F$in$W_{a}$is representable as: $$F(x)=F{(a)+}h(x)\left( \tfrac{1}{2}{D}^2F(a)\right) h(x)^{T}, \label{D1}$$ where$T$stands for a transposition. Moreover, the neighborhood$W_{a}$is fixed by \begin{equation*} W_{a}\subset \left\{ x\in V:\det B(x)\geq 0\text{ \ \ and \ } b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)>0\right\} . \end{equation*} where \begin{equation*} B(x) =\begin{pmatrix} b_{11}(x) & b_{12}(x) \\ b_{21}(x) & b_{22}(x) \end{pmatrix} =\left( \tfrac{1}{2}D^2F(a)\right) ^{-1}\int_0^{1}\int_0^{1}sD^2F\left( a+ts\left( x-a\right) \right) dt\,ds. \end{equation*} \end{theorem} \begin{proof} Let$a\in V$be such that${D}F(a)=0$and$\det {D}^2F(a)\neq 0$. First, for every$x\in V$we have \begin{equation*} F(x)-F(a)= {F}\left( a+s(x-a)\right) | _0^{1}=\int_0^{1}\tfrac{d}{ds}{F}% \left( a+s(x-a)\right) ds, \end{equation*} and due to$\tfrac{d}{ds}{F}\left( a+s(x-a)\right) =D{F}\left( a+s(x-a)\right) (x-a)^{T}$it will follow that \begin{equation*} F(x)=F{(a)+}\int_0^{1}D{F}\left( a+s(x-a)\right) ds(x-a)^{T}. \end{equation*} Analogously, we obtain \begin{equation*} DF(x)=DF(a){+}\int_0^{1}D^2F\left( a+t(x-a)\right) dt(x-a)^{T}, \end{equation*} where by assumption$DF(a)=0$. As a result, for every$x\in V$,$F$is representable in terms of$D^2F$as: \begin{equation*} F(x)=F{(a)+}(x-a)\int_0^{1}\int_0^{1}sD^2F\left( a+ts(x-a)\right) dt\,ds(x-a)^{T}, \end{equation*} or shortly $$F(x)=F{(a)+}(x-a)K(x)(x-a)^{T}. \label{D1.1}$$ Here$K(x)=\int_0^{1}\int_0^{1}sD^2F\left( a+ts(x-a)\right) dt\,ds$is a symmetric matrix. With this definition,$K(a)=\frac{1}{2}D^2F(a) $is symmetric and invertible ($\det {D}^2F(a)\neq 0$by assumption). Let us bring some intermediate results. (1) For every$x\in V$, there exists matrix$B$such that $$K(x)=K(a)B(x). \label{D1.2}$$ Indeed, since$K(a)$is invertible, the matrix$B(x)=K(a)^{-1}K(x)$is well-defined. We write \begin{equation*} B(x)=% \begin{pmatrix} b_{11}(x) & b_{12}(x) \\ b_{21}(x) & b_{22}(x)% \end{pmatrix}% , \end{equation*} Since$F\in C^{\infty }(V,\mathbb{R})$, so are$b_{ij}$,$i,j=1,2$. Note that$B(a)=I$. (2) Since$x\mapsto B(x)$is$C^{\infty }$in a neighborhood of$a$and$% B(a)=I$,$B(x)$is positive definite in a neighborhood of$a$, and hence allows a square root. In particular, it holds that $$C(x)=\sqrt{B(x)}:=\tfrac{1}{2\pi i}\oint_{\gamma }\sqrt{z}\left( Iz-B(x)\right) ^{-1}dz, \label{D1.11}$$ where$\gamma $is a Jordan curve in$\mathbb{C} $which goes around the eigenvalues$\lambda _1,\lambda _2\in \mathbb{C}$of$B(x)$and does not intersect$\mathop{\rm Re}(z)\leq 0$,$\mathrm{Im}(z)=0$. One may check that$C$, defined as follows \begin{equation*} C(x)=% \begin{pmatrix} \tfrac{b_{11}(x)+\sqrt{\det B(x)}}{\sqrt{b_{11}(x)+2\sqrt{\det B(x)} +b_{22}(x)}} & \tfrac{b_{12}(x)}{\sqrt{b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)} } \\ \tfrac{b_{21}(x)}{\sqrt{b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)}} & \tfrac{ b_{22}(x)+\sqrt{\det B(x)}}{\sqrt{b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)}}% \end{pmatrix}% , \end{equation*} is indeed such that $$C(x)^2=B(x). \label{D1.20}$$ With this definition,$C(a)=I$and$C(a)^2=I=B(a)$as required. Also, matrix$C$is well defined when $$\det B(x)\geq 0, \quad b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)>0. \label{D1.6}$$ (3) For those$xone finds \begin{align*} K(a)B(x) & = K(x)=K(x)K(a)^{-1}K(a) \\ &=K(x)^{T}\left( K(a)^{-1}\right)^{T}K(a) \\ & = \left( K(a)^{-1}K(x)\right) ^{T}K(a)=B^{T}(x)K(a). \end{align*} Due to this equality, we deduce the following \begin{equation*} \left( Iz-B(x)\right) K(a)^{-1}=K(a)^{-1}\left( Iz-B(x)\right) ^{T}, \end{equation*} and hence \begin{align*} K(a)\left( Iz-B(x)\right) ^{-1} & = \left( \left( Iz-B(x)\right) K(a)^{-1}\right) ^{-1}=\left( K(a)^{-1}\left( Iz-B(x)\right) ^{T}\right) ^{-1} \\ & = \left( \left( Iz-B(x)\right) ^{T}\right) ^{-1}K(a)=\left( \left( Iz-B(x)\right) ^{-1}\right) ^{T}K(a). \end{align*} Applying the integration (\ref{D1.11}) to the last identity we find $$K(a)C(x)=C(x)^{T}K(a). \label{D1.21}$$ Combining (\ref{D1.2}), (\ref{D1.20}) and (\ref{D1.21}) we have \begin{equation*} K(x)=K(a)B(x)=K(a)C(x)^2=C(x)^{T}K(a)C(x), \end{equation*} and therefore (\ref{D1.1}) for thosex$results in $$F(x)=F(a)+F{(a)+}h(x)K(a)h(x)^{T}, \label{D1.7}$$ where$K(a)=\tfrac{1}{2}{D}^2F(a)$and \begin{equation*} h(x)^{T}=C(x)(x-a)^{T}. \end{equation*} Note that by (\ref{D1.6}) the representation for$F$in (\ref{D1.7}) holds on a set$W_{a}\subset V$which is star-shaped with respect to$a$and such that $$W_{a}\subset \big\{ x\in V:\det B(x)\geq 0\text{ and } b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)>0\big\} . \label{D21}$$ \end{proof} \begin{remark} \rm For each pair$(F,a) $one can obtain an explicit estimate for$W_{a}$in (\ref{D21}). We will do this in the next subsection for the pair we are interested in. \end{remark} \subsection{The Morse Theorem applied} Let$Pbe the function given by formula (\ref{9}) and which is defined on \begin{equation*} V=\left\{ (\omega ,\lambda) :\left[ \tfrac{70}{180}\pi ,\tfrac{ 110}{180}\pi % \right] \times \left[ 2.900,5.100\right] \right\} . \end{equation*} Let us recall it here: \begin{aligned} P(\omega ,\lambda) &=\left( 1-\tfrac{\sqrt{2}}{2}\sin (2\omega )\right) ^{\lambda }+\left( 1+\tfrac{\sqrt{2}}{2}\sin (2\omega )\right) ^{\lambda } \\ &\quad +\left( \tfrac{1}{2}+\tfrac{1}{2}\cos ^2(2\omega )\right) ^{\frac{1}{2} \lambda }\Big[ 2\cos \left( \lambda \left( \arctan \left( \tfrac{\sqrt{2}}{2 }\tan (2\omega )\right) +\pi \right) \right) \\ &\quad -4\cos \left( \lambda \arctan \left( \tan ^2(\omega )\right) \right) \Big] . \end{aligned} \label{D2} The pointa=( \tfrac{1}{2}\pi ,4) $is such that$P(a)=0$and$DP(a)=0$. Theorem \ref{th1,ap.D} gives us the tool to study$P$in the vicinity of$a$% . In particular, the following holds. \begin{proposition}\label{Proposition.ap.D} Let$a$be as above. There is a closed ball$W_{R}(a)\subset V$of a radius$R$centered at$a$such that on$W_{R}(a)$we have: $$P(\omega ,\lambda )=-\tfrac{1}{2}h_2(\omega ,\lambda) \left( 16\sqrt{2}h_1(\omega ,\lambda) +\pi h_2(\omega ,\lambda) \right) . \label{D3.1}$$ Here$h_1,h_2\in C^{\infty }\left( W_{R}(a),\mathbb{R}\right) $are given by: \begin{gather} h_1(\omega ,\lambda) =\left( \omega -\tfrac{1}{2}\pi \right) c_{11}(\omega ,\lambda) +\left( \lambda -4\right) c_{12}(\omega ,\lambda) , \label{D18} \\ h_2(\omega ,\lambda) =\left( \omega -\tfrac{1}{2}\pi \right) c_{21}(\omega ,\lambda) +\left( \lambda -4\right) c_{22}(\omega ,\lambda) , \label{D19} \end{gather} with$c_{ij}\in C^{\infty }\left( W_{R}(a),\mathbb{R}\right) $,$i,j=1,2$are the entries of matrix$C$: $$C(\omega ,\lambda) =\begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix} (\omega ,\lambda) =\begin{pmatrix} \tfrac{b_{11}+\sqrt{\det B}}{\sqrt{b_{11}+2\sqrt{\det B}+b_{22}}} & \tfrac{ b_{12}}{\sqrt{b_{11}+2\sqrt{\det B}+b_{22}}} \\ \tfrac{b_{21}}{\sqrt{b_{11}+2\sqrt{\det B}+b_{22}}} & \tfrac{b_{22}+\sqrt{ \det B}}{\sqrt{b_{11}+2\sqrt{\det B}+b_{22}}} \end{pmatrix} (\omega ,\lambda) , \label{D20}$$ while$b_{ij}\in C^{\infty }(V,\mathbb{R})$,$i,j=1,2are as follows \begin{aligned} B(\omega ,\lambda) &=\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} (\omega ,\lambda) \\ &=\left( \tfrac{1}{2}D^2P(a)\right) ^{-1}\int_0^{1}\int_0^{1}sD^2P\left( a+ts\left( (\omega ,\lambda) -a\right) \right) dt\,ds. \end{aligned} \label{D15} Note thatB(a)=I$and$C(a)=I$. The ball$W_{R}(a)$is fixed by $$W_{R}(a):=\left\{ (\omega ,\lambda) \in V:| \left( \omega -\tfrac{1}{2}\pi ,\lambda -4\right) | \leq R\right\} , \label{D3.2}$$ with$R=-\tfrac{1}{120}\pi ^2-\tfrac{\sqrt{2}}{18}\pi +\tfrac{1}{120}\pi \sqrt{\pi ^2+\tfrac{40}{3}\sqrt{2}\pi +\tfrac{1568}{9}}\approx 0.078\dots$(In$\omega $-direction we have that$R\approx 4.5\dots^{\circ }$). \end{proposition} \subsubsection{Computational results I} It is straightforward for$a=\left( \frac{1}{2}\pi ,4\right) $that $$\begin{tabular}{ccc} \tfrac{\partial ^2P}{\partial \omega ^2}(a)=0,\quad \tfrac{\partial ^2P }{% \partial \omega \partial \lambda }(a)=-8\sqrt{2}\pi ,\quad \tfrac{\partial ^2P}{\partial \lambda ^2}(a)=-\pi ^2, & & \end{tabular} \label{D5}$$ and hence $$\det D^2P(a)=-128\pi ^2. \label{D5.1}$$ To simplify the notation, we use$x$instead of$(\omega ,\lambda) $when$% (\omega ,\lambda) $stands for a argument. Now let us bring two alternatives representations for the entries of matrix$B$given by (\ref{D15}), which we will use later on.\medskip \noindent\textit{Representation I.} We will need the explicit formula for the coefficients$b_{ij}$,$i,j=1,2$. Let us find them in a straightforward way from (\ref{D15}). We write down the integral term$\int_0^{1}\int _0^{1}sD^2P\left( a+ts\left( x-a\right) \right) dt\,ds$in (\ref{D15}) as follows $$\int_0^{1}\int_0^{1}sD^2P\left( a+ts\left( x-a\right) \right) dt\,ds=% \begin{pmatrix} r_1(x) & r_2(x) \\ r_2(x) & r_{3}(x)% \end{pmatrix} , \label{D1.45}$$ where$r_{j}$,$j=1,2,3$read as: \begin{gather} r_1(x)=\int_0^{1}\int_0^{1}s\tfrac{\partial ^2P}{ \partial \omega ^2}% (a+ts(x-a))dt\,ds, \label{D4.1} \\ r_2(x)=\int_0^{1}\int_0^{1}s\tfrac{\partial ^2P}{ \partial \omega \partial \lambda }(a+ts(x-a))dt\,ds, \label{D4.2} \\ r_{3}(x)=\int_0^{1}\int_0^{1}s\tfrac{\partial ^2P}{ \partial \lambda ^2}% (a+ts(x-a))dt\,ds. \label{D4.3} \end{gather} Then the entries$b_{ij}$,$i,j=1,2$of matrix$B$in terms of$r_{j}$,$% j=1,2,3$and due to (\ref{D5}), (\ref{D5.1}) will read: \begin{gather} b_{11}(x)=\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{\partial \lambda ^2}(a)r_1(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)r_2(x)\right) =\tfrac{1}{64}r_1(x)-\tfrac{\sqrt{2}}{8\pi } r_2(x), \label{D2.1} \\ b_{12}(x)=\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{\partial \lambda ^2}(a)r_2(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)r_{3}(x)\right) =\tfrac{1}{64}r_2(x)-\tfrac{\sqrt{2}}{8\pi } r_{3}(x), \label{D2.2} \\ b_{21}(x)=\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{\partial \omega ^2}(a)r_2(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)r_1(x)\right) =-\tfrac{\sqrt{2}}{8\pi }r_1(x), \label{D2.3} \\ b_{22}(x)=\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{\partial \omega ^2}(a)r_{3}(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)r_2(x)\right) =-\tfrac{\sqrt{2}}{8\pi }r_2(x). \label{D2.4} \end{gather} \noindent\textit{Representation II.} On the other hand, let us obtain for$r_{j}$,$j=1,2,3the following representations: \begin{equation*} r_1(x)=\tfrac{1}{2}\tfrac{\partial ^2P}{\partial \omega ^2} (a)+q_1(x), \quad r_2(x)=\tfrac{1}{2}\tfrac{\partial ^2P}{\partial \omega \partial \lambda }(a)+q_2(x), \quad r_{3}(x)=\tfrac{1}{2}\tfrac{\partial ^2P}{% \partial \lambda ^2} (a)+q_{3}(x), \end{equation*} where \begin{gather} \begin{aligned} q_1(x)&=\int_0^{1}\int_0^{1}\int_0^{1}ts^2 \left( \tfrac{\partial ^{3}P}{\partial \omega ^{3}}(a+ts\sigma (x-a)),\tfrac{ \partial ^{3}P}{\partial \omega ^2\partial \lambda }(a+ts\sigma (x-a))\right) \\ &\quad\times (x-a)^{T}d\sigma dt\,ds, \end{aligned} \label{D1.46} \\ \begin{aligned} q_2(x)&=\int_0^{1}\int_0^{1}\int_0^{1}ts^2 \left( \tfrac{\partial ^{3}P}{\partial \omega ^2\partial \lambda } (a+ts\sigma (x-a)),\tfrac{\partial ^{3}P}{\partial \omega \partial \lambda ^2}(a+ts\sigma (x-a))\right)\\ &\quad\times (x-a)^{T}d\sigma dt\,ds, \end{aligned} \label{D1.47} \\ \begin{aligned} q_{3}(x) &=\int_0^{1}\int_0^{1}\int_0^{1}ts^2 \left( \tfrac{\partial ^{3}P}{\partial \omega \partial \lambda ^2} (a+ts\sigma (x-a)),\tfrac{\partial ^{3}P}{\partial \lambda ^{3}}(a+ts\sigma (x-a))\right) \\ &\quad\times(x-a)^{T}d\sigma dt\,ds. \end{aligned} \label{D1.48} \end{gather} This will yield another representation formulas forb_{ij}$,$i,j=1,2$of matrix$B$, namely, \begin{gather} b_{11}(x)=1+\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{ \partial \lambda ^2}(a)q_1(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)q_2(x)\right) =1+\tfrac{1}{64}q_1(x)-\tfrac{\sqrt{2} }{8\pi }q_2(x), \label{D29} \\ b_{12}(x)=\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{\partial \lambda ^2}(a)q_2(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)q_{3}(x)\right) =\tfrac{1}{64}q_2(x)-\tfrac{\sqrt{2}}{8\pi } q_{3}(x), \label{D30} \\ b_{21}(x)=\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{\partial \omega ^2}(a)q_2(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)q_1(x)\right) =-\tfrac{\sqrt{2}}{8\pi }q_1(x), \label{D31} \\ b_{22}(x)=1+\tfrac{2}{\det {D}^2P(a)}\left( \tfrac{\partial ^2P}{ \partial \omega ^2}(a)q_{3}(x)-\tfrac{\partial ^2P}{\partial \omega \partial \lambda }% (a)q_2(x)\right) =1-\tfrac{\sqrt{2}}{8\pi }q_2(x). \label{D32} \end{gather} This representation, together with the estimates from above for$| q_{j}| $,$j=1,2,3$on$V$given below, will be useful in the proof of Proposition \ref% {Proposition.ap.D}.\medskip \noindent\textit{Estimates for$| q_{j}| $,$j=1,2,3$on$V$.} Let$q_{j}$,$j=1,2,3$be given by (\ref{D1.46}) -- (\ref{D1.48}). We will need the estimates for$| q_{j}| $,$j=1,2,3$on$V$. Let us also use the notations$\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}},% \tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}$for the corresponding differentiations in each$q_{j}$,$j=1,2,3$. For example for$% q_1$it will be$\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}=\tfrac{\partial ^{3}P}{\partial \omega ^{3}}$and$\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}=\tfrac{ \partial ^{3}P}{% \partial \omega ^2\partial \lambda }$, etc. For every$q_{j}$,$j=1,2,3we then in general have: \begin{aligned} | q_{j}(x)| &=\big|\int_0^{1}\int_0^{1}\int_0^{1}ts^2\left( \tfrac{ \partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}} (a+ts\sigma (x-a)),\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}(a+ts\sigma (x-a))\right) \\ &\quad\times(x-a)^{T}d\sigma dt\,ds\big| \\ &\leq \int_0^{1}\int_0^{1}\int_0^{1}ts^2\sup_{x\in V}| \left( \tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}(x),\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}(x)\right) | d\sigma dt\,ds| x-a| \\ &=\tfrac{1}{6}\sup_{x\in V}| \left( \tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}(x),\tfrac{\partial ^{3}P}{ \partial (\omega ,\lambda) ^{\beta _{j}}}(x)\right) | | x-a| \leq M_{j}| x-a| . \end{aligned} \label{D6} whereM_{j}$is an upper bound for function$\tfrac{1}{6}| \left(\tfrac{% \partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}} }(x),\tfrac{% \partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}(x)\right) | $on$% V $. In particular, one shows by explicit and tedious computations that for$% M_1=180$,$M_2=75$,$M_{3}=30$, the estimate (\ref{D6}) for the corresponding$| q_{j}| $,$j=1,2,3$holds true. We depict the results in Table \ref{table1.ap.D}. \begin{table}[ht] \renewcommand{\arraystretch}{1.5} \par \begin{center} \begin{tabular}{|c|c|c|} \hline$| q_1(x)| <180| x-a|$&$| q_2(x)| <75| x-a| $&$| q_{3}(x)| <30| x-a| $\\ \hline \end{tabular}% \end{center} \caption{Estimates from above for$| q_{j}| $,$j=1,2,3$on$V$} \label{table1.ap.D} \end{table} Now we proceed with a proof of Proposition \ref{Proposition.ap.D}. \subsubsection{Checking the range for Morse} \begin{proof}[Proof of Proposition \protect\ref{Proposition.ap.D}] Representation (\ref{D3.1}) is a consequence of Theorem \ref{th1,ap.D}. It is straightforward for$a=\left( \frac{1}{2}\pi ,4\right) $that$P(a)=0$. We also find that$DP(a)=0$, meaning$a$is a critical point of$P$. Due to ( \ref{D5.1}) we conclude that$a$is a non-degenerate critical point of$P$. By Theorem \ref{th1,ap.D} in a vicinity$W_{a}\subset V$of$awhich is defined in (\ref{D21}), we obtain \begin{aligned} P(x)&=\left( h_1(x),h_2(x)\right) \cdot \tfrac{1}{2}{D}^2P(a)\cdot \left( h_1(x),h_2(x)\right) ^{T} \\ &=-\tfrac{1}{2}h_2(x) \left( 16\sqrt{2}h_1(x) +\pi h_2(x) \right) , \end{aligned} \label{D36} whereh_1,h_2\in C^{\infty }\left( W_{a},\mathbb{R}\right) $. Their explicit formulas read as (\ref{D18}), (\ref{D19}). We show that$W_{a}$in our case can be taken as a closed ball$W_{R}(a)$centered at$a$of radius$R$and the numerical approximation for its range is given by (\ref{D3.2}). We will do this in two steps. (1) Let us solve the inequality$\det B(x)\geq 0$on$V. Due to (\ref{D29}) -- (\ref{D32}) we will get \begin{align*} \det B(x)&=b_{11}(x)b_{22}(x)-b_{12}(x)b_{21}(x) \\ &=1-\tfrac{\sqrt{2}}{4\pi }q_2(x)+\tfrac{1}{64}q_1(x)-\tfrac{1}{32\pi ^2 }% q_1(x)q_{3}(x)+\tfrac{1}{32\pi ^2}q_2^2(x) \\ &\geq 1-\tfrac{\sqrt{2}}{4\pi }| q_2(x)| -\tfrac{1}{64} | q_1(x)| -\tfrac{1}{% 32\pi ^2}| q_1(x)| | q_{3}(x)| \geq \dots \end{align*} we use estimates for| q_1(x)| $,$|q_2(x)| $and$| q_{3}(x)| $from Table % \ref{table1.ap.D} to get \begin{equation*} \dots\geq 1-\tfrac{75\sqrt{2}}{4\pi }| x-a| -\tfrac{180}{64} | x-a| -\tfrac{% 5400}{32\pi ^2}| x-a|^2. \end{equation*} The above expression is positive for all$x\in V$such that$| x-a| \leq R_1$% , with \begin{equation*} R_1=-\tfrac{1}{120}\pi ^2-\tfrac{\sqrt{2}}{18}\pi +\tfrac{1}{120}\pi \sqrt{% \pi ^2+\tfrac{40}{3}\sqrt{2}\pi +\tfrac{1568}{9}}. \end{equation*} The numerical approximation is$R_1\approx 0.078\dots$. Hence, the first estimate for a range of$W_{a}$is$|x-a| \leq R_1$. (2) Let us solve$b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)>0$on$V$. We have \begin{equation*} b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x)\geq b_{11}(x)+b_{22}(x)=\dots \end{equation*} due to formulas (\ref{D29}), (\ref{D32}) we obtain \begin{equation*} 2-\tfrac{\sqrt{2}}{4\pi }q_2(x)+\tfrac{1}{64}q_1(x)\geq 2-\tfrac{\sqrt{2 }}{% 4\pi }| q_2(x)| -\tfrac{1}{64}| q_1(x)| \geq \dots \end{equation*} we use the estimates for$| q_1(x)| $and$| q_2(x)| $\ from Table \ref% {table1.ap.D}\ to get \begin{equation*} \dots\geq 2-\tfrac{75\sqrt{2}}{4\pi }| x-a| -\tfrac{180}{64} | x-a| . \end{equation*} The above expression is strictly positive for all$x\in V$such that \begin{equation*} \begin{tabular}{ccc}$| x-a| 0, \\ 16\sqrt{2}\tfrac{\partial h_1}{\partial \lambda }(\omega ,\lambda) +\pi \tfrac{\partial h_2}{\partial \lambda }(\omega ,\lambda) >0, \end{gather*} and hence we can apply the Implicit Function Theorem proving that every function (\ref{D22}) and (\ref{D23}) allows its local parametrization $% \omega \mapsto \lambda (\omega )$ in $U$. This fact is used in Lemma \ref% {lemma5}. Now some preparatory technical steps are required. \subsubsection{Computational results II} \noindent\textit{Upper bounds for $| r_{j}|$, $j=1,2,3$ on $U$.} Let $r_{j}$, $j=1,2,3$ be given by (\ref{D4.1})--(\ref{D4.3}). We will find the upper bounds for $| r_{j}|$, $j=1,2,3$ on $U$. Setting again $\tfrac{% \partial ^2P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}$ for the corresponding differentiations in each $r_{j}$, $j=1,2,3$, we in general deduce that \begin{align*} | r_{j}(x)| &=|\int_0^{1}\int_0^{1}s\tfrac{\partial ^2P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}(a+ts(x-a))dt\,ds| \\ &\leq \int_0^{1}\int_0^{1}s\sup_{x\in U}| \tfrac{ \partial ^2P}{\partial (\omega ,\lambda) ^{\alpha _{j}}} (x)| dt\,ds \\ &=\tfrac{1}{2}\sup_{x\in U}| \tfrac{\partial ^2P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}(x)| \leq Q_{j}, \end{align*} where $Q_{j}$, $j=1,2,3$ is an upper bound for the function $\tfrac{1}{2}| \tfrac{\partial ^2P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}(x)|$ on $U$. In an analogous way we find the upper bounds for $| \frac{\partial r_{j}}{% \partial \omega }| , | \frac{\partial r_{j}}{\partial \lambda }|$ and $| \frac{\partial ^2r_{j}}{\partial \omega \partial \lambda }| , | \frac{% \partial^2r_{j}}{\partial \lambda ^2}|$, $j=1,2,3$ on $U$, we will need later on. Explicit bounds are given in Table \ref{table2.ap.D}. Note that we skip the derivatives $\frac{\partial ^2r_{j}}{\partial \omega ^2}$ since there will be no need for them. \begin{table}[ht] \begin{center} \begin{tabular}{|c|c|c|} \hline $| r_1(x)| <5$ & \parbox{43mm}{$|\frac{\partial r_1}{\partial \omega }(x)| <43$, $| \frac{\partial r_1}{\partial \lambda }(x)| <25$} & \parbox{48mm}{ $| \frac{\partial ^2r_1}{\partial \omega \partial \lambda }(x)| <175$, $| \frac{\partial ^2r_1}{\partial \lambda ^2}(x)|<65$} \rule[-2.5mm]{0mm}{6.5mm}\\ \hline\hline $| r_2(x)| <19$ & \parbox{43mm}{ $| \frac{\partial r_2}{\partial \omega }(x)| <25$, $| \frac{\partial r_2}{\partial \lambda }(x)| <6$} & \parbox{47mm}{$| \frac{\partial ^2r_2}{\partial \omega \partial \lambda } (x)| <65$, $| \frac{\partial ^2r_2}{\partial \lambda ^2}(x)|<33$} \rule[-2.5mm]{0mm}{6.5mm}\\ \hline\hline $| r_{3}(x)| <6$ & \parbox{43mm}{ $| \frac{\partial r_{3}}{\partial \omega }(x)| <6$, $| \frac{\partial r_{3}}{\partial \lambda }(x)| <4$} & \parbox{47mm}{ $| \frac{\partial ^2r_{3}}{\partial \omega \partial \lambda }(x)| <33$, $| \frac{\partial ^2r_{3}}{\partial \lambda ^2}(x)|<16$ } \rule[-2.5mm]{0mm}{6.5mm}\\ \hline \end{tabular}% \end{center} \caption{Estimates from above for the absolute value of $r_{j}$, $j=1,2,3$ and some higher order derivatives on $U$} \label{table2.ap.D} \end{table} \noindent\textit{Upper bounds for $| q_{j}|$, $j=1,2,3$\ on $U$.} Let $q_{j}$, $j=1,2,3$ by given by (\ref{D1.46}) -- (\ref{D1.48}). Earlier we found the estimates for $| q_{j}|$, $j=1,2,3$ on $V$ of the type $| q_{j}| \leq M_{j}| x-a|$, $j=1,2,3$ (see Table \ref{table1.ap.D}). Here, we will obtain the constants which are the upper bounds for $| q_{j}|$, $% j=1,2,3$ on $U$. Setting $\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}},% \tfrac{\partial ^{3}P}{\partial (\omega , \lambda) ^{\beta _{j}}}$ for the corresponding differentiations in each $q_{j}$, $j=1,2,3$, analogously to (% \ref{D6}), we have that \begin{aligned} | q_{j}(x)| &=| \int_0^{1}\int_0^{1}\int_0^{1}ts^2\left( \tfrac{ \partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}} (a+ts\sigma (x-a)),\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}(a+ts\sigma (x-a))\right) \\ &\quad \times (x-a)^{T}d\sigma dt\,ds| \\ &\leq \tfrac{1}{6}\sup_{x\in U}| \left( \tfrac{\partial ^{3}P}{ \partial (\omega ,\lambda) ^{\alpha _{j}}}(x),\tfrac{\partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}(x)\right) | \max_{U}| x-a| \leq M_{j}\max_{U}|x-a| , \end{aligned} \label{D99} where $M_{j}$ is an upper bound for the function $\tfrac{1}{6}|\left( \tfrac{% \partial ^{3}P}{\partial (\omega ,\lambda) ^{\alpha _{j}}}(x), \tfrac{% \partial ^{3}P}{\partial (\omega ,\lambda) ^{\beta _{j}}}(x)\right) |$ on $U$. In particular, it holds that \begin{aligned} \max_{U}| x-a|& =| x-a|\big| _{\partial U}\\ &= \sqrt{\left( \omega -\tfrac{1}{2} \pi \right) ^2+\left( \lambda -4\right) ^2} \big| _{(\omega ,\lambda ) =( \tfrac{1}{2}\pi -\tfrac{2}{180}\pi ,4-0.060)}\\ &=\sqrt{( \tfrac{1}{90}\pi ) ^2+0.060^2}, \end{aligned} \label{D104} and \begin{equation*} M_1=44, \quad M_2=27, \quad M_{3}=11. \end{equation*} The explicit upper bounds are given in Table \ref{table3.ap.D}. \begin{table}[ht] \renewcommand{\arraystretch}{1.5} \par \begin{center} \begin{tabular}{|c|c|c|} \hline $| q_1(x)| <3.1$ & $| q_2(x)| <1.9$ & $| q_{3}(x)| <0.8$ \\ \hline \end{tabular}% \end{center} \caption{Estimates from above for $| q_{j}|$, $j=1,2,3$ on $U$} \label{table3.ap.D} \end{table} \noindent\textit{Lower bounds for $F(x)=\det B(x)$ and $G(x)=b_{11}(x)+2\protect% \sqrt{\det B(x)}+b_{22}(x)$ on $U$.} Let us set $$\begin{gathered} F(x)=\det B(x), \\ G(x)=b_{11}(x)+2\sqrt{\det B(x)}+b_{22}(x), \end{gathered} \label{D102}$$ $b_{ij}$, $i,j=1,2$ are given in (\ref{D29})--(\ref{D32}) and find the lower bounds for $F$ and $G$ on $U$. By construction of the ball $W_{R}(a)\supset U$ from Proposition \ref{Proposition.ap.D} we know that $F\geq 0$ and $G>0$ on $W_{R}(a)$. More precisely, for every $x\in W_{R}(a)$ (and hence for every $x\in U$) it holds that \begin{gather*} F(x)\geq 1-\tfrac{\sqrt{2}}{4\pi }| q_2(x)| -\tfrac{1}{ 64}| q_1(x)| -\tfrac{% 1}{32\pi ^2}| q_1(x)| | q_{3}(x)| , \\ G(x)\geq 2-\tfrac{\sqrt{2}}{4\pi }| q_2(x)| -\tfrac{1}{ 64}| q_1(x)| . \end{gather*} Due to results of Table \ref{table3.ap.D} we finally obtain that on $U$, $$\begin{gathered} F(x)\geq 0.730\dots>0.7, \\ G(x)\geq 1.738\dots>1.7. \end{gathered} \label{D101}$$ \noindent\textit{Upper bounds for $| b_{ij}|$, $i,j=1,2$ on $U$.} Let $b_{ij}$, $i,j=1,2$ be given by (\ref{D2.1}) -- (\ref{D2.4}), namely, \begin{gather*} b_{11}(x)=\tfrac{1}{64}r_1(x)-\tfrac{\sqrt{2}}{8\pi }r_2(x), \\ b_{12}(x)=\tfrac{1}{64}r_2(x)-\tfrac{\sqrt{2}}{8\pi }r_{3}(x), \\ b_{21}(x)=-\tfrac{\sqrt{2}}{8\pi }r_1(x), \quad b_{22}(x)=-\tfrac{\sqrt{2}}{% 8\pi }r_2(x), \end{gather*} with $r_{j}$, $j=1,2,3$ as in (\ref{D4.1})--(\ref{D4.3}). Using the results of Table \ref{table2.ap.D} we will find the following upper bounds for the absolute values of $b_{ij}$, $i,j=1,2$ and some higher order derivatives on $U$ (see Table \ref{table4.ap.D}). \begin{table}[ht] \renewcommand{\arraystretch}{1.5} \par \begin{center} \begin{tabular}{|c|c|c|} \hline $| b_{11}(x)| <1.2$ & \parbox{47mm}{ $| \frac{\partial b_{11}}{\partial \omega }(x)| <2.1$, $| \frac{\partial b_{11}}{\partial \lambda }(x)| <0.8$} & \parbox{49mm}{ $| \frac{\partial ^2b_{11}}{\partial \omega \partial \lambda }(x)| <6.4$, $| \frac{\partial ^2b_{11}}{\partial \lambda ^2}(x)|<2.9$} \\ \hline $| b_{12}(x)| <0.7$ & \parbox{47mm}{ $| \frac{\partial b_{12}}{\partial \omega }(x)| <0.8$, $| \frac{\partial b_{12}}{\partial \lambda }(x)| <0.4$} & \parbox{49mm}{ $| \frac{\partial ^2b_{12}}{\partial \omega \partial \lambda }(x)| <2.9$, $| \frac{\partial ^2b_{12}}{\partial \lambda ^2}(x)|<1.5$ } \\ \hline $| b_{21}(x)| <0.3$ & \parbox{47mm}{ $| \frac{\partial b_{21}}{\partial \omega }(x)| <2.5$, $| \frac{\partial b_{21}}{\partial \lambda }(x)| <1.5$} & \parbox{49mm}{ $| \frac{\partial ^2b_{21}}{\partial \omega \partial \lambda }(x)| <9.9$, $| \frac{\partial ^2b_{21}}{\partial \lambda ^2}(x)|<3.7$} \\ \hline $| b_{22}(x)| <1.1$ & \parbox{47mm}{ $| \frac{\partial b_{22}}{\partial \omega }(x)| <1.5$, $| \frac{\partial b_{22}}{\partial \lambda }(x)| <0.4$} & \parbox{49mm}{ $| \frac{\partial ^2b_{22}}{\partial \omega \partial \lambda }(x)| <3.7$, $| \frac{\partial ^2b_{22}}{\partial \lambda ^2}(x)|<1.9$} \\ \hline \end{tabular}% \end{center} \caption{Estimates from above for the absolute value of $b_{ij}$, $i,j=1,2$ and some higher order derivatives on $U$} \label{table4.ap.D} \end{table} \noindent\textit{Upper bounds for $F(x)=\det B(x)$ and $G(x)=b_{11}(x)+2\protect% \sqrt{\det B(x)}+b_{22}(x)$ on $U$.} Recall that $F$ and $G$ are given by (\ref{D102}). They are positive functions on $U$ with lower bounds as in (\ref{D101}). Here we find their upper bounds together with the upper bounds for some higher order derivatives. In particular, in order to obtain the estimates for $F$, $| \frac{\partial F% }{\partial \omega }| , | \frac{\partial F}{\partial \lambda }|$ and $| \frac{\partial ^2F}{\partial \omega \partial \lambda }| , | \frac{\partial^2F% }{\partial \lambda ^2}|$ on $U$, we use the results of Table \ref% {table4.ap.D}. The estimates found are presented in the first row of Table % \ref{table5.ap.D}. To find the estimates for $| \frac{\partial G}{\partial \omega }| ,| \frac{% \partial G}{\partial \lambda } |$ and $| \frac{\partial ^2G}{\partial \omega \partial \lambda }| ,| \frac{\partial ^2G}{\partial \lambda ^2} |$ we use \begin{itemize} \item the lower bound for $F$on $U$, namely, $F(x)>0.7$, \item the results of Table \ref{table4.ap.D} and \item the results from the first row of Table \ref{table5.ap.D}. \end{itemize} E.g., for $| \tfrac{\partial G}{\partial \omega }|$ we will have that on $U$ \begin{align*} | \tfrac{\partial G}{\partial \omega }(x)| &=| \tfrac{\partial b_{11}}{% \partial \omega }(x)+\tfrac{\tfrac{\partial F}{ \partial \omega }(x)}{\sqrt{% F(x)}}+\tfrac{\partial b_{22}}{\partial \omega } (x)| \\ &\leq \sup_{U}| \tfrac{\partial b_{11}}{ \partial \omega }(x)| +\tfrac{% \sup_{U}| \tfrac{ \partial F}{\partial \omega }(x)| }{\inf_{U}\sqrt{F(x)}} +\sup_{U}| \tfrac{\partial b_{22}}{\partial \omega } (x)| \\ & <2.1+\tfrac{6.1}{\sqrt{0.7}}+1.5\approx 10.891\dots<11. \end{align*} The other estimates on $U$ for the derivatives of $G$ listed above are obtained in an analogous way and presented in Table \ref{table5.ap.D}. \begin{table}[ht] \begin{center} \begin{tabular}{|c|c|c|} \hline $F(x)<1.53$ & \parbox{24mm}{ $| \frac{\partial F}{\partial \omega }(x)| <6.10$, \hfill $| \frac{\partial F}{\partial \lambda }(x)| <2.53$} & \parbox{29mm}{ $| \frac{\partial ^2F}{\partial \omega \partial \lambda }(x)| <23.52$, \hfill $| \frac{\partial ^2F}{\partial \lambda ^2}(x)| <10.35$}\rule[-14pt]{0pt}{34pt} \\ \hline $-$ & \parbox{22mm}{$| \frac{\partial G}{\partial \omega }(x)| <11$, \hfill $| \frac{\partial G}{\partial \lambda }(x)| <4.3$} & \parbox{27mm}{ $| \frac{\partial ^2G}{\partial \omega \partial \lambda }(x)| <51.4$, \hfill $| \frac{\partial ^2G}{\partial \lambda ^2}(x)| <22.7$}\rule[-14pt]{0pt}{34pt} \\ \hline \end{tabular}% \end{center} \caption{Estimates from above for $F$ and $G$ on $U$} \label{table5.ap.D} \end{table} \noindent\textit{Upper bounds for $| c_{ij}|$, $i,j=1,2$ on $U$.} Let $c_{ij}$, $i,j=1,2$ be given by given by (\ref{D20}). It is convenient for further computations to set $$c_{ij}(x)=\tfrac{b_{ij}(x)+A\sqrt{F(x)}}{\sqrt{G(x)}}, \label{D103}$$ where $F$ and $G$ are as in (\ref{D102}) and \begin{equation*} A=% \begin{cases} 1 & \text{if }(i,j)=\{ (1,1),\text{ }(2,2)\} , \\ 0 & \text{if }(i,j)=\{ (1,2),\text{ }(2,1)\}.% \end{cases}% \end{equation*} To obtain the estimates for $| \frac{\partial c_{ij}}{ \partial \omega }| ,| \frac{\partial c_{ij}}{\partial \lambda }|$ and $| \frac{\partial ^2c_{ij}}{% \partial \omega \partial \lambda }| ,| \frac{\partial ^2c_{ij}}{ \partial \lambda ^2}|$, $i,j=1,2$ on $U$ we use \begin{itemize} \item the lower bound for $F$ and $G$ on $U$, defined by formula (\ref{D101}% ), \item the results of Table \ref{table4.ap.D} and \item the results of Table \ref{table5.ap.D}. \end{itemize} E.g., the estimate for $| \frac{\partial c_{ij}}{\partial \lambda } |$ on $% U$ is found in the following way: \begin{align*} | \tfrac{\partial c_{ij}}{\partial \lambda }(x)| &=\big| \tfrac{\tfrac{% \partial b_{ij}}{\partial \lambda }(x)}{\sqrt{G(x)} }+\tfrac{1}{2}A\tfrac{% \tfrac{\partial F}{\partial \lambda }(x)}{\sqrt{G(x)} \sqrt{F(x)}}-\tfrac{1}{% 2}\tfrac{\tfrac{\partial G}{\partial \lambda }(x)}{ G^{3/2}(x)}\left( b_{ij}(x)+A\sqrt{F(x)}\right) \big| \\ &\leq \big| \tfrac{\sup_{U}| \tfrac{\partial b_{ij}}{ \partial \lambda }(x)| }{\inf_{U}\sqrt{G(x)}}+\tfrac{1}{ 2}A\tfrac{\sup_{U}| \tfrac{\partial F}{% \partial \lambda } (x)| }{\inf_{U}\left( \sqrt{G(x)}\sqrt{F(x)}\right) } \\ &\quad +\tfrac{1}{2}\tfrac{\sup_{U}| \tfrac{\partial G}{ \partial \lambda }% (x)| }{\inf_{U}G^{3/2}(x)}\left( \sup_{U}| b_{ij}(x)| +A\sup_{U} \sqrt{F(x)}% \right) \big| . \end{align*} The other estimates on $U$ for the derivatives of $c_{ij}$, $i,j=1,2$ listed are obtained in an analogous way and listed in Table \ref{table6.ap.D} below. \begin{table}[ht] \renewcommand{\arraystretch}{1.5} \par \begin{center} \begin{tabular}{|c|c|} \hline $| \frac{\partial c_{11}}{\partial \lambda }(x)| <4.2$ & \parbox{27mm}{ $| \frac{\partial ^2c_{11}}{\partial \omega \partial \lambda } (x)| <99.4$ $| \frac{\partial ^2c_{11}}{\partial \lambda ^2}(x)|<35$} \\ \hline \parbox{24mm}{ $| \frac{\partial c_{12}}{\partial \omega }(x)| <2.4$ $| \frac{\partial c_{12}}{\partial \lambda }(x)| <1$} & \parbox{27mm}{ $| \frac{\partial ^2c_{12}}{\partial \omega \partial \lambda } (x)| <18.7$ $| \frac{\partial ^2c_{12}}{\partial \lambda ^2}(x)|<8.1$} \\ \hline $| \frac{\partial c_{21}}{\partial \lambda }(x)| <1.5$ & \parbox{27mm}{ $| \frac{\partial ^2c_{21}}{\partial \omega \partial \lambda } (x)| <20.1$ $| \frac{\partial ^2c_{21}}{\partial \lambda ^2}(x)|<8.4$} \\ \hline \parbox{24mm}{ $| \frac{\partial c_{22}}{\partial \omega }(x)| <9.8$ $| \frac{\partial c_{22}}{\partial \lambda }(x)| <3.8$} & \parbox{27mm}{ $| \frac{\partial ^2c_{22}}{\partial \omega \partial \lambda }(x)| <94.4$ $| \frac{\partial ^2c_{22}}{\partial \lambda ^2}(x)|<38.7$} \\ \hline \end{tabular}% \end{center} \caption{Estimates from above for the absolute value of some higher order derivatives of $c_{ij}$, $i,j=1,2$ on $U$} \label{table6.ap.D} \end{table} \noindent\textit{Upper bounds for $| h_{i}|$, $i=1,2$ on $U$.} Let $h_{i}$, $i=1,2$ be given by formulas (\ref{D18}), (\ref{D19}). First we compute the following derivatives of $h_{i}$, $i=1,2$ we will need below: \begin{gather} \tfrac{\partial h_1}{\partial \lambda }(x) =c_{12}(x) +\left( \tfrac{% \partial c_{11}}{\partial \lambda }(x) , \tfrac{\partial c_{12}}{\partial \lambda }(x) \right) \left( x-a\right) ^{T}, \label{D24} \\ \tfrac{\partial ^2h_1}{\partial \omega \partial \lambda }(x) =\tfrac{% \partial c_{12}}{\partial \omega }(x)+\tfrac{\partial c_{11}}{ \partial \lambda }(x)+\left( \tfrac{\partial ^2c_{11}}{\partial \omega \partial \lambda }(x) ,\tfrac{\partial ^2c_{12}}{\partial \omega \partial \lambda }% (x) \right) \left( x-a\right) ^{T}, \label{D25} \\ \tfrac{\partial ^2h_1}{\partial \lambda ^2}(x) =2\tfrac{ \partial c_{12}}{% \partial \lambda }(x)+\left( \tfrac{\partial ^2c_{11}}{ \partial \lambda ^2}% (x) ,\tfrac{\partial ^2c_{12}}{\partial \lambda ^2}(x) \right) \left( x-a\right) ^{T}, \label{D25.1} \\ \tfrac{\partial h_2}{\partial \lambda }(x) =c_{22}(x) +\left( \tfrac{% \partial c_{21}}{\partial \lambda }(x) , \tfrac{\partial c_{22}}{\partial \lambda }(x) \right) \left( x-a\right) ^{T}, \label{D26} \\ \tfrac{\partial ^2h_2}{\partial \omega \partial \lambda }(x) =\tfrac{% \partial c_{22}}{\partial \omega }(x)+\tfrac{\partial c_{21}}{ \partial \lambda }(x)+\left( \tfrac{\partial ^2c_{21}}{\partial \omega \partial \lambda }(x) ,\tfrac{\partial ^2c_{22}}{\partial \omega \partial \lambda }% (x) \right) \left( x-a\right) ^{T}, \label{D27} \\ \tfrac{\partial ^2h_2}{\partial \lambda ^2}(x) =2\tfrac{ \partial c_{22}}{% \partial \lambda }(x)+\left( \tfrac{\partial ^2c_{21}}{ \partial \lambda ^2}% (x) ,\tfrac{\partial ^2c_{22}}{\partial \lambda ^2}(x) \right) \left( x-a\right) ^{T}, \label{D27.1} \end{gather} where $c_{ij}$, $i,j=1,2$ are as in (\ref{D103}) and $a=\left( \frac{1}{2}% \pi ,4\right)$. Using the results of Table \ref{table6.ap.D} we find the estimates for $| \frac{\partial ^2h_{i}}{\partial \omega \partial \lambda } | ,| \frac{% \partial ^2h_{i}}{\partial \lambda ^2} |$, $i=1,2$ on $U$. E.g., for $| \frac{\partial ^2h_1}{\partial \omega \partial \lambda }|$ it holds that on $U$: \begin{align*} | \tfrac{\partial ^2h_1}{\partial \omega \partial \lambda }(x)| &=| \tfrac{% \partial c_{12}}{\partial \omega }(x)+ \tfrac{\partial c_{11}}{\partial \lambda }(x)+\left( \tfrac{\partial ^2c_{11}}{\partial \omega \partial \lambda }(x) ,\tfrac{ \partial ^2c_{12}}{\partial \omega \partial \lambda }% (x) \right) \left( x-a\right) ^{T}| \\ &\leq \sup_{U}| \tfrac{\partial c_{12}}{\partial \omega } (x)| +\sup_{U}| \tfrac{\partial c_{11}}{\partial \lambda }(x)| \\ &\quad +\sup_{U}| \left( \tfrac{\partial ^2c_{11}}{\partial \omega \partial \lambda }(x) ,\tfrac{\partial ^2c_{12}}{\partial \omega \partial \lambda } (x) \right) | \max_{U}| x-a| <\dots \end{align*} for $\max_{U}| x-a|$ see formula (\ref{D104}) and then \begin{equation*} \dots<2.4+4.2+\sqrt{99.4^2+18.7^2}\sqrt{\left( \tfrac{1}{90}\pi \right) ^2+0.060^2}\approx 13.621\dots<13.7. \end{equation*} Analogously, the other estimates on $U$ are obtained (see Table \ref% {table7.ap.D}). \begin{table}[ht] \renewcommand{\arraystretch}{1.5} \par \begin{center} \begin{tabular}{|c|c|} \hline $| \frac{\partial ^2h_1}{\partial \omega \partial \lambda }(x)| <13.7$ & $| \frac{\partial ^2h_2}{\partial \omega \partial \lambda }(x)| <18$ \\ \hline $| \frac{\partial ^2h_1}{\partial \lambda ^2}(x)|<4.5$ & $| \frac{\partial ^2h_2}{\partial \lambda ^2}(x)| <10.4$ \\ \hline \end{tabular}% \end{center} \caption{Estimates from above for the absolute value of some higher order derivatives of $h_{i}$, $i=1,2$ on $U$} \label{table7.ap.D} \end{table} \subsubsection{Strict positivity of the functions $\frac{\partial h_2}{% \partial \protect\lambda }(\protect\omega ,\protect\lambda)$ and $16\protect% \sqrt{2}\frac{\partial h_1}{\partial \protect\lambda } (\protect\omega ,% \protect\lambda) +\protect\pi \frac{\partial h_2}{\partial \protect\lambda }(% \protect\omega ,\protect\lambda)$ on $U$} Let us fix the notations which are common for two lemmas. \begin{notation}\label{Notation1,ap.D} \rm Let $U$ be as in (\ref{D100}), namely, \begin{equation*} U:=\left\{ (\omega ,\lambda) :\left[ \tfrac{88}{180}\pi ,\tfrac{ 92}{180}\pi \right] \times \left[ 3.940,4.060\right] \right\} . \end{equation*} By $\{ (\omega _{i},\lambda _{j})\} _{\substack{ i=0,\dots,n \\ j=0,\dots,m}}$ we mean a discretization of $U$ which is defined as follows \begin{equation*} \omega _{i}=\tfrac{88}{180}\pi +i\Delta \omega , \quad \lambda_{j}=3.94+j\Delta \lambda , \end{equation*} with \begin{equation*} \Delta \omega =\tfrac{\tfrac{4}{180}\pi }{n}, \quad \Delta \lambda = \tfrac{0.12}{m}. \end{equation*} \end{notation} Then we deduce the following two results. \begin{lemma}\label{lemma3,ap.D} It holds that $$\frac{\partial h_2}{\partial \lambda }(\omega ,\lambda) >0\quad\text{on }U. \label{D33}$$ \end{lemma} \begin{proof} Fix $n=m=2$ and consider the discretization $\{ (\omega _{i},\lambda _{j})\} _{i,j}$ of $U$ given by Notation \ref{Notation1,ap.D}. By straightforward computations it holds that $$\tfrac{\partial h_2}{\partial \lambda }(\omega _{i},\lambda _{j})>0, \label{D105}$$ for all $i=0,\dots,2$, $j=0,\dots,2$ and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in U} \tfrac{\partial h_2}{ \partial \lambda }(\omega _{i},\lambda _{j})=\tfrac{\partial h_2}{\partial \lambda }% (\omega _0,\lambda _0)\approx 0.952\dots\,. \label{D106}$$ Next to this, we compute \begin{equation*} a_1=\max \big\{ \Delta \omega ,\Delta \lambda \big\} =\max \big\{ \tfrac{% \tfrac{4}{180}\pi }{n},\tfrac{0.12}{m}\big\} =\tfrac{0.12}{2}=0.060, \end{equation*} and, by taking into account the results of Table \ref{table7.ap.D} and (\ref% {D106}), we also find \begin{equation*} a_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in U} \tfrac{ \partial h_2}{\partial \lambda }(\omega _{i},\lambda _{j})} {\sup_{U} | \left( \tfrac{% \partial ^2h_2}{\partial \omega \partial \lambda }(\omega ,\lambda ),\tfrac{% \partial ^2h_2}{\partial \lambda ^2} (\omega ,\lambda )\right) | }\approx 0.065\dots\,. \end{equation*} Since $a_10 \quad\text{on }U. \label{D60} \end{lemma} \begin{proof} Fix$n=7$,$m=12$and consider the discretization$\left\{ (\omega _{i},\lambda _{j})\right\} _{i,j}$of$U$given by Notation \ref% {Notation1,ap.D}. By straightforward computations it holds that $$16\sqrt{2}\tfrac{\partial h_1}{\partial \lambda }(\omega _{i},\lambda _{j})+\pi \tfrac{\partial h_2}{\partial \lambda }(\omega _{i},\lambda _{j})>0, \label{D107}$$ for all$i=0,\dots,7$,$j=0,\dots,12and, moreover, \begin{aligned} &\min_{(\omega _{i},\lambda _{j})\in U} \left( 16\sqrt{2}\tfrac{ \partial h_1}{\partial \lambda }(\omega _{i},\lambda _{j})+\pi \tfrac{ \partial h_2}{\partial \lambda }(\omega _{i},\lambda _{j})\right)\\ &=16\sqrt{2}\tfrac{\partial h_1}{\partial \lambda }(\omega _0,\lambda _0) +\pi \tfrac{\partial h_2}{\partial \lambda }(\omega _0,\lambda _0)\approx 2.936\dots\,. \end{aligned} \label{D108} Next to this, we compute \begin{equation*} b_1=\max \big\{ \Delta \omega ,\Delta \lambda \big\} =\max\big\{ \tfrac{% \tfrac{4}{180}\pi }{n},\tfrac{0.12}{m}\big\} =\tfrac{0.12}{12}=0.010, \end{equation*} and, by taking into account the results of Table \ref{table7.ap.D} and (\ref% {D108}), we also find \begin{align*} b_2&=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in U} \left( 16\sqrt{2}% \tfrac{\partial h_1}{\partial \lambda }(\omega _{i},\lambda _{j})+\pi \tfrac{% \partial h_2}{\partial \lambda }(\omega _{i},\lambda _{j})\right) }{% \sup_{U}| \left( 16\sqrt{2}\tfrac{ \partial ^2h_1}{\partial \omega \partial \lambda }(\omega ,\lambda )+\pi \tfrac{\partial ^2h_2}{\partial \omega \partial \lambda }(\omega ,\lambda ),16\sqrt{2}\tfrac{\partial ^2h_1}{% \partial \lambda ^2} (\omega ,\lambda )+\pi \tfrac{\partial ^2h_2}{\partial \lambda ^2} (\omega ,\lambda )\right) | } \\ &\approx 0.011\dots\,. \end{align*} Sinceb_10, \label{D111} for all $i=0,\dots,14$, $j=0,\dots,120$ and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_1} \left( -\tfrac{\partial P}{\partial \omega }(\omega _{i},\lambda _{j})\right) =-\tfrac{\partial P}{ \partial \omega }(\omega _{9},\lambda _0)\approx 1.022\dots\,. \label{D112}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{6}{180}\pi }{n},\tfrac{0.94}{m}\big\} =\tfrac{0.94}{120}\approx 0.00783\dots\,, \end{equation*} and, by taking into account (\ref{D110}) and (\ref{D112}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_1} \left( -\tfrac{% \partial P}{\partial \omega }(\omega _{i},\lambda _{j})\right) }{\sup_{H_1} % \big| \big( \tfrac{\partial ^2P }{\partial \omega ^2}(\omega ,\lambda ),% \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda )% \big) \big| }\approx 0.00860\dots\,. \end{equation*} Since $c_10$ on $H_2$. \end{claim} \begin{proof} First we find the following estimates: $$\begin{tabular}{llll} | \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda) | <48,\quad | \tfrac{\partial ^2P }{\partial \lambda ^2}(\omega ,\lambda )| <25\quad\text{on }H_2. & & & \end{tabular} \label{D120}$$ Then we fix $n=40$, $m=55$ and consider the discretization $\left\{ (\omega_{i},\lambda _{j})\right\} _{i,j}$, $i=0,\dots,40$, $j=0,\dots,55$ of $H_2$ such that \begin{equation*} \omega _{i}=\tfrac{87}{180}\pi +i\Delta \omega , \quad \lambda_{j}=4.750+j\Delta \lambda , \end{equation*} with \begin{equation*} \Delta \omega =\tfrac{\tfrac{14}{180}\pi }{n}, \quad \Delta \lambda = \tfrac{% 0.35}{m}. \end{equation*} By straightforward computations it holds that $$\tfrac{\partial P}{\partial \lambda }(\omega _{i},\lambda _{j})>0, \label{D121}$$ for all $i=0,\dots,40$, $j=0,\dots,55$ and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_2} \tfrac{\partial P}{ \partial \lambda }(\omega _{i},\lambda _{j})=\tfrac{\partial P}{\partial \lambda }% (\omega _0,\lambda _0)\approx 0.245\dots\,. \label{D122}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{14}{180}\pi }{n},\tfrac{0.35}{m}\big\} =\tfrac{0.35}{ 55}\approx 0.00636\dots\,, \end{equation*} and, by taking into account (\ref{D120}) and (\ref{D122}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_2} \tfrac{\partial P% }{\partial \lambda }(\omega _{i},\lambda _{j})} {\sup_{H_2}\big| \big( \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda ),% \tfrac{\partial ^2P}{\partial \lambda ^2}(\omega ,\lambda )\big) \big| }% \approx 0.00641\dots\,. \end{equation*} Since $c_10$ on $H_{3}$. \end{claim} \begin{proof} First we find the following estimates: $$| \frac{\partial ^2P}{\partial \omega ^2}(\omega ,\lambda) | <166, \quad | \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda )| <37\quad\text{on }H_{3}. \label{D130}$$ Then we fix $n=10$, $m=60$ and consider the discretization $\left\{ (\omega_{i},\lambda _{j})\right\} _{i,j}$ of $H_{3}$ such that \begin{equation*} \omega _{i}=\tfrac{100}{180}\pi +i\Delta \omega , \quad \lambda_{j}=4.000+j\Delta \lambda , \end{equation*} with \begin{equation*} \Delta \omega =\tfrac{\tfrac{8}{180}\pi }{n},\quad \Delta \lambda = \tfrac{% 0.85}{m}. \end{equation*} By straightforward computations it holds that $$\tfrac{\partial P}{\partial \omega }(\omega _{i},\lambda _{j})>0, \label{D131}$$ for all $i=0,\dots,10$, $j=0,\dots,60$ and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{3}} \tfrac{\partial P}{ \partial \omega }(\omega _{i},\lambda _{j})=\tfrac{\partial P}{\partial \omega }% (\omega _0,\lambda _{11})\approx 1.885\dots\,. \label{D132}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{8}{180}\pi }{n},\tfrac{0.85}{m}\big\} =\tfrac{0.85}{60}\approx 0.0142\dots\,, \end{equation*} and, by taking into account (\ref{D130}) and (\ref{D132}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{3}} \tfrac{% \partial P}{\partial \omega }(\omega _{i},\lambda _{j})} {\sup_{H_{3}}\big| % \big( \tfrac{\partial ^2P}{\partial \omega ^2} (\omega ,\lambda ),\tfrac{% \partial ^2P}{\partial \omega \partial \lambda } (\omega ,\lambda )\big) % \big| }\approx 0.0157\dots\,. \end{equation*} Since $c_10, \label{D141} for all$i=0,\dots,50$,$j=0,\dots,36$and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{4}} \left( -\tfrac{\partial P}{% \partial \lambda }(\omega _{i},\lambda _{j})\right) =-\tfrac{\partial P}{ \partial \lambda }(\omega _0,\lambda _0)\approx 0.118\dots\,. \label{D142}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{11}{180}\pi }{n},\tfrac{0.15}{m}\big\} =\tfrac{0.15}{36}% =0.00416\dots\,, \end{equation*} and, by taking into account (\ref{D140}) and (\ref{D142}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{4}} \left( -\tfrac{% \partial P}{\partial \lambda }(\omega _{i},\lambda _{j})\right) }{% \sup_{H_{4}} \big| \big( \tfrac{\partial ^2P }{\partial \omega \partial \lambda }(\omega ,\lambda ),\tfrac{\partial ^2P }{\partial \lambda ^2}% (\omega ,\lambda )\big) \big| }\approx 0.00423\dots\,. \end{equation*} Since$c_10$on$H_{5}$. \end{claim} \begin{proof} First we find the following estimates: $$| \frac{\partial ^2P}{\partial \omega ^2}(\omega ,\lambda) | <105, \quad | \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda )| <37\quad\text{on } H_{5}. \label{D150}$$ Then we fix$n=11$,$m=94$and consider the discretization$\left\{ (\omega_{i},\lambda _{j})\right\} _{i,j}$,$i=0,\dots,11$,$j=0,\dots,94$of$H_{5} $such that \begin{equation*} \omega _{i}=\tfrac{90}{180}\pi +i\Delta \omega , \quad \lambda_{j}=3.030+j\Delta \lambda , \end{equation*} with \begin{equation*} \Delta \omega =\tfrac{\tfrac{6}{180}\pi }{n}, \quad \Delta \lambda = \tfrac{% 0.94}{m}. \end{equation*} By straightforward computations it holds that $$\tfrac{\partial P}{\partial \omega }(\omega _{i},\lambda _{j})>0, \label{D151}$$ for all$i=0,\dots,11$,$j=0,\dots,94$and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{5}} \tfrac{\partial P}{\partial \omega }(\omega _{i},\lambda _{j}) =\tfrac{\partial P}{\partial \omega }% (\omega _0,\lambda _0) \approx 0.807\dots\,. \label{D152}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{6}{180}\pi }{n},\tfrac{0.94}{m}\big\} =\tfrac{0.94}{94}\approx 0.0100, \end{equation*} and, by taking into account (\ref{D150}) and (\ref{D152}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{5}} \tfrac{% \partial P}{\partial \omega }(\omega _{i},\lambda _{j})} {\sup_{H_{5}} \big| % \big( \tfrac{\partial ^2P}{\partial \omega ^2} (\omega ,\lambda ),\tfrac{% \partial ^2P}{\partial \omega \partial \lambda } (\omega ,\lambda )\big) % \big| }\approx 0.0102\dots\,. \end{equation*} Since$c_10, \label{D161} for all $i=0,\dots,33$, $j=0,\dots,41$ and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{6}} \left( -\tfrac{\partial P}{% \partial \lambda }(\omega _{i},\lambda _{j})\right) =-\tfrac{\partial P}{ \partial \lambda }(\omega _{33},\lambda _{41}) \approx 0.227\dots\,. \label{D162}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{15}{180}\pi }{n},\tfrac{0.33}{m}\big\} =\tfrac{0.33}{ 41}% =0.00805\dots\,, \end{equation*} and, by taking into account (\ref{D160}) and (\ref{D162}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{6}} \left( -\tfrac{% \partial P}{\partial \lambda }(\omega _{i},\lambda _{j})\right) }{% \sup_{H_{6}} \big| \big( \tfrac{\partial ^2P }{\partial \omega \partial \lambda }(\omega ,\lambda ),\tfrac{\partial ^2P }{\partial \lambda ^2}% (\omega ,\lambda )\big) \big| }\approx 0.00820\dots\,. \end{equation*} Since $c_10, \label{D171} for all$i=0,\dots,5$,$j=0,\dots,30$and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{7}} \left( -\tfrac{\partial P}{% \partial \omega }(\omega _{i},\lambda _{j})\right) =-\tfrac{\partial P}{ \partial \omega }(\omega _{5},\lambda _{27})\approx 2.663\dots\,. \label{D172}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{8}{180}\pi }{n},\tfrac{0.85}{m}\big\} =\tfrac{0.85}{ 30}\approx 0.0283\dots\,, \end{equation*} and, by taking into account (\ref{D170}) and (\ref{D172}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{7}} \left( -\tfrac{% \partial P}{\partial \omega }(\omega _{i},\lambda _{j})\right) }{\sup_{H_{7}}% \big| \big( \tfrac{\partial ^2P }{\partial \omega ^2}(\omega ,\lambda ),% \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda )% \big) \big| }\approx 0.0322\dots\,. \end{equation*} Since$c_10$on$H_{8}$. \end{claim} \begin{proof} First we find the following estimates: $$| \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda) | <36, \quad | \tfrac{\partial ^2P }{\partial \lambda ^2}(\omega ,\lambda )| <10 \quad\text{on }H_{8}. \label{D180}$$ Then we fix$n=40$,$m=30$and consider the discretization$\{ (\omega_{i},\lambda _{j})\} _{i,j}$,$i=0,\dots,40$,$j=0,\dots,30$of$% H_{8} $such that \begin{equation*} \omega _{i}=\tfrac{78}{180}\pi +i\Delta \omega , \quad \lambda_{j}=3.900+j\Delta \lambda , \end{equation*} with \begin{equation*} \Delta \omega =\tfrac{\tfrac{11}{180}\pi }{n}, \quad \Delta \lambda =\tfrac{% 0.15}{m}. \end{equation*} By straightforward computations it holds that $$\tfrac{\partial P}{\partial \lambda }(\omega _{i},\lambda _{j})>0, \label{D181}$$ for all$i=0,\dots,40$,$j=0,\dots,30$and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{8}} \tfrac{\partial P}{\partial \lambda }(\omega _{i},\lambda _{j}) =\tfrac{\partial P}{\partial\lambda }% (\omega _{40},\lambda _{30}) \approx 0.134\dots\,. \label{D182}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{11}{180}\pi }{n},\tfrac{0.15}{m}\big\} =\tfrac{0.15}{30}=0.00500, \end{equation*} and, by taking into account (\ref{D180}) and (\ref{D182}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{8}} \tfrac{% \partial P}{\partial \lambda }(\omega _{i},\lambda _{j})} {\sup_{H_{8}}{\sup }\big| \big( \tfrac{\partial ^2P}{\partial \omega \partial \lambda }(\omega ,\lambda ),\tfrac{\partial ^2P}{\partial \lambda ^2}(\omega ,\lambda )\big) % \big| }\approx 0.00508\dots\,. \end{equation*} Since$c_10$on$H_0$. \end{claim} \begin{proof} First we find the following estimates: $$| \tfrac{\partial G}{\partial \omega }(\omega ,\lambda) | <25000, \quad | \tfrac{\partial G}{\partial \lambda }(\omega ,\lambda )| <14000 \quad\text{% on }H_0. \label{D190}$$ Then we fix$n=600$,$m=300$and consider the discretization$\left\{(\omega _{i},\lambda _{j})\right\} _{i,j}$,$i=0,\dots,600$,$j=0,\dots,300$of$H_0$such that \begin{equation*} \omega _{i}=\tfrac{84}{180}\pi +i\Delta \omega , \quad \lambda_{j}=2.960+j\Delta \lambda , \end{equation*} with \begin{equation*} \Delta \omega =\tfrac{\tfrac{10}{180}\pi }{n},\quad \Delta \lambda = \tfrac{% 0.1}{m}. \end{equation*} By straightforward computations it holds that $$G(\omega _{i},\lambda _{j})>0, \label{D191}$$ for all$i=0,\dots,600$,$j=0,\dots,300$and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_0} G(\omega _{i},\lambda _{j})=G(\omega _0,\lambda _0)\approx 8.380\dots\,. \label{D192}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{10}{180}\pi }{n},\tfrac{0.1}{m}\big\} =\tfrac{0.1}{ 300}\approx 0.000(3), \end{equation*} and, by taking into account (\ref{D190}) and (\ref{D192}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_0} G(\omega _{i},\lambda _{j})}{\sup_{H_0}\big| \big( \tfrac{ \partial G}{\partial \omega }(\omega ,\lambda ),\tfrac{\partial G}{\partial \lambda }(\omega ,\lambda )\big) \big| }\approx 0.000414\dots\,. \end{equation*} Since$c_10, \label{D201} for all $i=0,\dots,70$, $j=0,\dots,1100$ and, moreover, $$\min_{(\omega _{i},\lambda _{j})\in H_{\star }} \left( -F(\omega _{i},\lambda _{j})\right) =-F(\omega _{70},\lambda _{1100})\approx 0.0773\dots \,. \label{D202}$$ Next to this, we compute \begin{equation*} c_1=\max \left\{ \Delta \omega ,\Delta \lambda \right\} =\max \big\{ \tfrac{% \tfrac{2}{180}\pi }{n},\tfrac{0.57}{m}\big\} =\tfrac{0.57}{ 1100}% =0.000518\dots, \end{equation*} and, by taking into account (\ref{D200}) and (\ref{D202}), we also find \begin{equation*} c_2=\sqrt{2}\dfrac{\min_{(\omega _{i},\lambda _{j})\in H_{\star }} \left( -F(\omega _{i},\lambda _{j})\right) } {\sup_{H_{\star }}\big| \big( \tfrac{% \partial F}{\partial \omega }(\omega ,\lambda ),\tfrac{\partial F}{\partial \lambda }(\omega ,\lambda )\big) \big| }\approx 0.000555\dots\,. \end{equation*} Since \$c_1