Remove the common nonzero factor $e^{-i\omega t}$ and expand in the small parameter $ϵ$. To first order, the modified Laplacian contributes only its $y$ derivatives: its $x$ derivative term has the extra factor ${ϵ}^2$. The regular perturbation expansion therefore givesThus the two residuals are $P+ϵQ+O({ϵ}^2)$ and $R+ϵS+O({ϵ}^2)$, respectively. Subscripts after a comma denote partial derivatives. We retain $\omega \ne 0$, as appropriate to the oscillatory wave and the elimination below.

For the two sine modes, orthogonality givesThe diagonal potential integrals vanish, while the mixed one isThus ${\int }_0^1\cos (\pi x){\tilde{\psi }}^{2}dx=\alpha /2$, giving the two-sine variational bound for a Dirichlet Mathieu equation:Its derivative vanishes when $1+6{\pi }^2\alpha -{\alpha }^2=0$. Hence the two stationary parameters are $3{\pi }^2\pm \sqrt{9{\pi }^4+1}$. To identify the minimum, use the normalized sine basis: the restricted self-adjoint operator has real symmetric matrixIts smaller eigenvalue is the minimum of the Rayleigh quotient, attained at the negative stationary parameter. ThereforeThe estimate improves on the single-sine value ${\pi }^2$. The Rayleigh-Ritz variational principle remains valid here even though $q(x)=\cos (\pi x)$ changes sign: on this finite interval the Sturm-Liouville operator is self-adjoint and bounded below, which is sufficient for the upper bound.

Use the normalized eigenfunctions and their weighted inner product ${\int }_a^{b}w\overline{{\psi }_m}{\psi }_{n}dx={\delta }_{mn}$. Orthogonality givesFor complex trials, $F$ means the corresponding Hermitian quadratic form with $|{\psi }^{\prime }{|}^2,|\psi {|}^2$; it agrees with the displayed functional on real trials. If $a_0\ne 0$, division by $|a_0{|}^2$ yieldsSince ${\lambda }_n\ge {\lambda }_0$, this is at least ${\lambda }_0$. The Rayleigh-Ritz variational principle therefore gives an upper bound from any admissible trial, improved by minimizing over trial parameters or enlarging the trial space. Equality holds when the trial is entirely in the lowest eigenspace.

The displayed ratio using $b_n$ presupposes $a_0\ne 0$. If $a_0=0$, use the undivided quotient $\sum {\lambda }_n|a_n{|}^2/\sum |a_n{|}^2$ instead; the same upper-bound argument still applies. No ground-state coefficient can be divided out in that case.

Choose homogeneous Dirichlet boundary conditions $\psi (a)=\psi (b)=0$, and take a nonzero real admissible trial $\psi$. Write the numerator as $F[\psi ]$ and denominator as $G[\psi ]>0$. For an endpoint-zero variation $\eta$, integration by parts givesThus the first variation of the Rayleigh quotient isBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiplying an eigenfunction equation by $\psi$ and integrating gives $F=\lambda G$, so its first variation vanishes. This establishes the equivalence for the chosen Sturm-Liouville problem. Homogeneous Neumann boundary conditions, or a mixed Dirichlet-Neumann pair, also cancel the boundary term; general Robin boundary conditions require adding the corresponding boundary contribution to the numerator.

Since $\mathbf{F}=\nabla \Omega$, its divergence is the Laplacian of $\Omega$:The divergence theorem changes the outward flux through the whole closed boundary into $I={\int }_V(6x+4z)dV$. Rotational symmetry makes the $x$ term vanish. With $z=(u^2-v^2)/2$ and the Jacobian determinant from part (c),Both the curved face and the top disk are included in this surface integral through the divergence theorem.

The Jacobian determinant is the oriented scalar triple product of the three coordinate basis vectors:Here ${\mathbf{h}}_u\times {\mathbf{h}}_v=(u^2+v^2)(-\sin \varphi ,\cos \varphi ,0)$, soThe sign is nonnegative on the stated range, and the absolute Jacobian determinant also equals $h_u{h}_v{h}_{\varphi }$ as expected for orthogonal coordinates.

Differentiate the position vector with respect to the three coordinates to obtain the coordinate basis vectors:The chain rule gives $d\mathbf{x}={\mathbf{h}}_{u}du+{\mathbf{h}}_{v}dv+{\mathbf{h}}_{\varphi }d\varphi$. The scale factors of orthogonal coordinates are the corresponding norms:Direct dot products give ${\mathbf{h}}_u\cdot {\mathbf{h}}_v=uv-uv=0$, ${\mathbf{h}}_u\cdot {\mathbf{h}}_{\varphi }=0$, and ${\mathbf{h}}_v\cdot {\mathbf{h}}_{\varphi }=0$. Thus the parabolic coordinates are orthogonal wherever the coordinate vectors are nondegenerate. Their degeneracy on $uv=0$ is the usual axis degeneracy of the azimuthal coordinate.

Let $\rho =\sqrt{x^2+y^2}=uv$. The condition $u\le v$ gives $z\le 0$. On the outer coordinate surface $v=1$, elimination of $u$ gives $z=({\rho }^2-1)/2$. On $u=v$, the surface is the plane $z=0$, with $\rho =u^2\le 1$. Thus the closure of the body isIt is a solid circular cap bounded below by an elliptic paraboloid of revolution and above by the unit disk in $z=0$. Its vertex is $\boxed{(0,0,-1/2)}$, its height is $\boxed{1/2}$, and its rim is the circle $\boxed{x^2+y^2=1,z=0}$. The symmetry axis is the $z$ axis. The coordinate surface $u=0$ maps to that axis rather than to an additional two-dimensional boundary. These are parabolic coordinates.

Let $n=p\ge 3$ be prime. Any subgroup contained in $⟨R⟩\cong C_p$ is either $\{I\}$ or the whole rotation subgroup. Both are normal: the trivial subgroup always is, and the rotation subgroup has index two, or equivalently is preserved by $mR{m}^{-1}=R^{-1}$.

If a normal subgroup contains any reflection, it contains its entire conjugacy class. Part (b) shows that this is every reflection for odd $p$. Products of reflections include a generator of the rotations, since $(Rm)m=R$. Such a normal subgroup must therefore be the whole dihedral group.

Thus the complete list of proper normal subgroups, including the trivial subgroup, isThis also agrees with classification of subgroups of a dihedral group: the reflection subgroups of order two are proper but are not normal when $p$ is odd.

Rotations commute with one another, while reflection conjugation sends $R^j$ to $R^{-j}$. Thus the rotational conjugacy classes are pairs $\{R^j,R^{-j}\}$, except when the two entries coincide. For reflections, direct use of the relations givesIf $n$ is odd, multiplication by two permutes residues modulo $n$, so all reflections are conjugate. The complete class list isTheir sizes are $1$, then $(n-1)/2$ classes of size two, then one class of size $n$, giving $(n+3)/2$ classes and $2n$ elements in total.

If $n$ is even, the half-turn $R^{n/2}$ is central and forms its own singleton. Conjugation changes a reflection exponent by an even amount or negates it, so it preserves parity and reaches every exponent of that parity. The list isThere are $n/2+3$ classes; their sizes are $1,1$, then $n/2-1$ pairs, then two classes of size $n/2$. Geometrically the two reflection classes correspond to vertex-axis and edge-axis reflections, in either order depending on the choice of $m$. These are the conjugacy classes of a finite dihedral group.

The relation $mR=R^{-1}m$ moves every reflection factor to the right of every rotation factor. Canceling pairs $m^2=I$ and reducing rotation powers modulo $n$ shows that every word has one of the formsWithin either family the elements are distinct, since equality would imply equality of two powers of $R$, which has order $n$. No rotation equals a reflection: in the polygon action, rotations preserve planar orientation and reflections reverse it. Thus this complete list has exactly $2n$ elements.

Geometrically, $R$ rotates the polygon by $2\pi /n$ about its centre, and $m$ reflects it in a symmetry axis through the centre. The elements $R^j$ are all rotations; $R^{j}m$ are all reflections. We use the paper's convention that $D_n$ has order $2n$, since another common convention calls this same group $D_{2n}$.

The subgroup generated by $T,P$ is $H=\{I,T,P,-I\}$, a Klein four-group. It is not normal in $G^{\prime }$. Indeed,For $x\ne 0$ the off-diagonal entries are nonzero, whereas every member of $H$ is diagonal. Thus this conjugate of $T$ is outside $H$, givingThe normal subgroup in part (c) is the continuous boost subgroup; choosing one representative from each of its cosets does not make those representatives a normal subgroup.

Direct multiplication gives $T^2=P^2=I$, $TP=PT=-I$, andThus conjugation by each generator of $G^{\prime }$ preserves $G$; so does conjugation by every product of the generators. Therefore $\boxed{G⊴G^{\prime }}$.

Using $TB(x)=B(-x)T$ and its analogue for $P$, every group word can be reduced to $B(x)S$ with $S\in \{I,T,P,TP\}$. Consequently the quotient has at most four cosets. They are distinct: a nonidentity diagonal matrix among $T,P,-I$ cannot equal $B(x)$, since a diagonal $B(x)$ requires $\sinh x=0$ and hence $B(x)=I$. ThereforeThe three nonidentity cosets all square to $G$, and $TG$ and $PG$ commute because $T,P$ commute. HenceEquivalently, $G^{\prime }$ is the full real $O(1,1)$ group and $G$ its identity component. This is the component quotient of the real Lorentz group in two dimensions; the quotient records the four components, not the continuous boost parameter.

For a subgroup $N\le H$ and $h\in H$, a left coset is $hN=\{hn:n\in N\}$ and a right coset is $Nh=\{nh:n\in N\}$. A normal subgroup satisfies $hN{h}^{-1}=N$ for every $h\in H$, equivalently $hN=Nh$.

If $N$ is normal, the quotient group is the set $H/N$ of cosets, with multiplicationTo check independence of representatives, replace $h,k$ by $h{n}_1,k{n}_2$. Thenbecause normality puts $k^{-1}{n}_{1}k$ in $N$. Thus the operation is well defined. Associativity follows from that in $H$, the identity is $N$, and the inverse of $hN$ is $h^{-1}N$. This proves the group axioms for $H/N$ and explains the role of coset multiplication is well-defined exactly for normal subgroups.

Write $B(x)$ for the displayed matrix. The hyperbolic addition identities giveThus matrix multiplication is closed in $G$. It is associative because ordinary matrix multiplication is associative. The identity is $B(0)=I$, and $B(x)B(-x)=I$ gives the inverse $B(x{)}^{-1}=B(-x)$. Hence $\boxed{G\text{ is a group}}$.

Moreover, $x\mapsto B(x)$ is an injective homomorphism from the additive real numbers, since $\sinh x$ determines $x$. Therefore $\boxed{G\cong (\mathbb{R},+)}$; in particular it is abelian. These matrices are the connected boost subgroup relevant to the component quotient of the real Lorentz group in two dimensions.