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The nullclines are $y=\mu x$ and $y=x^2/[\nu (1+x^2)]$. At a nonzero fixed point, cancellation of $x$ yields $\mu \nu =x/(1+x^2)$. The latter function attains its maximum $1/2$ at $x=1$. HenceFor $0<\mu <{\mu }_c$, $x_{-}<1<x_{+}$ and $x_{-}{x}_{+}=1$.

The Jacobian matrix isAt $O$ the distinct eigenvalues are $-\mu ,-\nu$, so the origin is a stable node. At a nonzero fixed point its trace is $-(\mu +\nu )<0$ andThe discriminant is $(\mu -\nu {)}^2+8x/(1+x^2{)}^2>0$. Thus $P_{-}$ is a saddle and $P_{+}$ is a stable node, never a focus.

Above the straight nullcline $\dot{x}>0$; below it $\dot{x}<0$. Below the curved nullcline $\dot{y}>0$; above it $\dot{y}<0$. At the saddle, an eigenvector can be chosen as $(1,\mu +\lambda )$: the stable eigenvalue gives negative slope, and the unstable eigenvalue gives positive slope. The stable manifold is the basin boundary. The two branches of the unstable manifold lead respectively to the origin and the positive stable node, as illustrated for representative parameters.

The nonnegative quadrant is forward invariant. Since $\dot{y}\le 1-\nu y$, $y$ stays bounded, and then $\dot{x}=-\mu x+y$ also bounds $x$. Moreover the divergence is the negative constant $-(\mu +\nu )$, excluding periodic or homoclinic loops by the Bendixson-Dulac criterion. The Poincaré-Bendixson theorem then leaves convergence to a fixed point. Initial data on the saddle's stable manifold approach the saddle; data in the two open basins approach extinction at $O$ or persistence at $P_{+}$, respectively.

The rotation unitary operator is $U(\mathbf{n},\theta )=\exp (-i\theta \mathbf{n}\cdot \mathbf{J}/\hslash )$. A state with zero orbital angular momentum has a rotationally invariant orbital factor; a spinless $j=0$ state is wholly unchanged. Zero orbital angular momentum alone does not rule out a nontrivial rotation of intrinsic spin angular momentum.

For spin one-half, $\mathbf{J}=\hslash \mathbf{\sigma }/2$. Direct multiplication of the Pauli matrices gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$. The antisymmetric part vanishes when contracted with $a_i{a}_j$, henceFor $\mathbf{a}\ne 0$, the operator has trace zero, so its eigenvalues are $\pm \hslash |\mathbf{a}|/2$; for $\mathbf{a}=0$ both are zero. With unit $\mathbf{n}$, the even and odd powers in the exponential give

Take ${\mathbf{n}}^{\prime }=(\sin \theta ,0,\cos \theta )$. The rotated state $|m^{\prime }{⟩}_{\theta }=U(\mathbf{y},\theta )|m^{\prime }⟩$ satisfies $({\mathbf{n}}^{\prime }\cdot \mathbf{J})|m^{\prime }{⟩}_{\theta }=m^{\prime }\hslash |m^{\prime }{⟩}_{\theta }$, because $U{J}_3{U}^{†}={\mathbf{n}}^{\prime }\cdot \mathbf{J}$. By the Born rule, the probability of this outcome in the initial state $|m⟩$ is $|{}_{\theta }⟨m^{\prime }|m⟩{|}^2=|⟨m|U(\mathbf{y},\theta )|m^{\prime }⟩{|}^2$. In the $J_3$ eigenbasis,The transition from $m=1/2$ to $m^{\prime }=-1/2$ thus has probability ${\sin }^2(\theta /2)=1-{\cos }^2(\theta /2)$. $\boxed{A=1,B=-1}$.

Let $v=T(0)$ and $B(x)=T(x)-v$. The Euclidean isometry $B$ fixes the origin and preserves norms. The polarization identity therefore gives $⟨Bx,By⟩=⟨x,y⟩$. In particular, $B{e}_1,B{e}_2$ are an orthonormal basis. The coefficients of $Bx$ in this basis are $⟨Bx,B{e}_j⟩=x_j$, so $Bx=x_{1}B{e}_1+x_{2}B{e}_2$. Thus $B$ is a linear map represented by an orthogonal matrix, and $T(x)=Bx+v$. Conversely, every such map preserves distances because $\Vert B(x-y)\Vert =\Vert x-y\Vert$.

When $\det B=-1$, the real eigenvalues of the two-dimensional orthogonal matrix are $1,-1$. Choose corresponding unit eigenvectors $e,n$, and write $v=ae+bn$. In these coordinates $T(s,t)=(s+a,-t+b)$. This is reflection in the line $t=b/2$, followed by translation $ae$ along that line. It is a Euclidean reflection if $a=0$, and a glide reflection if $a\ne 0$.

The only equilibrium point is $(0,0)$. Its Jacobian matrix is $\left(\begin{array}{cc}-\mu & 1\\ -1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2+\mu \lambda +1$. For $\mu >0$ both eigenvalues have negative real parts, so the origin is locally asymptotically stable.

Use the Lyapunov function $V=(x^2+y^2)/2$. Along a solution,A Lyapunov function is positive away from the equilibrium and nonincreasing along trajectories; its sufficiently small sublevel sets prove stability. Asymptotic stability additionally means nearby trajectories converge to the equilibrium. The basin of attraction is the set of initial states whose forward solutions exist for all positive time and converge to it. Every closed disc of radius at most $\sqrt{3}$ is positively invariant, since $\dot{V}\le 0$ there. The LaSalle invariance principle says that a solution in a compact positively invariant set where $\dot{V}\le 0$ approaches the largest invariant subset of $\{\dot{V}=0\}$. Here that subset is only the origin: on $x=0$, remaining on the line would require $y=0$; the isolated points $x=\pm \sqrt{3},y=0$ on the outer disc do not remain there because $\dot{y}=-x\ne 0$. Hence the closed disc $x^2+y^2\le 3$ lies in the basin of attraction.

This estimate does not assert that the basin is exactly that disc, or the whole plane. For example, choose $R$ so large that $R^2\ge 6(\mu +1)/\mu$ and $R^2>3(\mu +2)/\mu$. The region $x\ge R,y\ge -x$ is positively invariant: on $x=R$ one has $\dot{x}>0$, and on $y=-x$ one has $(x+y{)}^{\cdot }=\mu x^3/3-(\mu +2)x>0$. Within it $\dot{x}\ge \mu x^3/6$, which forces finite-time escape. Thus the attraction is genuinely local. The Lyapunov function gives a rigorous sufficient domain for attraction without identifying its entire boundary.

For $T=\frac{1}{2}{\dot{q}}^{T}T\dot{q}$ and $V=\frac{1}{2}{q}^{T}Vq$, where the matrices are symmetric and constant, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$. A normal mode $q=v{e}^{i\omega t}$ therefore solves $(V-{\omega }^{2}T)v=0$.

A string segment of horizontal length $b$ has extension $\sqrt{b^2+(\Delta q{)}^2}-b=(\Delta q{)}^2/(2b)+O((\Delta q{)}^4)$. Multiplying by the tension $S$ and using $m=1$ and $S/b=1$, the quadratic kinetic energy and potential energy areThus $T=I$ and $V=\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right)$. Taking $v_i=\sin (ij\pi /4)$, $i=1,2,3$, gives the three eigenvalues $2-2\cos (j\pi /4)$, $j=1,2,3$. The normal mode frequencies are

For parameter-dependent centre manifold theorem, append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the eigenvalues are $0,-1,0$: the $y$ direction is strongly stable and $(x,\mu )$ span the extended centre directions. An extended centre manifold for a parameter is an invariant graph of a function $y=h(x,\mu )$ tangent to $y=0$; nearby trajectories approach its reduced dynamics along stable fibers. The stable eigenvalue remains separated from zero, so local bifurcation and stability are governed by this reduction. This is the parameter-extension content of the requested stable/centre-manifold discussion.

The centre-manifold invariance equation isReflection symmetry permits an invariant graph of a function with $h(-x,\mu )=h(x,\mu )$ and $h(0,\mu )=0$. At quadratic order the left side vanishes and the right side is $-h-x^2$, giving $h=-x^2+O(|\mu |x^2+x^4)$. More explicitly its first mixed coefficient is $2\mu x^2$, found by equating the $\mu x^2$ terms. The reduced equation isThus the origin is locally asymptotically stable for $\mu <0$, becomes a saddle for $\mu >0$, and creates two stable equilibria $x=\pm \sqrt{\mu /2}+O({\mu }^{3/2})$, $y=-\mu /2+O({\mu }^2)$. This is a supercritical pitchfork bifurcation. The exact nearby nonzero branches provide a check: the equilibrium equations givewhich are real and nonzero for small positive $\mu$. The omitted branch $y=1+\sqrt{1+\mu }$ is not near the origin. At $\mu =0$ the leading $-2x^3$ also attracts, although at an algebraic rather than exponential rate.

We use the standard facts that finite complex representations are completely reducible, irreducible characters are orthonormal, the regular character is ${\sum }_{\chi }\chi (1)\chi$, and the number of irreducible characters equals the number of conjugacy classes. We also use Schur's lemma, Sylow's theorem and the nontrivial center of a nontrivial finite $p$-group.

For an irreducible character $\chi$ of degree $d$, a conjugacy-class sum $z_C$ acts as the scalar ${\lambda }_C=|C|\chi (g)/d$ by Schur's lemma and taking traces. This scalar is an algebraic integer: class sums form an integral basis of the center of the integral group algebra, their multiplication has integer structure constants, and multiplication by $z_C$ satisfies its monic integer characteristic polynomial. That same polynomial annihilates its image in any irreducible representation. Character values are also algebraic integers, since they are sums of roots of unity. Character orthogonality now givesThis is a rational algebraic integer, hence an integer. Therefore $\chi (1)$ divides $|G|$.

The Burnside's pa qb theorem states that every finite group of order $p^a{q}^b$ is soluble. We prove its character-theoretic ingredient. If a class has size $m$ coprime to $d$, integers $u,v$ with $ud+vm=1$ givean algebraic integer. Every algebraic conjugate of this number is the average of $d$ roots of unity and has modulus at most one. If it is nonzero, its algebraic norm is a nonzero integer of absolute value at least one; since every factor has modulus at most one, all must have modulus one. In particular $|\chi (g)|=d$. Equality in the triangle inequality for the unit-modulus eigenvalues implies that $\rho (g)$ is scalar. Thus if $(m,d)=1$, either $\chi (g)=0$ or $\rho (g)$ is scalar.

Suppose a group of order $p^a{q}^b$ were nonabelian simple, with both primes present. Choose a nonidentity $g$ in the center of a Sylow $p$-subgroup. Its centralizer contains that Sylow subgroup, so its class size is a power of $q$. Every nontrivial irreducible representation is faithful by simplicity. If it sent $g$ to a scalar, faithfulness would make $g$ central in the group, impossible in a nonabelian simple group. The ingredient above therefore gives $\chi (g)=0$ for every nontrivial character with degree coprime to $q$. Evaluating the regular character at $g\ne 1$ givesThe right side is an algebraic integer, but $1/q$ is not, a contradiction. A group of prime-power order cannot be nonabelian simple either, by its nontrivial center. Induct on the order: every nonsimple group has a proper nontrivial normal subgroup, and that subgroup and the quotient have smaller orders with at most these two prime divisors. By induction both are soluble; their derived-series extensions make the original group soluble. Simple abelian groups have prime order and start the induction. This proves Burnside's theorem.

For the final divisibility, let $|G|=n$ be odd and let $k$ be its number of conjugacy classes. All irreducible degrees are odd by the divisibility just proved. We use the standard Frobenius-Schur indicator theorem: $\nu (\chi )=n^{-1}{\sum }_g\chi (g^2)$ is zero precisely for an irreducible character not equal to its conjugate, and is $\pm 1$ otherwise. Squaring is a bijection on an odd-order group: for each $g$, its unique square root is $g^{(n+1)/2}$, since every element order divides $n$. For every nontrivial irreducible character, orthogonality to the trivial character consequently gives $\nu (\chi )=n^{-1}{\sum }_g\chi (g)=0$. The nontrivial characters therefore occur in distinct conjugate pairs of equal odd degree. Since an odd square is one modulo eight,is a sum of pairs $2(d^2-1)$, each divisible by sixteen; the trivial character contributes zero. Hence

The shape operator is the self-adjoint tangent map $S_p=-d{N}_p$, using a locally chosen unit normal. Its eigenvalues ${\kappa }_1,{\kappa }_2$ are the principal curvatures, and their eigenvectors give the principal directions. At an umbilical point they coincide, so $S_p=\kappa I$ and every tangent direction is principal. Reversing the normal changes both curvature signs but does not change the principal directions or umbilicity.

The canonical commutation relations are $[\hat{x},\hat{p}]=i\hslash I$ (in several dimensions $[x_i,p_j]=i\hslash {\delta }_{ij}I$, with the position-position and momentum-momentum commutators zero). For Hermitian $x,p$, the commutator is anti-Hermitian, just as $i\hslash I$ is; there is no conflict with their self-adjointness. The annihilation operator and its adjoint obeyThus the quantum harmonic oscillator Hamiltonian is $\boxed{H=\hslash (N+1/2)}$. The commutators $[N,a]=-a$, $[N,a^{†}]=a^{†}$ and the normalized ground state constructNormalization follows inductively from $a{a}^{†}=N+1$. Positivity of $N$ and repeated lowering exclude a noninteger eigenvalue: lowering it until it becomes negative would produce a nonzero vector of negative $N$ eigenvalue. Every eigenstate therefore lowers to the unique ground state, so these normalized ladder states exhaust the oscillator eigenstates.

For two independent oscillators, the tensor products $|n_1,n_2⟩$ have $E=\hslash (n_1+n_2+1)$. The level $n=n_1+n_2$ has degeneracy $n+1$. In its degenerate subspace the perturbation $\lambda x_1{x}_2$ reduces tosince terms that raise or lower both numbers leave the subspace. For $n=0$ its matrix is zero. For $n=1$, in $|1,0⟩,|0,1⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. For $n=2$, in $|2,0⟩,|1,1⟩,|0,2⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right)$. Their eigenvalues giveThe $n=1$ states are the symmetric and antisymmetric combinations. At $n=2$, the shifted states are $(|2,0⟩\pm \sqrt{2}|1,1⟩+|0,2⟩)/2$, and the zero-shift state is $(|2,0⟩-|0,2⟩)/\sqrt{2}$.

Take the boundary-fitted grid $h=1/(M+1)$ and assume the standard consistency of a numerical method hypothesis $u\in C^4(\overline{\Omega })$ with bounded fourth partial derivatives. Taylor expansion along each coordinate, including a bounded fourth-order remainder, gives the five-point Laplacian truncation errorThe numerical equation and exact grid values therefore imply $A_h\mathbf{e}=\mathbf{\tau }$, where $A_h=-{\Delta }_h$ with zero boundary values. Its discrete sine transform eigenvectors are $\sin (\pi rih)\sin (\pi sjh)$, $1\le r,s\le M$, and its eigenvalues areThe one-dimensional sine vectors form an orthogonal basis, hence so do their tensor products. The least eigenvalue occurs at $r=s=1$. Since $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8$, so the Euclidean operator norm of $A_h^{-1}$ is at most $1/8$. There are $M^2$ residual components, and $Mh<1$, givingThe factor $M$ matters: pointwise second-order consistency of a numerical method becomes first order in the unweighted Euclidean grid norm.

The PDF does not explicitly impose regularity on the exact solution. Without such a hypothesis its unrestricted error claim is false. Here is a counterexample, with a classical solution smooth in the open square and continuous with zero boundary values:The sampled numerical solution separates as $v_i\sin (\pi jh)$. Its error $e_i=v_i-g(ih)$ solves $Ae=\tau$, where $A=-{\Delta }_{h,x}+{\lambda }_{y}I$ and ${\lambda }_y=4h^{-2}{\sin }^2(\pi h/2)$. The one-dimensional residual is ${\tau }_i={\Delta }_{h,x}g(ih)-g^{\prime \prime }(ih)+({\pi }^2-{\lambda }_y)g(ih)$. At the first node,The leading coefficient is strictly negative. Moreover $g^{\prime \prime \prime \prime }<0$ on $(0,1)$, so the triangular-average formula for the second difference and concavity of $g^{\prime \prime }$ give ${\Delta }_{h,x}g\le g^{\prime \prime }$. Thus the positive part of every residual is at most $C{h}^2$. The matrix $A$ has a nonnegative inverse, and its first diagonal inverse entry is at least $1/(2h^{-2}+{\lambda }_y)$, as follows from the nonnegative Neumann expansion of its tridiagonal matrix. The discrete comparison principle bounds the inverse image of the residual's positive part by $C{h}^2$, while the negative first residual gives $e_1\le -c\sqrt{h}+C{h}^2$. Hence $|e_1|\ge c^{\prime }\sqrt{h}$ for small $h$. Since ${\sum }_{j=1}^M{\sin }^2(\pi jh)=1/(2h)$, the full Euclidean error is bounded below by a positive constant. It cannot be $O(h)$. Thus the boxed estimate holds for the intended smooth solution class; a regularity condition is genuinely necessary.

The circle group is $U(1)=\{z\in \mathbb{C}:|z|=1\}$ under multiplication. For continuous finite-dimensional complex representations, its irreducibles are the one-dimensional charactersA group representation is a continuous homomorphism to the invertible linear maps; it is irreducible when it has no proper nonzero invariant subspace. Compactness permits an invariant Hermitian inner product by averaging. Since the resulting unitary operators commute for $U(1)$, simultaneous diagonalization reduces an irreducible to a character, and the continuous characters of the circle have the stated integer winding number.

The group $SU(2)$ consists of unitary two-by-two complex matrices of determinant one. Each has the unique formidentifying it homeomorphically with the unit 3-sphere in ${\mathbb{R}}^4$. This is the Spin group $\mathrm{Spin}(3)$. A unitary matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$, and its conjugacy class is determined by $0\le \theta \le \pi$, or equivalently its trace $2\cos \theta$.

The irreducible representations of SU2 are $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ for integers $n\ge 0$, with dimension $n+1$. The symmetric power can be realized as homogeneous degree-$n$ polynomials in two variables, with the induced linear action. On the diagonal torus its eigenvalues are $e^{i(n-2j)\theta }$ for $j=0,\dots ,n$, so its character, the trace of the representing operator, isAt the endpoints use the continuous values ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The highest-weight classification says this list is complete.

For the final action, multiplication of block matrices gives $(AB{)}_1=A_1{B}_1$, so conjugation is a linear group representation. Decompose the underlying three-dimensional space as $V_1\oplus V_0$. Its endomorphism space isThe two off-diagonal blocks are each $V_1$, since the invariant alternating form identifies $V_1^{\ast }$ with $V_1$. The scalar part of $\mathrm{End}(V_1)$ is $V_0$ and its traceless part is $V_2$: the same alternating form identifies traceless endomorphisms with symmetric tensors of degree two. The lower scalar block is a second $V_0$. ThusThe dimensions add to $2+4+3=9$, and its character is $2{\chi }_0+2{\chi }_1+{\chi }_2$.

A bounded linear operator is compact if it maps the unit ball to a set with compact closure; equivalently, every bounded sequence has a subsequence whose images converge.

If $S,T$ are compact operators, select a subsequence on which $S{x}_n$ converges and then a further subsequence on which $T{x}_n$ converges. This proves compactness of any linear combination. For norm closure, suppose compact $T_j$ converge to $T$ in operator norm. Given $\epsilon >0$, choose $j$ with $\Vert T-T_j\Vert <\epsilon /3$. A finite $\epsilon /3$-net for $T_j$ of the unit ball is then a finite $2\epsilon /3$-net for its image under $T$. This image is totally bounded, and its closure is compact because $X$ is complete. Thus $B_0(X)$ is a closed linear subspace.

For bounded $S$ and compact $T$, the image of a bounded sequence under $S$ is bounded, so $TS$ is compact; and a convergent subsequence of $T{x}_n$ remains convergent after applying the continuous map $S$, so $ST$ is compact. Therefore the compact operators form a closed two-sided ideal in $B(X)$.

For the weighted backward shift, truncate its output after coordinate $N$, obtaining a finite-rank operator $T_N$. The exact norm estimate isHence $T$ is compact. Its iterates areThus for every nonzero complex $\lambda$ the Neumann seriesconverges absolutely in operator norm. Zero is an eigenvalue, since $T{e}_1=0$, whereas every nonzero value is in the resolvent set. Consequently

There are two conventions for loxodromic Möbius transformations. In the inclusive convention, a transformation conjugate to $z\mapsto \lambda z$ with $|\lambda |\ne 1$ is loxodromic, including a hyperbolic Möbius transformation with $\lambda >0$. Under that convention the assertion about discs is false: $z\mapsto 2z$ preserves the upper half-plane, and a Cayley conjugation gives an invariant ordinary disc.

The question uses the classification in which hyperbolic Möbius transformations and genuinely spiralling loxodromic Möbius transformations are separate. Thus here the multiplier satisfies $|\lambda |\ne 1$ and $\lambda \notin (0,\infty )$. Such a Möbius transformation has two distinct fixed points; after sending them to zero and infinity it acts by the stated dilation and rotation.

For a determinant-one representative, put $\tau =a+d$. Its eigenvalues are $\eta ,{\eta }^{-1}$, with $\eta +{\eta }^{-1}=\tau$, and the normalized multiplier is ${\eta }^2$ or its reciprocal. If $\tau$ is real, the eigenvalues are either real or conjugate unit-modulus numbers. These give the hyperbolic, elliptic, parabolic or identity cases, not a genuine spiral. Conversely, a real eigenvalue or a unit-modulus eigenvalue makes $\tau$ real. Hence the practical criterion in the question's convention isChanging the determinant-one representative by its sign does not affect this criterion.

Finally, conjugate an invariant disc to the upper half-plane. A Möbius transformation preserving that half-plane has a representative in $S{L}_2(\mathbb{R})$: preservation of the boundary gives real coefficients up to a common scalar, and preservation of the chosen side makes the real determinant positive, allowing normalization to one. Its trace is real. Trace is invariant under conjugacy, up to the harmless sign of the representative, so a disc-preserving transformation cannot be genuinely loxodromic. This proves the intended assertion while specifying its necessary terminology.