For integrable boundary data $g$ on the unit circle, this follows from the Poisson kernel bound $C_A\delta /({\delta }^2+d^2)$ in a fixed Stolz region. The arc of radius $\delta$ and successive dyadic annuli contribute at most $C_A{2}^{-k}Mg(\xi )$, which sum to the displayed estimate. This links harmonic extension to the Hardy-Littlewood maximal function and its strong $L^p$ bounds for $p>1$.

Shear banding is spatial separation into regions with different local shear rates. In simple inertialess planar shear, equal shear stress can coexist with different rates when the homogeneous constitutive curve is nonmonotonic. Gradient terms and boundary conditions can determine the interface and selected stress; a local multivalued flow curve alone does not provide that selection. Shear banding need not be confused with the ordinary smooth shear-rate variation in a pressure-driven channel.

A simple fluid has local stress determined by the history of deformation of a material element, rather than by spatial derivatives of that history. An isotropic incompressible simple fluid has an arbitrary isotropic pressure and an objective constitutive functional of the relative deformation gradients. Internal-variable equations can represent such a functional when the initial material state is fixed by the past history.

For a surjective module endomorphism of a Noetherian module, the chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes at some $N$. If $f(x)=0$, choose $y$ with $f^N(y)=x$. Then $y\in \ker f^{N+1}=\ker f^N$, giving $x=0$. Thus the endomorphism is also injective and is an isomorphism.

For a short exact sequence $0\to N\to M\to M/N\to 0$, $M$ is a Noetherian module exactly when both the other terms are Noetherian modules. For the extension direction, an ascending chain $M_i$ has eventually constant intersections $M_i\cap N$ and images $(M_i+N)/N$. Equality of both forces equality of consecutive $M_i$: subtract an element with the same image and then use the common intersection.

For a finite integral inclusion $A=k[t_1,\dots ,t_r]\subset B$ of domains, a nonzero $b\in B$ has a monic minimal polynomial over the fraction field with nonzero constant term $a_0\in A$. Its equation gives $a_0\in bB\cap A$. Consequently the image of $V_B(b)$ under the finite morphism to ${\mathbb{A}}^r$ is contained in the proper Zariski-closed set $V_A(a_0)$. Finite morphisms preserve the dimension of a closed subvariety and its image, so this proves dimension drops for proper closed subsets after Noether normalization. A power of $a_0$, up to sign, is the field norm of $b$.

For $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$, the fixed-coordinate field variation is $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$. Suppose action invariance is expressed as ${\delta }_0\mathcal{L}/\epsilon +{\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })={\partial }_{\mu }{K}^{\mu }$. Writing ${\Pi }_a^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)$ and $E_a={\mathcal{L}}_{{\varphi }_a}-{\partial }_{\mu }{\Pi }_a^{\mu }$, the product rule gives ${\partial }_{\mu }{j}^{\mu }=-{\sum }_a{E}_a{Q}_a$. Thus $j^{\mu }$ is a conserved current on shell. A vanishing spatial boundary flux is additionally required for its Noether charge to be conserved. This formula includes internal transformations, translations and Lorentz transformations.

A virial current $V^{\mu }$ represents the trace of a conserved stress-energy tensor as a divergence. Then the dilation Noether current $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$ is conserved, since its divergence is $T^{\mu }{}_{\mu }-{\partial }_{\mu }{V}^{\mu }$. This establishes scale invariance; a traceless stress-energy tensor improvement is the extra condition needed for conformal currents.

For a symmetric conserved stress-energy tensor $\Theta$ and a Conformal Killing vector field $\omega$ in dimension $d$, the displayed conformal current has divergence $(\partial \cdot \omega ){\Theta }^{\mu }{}_{\mu }/d$. This follows by contracting the Conformal Killing equation with $\Theta$. A traceless tensor therefore gives a conserved current for every conformal generator.

The no-scale supergravity identity cancels the negative gravitino contribution, leaving $V=e^{|C{|}^2}|3C^2+\bar{C}(C^3+B){|}^2/(T+\bar{T}{)}^3$. Zero-energy minima occur at $C=0$ and, for $B\ne 0$, three phase-related nonzero solutions with $|B|=|C|(3+|C{|}^2)$. Their $T$ supergravity auxiliary field is nonzero if $B\ne 0$, while both real components of $T$ are flat on each vacuum branch. For $B=0$, the $C=0$ branch is instead supersymmetric and has zero gravitino mass from a superpotential.

For $s=S+\bar{S}>0$, $t=T+\bar{T}-|C{|}^2>0$, $K=-\log s-3\log t$ and a holomorphic superpotential independent of $T$, the exact F-term potential is $V=e^K[s^2|D_{S}W{|}^2+(t/3)|W_C{|}^2]$. Inverting the mixed $T,C$ metric gives $K^{T\bar{T}}=t(t+|C{|}^2)/3$, $K^{T\bar{C}}=t\bar{C}/3$, $K^{C\bar{T}}=tC/3$, and $K^{C\bar{C}}=t/3$. Substitution of $D_{T}W=-3W/t$ and $D_{C}W=W_C+3\bar{C}W/t$ cancels all mixed terms, leaving $3|W{|}^2+(t/3)|W_C{|}^2$; the first term cancels the supergravity negative term.

Let $b$ be the undiscounted annual premium and $d_i$ the discount fraction at level $i$. Compare the next two levels $i_r^0$ after no current claim with $i_r^1$ after a current claim, assuming no later accidents. Their total additional premium is the displayed threshold. A repair cost above it makes a claim advantageous under this two-year rule. For discounts $0,\alpha ,\beta$ and one-level moves with reflecting endpoints, the thresholds at levels one, two, and three are respectively $b\beta$, $b(2\beta -\alpha )$, and $b(\beta -\alpha )$. The premium scale must be specified or fixed as the monetary unit.

For a three-level birth-death chain whose down-or-stay claim probabilities are $p_1,p_2,p_3$, with each in $(0,1)$, detailed balance is ${\pi }_1(1-p_1)={\pi }_2{p}_2$ and ${\pi }_2(1-p_2)={\pi }_3{p}_3$. Normalizing gives the displayed stationary distribution. The finite chain is irreducible and has endpoint holding probabilities, so it is aperiodic and its level proportions converge to this distribution from any initial level.

Suppose annual claim counts are independent with geometric distribution mass $p{q}^j$, $j\ge 0$. In a no claims discount system with discounts $0,\alpha ,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The transition matrix is $\left(\begin{array}{ccc}q& p& 0\\ q& 0& p\\ q^2& pq& p\end{array}\right)$. Solving $\pi P=\pi$ gives the displayed stationary distribution. With full premium $c$, the stationary expected value of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^{2}q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.

In a no claims discount system with discounts $0,\alpha ,\beta$, a reported claim resets next year's discount to zero. With no later accidents, the next two premiums after reporting are $C+(1-\alpha )C$. At the zero level, not reporting instead gives $(1-\alpha )C+(1-\beta )C$; at either other level it gives $2(1-\beta )C$. Their differences give the displayed repair-cost thresholds. A fully reimbursed loss above its threshold should be reported under this undiscounted two-year rule. Current premiums cancel. Different thresholds generally produce state-dependent submitted-claim probabilities, even with a common accident mechanism.

At the positive Nicholson-Bailey model equilibrium, $\mathrm{tr}{J}_{\ast }=1+\log r/(r-1)$ lies between one and two, and the displayed determinant exceeds one. These follow from $(r-1)/r<\log r<r-1$ for $r>1$. The discriminant is therefore negative, and the conjugate eigenvalues have moduli greater than one. Nearby annual samples spiral outward rather than approach coexistence.

Remove a spherical void centered at $c$ from a uniform sphere centered at the origin. At points inside the void, superposition combines the sphere's interior field $-4\pi G\rho x/3$ with the removed sphere's field $+4\pi G\rho (x-c)/3$. Their sum is the displayed constant Newtonian gravitational field. A second removed mass outside this void adds a nonuniform contribution.

Let $f\in C^2([a,b])$ have $f^{\prime },f^{\prime \prime }>0$ and $f(a)<0<f(b)$. Its unique root $r$ lies in $(a,b)$. Starting Newton root-finding iteration at $b$, strict convexity puts the tangent root strictly between $r$ and the current iterate, so the iterates decrease to $r$. Taylor expansion about the current iterate gives $e_{n+1}=f^{\prime \prime }({\xi }_n)e_n^2/[2f^{\prime }(x_n)]$, where $e_n=x_n-r$ and ${\xi }_n\in (r,x_n)$. Continuity proves the displayed positive asymptotic constant and quadratic convergence.

For a $C^2$ map $\Phi$ on a Banach space, define $N(p)=p-[D\Phi (p){]}^{-1}\Phi (p)$ wherever the bounded derivative is invertible. It solves the local affine approximation to $\Phi =0$. Every zero is a fixed point of $N$. Near a zero with invertible derivative, bounded inverses and a locally Lipschitz derivative give the quadratic Newton error bound in a Banach space.

// Target: numerical-analysis.bigb