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Choose homogeneous Dirichlet boundary conditions, $y(a)=y(b)=0$, and nonzero admissible functions with finite $F$ and positive $G$. Variations obey the same zero endpoint values. At a stationary Rayleigh quotient, set $\lambda =F[y]/G[y]$. ThenThe boundary term vanishes, and the fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Thus stationary functions are eigenfunctions. Conversely, multiply that equation by $y$ and integrate by parts to obtain $F=\lambda G$; its first variation then vanishes by the same calculation. This proves that stationary values are the corresponding eigenvalues. Homogeneous Neumann boundary conditions also make the displayed boundary terms vanish; general Robin boundary conditions can require a boundary contribution to the quadratic form.

For $p=1$, $q=0$, $w=x$ on $(0,\pi )$, the polynomial trial function $y=x(\pi -x)$ satisfies the endpoint conditions. Its Rayleigh quotient usesso the polynomial estimate is $\boxed{{\lambda }_{\mathrm{poly}}=20/{\pi }^3\simeq 0.64503}$. The trigonometric trial function $y=\sin x$ givesand hence $\boxed{{\lambda }_{\mathrm{trig}}=2/\pi \simeq 0.63662}$. Both are upper bounds on the lowest eigenvalue, by the Rayleigh-Ritz method; neither trial is an exact eigenfunction, so the inequalities are strict.

Their closeness has a useful explanation beyond their similar single-arch shapes. Both trial functions are symmetric under $x\mapsto \pi -x$, forcing ${\int }_0^{\pi }x{y}^{2}dx=(\pi /2){\int }_0^{\pi }{y}^{2}dx$. The reflection-symmetric trials for a linearly weighted Rayleigh quotient therefore reduce this problem to a constant multiple of the ordinary Dirichlet energy quotient. The sine is the best trial within this symmetric class, and the polynomial's value is only a factor $10/{\pi }^2\simeq 1.0132$ larger. The unrestricted lowest eigenfunction can be asymmetric because the weight $x$ is asymmetric.

For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are hyperbolic cylinders. The degenerate $u=0$ surface is $y=0$, $|x|\le a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi /2,3\pi /2$ they are vertical half-axes.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. Away from the focal lines, divide the coordinate tangent vectors by their lengths to obtain the local orthonormal unit vectors $e_u,e_v,e_z$. For physical vector field components $A_u,A_v,A_z$, the formulas for vector calculus in elliptic cylindrical coordinates areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

Conjugacy classes. For $a\in G$, its conjugacy class isEvery element belongs to its own conjugacy class, by taking $g=e$. To prove uniqueness, suppose $x=ga{g}^{-1}=hb{h}^{-1}$ belongs to both $\mathrm{Cl}(a)$ and $\mathrm{Cl}(b)$. ThenFor every $t\in G$, this makes $tb{t}^{-1}$ a conjugate of $a$ as well. Hence $\mathrm{Cl}(b)\subseteq \mathrm{Cl}(a)$, and reversing the argument gives the other inclusion. Two conjugacy classes that meet are therefore equal, proving that the classes partition $G$.

Abelian case. If $G$ is an abelian group, then $ga{g}^{-1}=ag{g}^{-1}=a$ for every $g$, so $\boxed{\mathrm{Cl}(a)=\{a\}}$ for each $a$.

Centre. An element has a singleton conjugacy class precisely when $gz{g}^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual centre of a group:The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=g{z}^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a subgroup. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.

Square symmetry. Use the paper's notation $D_4$ for the eight-element dihedral group of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^{j}s$ commutes with $r$, becauseand $r^{j-1}\ne r^{j+1}$. Therefore

Cosets and counting. A left coset of a subgroup $H$ in a group $G$ is $gH=\{gh:h\in H\}$. The left cosets partition $G$ and each has $|H|$ elements. Lagrange's theorem states that for a finite group, $\boxed{|G|=[G:H]|H|}$; in particular, the subgroup order divides the group order.

The matrix group. Work throughout in the finite field ${\mathbb{F}}_5$, so every nonzero entry has a multiplicative inverse. Denote a matrix by $A(a,b,c)$, with $a,c\ne 0$. Ordinary matrix multiplication, interpreted using modular arithmetic, givesIts diagonal entries remain nonzero, proving closure. Associativity is inherited from matrix multiplication, $A(1,0,1)$ is the identity element, andThis lies in the same set and multiplication checks the inverse in either order. Thus these matrices form a group. There are four choices each for $a$ and $c$ and five for $b$, giving $\boxed{|G|=4\cdot 5\cdot 4=80}$.

The subgroup. Write $h(a,c)=A(a,a-c,c)$. ThenIt contains the identity $h(1,1)$, is closed, and has inverse $h(a^{-1},c^{-1})$. Interchanging the two factors leaves the product unchanged because multiplication in ${\mathbb{F}}_5$ is commutative. Therefore $H$ is an abelian subgroup, withEquivalently $H$ is isomorphic to ${\mathbb{F}}_5^{\times }\times {\mathbb{F}}_5^{\times }$ by $h(a,c)\mapsto (a,c)$, which also explains its abelian property. Its order divides $80$, verifying Lagrange's theorem.

There are five distinct left cosets. Representatives can be taken as $U_t=A(1,t,1)$, $t\in {\mathbb{F}}_5$: indeedEvery matrix lies in exactly the coset specified by this unique $t$.

Elements whose square is the identity. Squaring givesThus $a,c\in \{1,-1\}=\{1,4\}$ and $b(a+c)=0$. For equal signs, $a+c\ne 0$ in ${\mathbb{F}}_5$, so $b=0$. For opposite signs, $a+c=0$ and all five $b$ values work. The complete set is thereforeIt contains $2+5+5=12$ elements, including the identity; the other eleven are involutions. Since $12$ does not divide $80$, Lagrange's theorem proves that this set is not a subgroup.

Definitions. In a punctured neighborhood of $z_0$, let the Laurent series be $f(z)={\sum }_{j=-\infty }^{\infty }{c}_j(z-z_0{)}^j$. The point is a pole of order $m\ge 1$ if $c_{-m}\ne 0$ and every coefficient with index less than $-m$ vanishes. Its residue is $\boxed{{\mathrm{Res}}_{z_0}f=c_{-1}}$. A simple pole has $m=1$.

Poles for $0<a<1$. Denote the bracket in the denominator by $D(z)$. Factoring givesPut $q=\sqrt{(1-a)/(1+a)}$, so $0<q<1$. The finite poles are $0,\pm q,\pm q^{-1}$, all simple; the numerator is nonzero at each. At zero,At a nonzero simple root $r$ of $D$, the residue formula givesFor $r^2=q^2$, $D^{\prime }(r)=8ar$; for $r^2=q^{-2}$, $D^{\prime }(r)=-8ar$. Substitution yieldsThe two zeros of a function at $z=\pm i$ do not cause additional poles: there $D=-4$ and the numerator vanishes.

The case $a=1$. Simplify before using any formulas that divide by $1-a^2$:Thus its only finite pole isAs $a\uparrow 1$, the two inner poles merge with zero while the outer two escape to infinity. If one also considers the Riemann sphere, $f$ at $a=1$ has a simple pole at infinity because $f(z)\sim z/4$; the requested unit-circle calculation only uses finite poles.

Unit-circle integral. Set $z=e^{i\theta }$. ThenAt the two angles where the tangent is undefined, the real integrand has continuous limiting value zero. Thus the substitution is valid for the improper integral, or after that continuous extension. No pole lies on $|z|=1$ for any allowed $a$.

For $0<a<1$, the sum of the enclosed residues isThe residue theorem therefore givesAt $a=1$ the enclosed residue is $1/2$, giving $I=\pi$, exactly the same formula. The direct simplification $1/(1+{\tan }^2\theta )={\cos }^2\theta$ provides an independent check of that endpoint.

Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

Integral identity and representation. Define $L={\nabla }^2+k^2$ and take $n$ to be the outward unit normal. The product rule givesThe $k^2\varphi \psi$ terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:

Choose the Dirichlet Green function with the sign conventionwhere $\delta$ is the three-dimensional Dirac delta. Substitute $\psi =G$, use $L\varphi =f$, and evaluate the Dirac delta integral. The boundary term containing $G$ vanishes and $\varphi =g$ on $S$, leavingHenceThe choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen $k^2$ is not a Dirichlet eigenvalue of $-{\nabla }^2$; at such an eigenvalue a Green inverse for arbitrary data does not exist.

Infinite space. Put $r=|x-x^{\prime }|$. Away from the source, spherical symmetry reduces the Helmholtz equation toFor $k\ne 0$ its general radial solution isIntegrate $LG=\delta$ over a ball of radius $\epsilon$ about the source. The divergence theorem yieldsSince $G=A/r+Bk+O(r)$, the volume term is $O({\epsilon }^2)$ and the flux tends to $-4\pi A$. Therefore $A=-1/(4\pi )$, while $B$ remains free:The $B$ term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence $e^{-i\omega t}$, the outgoing Green function for the three-dimensional Helmholtz equation is $-e^{ikr}/(4\pi r)$, corresponding to $B=-i/(4\pi )$.

Zero wavenumber. For fixed finite $B$, $\sin (kr)/r\to 0$ as $k\to 0$, andThis is the decaying Green function of the Laplacian. Directly at $k=0$, the general radial solution is $A/r+C$; source normalization fixes $A=-1/(4\pi )$, and decay at infinity removes $C$. If the free parameter depends on $k$ as $B(k)\sim C/k$, the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.

Write the two-dimensional character as $(2,x,y)$. Character orthogonality with the first two rows givesso $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete character table is thereforeThe last row belongs to the standard representation of the symmetric group.

To decompose the given representation, use the character inner productIts multiplicities areThusConcretely the invariant line is spanned by $(1,1,1{)}^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from character orthogonality.

Apply the Laplacian in polar coordinates twice to the proposed separated function. For any real exponent $m$,so the biharmonic equation becomesIts four distinct roots are $m=-2,0,2,4$. The separated polar solutions of the biharmonic equation therefore give the complete radial family in this angular sector,Equivalently, its radial equation is a fourth-order Cauchy-Euler equation; four distinct powers supply four linearly independent solutions, so no logarithmic terms are required.

Interior. Continuity with value zero at the origin removes both $r^{-2}\cos 2\theta$ and $\cos 2\theta$. The latter depends on the approach direction even though it is bounded. The two boundary conditions at unit radius then give $C+D=0$ and $2C+4D=1$. ConsequentlyThis extends smoothly through the origin: $r^2\cos 2\theta =x^2-y^2$ and $r^4\cos 2\theta =(x^2+y^2)(x^2-y^2)$ are polynomials. Its radial derivative at $r=1$ is $(2r^3-r)\cos 2\theta =\cos 2\theta$.

Exterior. The condition on the radial derivative at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos 2\theta$ is permitted, since the required decay concerns the radial derivative. Now $A+B=0$ and $-2A=1$, givingIndeed ${\partial }_r{\psi }_{\mathrm{out}}=r^{-3}\cos 2\theta \to 0$, and its value at unit radius is zero. These are particular interior and exterior solutions for the biharmonic streamfunction; no condition requiring $\psi$ itself to vanish at infinity was imposed.