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By the law of mass action, the net changes in the intermediate concentrations areThe autocatalytic reaction increases $X$ by one and decreases $Y$ by one; its reaction-rate convention is $k_3{X}^{2}Y$. With the specified common concentration scale and time scale, chooseSubstitution gives the nondimensional Brusselator system. Adding its two steady equations gives $u=1$, and then the second gives $v=b/a$. Its Jacobian matrix there isThe equilibrium is asymptotically stable for $b<1+a$ and unstable for $b>1+a$. At $b_c=1+a$ the eigenvalues are $\pm i\sqrt{a}$, crossing the imaginary axis with nonzero speed $\frac{d}{db}\mathrm{Re}\lambda =1/2$.

The nonlinear Hopf bifurcation is nondegenerate. At the threshold, put $x=u-1$ and $y=-\sqrt{a}[(u-1)+(v-b/a)]$. The translated equations areThe Hopf coefficient of the Brusselator in the planar radial normalization is $-(a+2)/8<0$: the cubic contribution is $-6a/16$ and the quadratic contribution is $[-2\sqrt{a}2(1-a)]/(16\sqrt{a})$. Consequently the bifurcation is supercritical, producing small stable oscillations on the unstable-equilibrium side. Their limiting period is

An exponential dispersion family has density or mass functionwith support independent of the natural parameter $\theta$. Usually $a(\varphi )=\varphi /w$ for a known weight $w$. For the Poisson distribution, identifyThe dispersion is fixed at one. Differentiating the normalizing identity gives $\mathbb{E}Y=b^{\prime }(\theta )$; differentiating again gives $\mathrm{Var}Y=a(\varphi )b^{\prime \prime }(\theta )$. Here both derivatives equal $e^{\theta }=\lambda$, soThe variance function expresses $b^{\prime \prime }(\theta )$ in terms of the mean $\mu$. The canonical link function maps the mean to the natural parameter, so the Poisson canonical link is $g(\mu )=\log \mu$.

The supplied congruence is a congruence of squares, since $25=5^2$. HenceA nontrivial square root modulo a product of two odd primes may agree with $+5$ modulo one prime and $-5$ modulo the other. Taking greatest common divisors separates these factors. The Euclidean algorithm givesFor example the successive remainders for the first computation are $1072,737,335,67,0$; for the second they are $1062,767,295,177,118,59,0$. ThereforeBoth numbers are prime by trial division by primes at most their square roots. More generally, given $a^2\equiv b^2(\mathrm{mod}N)$ with $a\not\equiv \pm b(\mathrm{mod}N)$, computing $\gcd (N,a-b)$ often produces a proper factor. The method requires no prior knowledge of which prime gives which sign, and is the basic factor-extraction step in congruence-of-squares algorithms.

In the upper-half-plane model, with metric $d{s}^2=(d{x}^2+d{y}^2)/y^2$, the hyperbolic geodesics are vertical lines and semicircles orthogonal to the real boundary. Two geodesics with disjoint endpoints and no intersection are ultraparallel and have a common perpendicular $h$. To construct it, send one geodesic to the imaginary axis by an isometry. The second then has endpoints $0<a<b$ after a possible reflection. The semicircle centered at zero with radius $\sqrt{ab}$ is perpendicular to the imaginary axis and to the second circle: its squared radius plus $((b-a)/2{)}^2$ equals $((a+b)/2{)}^2$, the orthogonality condition for the two circles.

Let the intersection points of $h$ with the given geodesics have signed arclength coordinates $s_1,s_2$ on $h$. Each hyperbolic reflection preserves $h$ and restricts there to $s\mapsto 2s_i-s$: the perpendicular intersection point is fixed and the tangent direction along $h$ is reversed. Thus their composition restricts toThe displacement is nonzero since the two perpendicular intersection points differ. Its $N$th iterate translates by $2N(s_1-s_2)$, which is never zero for $N>0$. The composite is a hyperbolic translation of infinite order. The assumption excluding a common endpoint ensures the common perpendicular used here exists.

The displacement equation is $\rho {\mathbf{u}}_{tt}=(\lambda +\mu )\nabla (\nabla \cdot \mathbf{u})+\mu \Delta \mathbf{u}$. With $\mathbf{u}=\nabla \varphi +\nabla \times \mathbf{\psi }$ and divergence-free vector-potential gauge, divergence and curl give wave equations withThus ${\varphi }_{tt}=c_P^2\Delta \varphi$ and ${\mathbf{\psi }}_{tt}=c_S^2\Delta \mathbf{\psi }$, up to gauge constants. The P wave displacement is parallel to its propagation direction; S wave displacement is perpendicular, with two transverse polarizations.

Take the plane of incidence to be $xy$. Let $k$ be the shared tangential wave number, $h=({\omega }^2/c_P^2-k^2{)}^{1/2}$ and $s=({\omega }^2/c_S^2-k^2{)}^{1/2}$. Incident P, reflected P and reflected SV waves have respectively wave vectors $(k,h)$, $(k,-h)$ and $(k,-s)$. Thus the reflected P angle equals $\theta$ and $\sin {\theta }_S=(c_S/c_P)\sin \theta$. No SH wave is excited by this planar P incidence.

Use scalar potentials $\mathbf{u}=({\varphi }_x+{\psi }_y,{\varphi }_y-{\psi }_x)$, with incident potential amplitude one, reflected P amplitude $R$, and reflected SV amplitude $Q$. At the surface their phases are common. Vanishing $xy$ and $yy$ tractions gives, with $D=s^2-k^2$,Eliminating $Q$ givesThere is no reflected P wave precisely when $4k^{2}hs=D^2$. If $\beta =c_S^2/c_P^2$ and $\sigma =k^2{c}_S^2/{\omega }^2=\beta {\sin }^2\theta$, then $h=(\omega /c_S)\sqrt{\beta -\sigma }$, $s=(\omega /c_S)\sqrt{1-\sigma }$ and $D=({\omega }^2/c_S^2)(1-2\sigma )$. Substitution provesThe condition applies only to incidence angles in the physical range $0\le \sigma \le \beta$; it need not have a solution for every elastic-modulus ratio.

Incompressibility gives $r^{-1}(ru{)}_r+r^{-1}{v}_{\varphi }=0$, permitting a stream function with $u={\psi }_{\varphi }/r$, $v=-{\psi }_r$. Its axial vorticity is $-\Delta \psi$. Taking the curl of the Stokes flow momentum equation eliminates pressure and gives $\Delta$ of that vorticity zero, hence ${\Delta }^2\psi =0$.

At an impermeable constant-angle plane, $v=0$ means $\psi$ is constant along that boundary. Fix that constant to zero. A rigid boundary also requires ${\psi }_{\varphi }=0$. The tangential stress is $\mu (-{\psi }_{rr}+{\psi }_r/r+{\psi }_{\varphi \varphi }/r^2)$; along an impermeable boundary its first two terms vanish, so a stress-free boundary requires ${\psi }_{\varphi \varphi }=0$.

Write $\psi =(S/\mu )r^{2}F(\varphi )$. The biharmonic equation becomes $F^{\prime \prime \prime \prime }+4F^{\prime \prime }=0$. The boundary conditions are $F(-\alpha )=F^{\prime }(-\alpha )=F(0)=0$ and $F^{\prime \prime }(0)=1$. ConsequentlyThe requested $f$ is $F/\mu$. Evaluating $u(r,0)=(Sr/\mu )F^{\prime }(0)$ givesIts small-angle ratio is $\alpha /4+O({\alpha }^3)$, so $U$ and $S$ have the same sign for sufficiently small positive angle. At the first nonzero critical root, $2{\alpha }_c\simeq 4.493409$, the denominator vanishes while the numerator does not. The pure $r^2$ similarity solution is then resonant and cannot satisfy the forcing; logarithmic radial terms or additional global boundary data are needed. Just beyond this angle, and up to $\alpha <\pi$, the denominator is negative and numerator positive, so the formal surface flow reverses relative to $S$. This is not a contradiction of viscous dissipation: the wedge is radially unbounded and its radial-end energy flux has not been prescribed. A physical finite-domain flow needs those outer conditions, and the Stokes approximation also fails sufficiently far out when the velocity grows with $r$.

For $t>0$ the nonzero Christoffel symbols, apart from lower-index symmetry, arePut $S=\sum p_i$ and $Q=\sum p_i^2$. Their substitution in the Ricci tensor gives $R_{00}=(S-Q)/t^2$ and $R_{ii}=p_i(S-1)t^{2p_i-2}$, with all off-diagonal entries zero. Thus either every $p_i=0$, or some exponent is nonzero andThe latter are the Kasner metric conditions; the all-zero static Minkowski solution is a separate vacuum possibility.

In an orthonormal frame the independent curvature components are proportional to $p_i(p_i-1)/t^2$ and $p_i{p}_j/t^2$. They all vanish precisely when the exponents are all zero or a permutation of $(1,0,0)$. The latter flat metric is Minkowski space in a Milne-type coordinate chart.

All three exponents cannot be strictly positive: in the nonstatic branch, $S^2-Q=2{\sum }_{i<j}{p}_i{p}_j=0$, contradicting positive pairwise products. If all three are nonzero, there is exactly one negative exponent: two negative ones would force the positive one above one, violating $Q=1$. Therefore one scale factor contracts and two expand as $t$ increases in that branch.

Assume the physical potential is real, as required for the stated conjugation identity, and use the regular radial solution and its analytic continuation. The differential equation depends on $k^2$, so exchanging $k$ and $-k$ exchanges the outgoing and incoming coefficients of the same regular solution. Their ratio consequently gives $S(-k)=S(k{)}^{-1}$. Complex conjugation gives a solution at $k^{\ast }$ and exchanges those coefficients, so $S(k{)}^{\ast }S(k^{\ast })=1$.

For real $k$, $|S(k)|=1$, hence $S=e^{2i{\delta }_0(k)}$ with real phase. The first identity makes the phase odd modulo $\pi$; one can choose odd branches on paired nonzero intervals. A continuous odd branch through zero additionally requires $S(0)=1$, excluding a threshold-resonance exception. For the given rational expression, which has $S(0)=1$, chooseIts small-$k$ behavior is ${\delta }_0(k)=-4k/(3\lambda )+O(k^3)$. ThusA zero of the continued $S$ means that the outgoing coefficient vanishes, leaving an incoming-only asymptotic solution. It is paired by $S(k)S(-k)=1$ with a pole at the opposite wave number. The displayed zeros are $-i\lambda$ and $-3i\lambda$. Genuine bound-state poles on the positive imaginary axis correspond to negative energies and decaying regular solutions; their opposite zeros encode the same spectral information. A bare continued pole outside the physical analyticity region need not be a bound state, so no additional classification of the two poles follows merely from the rational expression without further potential assumptions.

Putting the unit vector along a coordinate axis gives ${\sigma }_i^2=I$. Putting it along $(e_i+e_j)/\sqrt{2}$ gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$ for $i\ne j$. Combining the resulting anticommutators with the given commutators yields ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, and henceThe operators $S_i=(\hslash /2){\sigma }_i$ obey angular-momentum commutators and ${\mathbf{S}}^2=3{\hslash }^{2}I/4$, the spin-one-half value. In a spin-half irreducible sector they are the intrinsic spin operators; the algebra also permits direct sums of such sectors.

Since $Q=\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian with $Q^2=I$, its operators $P_{\pm }=(I\pm Q)/2$ are Hermitian and satisfy $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Moreover $Q{P}_{\pm }=\pm P_{\pm }$. Thus $|\chi ⟩=P_{+}|\chi ⟩+P_{-}|\chi ⟩$ is its orthogonal eigenstate decomposition, omitting any zero component. The Born probabilities for a normalized state areFor spin up along $\mathbf{m}$, the anticommutator $\{\mathbf{m}\cdot \mathbf{\sigma },{\sigma }_j\}=2m_{j}I$ gives $⟨{\sigma }_j⟩=m_j$. Therefore the probability of spin up along $\mathbf{n}$ is $(1+\mathbf{n}\cdot \mathbf{m})/2$.

Use a common invariant operator domain on which the stated differentiations and commutator are valid. For a differentiable normalized eigenfunction, the Hellmann-Feynman calculation and self-adjointness giveThe Lax equation preserves the spectrum. Differentiating $Lf=\lambda f$ now gives $(L-\lambda )(f_t+Af)=0$, so $f_t+Af$ is in the same eigenspace, or is zero. For a nondegenerate eigenvalue it equals $c(t,\lambda )f$. Multiplying $f$ by $\exp [-\int c(t,\lambda )dt]$ produces $\hat{f}$ with ${\hat{f}}_t+A\hat{f}=0$ while preserving its eigenvalue.

Write $b=u_x+a_0$ and $d=u-\lambda +a_1$. The spatial eigenvalue equation gives ${\hat{f}}_{xx}=(u-\lambda )\hat{f}$, so $A\hat{f}=b\hat{f}+d{\hat{f}}_x$. For $F=(\hat{f},{\hat{f}}_x{)}^T$ the two first-order systems are thereforeThe second row of $V$ follows by differentiating the first time equation in $x$ and using the spatial equation once more. Conversely these matrix equations imply both scalar equations, so the formulations are equivalent.