Suppose $D\subseteq \mathbb{C}$ is open and admits an injective holomorphic map $\Phi :D\to \mathbb{C}$ whose image is bounded. Every entire function $f:\mathbb{C}\to D$ is constant: the function composition $\Phi \circ f$ is a bounded entire function, hence constant by Liouville theorem, and injectivity of $\Phi$ then forces $f$ to be constant. Surjectivity of $\Phi$ onto a disk is unnecessary. For example $z\mapsto z/(1-z)$ is a conformal map from $\mathbb{C}\setminus [0,1]$ onto $(\mathbb{C}\setminus [0,\infty ))\setminus \{-1\}$. A square-root branch with argument in $(0,2\pi )$ followed by the Cayley transform between the half-plane and disk maps this to the punctured unit disk. It therefore supplies a bounded injective holomorphic coordinate on the interval complement, giving the same rigidity even though that domain is not simply connected.

Let $f,g$ be entire functions, with $g$ not identically zero. If $|f(z)|\le C|g(z)|$ everywhere, then $f/g$ is bounded and holomorphic away from the isolated zeros of $g$. The removable singularity extends it across each zero, giving a bounded entire function. Liouville's theorem makes it constant, so $f=cg$ with $|c|\le C$. If $g$ is identically zero the inequality forces $f=0$. For example, domination by $|\cos z|$ forces a constant multiple of $\cos z$.

For a discrete dynamical system, eigenvalues of the linearization strictly inside the unit circle imply local asymptotic stability; an eigenvalue outside implies instability. Eigenvalues on the unit circle require more information. A nontrivial Jordan block there can already make the linearized iterates unbounded.

Even a bounded linearization leaves nonlinear stability undecided. The two displayed maps of the complex plane have the same derivative, a rotation matrix. Their radial updates are $r\mapsto r(1-r^2)$ and $r\mapsto r(1+r^2)$. For small positive radius, the first decreases to zero and the second moves out of every sufficiently small fixed neighborhood. Thus identical unit-modulus linear multipliers can accompany attraction or instability.

For a second-order real linear partial differential equation in two variables, an invertible local change of variables simplifies its principal part. At regular points, the hyperbolic case has mixed-derivative principal part $U_{\xi \eta }$, the elliptic case has principal part $U_{\xi \xi }+U_{\eta \eta }$, and the parabolic case has rank-one principal part $U_{\xi \xi }$, after multiplication by nonzero factors. Lower-order terms and forcing must also be transformed with the chain rule; simplifying only the principal part does not in general give the full canonical equation. In the hyperbolic partial differential equation case the new variables can be chosen constant along the two characteristic families.

An annihilating polynomial of a linear operator is a nonzero polynomial $p$ for which the displayed operator is zero. The minimal polynomial is its unique monic annihilating polynomial of least degree and divides every other annihilating polynomial. For a diagonalizable operator with distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_s$, the product ${\prod }_j(t-{\lambda }_j)$ annihilates it. For an individual vector it suffices to annihilate the eigenvalues contributing to that vector; this gives exact finite-step termination bounds for a Krylov subspace method.

For an ordinary differential equation $y^{\prime }=\Phi (t,y)$, this linear multistep method uses a centered derivative across the two-step interval $[t_{i-1},t_{i+1}]$, evaluated at its midpoint $t_i$. It requires two starting values, $y_0$ and $y_1$. For a sufficiently smooth exact solution, Taylor's theorem givesThe unscaled local residual is therefore $O(h^3)$ and the method has order two. Its first characteristic polynomial is $z^2-1$, with the simple roots $1,-1$, so it satisfies the root condition for a multistep method; the second root is a parasitic amplification root. This two-step method differs from the one-step midpoint method and the implicit midpoint rule.

For the special equation $G^{\prime }=g(t)$ and $G(a)=0$, an even number $n=2r$ of steps gives $y_n=2h{\sum }_{j=1}^{r}g(t_{2j-1})$, independently of the odd-index starting value. It is the composite midpoint quadrature rule on $r$ intervals of width $2h$. An odd final index depends on $y_1$, which must be specified separately.

Fix a positive shear modulus $\mu$ and let the other Lamé parameter tend to infinity in $S_{\lambda }=\lambda \mathrm{tr}(e_{\lambda })I+2\mu e_{\lambda }$. If strain and stress converge to finite limits and differentiation commutes with the limit, taking the trace shows $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$. Thus $\mathrm{div}u=0$ and $S=-pI+2\mu e$ for $p=-\lim \lambda \mathrm{tr}{e}_{\lambda }$. Force balance becomes the displayed Stokes equation, with an elastic shear modulus in place of fluid dynamic viscosity. The vanishing volumetric strain is the linearization of elastic incompressibility $\det Dy=1$.

Write a quadratic small-strain energy as $\tilde{W}(F)=Q(\mathrm{sym}(F-I))$ with $Q$ homogeneous of degree two. If it were fully frame indifferent, the rotation $R=\mathrm{diag}(1,-1,-1)$ would have zero energy. Quadratic homogeneity would then make every $F_s=I+s(R-I)$ have zero energy. At $s=1/4$, $F_s=\mathrm{diag}(1,1/2,1/2)$ has determinant $1/4$, yet zero energy. Thus a nondegenerate linear elasticity energy cannot also be used as a fully objective finite-deformation energy. Its invariance concerns infinitesimal rotations around the reference state.

A linear map from symmetric three-dimensional infinitesimal strain tensors to symmetric stress tensors acts between two six-dimensional spaces. Choosing the minor index symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$ leaves $6\times 6=36$ independent constitutive coefficients. Symmetry of the two second-rank tensors alone does not force pair interchange. If a strain-energy density exists and stress is its derivative, equality of mixed second derivatives gives the major symmetry $C_{ijkl}=C_{klij}$. The resulting symmetric six-by-six coefficient matrix has $6\cdot 7/2=21$ independent coefficients. This distinguishes minor tensor symmetry from the extra reciprocity supplied by an energy potential.

On a connected isotropic linear elastic body, assume $\mu >0$ and $3\lambda +2\mu >0$. Then $W(e)=\mu |e^{\mathrm{dev}}{|}^2+(3\lambda +2\mu )(\mathrm{tr}e{)}^2/6$ is positive for every nonzero symmetric strain tensor. If two equilibria have the same body force and boundary traction, their difference $v$ satisfies zero loading. Multiplication of equilibrium by $v$ and the divergence theorem yield ${\int }_{B}2W(e(v))dV=0$. Hence $e(v)=0$, and the kernel of the symmetric gradient consists of infinitesimal translations and rotations: $v=a+\omega \times x$.

The conclusion is uniqueness modulo these modes, not existence for arbitrary tractions. Existence additionally requires zero total resultant force and torque. On a disconnected body each connected component has its own rigid modes; fixing enough displacements removes those modes and can produce full uniqueness.

For uniform twist $\Omega$ and displacement $(-\Omega yz,\Omega xz,w(x,y))$, linear elasticity gives ${\tau }_{xz}=\mu (w_x-\Omega y)$ and ${\tau }_{yz}=\mu (w_y+\Omega x)$. Their planar divergence vanishes by axial equilibrium. On a simply connected section there is therefore a stress function $\varphi$ with ${\tau }_{xz}={\varphi }_y$ and ${\tau }_{yz}=-{\varphi }_x$.

The mixed derivatives of $w$ imply $\Delta \varphi =-2\mu \Omega$. The lateral axial traction is the tangential derivative of $\varphi$, so a traction-free connected boundary makes $\varphi$ constant; choose it to be zero. Finally the torque is $M={\int }_D(x{\tau }_{yz}-y{\tau }_{xz})dA=-{\int }_D(x{\varphi }_x+y{\varphi }_y)dA=2{\int }_D\varphi dA$, since its boundary term vanishes. Thus one scalar Poisson equation determines both the shear field and torsional stiffness. Multiply connected sections require additional constants and compatibility conditions on the separate boundary components.

For a infinitesimal strain tensor $e=(\nabla u+\nabla u^T)/2$, commuting displacement derivatives gives the symmetric incompatibility tensor $C=0$. Its six component equations are the Saint-Venant compatibility equations. They satisfy three differential identities because ${\partial }_i{\partial }_k$ is symmetric while ${ϵ}_{ik\ell }$ is antisymmetric. Thus there are three independent differential constraints, although keeping all six equations is a coordinate-symmetric way to state them. At a nonzero Fourier wavevector along the third axis the constrained components are $e_{11},e_{22},e_{12}$, while $e_{13},e_{23},e_{33}$ arise from the three displacement components. On a simply connected region compatibility gives local integrability up to a rigid motion; on a multiply connected region circulation conditions must also be imposed.

For an autonomous ordinary differential equation, the tangent to a periodic orbit gives a unit Floquet multiplier. A hyperbolic attracting cycle has that multiplier simple and all transverse multipliers strictly inside the unit disk. Thus the periodic variational operator has a one-dimensional null space spanned by the tangent. Nonlinear asymptotic attraction without hyperbolicity does not suffice: $\dot{r}=-(r-1{)}^3$, $\dot{\phi }=2\pi$ attracts to a unit-period circle but its radial variational equation is neutral.

Near a hyperbolic attracting limit cycle, a weak perturbation can be projected onto the neutral phase direction while transverse perturbations decay. Solving the periodic adjoint variational equation gives a phase response function. Averaging its inner product with weak forcing or coupling yields the first-order phase dynamics.

Electric current in the ground after a nearby lightning discharge. A person can bridge points of different electric potential, creating indirect exposure without the main channel striking the body.

The support or pivot about which a lever turns.

A first-class lever places the fulcrum between the applied effort and the load. The two loads produce opposite torques about the pivot, with equilibrium condition $E{d}_E=L{d}_L$ for the perpendicular lever arms. Its mechanical advantage can be greater than, equal to or less than one depending on the arm ratio.

A balanced seesaw is a direct test: equal loads balance at equal arms, while moving one load closer to the pivot requires a larger load there to maintain balance. For a small rotation, the two ends move in opposite directions and their travel distances scale with their distances from the pivot. The force ratio and displacement ratio therefore compensate in an ideal work balance. The defining feature is the pivot's position between effort and load, rather than a guaranteed force multiplication.

A second-class lever places the load between the fulcrum and the applied effort. In the simple straight-lever geometry with perpendicular forces, the effort lever arm is longer than the load arm. Static torque balance then gives mechanical advantage $L/E=d_E/d_L>1$.

A wheelbarrow illustrates the geometry when the wheel axle acts as the pivot, the carried load lies between axle and handles, and the hands supply the effort. The handles must move through a greater distance than the load's corresponding vertical motion. That motion trade follows directly from a common small rotation and explains why a reduced effort force does not imply reduced ideal work. Actual geometry and force directions determine the perpendicular arms; distances measured along an inclined body cannot always be substituted unchanged.

A third-class lever places the applied effort between the fulcrum and the load. For perpendicular loads on a straight ideal lever, the effort lever arm $d_E$ is shorter than the load arm $d_L$. Static torque balance gives $E{d}_E=L{d}_L$, so its mechanical advantage $L/E=d_E/d_L$ is less than one.

The arrangement trades a larger required force for a larger load-end travel and speed. Tweezers provide a familiar model: the fingers press between the joined end and the gripping tips. For a small rotation, tip displacement is proportional to its distance from the pivot, so the displacement ratio is inverse to the force ratio. An ideal lever therefore conserves work while changing the force and motion scales.

Historical term for an excess of white blood corpuscles, especially the persistent organ-associated disorders subsequently classified as leukaemia. The old term does not make every raised white count malignant.