An anti-dynamo theorem gives a geometric or analytic restriction excluding self-sustained dynamo action under stated magnetic diffusivity, domain and boundary conditions. It excludes regeneration against resistive decay, not all transient amplification or field induced by an external source. Important restrictions include Cowling anti-dynamo theorem, the toroidal-velocity anti-dynamo theorem, and the planar anti-dynamo theorem. Backus' necessary condition for dynamo action instead bounds the stretching required for possible regeneration.

An axisymmetric magnetic mode is unchanged by axial rotations, while an equatorial magnetic mode and a corresponding velocity mode transform as $A,V\mapsto e^{i\theta }(A,V)$. Magnetic reversal sends $(B,A,V)$ to $(-B,-A,V)$. For magnetic evolution linear in magnetic amplitudes, rotational equivariance requires $\dot{B}=Bf(|V{|}^2)+2\mathrm{Re}[A^{\ast }Vg(|V{|}^2)]$ and $\dot{A}=Ah(|V{|}^2)+BVj(|V{|}^2)+A^{\ast }{V}^2\ell (|V{|}^2)$, with $f$ real and the other functions complex. Velocity monomials $B^p{A}^q(A^{\ast }{)}^r{V}^s(V^{\ast }{)}^t$ must have even $p+q+r$ and weight $q-r+s-t=1$. These conditions enumerate the cubic normal form without assuming reflection symmetry or nonlinear magnetic self-interactions.

Dynamo quenching is the reduction or modification of magnetic-field generation by magnetic feedback on the generating flow and its correlations. The Lorentz force can change shear or turbulence, while magnetic helicity affects mean-field regeneration. A phenomenological scalar example is $\alpha (B)={\alpha }_0/(1+B^2/B_{\ast }^2)$. In the illustrative amplitude equation $\dot{B}=[c\alpha (B)-d]B$ with $c,d>0$, a growing field saturates at $B^2=B_{\ast }^2(c{\alpha }_0/d-1)$ when $c{\alpha }_0>d$. This demonstrates the negative feedback mechanism without claiming that algebraic quenching is a complete solar closure.

Let a solenoidal flow be confined to a sphere $V$, with the insulating boundary condition for the radial magnetic scalar. Define $N={\int }_V{P}^2$, $D={\int }_{{\mathbb{R}}^3}|\nabla P{|}^2$, $M={\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^2$ and $q={\max }_V|\mathbf{u}\cdot \mathbf{x}|$. Integration by parts inside and outside the sphere givesThe Cauchy-Schwarz inequality therefore makes $q^2\ge {\eta }^{2}D/M$ necessary whenever $N^{\prime }\ge 0$ and $D>0$. A uniform strict violation with a positive gap forces decay, using a Sobolev inequality to bound $N$ by a constant times $D$. A bounded statistically steady field instead requires $q_{\ast }^2\ge {\eta }^2⟨D⟩/⟨M⟩$, with $q_{\ast }={\sup }_{t}q(t)$. This is a necessary condition for dynamo action, not a sufficient criterion or a pointwise assertion at every instant of an arbitrary time-dependent solution.

// Target: analysis.bigb

For a smooth heat equation solution $w_t=\Delta w$ with $w_t+\alpha {\partial }_{n}w=0$ on the boundary and nonnegative $\alpha$, Green's first identity gives the displayed dissipation identity for $E={\int }_V|\nabla w{|}^2$. The outward normal derivative is used. Writing the boundary loss as $\alpha ({\partial }_{n}w{)}^2$ avoids division by $\alpha$ and remains valid where $\alpha =0$. A stationary solution is harmonic with $\alpha {\partial }_{n}w=0$; it need not be a particular fixed-data Dirichlet problem solution.

Assume $a=c/r$ decreases outward, tends to zero at $R$, and define $h(a)=-d\ln r/da$. Differentiation of the Duvall law gives $F^{\prime }(w)={\int }_0^{w}h(b)b/[w^2\sqrt{w^2-b^2}]db$. Interchanging integrals and using ${\int }_b^{a}dw/[w\sqrt{w^2-b^2}\sqrt{a^2-w^2}]=\pi /(2ab)$ proves the displayed Abel transform inversion. For a finite outer reference value $w_s$, both lower limits become $w_s$, after removing the known outer-layer contribution to the travel time. Inversion requires a monotone branch and amplifies errors in differentiated frequency data.

For saturated Darcy flow through a long dam occupying a channel of width $\alpha z^2$, let hydraulic conductivity be $K=k\rho g/\mu$ and free-surface height $h(x)$. The Dupuit approximation gives $Q=-K\alpha h^3{h}_x/3$. Fixed heads $h(0)=H$ and $h(L)=\eta$ give $h^4=H^4-(H^4-{\eta }^4)x/L$ and the displayed steady discharge. Porosity relates pore velocity to superficial Darcy velocity but does not independently multiply this discharge when $k$ is intrinsic permeability. For depth-dependent permeability, integrate the horizontal conductivity over the saturated cross-section first.

For $n\ge 1$, any distinct node set in $[-1,1]$ different from the $n+1$ Chebyshev polynomial extrema admits one degree-$n$ polynomial $q$, bounded by one at all those nodes, whose every nonconstant derivative violates the corresponding sharp endpoint bound. In decreasing order set $q(t_i)=(-1{)}^i$. The endpoint derivative signs of Lagrange cardinal polynomials give $q^{(k)}(1)={\sum }_i|L_i^{(k)}(1)|$. Interpolating $T_n$ on the same nodes showsEach term is nonnegative and at least one is positive, because the other node set includes a nonextremal point. Uniqueness is a claim about node sets: permuting the same nodes does not change their norming property.

Start with cash one, switch to $X/a$ when a nonnegative supermartingale first reaches at most $a$, and switch to cash $q=b/a$ when it next reaches at least $b$. Repeat with coefficients multiplied by $q$ after every completed upcrossing. At each switch the new value is at most the old one, so pasting supermartingales with downward jumps gives a supermartingale when stopped after any fixed number of switches. Its initial value is $\min (X_0/a,1)$, and its value at the $k$th completed crossing is exactly $q^k$. Applying its expected value bound proves the Dubins upcrossing inequality.

For an irreducible projective variety $Y\subset \mathbb{P}(V)$ over the complex numbers, its projective dual variety is the closure in the Zariski topology of the projective hyperplanes containing the embedded tangent space at some smooth point of $Y$. The projective biduality theorem recovers $Y$ as $(Y^{\ast }{)}^{\ast }$: the family of tangent projective hyperplanes determines the original projective variety.

For a nondegenerate smooth projective curve $C\subset {\mathbb{P}}^n$ with $n\ge 2$, the incidence pairs $(p,H)$ with $T_{p}C\subset H$ form a projective bundle over $C$ with fiber ${\mathbb{P}}^{n-2}$, hence dimension $n-1$. The general contact locus is a projective subspace of $C$; since $C$ is not a projective line, it is a single point. Thus $C^{\ast }$ is an irreducible projective hypersurface, and a general member represents one ordinary tangency. For a smooth plane conic, the tangent-line equation gives another smooth plane conic in the dual projective plane.

The ambient-dimension hypothesis excludes $C={\mathbb{P}}^1\subset {\mathbb{P}}^1$: no projective hyperplane contains its whole tangent space, so its dual is empty. The stated reflexivity and general tangency assertions also use characteristic zero; inseparable behavior in positive characteristic requires additional hypotheses.^[ref]

For $A=k[\epsilon ]/({\epsilon }^2)$ and $k=A/(\epsilon )$, the kernel and image of multiplication by $\epsilon$ both equal $(\epsilon )$. Thus the displayed sequence is a free resolution. Applying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. The module $k$ therefore has infinite projective dimension.

In bond percolation on the square lattice, the outer boundary of a finite percolation cluster, after filling its holes, consists of closed bonds. Their crossing edges in the planar dual graph form a connected enclosing contour, and hence belong to a single dual bond percolation cluster $K$. If $|K|=m$, both its coordinate widths are at most $m-1$, so the enclosed primal cluster contains at most $m^2$ vertices. A contour enclosing the origin also places every vertex of $K$ within distance $m+1$ of the origin. An exponential volume tail for the subcritical dual bond percolation clusters thus gives a bound $C{e}^{-c\sqrt{n}}$ for a finite primal cluster of size at least $n$.

A planar finite network with three boundary terminals induces one of five connection partitions: all three connected, all three separate, or one connected pair and an isolated third terminal. Put three dual terminals on the intervening boundary arcs. Passing to the planar dual graph interchanges the all-connected and all-separate partitions and permutes the three pair partitions. If the network is symmetric under all terminal permutations, equality $P_3=P_0$ of the first two probabilities makes its whole partition law dual-invariant. Independent cells on a self-dual triangular terminal arrangement then give identical primal and dual coarse connection laws. Infinite connectivity is preserved because cells are uniformly finite. For a martini lattice cell with six independent bonds of common parameter $p$, $P_3=3p^5-2p^6$ and $P_3-P_0=(2p^2-1)(p^4-3p^3+2p^2+1)$. The unique zero in $[0,1]$ is $1/\sqrt{2}$.

For a down-and-out European call with barrier equal to strike, zero interest and no dividends, hold one share and borrow the strike until barrier hitting or maturity. At continuous hitting the share price is exactly the debt, so liquidating leaves zero; on survival the portfolio pays $S_T-K$. Hence the contract has value $S-K$ and delta one before knock-out. The pathwise replication works beyond constant-volatility Black-Scholes model, but can fail through downward jumps or execution gaps, and nonzero carry, transaction costs or discrete monitoring require changes.

Suppose the generators of a binary linear code have Hamming weights divisible by four and have pairwise zero binary inner products. Their linear span is doubly even: for any two vectors,The binary inner product says the intersection size is even. Induction on the number of generators in a sum proves divisibility by four.

// Target: algebra.bigb

The motion of a star around the astronomical barycenter caused by an orbiting companion. In an isolated two-body problem with relative semi-major axis $a$, the stellar semi-major axis is $a_{\star }=M_{p}a/(M_{\star }+M_p)$. Doppler spectroscopy measures its line-of-sight velocity and astrometry measures its projected position on the sky.

For independent zero-mean square-integrable increments, their partial sums form a martingale, and their squares form a nonnegative submartingale. Apply the Doob maximal inequality to the squared process at threshold $x^2$. Independence and zero means make its terminal second moment the sum of the increment variances. Identical distributions are not required.

For a square-integrable martingale with $M_0=0$ and $V=\mathbb{E}{M}_n^2$, apply the Doob maximal inequality to $(M_k+c{)}^2$ for $c\ge 0$. An upper crossing of $x$ implies a square crossing of $(x+c{)}^2$, giving $(V+c^2)/(x+c{)}^2$. Minimization at $c=V/x$ gives the displayed bound. A one-step martingale taking $x$ with probability $V/(V+x^2)$ and $-V/x$ otherwise attains equality when $V>0$.

For a nonnegative submartingale with maximum $M_t$, the first-moment maximal estimate is $\lambda \mathbb{P}(M_t\ge \lambda )\le \mathbb{E}[X_t{1}_{\{M_t\ge \lambda \}}]$. Integrate $2\lambda$ times the tail probability only up to $R$, obtaining $\mathbb{E}(M_t\wedge R{)}^2\le 2\mathbb{E}[X_t(M_t\wedge R)]$. The Cauchy-Schwarz inequality gives $\Vert M_t\wedge R{\Vert }_2\le 2\Vert X_t{\Vert }_2$, and monotone convergence gives the result as $R\to \infty$. Truncation avoids assuming beforehand that the maximum has a finite second moment.

// Target: brownian-motion.bigb

For centered unit-variance independent and identically distributed increments with $m_4=\mathbb{E}{\xi }^4<\infty$, an interpolated increment is $N^{-1/2}{\sum }_j{c}_j{\xi }_j$. Here $c_j$ is the length of the overlap of $(Ns,Nt)$ with the $j$th unit cell, so $0\le c_j\le 1$ and $\sum c_j=N(t-s)$. Independence gives its fourth moment as $N^{-2}[3(\sum c_j^2{)}^2+(m_4-3)\sum c_j^4]$. This is at most $\max (3,m_4)N^{-2}(\sum c_j^2{)}^2$, and $\sum c_j^2\le N(t-s)$ gives the bound, including intervals crossing a mesh point. A uniform fourth-moment continuity criterion yields tightness of probability measures on $C_0[0,1]$.