For $B_{\varphi }\propto r^p$, constant density and Keplerian ${\kappa }^2={\Omega }^2$, the Michael criterion for axisymmetric toroidal-field interchange becomes ${\Omega }^2-2(p-1)v_A^2/r^2$. Instability therefore requires $p>1$ and $v_A^2>r^2{\Omega }^2/[2(p-1)]$. For a field varying on the radial scale $r$, this is an orbital-speed field, much stronger than a gas-pressure-scale field with $v_A\lesssim c_s\sim H\Omega$ in a thin disk. Steep radial field gradients require the exact derivative criterion rather than this scale estimate. The instability is powered by the toroidal field distribution and is distinct from weak vertical-field magnetorotational instability.

// Target: fluid-mechanics.bigb

A Metropolis-within-Gibbs sweep updates one coordinate or block at a time by a Metropolis–Hastings algorithm transition whose invariant law is that block's full conditional distribution. For a proposal changing only block $j$, the acceptance ratio uses the full joint target with the other coordinates fixed, together with the reverse/forward proposal ratio. Each block kernel preserves the joint posterior, so their composition does as well. Conjugate blocks can instead use exact Gibbs sampling draws. Systematic composition need not be reversible, even when every block kernel is reversible.

For the unit ball of $C^p({\mathbb{T}}^n)$ measured in the uniform metric, $c{\epsilon }^{-n/p}\le H(\epsilon )\le C{\epsilon }^{-n/p}\log (1/\epsilon )$. The lower bound uses disjoint bump functions of spatial scale $h$, height proportional to $h^p$ and independently chosen zero/one coefficients: there are $\exp (c{h}^{-n})$ functions separated at scale $h^p$. For the upper bound, Multivariable Jackson approximation gives a trigonometric polynomial of degree $O({\epsilon }^{-1/p})$. Its $O({\epsilon }^{-n/p})$ coefficients are bounded and can be rounded with coefficient accuracy proportional to $\epsilon$ divided by the coefficient count. Counting those rounded coefficient arrays supplies the logarithmic factor. The continuous-function unit ball without derivative bounds has infinite entropy for $\epsilon <1/2$, by an infinite sequence of disjoint height-one bumps.

Suppose outer functions in a family $B$ obey $|f(x)-f(y)|\le L{\max }_j|x_j-y_j|$, and compose them with $m$ inner functions drawn from $E$. Approximate the outer function within $\epsilon /2$ and each inner function within $\epsilon /(2L)$. The triangle inequality bounds the composed error by $\epsilon$. Multiplying the numbers of possible net choices and taking logarithms gives the stated inequality. A uniform bound on first partial derivatives supplies $L=m$ for the $C^p$ unit ball, so the exponent of a common entropy bound for the outer and inner classes is retained under function composition.

For a constant-coefficient linear differential equation $P(D)y=e^{rx}$, where $D$ denotes differentiation, a simple root $P(r)=0$ forces a resonant trial. Factor $P(z)=(z-r)Q(z)$; since $(D-r)(x{e}^{rx})=e^{rx}$, commuting the differential operators gives $P(D)(x{e}^{rx})=Q(r)e^{rx}=P^{\prime }(r)e^{rx}$. Thus the displayed expression is a particular solution. More generally, if the root has multiplicity $m$, the same factorization gives $P(D)(x^m{e}^{rx})=P^{(m)}(r)e^{rx}$, so $y_p=x^m{e}^{rx}/P^{(m)}(r)$. This is the extra polynomial factor required by resonance in a differential equation.

For $y^{\prime \prime }+ϵ(y^{\prime }{)}^n+y=0$, the method of multiple scales gives no first-order phase drift. If $n$ is odd, the amplitude obeys the displayed damping law with $C_n=2^{-(n+1)}\left(\genfrac{}{}{0ex}{}{n+1}{(n+1)/2}\right)$. If $n$ is even, the average is zero and the amplitude is unchanged on the first slow time $T=ϵt$. The odd case dissipates mechanical energy, whereas the even case is reversible and its force is not damping of both velocity signs.

For $x^{\prime \prime }+3ϵy{x}^{\prime }+x=2(y^{\prime }{)}^2$ and $y^{\prime }=ϵ(1+x-y)$ with leading oscillation $x=A(\tau )\sin t$, $\tau =ϵt$, averaging the auxiliary equation gives $Y^{\prime }=1-Y$. Cancellation of resonant forcing in the oscillator gives $2A^{\prime }+3YA=0$. With $Y(0)=3$ and $A(0)=1$, $Y=1+2e^{-\tau }$ and the displayed amplitude follows. Bounded oscillatory corrections repair the instantaneous auxiliary equation; secular terms in a fixed-time expansion show why the slow scale is needed.

For homogeneous Dirichlet endpoints, the centered first-difference matrix is skew-Hermitian and the centered second-difference matrix is negative definite. Thus $\dot{U}=(D_{xx}-\alpha D_x)U$ is contractive in the discrete L2 norm for every real $\alpha$ and every mesh width. The displayed identity follows by summation by parts. Nonnegative off-diagonal entries additionally require $|\alpha |h\le 2$; that maximum-principle condition is not necessary for the L2 energy estimate. numerical consistency and Duhamel's formula then give second-order numerical convergence for smooth solutions and compatible initial approximations.

For periodic indexing let $S$ be the unitary forward shift. The operator is $A=-3I/2+2S-S^2/2$. A discrete Fourier transform diagonalizes it with eigenvalue $\lambda (\theta )=-3/2+2e^{i\theta }-e^{2i\theta }/2$ and real part $-(1-\cos \theta {)}^2$. Consequently $\Vert e^{tA/h}{\Vert }_2\le 1$ for every $t\ge 0$, uniformly in the grid. Equivalently $A+A^{\ast }=-\frac{1}{2}(2I-S-S^{\ast }{)}^2$, proving decay of the discrete energy directly. The result is semidiscrete stability; a further time discretization must also have adequate absolute stability.

If initial data are parameterized by $s$ and their characteristic invariant is $D(s)$, a solution constant on characteristics requires $f(s)=F(D(s))$. At a fold $D^{\prime }(s_0)=0$, equal invariant values must have equal data, and $f^{\prime }(s)/D^{\prime }(s)$ must extend continuously if $F$ is continuously differentiable. The condition $f^{\prime }(s_0)=0$ alone is insufficient: for $D(s)=1/4-(s-1/2{)}^2$, the data $|s-1/2{|}^{3/2}$ would require a derivative of $F$ blowing up at $1/4$.

For $x{f}_x+\beta y{f}_y=\gamma f$, the curves in the method of characteristics are $x=x_0{e}^s$, $y=y_0{e}^{\beta s}$ and $f=f_0{e}^{\gamma s}$. On either component $x\ne 0$, the invariant is $y|x{|}^{-\beta }$, giving $f=|x{|}^{\gamma }F(y|x{|}^{-\beta })$. A coordinate change that includes $y/x$ may be singular on an axis, so smooth continuation there must be checked against the original partial differential equation.

Dependence of a system's observed state on the path by which its control parameters were reached. In a first-order phase transition, a metastable local minimum may persist after another phase becomes globally stable, so increasing and decreasing the control parameter give different switching points. Nucleation and sweep rate determine actual switching; a mean-field spinodal point is a local-stability limit, not the equilibrium phase coexistence condition.

Nucleation forms a sufficiently large region of a new phase within an old metastable phase. In classical nucleation theory, positive surface energy competes with bulk free-energy gain, producing a critical nucleus and a barrier to conversion. The Arrhenius nucleation time depends exponentially on this barrier; boundaries and impurities can alter it. This explains how metastability and hysteresis can persist despite a different globally stable phase.

The fraction of total mass in chemical elements heavier than helium, so the hydrogen mass fraction, helium mass fraction and metal mass fraction satisfy $X+Y+Z=1$. Despite the astronomical name, this includes elements such as carbon and oxygen that are not chemical metals.

For steady stellar meridional circulation, $T\nabla s=\nabla (h+\varphi )$ and $\nabla \cdot (\rho \mathbf{v})=0$. Thus $\rho \mathbf{v}\cdot T\nabla s=\nabla \cdot [\rho \mathbf{v}(h+\varphi )]$. In the leading uniformly rotating stellar hydrostatic equilibrium, $h+\varphi$ is constant on an equipotential surface; its total advected flux vanishes because the net mass flux vanishes. Integrating the energy equation then identifies the outward luminosity with enclosed nuclear energy generation, even though local heat advection is nonzero.

The thermal imbalance of a slowly rotating radiative star can drive stellar meridional circulation. Its fractional centrifugal distortion is ${ϵ}_{\Omega }\sim {\Omega }^2{R}^3/(GM)$, so balancing an order-${ϵ}_{\Omega }$ fraction of the luminosity by thermal advection estimates $t_{\mathrm{ES}}\sim t_{\mathrm{KH}}/{ϵ}_{\Omega }$, where $t_{\mathrm{KH}}\sim G{M}^2/(RL)$. Stable composition gradients can further impede the circulation. This radiative thermal estimate is different from rapid torque-driven circulation in a convective envelope.

To isolate finite size from quantum vortex penetration, keep the same weak-field London equations but replace the half-space by an infinite slab $-d/2\le z\le d/2$. Put equal tangential magnetic fields $B_0\hat{\mathbf{x}}$ at its two faces. With $z=0$ now at the slab center, the even solution and its signed current density areThe hyperbolic cosine profile has equal boundary values, while the electric currents on the two sides have opposite signs. In particular,For the stipulated thickness $d=2{\lambda }_L$, the center retains about $0.648B_0$ and the mean is about $0.762B_0$, even though this is a Meissner state. For $d\gg {\lambda }_L$, the center magnetic field is exponentially small and the mean is about $2{\lambda }_L{B}_0/d$, reflecting the two surface layers. For $d\ll {\lambda }_L$, the mean tends to $B_0$: little screening is possible across such a short distance. These are analytic model examples, not material measurements. Real finite objects also need their demagnetizing field and edge geometry accounted for; the slab calculation only isolates the effect of thickness.

Take a homogeneous material exhibiting superconductivity in the half-space $z\ge 0$, bounded by vacuum at $z=0$. Fix the temperature below the superconducting transition temperature, take a constant positive London penetration depth ${\lambda }_L$, and assume a local, linear London equation description in a Meissner state. The applied magnetic field is weak enough that it neither nucleates a quantum vortex nor appreciably suppresses the material's superconductivity. Assume translation invariance in $x,y$, no imposed transport electric current, and a tangential boundary magnetic field $\mathbf{B}(0)=B_0\hat{\mathbf{x}}$. Here $B_0$ is the local surface magnetic flux density, equal to the imposed vacuum magnetic field in this infinite planar geometry; it is not automatically the far-away applied value for a finite object.

For a bounded, decaying interior solution, the London equations, magnetostatic Ampère law and Gauss's law for magnetism giveThe solution isAt one London penetration depth the magnetic field is $e^{-1}\simeq 0.368$ of its surface value; at five it is $e^{-5}\simeq 0.00674$. There is no finite depth where this exponential is exactly zero. The differential equation excludes the growing exponential through the deep-interior boundary condition. A constant normal component cannot decay in this geometry: Gauss's law for magnetism would require ${\partial }_z{B}_z=0$. A perpendicular applied magnetic field on a finite plate therefore requires a different spatial problem.^[ref]

On a square grid, let $C_h$ sum the four axial neighbors and $K_h$ the four diagonal neighbors. With $M_h=2I/3+C_h/12$ and $L_h=(-10I/3+2C_h/3+K_h/6)/h^2$, the method of lines has fourth-order normalized spatial numerical consistency. Its mass symbol is $(4+\cos \xi +\cos \eta )/6\ge 1/3$, and its diffusion symbol is $[-10+4(\cos \xi +\cos \eta )+2\cos \xi \cos \eta ]/(3h^2)\le 0$. Every continuous-time mode therefore contracts, and Parseval identity gives mesh-independent discrete L2 stability. No time-step restriction is involved until a time integrator is selected.

A graph embedding into an ordered tournament is half-sparse when the displayed inequality holds for every proper final interval $I$. Reserving two extra tournament vertices allows extension at any outward-oriented leaf: the median order of a tournament supplies at least half of the later vertices as outneighbours of the parent's image, while half-sparsity leaves at least one of them unused. Adding one image while extending every old final interval by two preserves the inequality.