A codimension-two bifurcation requires two independent parameter conditions to hold simultaneously. Simultaneous stationary and Hopf bifurcation thresholds require both kinds of critical mode in the reduced amplitude equation. A crossing of modes at different wavenumbers is different from a zero-frequency double-zero degeneration of one fixed-wave-number cubic.

A supercritical branch emerges on the parameter side where the original state loses stability. In the cubic Landau amplitude equation $\dot{A}=\mu A-g|A{|}^{2}A$ with $g>0$, the nonzero branch $|A{|}^2=\mu /g$ lies at $\mu >0$ and is stable to radial amplitude perturbations. Continuous phase symmetry still leaves a neutral phase direction. The direction of a branch and its stability should both be checked in a general problem of bifurcation theory.

A subcritical branch emerges on the parameter side where the original state remains stable. In the cubic Landau amplitude equation $\dot{A}=\mu A-g|A{|}^{2}A$ with $g<0$, the small nonzero branch lies at $\mu <0$ and is unstable to radial amplitude perturbations. Higher-order saturation can generate a finite-amplitude stable branch and hysteresis, but this does not follow from a cubic truncation alone.

The Hankel functions are the combinations $H_{\nu }^{(1)}=J_{\nu }+i{Y}_{\nu }$ and $H_{\nu }^{(2)}=J_{\nu }-i{Y}_{\nu }$ of the Bessel function of the first kind and Bessel function of the second kind. They form a standard independent pair for the Bessel differential equation.

For real $k,r>0$ and time dependence $e^{-i\omega t}$, their leading radial factors are proportional to $e^{ikr}$ and $e^{-ikr}$. Constant wave phase $kr-\omega t$ moves outward with increasing time, whereas $-kr-\omega t$ moves inward. Reversing the time-sign convention reverses these labels. This makes the combinations useful when imposing a Sommerfeld radiation condition, rather than regularity at the origin.^[ref]

For complex arguments, the Hankel asymptotic sectors and the chosen branch of a multivalued function are part of the approximation. A formula continued around zero must not be treated as one global exponential on every argument sheet.

A Dulac function is a multiplier $B(x,y)$ chosen so that the rescaled planar vector field $BF$ has a divergence of one sign. The Bendixson-Dulac theorem then excludes periodic orbits in a suitable simply connected region on which the multiplier is continuously differentiable.

For example, considerwith $b>0$, on the positive quadrant. Choosing $B=1/(xy)$ givesNo periodic orbit can stay wholly in the positive quadrant. By contrast, setting $b=0$ makes this divergence zero, so this multiplier no longer excludes the closed-orbit families of the Lotka-Volterra equations. The axes are outside the multiplier's regular domain, so the argument is explicitly restricted to positive populations.

For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

For iid $N(\mu ,v)$ observations, the prior density $1/v$ relative to $d\mu dv$ gives posterior kernel $v^{-n/2-1}\exp [-\{S+n(\mu -\bar{x}{)}^2\}/(2v)]$. It is proper exactly when $n>1$ and $S=\sum (x_i-\bar{x}{)}^2>0$. The conditionals are $\mu \mid v,x\sim N(\bar{x},v/n)$ and $v\mid \mu ,x\sim \mathrm{IG}(n/2,{\sum }_i(x_i-\mu {)}^2/2)$. Integrating out $\mu$ changes the inverse-gamma shape to $(n-1)/2$.

The ${\Lambda }^0$ in the baryon octet is an isospin singlet with strangeness $-1$ and zero electric charge. Its ordinary valence quark content is $uds$.

The baryon octet contains an isospin triplet ${\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-}$. Each has strangeness $-1$; their valence quark contents are $uus$, $uds$, $dds$.

The baryon octet contains the isospin doublet ${\Xi }^0,{\Xi }^{-}$. Each has strangeness $-2$, with valence quark content $uss$, $dss$.

The invariant colour state of three fundamental quarks is the normalized completely antisymmetric tensor ${ϵ}_{abc}/\sqrt{6}$. With this colour factor, the Pauli exclusion principle requires the remaining spatial-spin-flavour wavefunction to be symmetric. Colour and approximate flavour symmetry are distinct SU(3) actions.

In a three-quark colour singlet, a symmetric spatial state requires symmetric spin-flavour symmetry. Completely symmetric decuplet flavour couples to symmetric spin $3/2$; mixed octet flavour couples to mixed spin $1/2$ in the symmetric permutation combination. Antisymmetric singlet flavour cannot have a symmetric S-wave orbital state, since ${\bigwedge }^3{\mathbb{C}}^2=0$ forbids a completely antisymmetric state of three spin-$1/2$ quarks. Mixed orbital and spin states can instead realize the required antisymmetry.

A baryon with constituent content four quarks and one antiquark, whose net baryon number is one. Such content admits more flavour representations than three quarks alone. A representation in the tensor product is not evidence by itself of a bound state.

For a barotropic fluid, an energy per mass $e$ can be defined by $de/d\rho =p/{\rho }^2$. For the isothermal equation of state, $e=c_s^2\ln (\rho /{\rho }_0)$. Adding kinetic, gravitational and magnetic energy gives the ideal magnetic energy flux $\mathbf{u}(E+p+B^2/(2{\mu }_0))-\mathbf{B}(\mathbf{u}\cdot \mathbf{B})/{\mu }_0$. This is a mechanical barotropic energy identity; it does not say that an isothermal gas is thermally isolated.

For a barotropic fluid with pressure $P=P(\rho )$, choose an energy density $U$ satisfying $\rho U^{\prime }(\rho )-U=P$. The continuity equation then gives ${\partial }_{t}U+\nabla \cdot (U\mathbf{u})=-P\nabla \cdot \mathbf{u}$. For an isothermal equation of state $P=c_s^2\rho$, one choice is $U=c_s^2\rho \ln (\rho /{\rho }_{\mathrm{ref}})$. A fixed reference density changes $U$ only by a conserved mass term. The same construction holds with surface density and vertically integrated pressure. This closure energy need not equal the microscopic internal energy of a gas maintained at fixed temperature.

For angular velocity depending only on cylindrical radius, the centrifugal acceleration is a gradient. Steady barotropic balance is $\nabla p/\rho =-\nabla \Psi$, hence specific enthalpy satisfies $h(\rho )+\Psi =\text{constant}$. On an invertible equation-of-state branch, mass density and pressure are functions of $\Psi$. Cylindrical rotation is essential to this scalar centrifugal-potential representation.

Suppose $a,b,L_0>0$, $L(x)$ is smooth and positive near zero with $L(0)=L_0$, and $d=O(\eta )$ as $\eta \to 0^{+}$. The leading contribution tois obtained by scaling $x=\eta s$ and replacing $L$ by $L_0$. Integration of the resulting reciprocal quadratic gives the displayed formula. For constant $L$ the exact expression isThis proves the half-peak value at $d=0$ and the transition between an interior peak and a pole outside the interval. If $L$ varies, smoothness and positivity on the contributing neighborhood justify its leading replacement; other zeros or singular contributions elsewhere must be excluded.

For a polynomial depending on a small parameter, substitute $t={ϵ}^{p}s$ and compare the exponents of the parameter. Candidate root scales arise when two or more lowest exponents balance. Solve the leading scaled equation and expand about each of its nonzero roots. Different balances can reveal different families of roots; real-root counts must be checked separately.

Optimal truncation chooses how many terms to retain for a useful finite approximation. In a divergent asymptotic expansion, increasing the truncation order eventually makes the terms larger rather than smaller. Stopping near a smallest term often gives the best direct approximation, but the actual remainder must be checked. A small last term alone is not a universal error bound.^[ref]

For a divergent factorial series can approximate a convergent integral, $T_k=k!/z^k$ decreases while $k+1<z$ and increases while $k+1>z$. Thus the useful cutoff is near $k=z$, rather than at arbitrarily large order. At $z=10$, $T_9=T_{10}=0.00036288$ are the two smallest terms. Retaining $N=10$ terms givesThe actual error is about $0.0001770$. Retaining twenty terms instead gives $S_{20}(10)\approx 0.9076137084$, with error about $0.008020$: more work produces a worse answer.

For this example the bound $N!/z^N$ is explicit, so a requested tolerance can be tested before using a truncation. For a different integral, a nearby saddle point or simple pole can change the relevant remainder scale. This is why the smallest-term rule is a useful guide, rather than a replacement for remainder analysis.

The cutoff for factorial terms $n!z^{-n}$ has $n=|z|+O(1)$. Better analysis can expose contributions smaller than every retained inverse power, linking optimal truncation to the Stokes phenomenon. Such contributions can be important even when many asymptotic coefficients agree. Directly summing ever more terms does not automatically recover them.

For the formal power series in inverse powers ${\sum }_{n\ge 0}{a}_n{z}^{-n-1}$, its Borel transform isTo define a Borel summation along the positive real ray, this series must first converge near zero, admit a suitable analytic continuation along that ray, and have at most exponential growth there. If $|\mathcal{B}(t)|\le C{e}^{At}$, the Laplace transformconverges absolutely for $\mathrm{Re}z>A$. Local expansion near zero, the Gamma integral and a controlled tail then recover the original asymptotic expansion. Dividing coefficients by factorials alone does not establish these continuation and growth conditions.

For example, $a_n=(-1{)}^{n}n!$ gives $\mathcal{B}(t)=1/(1+t)$. The divergent factorial series consequently has the well-defined sum ${\int }_0^{\infty }{e}^{-zt}/(1+t)dt$ for $z>0$. Scaling $t=s/z$ shows that this equals $F(z)/z$ for a divergent factorial series can approximate a convergent integral; the finite geometric remainder there proves its asymptotic expansion directly.

A singularity on the chosen ray requires a contour prescription. Passing it on opposite sides can change the sum by an exponentially small residue contribution, as in Borel integral with a simple pole. That ambiguity links Borel summation to the Stokes phenomenon.