A measurable set $E$ has finite perimeter in $\Omega$ when ${\chi }_E$ belongs to the BV space. Its relative perimeter is $\mathrm{Per}(E;\Omega )=|D{\chi }_E|(\Omega )$. Relative perimeter excludes the exterior image signal boundary and measures the interior interface in Caccioppoli partitions.

With constant volume flux per unit width $Hu=q_0>0$, define $\beta =l/\lambda$, $\mathcal{H}=H/\beta$, and $\mathcal{Q}=\nu q_0/(g{\beta }^3)$. The friction closure gives $H_x=\alpha -q_0\nu /(\beta g{H}^2)$. Differentiating constant flux and applying the unbuttressed Newtonian grounding-line stress condition gives $q_0{H}_x/H+g^{\prime }{H}^2/(8\nu )=0$. Substitution proves the displayed relation. For $\alpha \ge 0$, $g^{\prime }>0$ and $\mathcal{Q}>0$, the polynomial $(g^{\prime }/8g){\mathcal{H}}^5+\alpha \mathcal{Q}{\mathcal{H}}^2-{\mathcal{Q}}^2$ increases strictly on the positive half-line, begins negative and tends to infinity; hence it has exactly one positive root. Its root satisfies $\mathcal{Q}>\alpha {\mathcal{H}}^2$, so the thickness decreases seaward at that steady grounding line.

In ZFC, the Freiling axiom of symmetry is equivalent to failure of the Continuum hypothesis. Under the hypothesis, countable initial segments of a well-order of the reals violate symmetry. When the continuum exceeds ${\aleph }_1$, the union of the countable values on an ${\aleph }_1$-sized set leaves a real outside; avoiding that real's countable value gives the two witnesses.

A free product $A\ast B$ is nontrivial as a splitting when both factors differ from the trivial group. Alternating words $(ab{)}^r$ with $a,b\ne 1$ in different factors show that every such splitting gives an infinite group. The distinct normal forms $ab$ and $ba$ also show that the group is not an abelian group.

Given group homomorphisms $f_A:A\to H$ and $f_B:B\to H$, there is exactly one group homomorphism $A\ast B\to H$ extending both. Its value on an alternating normal form is the product of the factor-map values. Thus the free product is the coproduct in the category of groups.

Let $D_h$ be the scaled Dirichlet discrete Laplacian and $V_h$ a bounded diagonal reaction matrix. If $0\le k/h_x^2\le 1/2$, the heat update $H_h=I+k{D}_h$ is contractive in both the maximum norm and the mesh-weighted L2 norm. For $\Vert V_h\Vert \le A$,This proves finite-time mesh-uniform stability without requiring the full update to have nonnegative entries or to be contractive. Negative reaction terms can destroy nonnegativity at the endpoint $k/h_x^2=1/2$, so it is the heat part alone that is treated as a nonnegative contraction.

For each formula, its forcing relation on names and conditions is uniformly first-order definable inside the ground model. This permits ground-model separation and replacement using forcing predicates.

A formula holds in a generic extension exactly when some condition in the generic filter forces it for names of its parameters.

A generic extension of a transitive model has exactly the same ordinals as its ground model. Rank induction on names bounds every value; transitivity and agreement on membership make ordinalhood absolute.

For a name with finitely many possible values, record which value each condition forces, using an extra marker for undecided conditions. Equal patterns mean that any condition deciding one entry decides all entries with that pattern identically.

If the prescribed gas velocity is itself a free epicyclic motion solution, subtract its two momentum equations from the forced dust epicycle with aerodynamic drag equations. The velocity difference obeys a damped epicycle and decays as $e^{-ϵ\Omega t}$. Therefore dust approaches the gas velocity with identical phase and amplitude for every positive drag rate. At zero drag the difference does not decay, so the late-time limit is nonuniform as drag vanishes. A finite-wavelength inertial-acoustic wave is detuned from the free epicycle; matching its phase and amplitude in the weak-drag limit additionally requires $|\omega -\Omega |\ll ϵ\Omega$. For a gas density wave of frequency $\omega$, the exact radial transfer amplitude isobtained by solving the two dust equations and using $U_y=-i\Omega U_x/(2\omega )$. At $\omega =\Omega$ it is exactly one.

A force-free, torque-free straight rod contributes a particle stresslet tensor $(C{L}^3/3)(\mathbf{p}\cdot E\mathbf{p})\mathbf{pp}$. Its extra viscous dissipation is $(C{L}^3/3)(\mathbf{p}\cdot E\mathbf{p}{)}^2$. In a simple shear flow, $\dot{\theta }=-\gamma {\sin }^2\theta$; the dissipation is greatest along the extensional or compressional axes and vanishes at flow and gradient alignment.

For a flux tube with $B_{\varphi }=\gamma (s/s_0)B_z$, regularity at the axis and cylindrical magnetostatic pressure balance give $B_z(0{)}^2=B_p^2-2{\mu }_{0}p(0)+{\gamma }^2⟨B_z^2⟩$. The flux-tube axial field virial identity and $p\ge 0$ give the bound. A twisted toroidal magnetic field can support a central axial field above the untwisted magnetic pressure limit even though its cross-sectional mean obeys that limit.

For symmetric finite-mass Marangoni spreading on a liquid film, the leading outer profiles are $h=2h_0|x|/x_N$ and $C=(M/x_N)(1-|x|/x_N)$ inside $|x|<x_N$, where $x_N=(6AM{h}_{0}t/\mu {)}^{1/3}$. Outside, $h=h_0$ and $C=0$. The similarity solution follows from $H{\Gamma }^{\prime }=-\eta /3$, $\eta H^{\prime }=H$, $2\int \Gamma d\eta =1$ and $\int (H-1)d\eta =0$. The film reaches height $2h_0$ immediately behind the front, leaving an outer height jump that gravity or surface tension must smooth. These are ideal leading outer profiles, not a description of the microscopic central dry region or the smoothed edge.

The height jump in linear similarity profiles for surfactant spreading is smoothed by surface tension. If $C_x\sim M/x_N^2$ through a region of width $\Delta$, the Marangoni stress flux is $AM{h}_0^2/(\mu x_N^2)$ and the capillary flux is ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$. Their balance gives $\Delta \sim ({\gamma }_0{h}_0^2{x}_N^2/(AM){)}^{1/3}\propto t^{2/9}$. This estimate assumes gravity is negligible and the smoothing region remains narrow compared with the pool.

When the smoothing width exceeds the capillary length, gravity dominates capillarity. Balancing hydrostatic flux $\rho g{h}_0^4/(\mu \Delta )$ against the Marangoni stress flux gives $\Delta \sim \rho g{h}_0^2{x}_N^2/(AM)\propto t^{2/3}$. The condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$ in physical units. The smoothing region catches up with the spreading pool at $x_N\sim AM/(\rho g{h}_0^2)$ and $t\sim \mu (AM{)}^2/[(\rho g{)}^3{h}_0^7]$, after which the sharp-front outer similarity solution ceases to apply.

The E-step computes mixture responsibilities ${\tau }_{ij}$ from the old parameters. Put $N_j={\sum }_i{\tau }_{ij}$. The M-step updates ${\pi }_j=N_j/n$, ${\mu }_j={\sum }_i{\tau }_{ij}{y}_i/N_j$ and ${\sigma }^2={\sum }_{i,j}{\tau }_{ij}(y_i-{\mu }_j^{\mathrm{new}}{)}^2/n$. The variance uses new means and old responsibilities. EM likelihood monotonicity guarantees nondecrease of the observed likelihood, not a global optimum.

Residual mixture clustering first removes a fitted covariate-dependent mean and then explores latent groups in the regression residuals. It seeks differences relative to the adjustment rather than raw-response differences. Mixture responsibilities express uncertain membership. The fitted mixture alone does not establish distinct real-world populations, and two-stage residual mixture fitting ignores some adjustment uncertainty.

A mixture regression with shared slopes models $Y_i\mid Z_i=j,x_i\sim N({\alpha }_j+x_i^T\beta ,{\sigma }^2)$ with common slopes and component-specific intercepts. It jointly estimates mean adjustment and latent groups, rather than fitting a mixture to fixed residuals. If both a global intercept and component offsets are included, an identifying constraint is required. Constant weights assume latent membership proportions do not depend on the covariates.

The fast-rate Lasso prediction bound under compatibility implies $\Vert {\hat{\beta }}_S-{\beta }_S^0{\Vert }_1\le (1+c)\lambda s/{\varphi }^2$. Each true coefficient larger in absolute value than this bound has the correct estimated sign. This conclusion alone does not ensure absence of false positives.