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A regular hyperbolic octagon with opposite sides paired gives a genus-two surface when each angle is $\pi /4$. The eight vertices then form one smooth quotient point because their angles sum to $2\pi$.

Identifying the $2n$ sides of one polygon in pairs gives a cell structure with one face, $n$ edges and some number $V$ of vertex classes. For a closed orientable surface of genus $g$,Since $V\ge 1$, every such schema satisfies $n\ge 2g$.

For a complex sesquilinear form $B$, the values $B(x,x)$ determine $B(x,y)$. In particular, expanding at $x+y$ and $x+iy$ shows that $B(x,x)=0$ for every $x$ implies $B=0$. Over the reals this conclusion can fail for nonsymmetric bilinear forms, as a nonzero skew-symmetric form has zero diagonal.

For an M-M-infinity queue with empty initial state, Poisson arrivals of rate $\lambda$ and exponential lifetimes of rate $\mu$ produce a Poisson occupancy of mean $\lambda (1-e^{-\mu t})/\mu$ at time $t$.

Conditioned on a total of $n$ Poisson points, the number falling in a region of relative intensity $p$ is $\mathrm{Binomial}(n,p)$.

The Poisson integral convolves boundary data with the Poisson kernel to produce a harmonic function in a half-space. In the upper half-plane,

For testing $H_0:\theta =1$ against a larger alternative, approximate size-$0.05$ upper-tail tests reject whenor, respectively,

For $S_r={\sum }_i{x}_i^r$, unweighted least squares through the origin and maximum likelihood giveBoth estimators are unbiased.

For sufficiently smooth curves, $Q(\gamma ,s)={\kappa }^{\prime }(s)$ obeys Euclidean and parameter-translation covariance but is not determined by the pair $(\kappa (s),\tau (s))$. Under the bare covariance definition, nonlocal examples such as $Q(\gamma ,s)=\kappa (s+1)$ work as well.

Projecting a sign-corrected displacement map back to a square turns Brouwer's fixed point into a zero of the displacement, proving that paths joining opposite side pairs intersect.

If $U$ is bounded and connected and a boundary portion ${\Gamma }_D$ has positive surface measure, thenfor every $v\in H^1(U)$ whose trace vanishes on ${\Gamma }_D$. Otherwise a normalized counterexample sequence, the Rellich-Kondrashov compactness theorem, and continuity of the trace would produce a nonzero constant with zero trace on ${\Gamma }_D$.

On a bounded connected Lipschitz domain $U$, every $u\in H^1(U)$ satisfiesEquivalently, the gradient norm is equivalent to the $H^1$ norm on the mean-zero Sobolev space.

For a rectangle in a planar lattice, exactly one of an open primal left-to-right crossing and a closed dual top-to-bottom crossing occurs.

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

For an effective enumeration of computably enumerable languages $W_v$, the totality problem isIt is ${\Pi }_2$-complete: a ${\Pi }_2$ condition $w\in P⟺\forall xf(w,x)\downarrow$ maps effectively to a program whose domain is $\{x:f(w,x)\downarrow \}$.

In Newtonian gauge, ${\Theta }_0+\Psi$ combines the intrinsic photon temperature perturbation with the gravitational redshift at emission. With constant potentials and negligible baryon loading it oscillates as $\cos (k{r}_s)$ for adiabatic initial conditions.

The Debye model replaces the crystal's vibrational spectrum by a continuum of phonon modes up to a cutoff chosen to preserve the total number of vibrational degrees of freedom.

A phonon dispersion relation gives its angular frequency $\omega$ as a function of its wavevector $k$. For an isotropic power law, $\omega =C|k{|}^{\alpha }$.

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

At a first-order phase transition, the Gibbs free energy is continuous while at least one first derivative, such as entropy or volume, jumps. A nonzero entropy jump produces latent heat $L=T\Delta S$.