In plane strain, the deformation map has $x_1=x_1(X_1,X_2)$, $x_2=x_2(X_1,X_2)$ and $x_3=X_3$. The out-of-plane stretch is one and out-of-plane shear vanishes. This kinematic condition need not make the out-of-plane stress vanish.

With the right Cauchy-Green deformation tensor $C=F^{T}F$, the Green-Lagrange strain tensor measures the change in the material metric. Since ${\dot{E}}^{(2)}=F^{T}DF$, its work-conjugate stress and strain partner is the second Piola-Kirchhoff stress tensor $F^{-1}\tau F^{-T}$. The formula follows by cycling factors in $T:{\dot{E}}^{(2)}$, not by identifying nominal and symmetric material stress.

For $C=F^{T}F$, differentiation of $C^{-1}$ gives ${\dot{E}}^{(-2)}=F^{-1}D{F}^{-T}$. Its reference-power conjugate is $T^{(-2)}=F^T\tau F$. In a convected coordinate basis $g_I=F{e}_I$, these are the covariant components of the Kirchhoff stress tensor. The second Piola-Kirchhoff stress tensor instead gives its contravariant components, since $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$.

For a right stretch tensor $U={\sum }_a{\lambda }_a{n}_a\otimes n_a$, define $E^f=f(U)={\sum }_{a}f({\lambda }_a)n_a\otimes n_a$. A standard normalized choice has $f\in C^1(0,\infty )$, $f(1)=0$, $f^{\prime }(1)=1$ and $f$ strictly increasing; requiring $f^{\prime }>0$ gives a nonsingular stretch-to-strain parameterization. This strain vanishes exactly for a rigid rotation, is independent of superposed spatial rotations, and agrees to first order with the infinitesimal symmetric displacement gradient. Indeed $F=I+H$ gives $U=I+(H+H^T)/2+O(\Vert H{\Vert }^2)$, and Taylor expansion of $f$ proves the last statement.

An automaton transition changes an automaton state on reading a symbol. A DFA specifies one destination; an NFA may specify several or none.

An automaton state is one possible memory configuration of a finite-state machine. The initial automaton state is the starting configuration; accepting automaton states indicate successful recognition when the whole word has been read.

For a relational structure, the Gaifman graph has the same universe. Two distinct elements are adjacent if they occur together in some tuple of some interpreted relation. A tuple's radius-$r$ neighbourhood is the induced structure on the union of its radius-$r$ graph balls, with the tuple marked by constants.

Extend first-order logic by iterating definable relation operators from the empty relation. If iteration reaches a fixed point, use that relation; if it enters a nontrivial cycle, the partial fixed point is the empty relation. For a fixed arity on a finite universe, a relation requires polynomially many bits and there are exponentially many states, so iteration is evaluable in PSPACE. On finite ordered structures this logic captures PSPACE.

Two formulas with the same free variables are asymptotically equivalent when the probability of their universal equivalence tends to one over the chosen finite-structure distribution. If a quantifier-eliminating extension theory has all its axioms of asymptotic probability one, compactness makes every formula asymptotically equivalent to a quantifier-free one: each particular equivalence follows from finitely many axioms, whose conjunction still has limiting probability one.

A group between a nonabelian simple group and its automorphism group. Finite two-transitive groups are either of this type or affine two-transitive groups, a structural theorem rather than an order-divisibility test.

A group of order ten is either a cyclic group or the dihedral group of the pentagon. Lagrange's theorem and the fact that a group of exponent two has power-of-two order force an element of order five or ten. In the latter case the group is cyclic. In the former case its order-five cyclic subgroup is normal by index two; an outside involution acts on it either trivially or by inversion. The trivial action gives $C_5\times C_2\cong C_{10}$, and inversion gives $D_{10}$.

The largest nilpotent normal subgroup of a finite group $G$. If $O_p(G)$ denotes its p-core, distinct p-cores commute, and $F(G)={\prod }_p{O}_p(G)$. Every normal nilpotent group has Sylow subgroups which are characteristic subgroups, so it is contained in this product.

A finite game is symmetric when all players have the same pure strategies and permuting the players permutes their payoffs. In the anonymous special case, a player's payoff depends on their own action and the multiset of opposing actions. A symmetric equilibrium exists even when there are more than two players, by the gain-map proof of symmetric equilibrium.

For ${\mathbb{F}}_{q^2}/{\mathbb{F}}_q$, the field norm maps the nonzero elements onto ${\mathbb{F}}_q^{\ast }$: the multiplicative group of a finite field is cyclic, and the power $q+1$ has image of order $q-1$. The field trace is also onto. Indeed it is an ${\mathbb{F}}_q$-linear map into ${\mathbb{F}}_q$, and is not zero because its polynomial $X+X^q$ has degree $q$ and cannot vanish at all $q^2$ field elements. Consequently $a\mapsto \mathrm{Tr}(ac)$ is onto whenever $c\ne 0$.

For arbitrary maps $\varphi :{\mathbb{R}}^2\to S^2$, integrability of the Dirichlet energy does not imply continuous extension to the one-point compactification. For large $r$, takeand interpolate smoothly to a constant on a bounded disk. Its exterior energy isThe field has no limit because $a(r)$ keeps winding. A continuous extension at the point at infinity instead requires $\varphi (x)\to {\varphi }_{\infty }$ uniformly outside sufficiently large disks. This fixed-limit boundary condition is standard for a sigma-model lump, but is extra information beyond finite energy for arbitrary fields.

The Rayleigh-Ritz method minimizes an energy functional over a finite-dimensional trial space. For a real Hilbert space $V$, a symmetric coercive bilinear form $a$ and a continuous linear functional $\ell$, takeWriting $u_h={\sum }_k{c}_k{\varphi }_k$ in a trial basis and differentiating with respect to each coefficient givesThe same equations follow from testing the variational problem against each ${\varphi }_j$, which is the Ritz-Galerkin equivalence for a symmetric coercive form. Nonsymmetric problems need a separate argument because differentiating $a(v,v)$ sees only its symmetric part.

For $-u^{\prime \prime }=1$ on $(0,1)$ with zero Dirichlet boundary conditions, choose $a(v,w)={\int }_0^1{v}^{\prime }{w}^{\prime }dx$, $\ell (v)={\int }_0^{1}vdx$ and the one-dimensional trial space spanned by $\varphi =x(1-x)$. Then $K=1/3$ and $F=1/6$, giving $u_h=x(1-x)/2$. Here the exact solution lies in the trial space, so minimization recovers it exactly; in general the Céa lemma controls the approximation error.

Consider the Poisson equation on an interval with homogeneous Dirichlet boundary conditions:Two integrations and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a test function $v$ that vanishes at both endpoints. Integration by parts givesThe boundary term is zero. Thus the weak formulation is to find $u\in V=H_0^1(0,1)$ such thatThe zero-trace Sobolev space $H_0^1(0,1)$ consists of square-integrable functions with square-integrable weak derivatives and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous piecewise linear functions are admissible despite corners. The Poincaré inequality makes $a$ a coercive bilinear form on $V$. Strang derives this one-dimensional weak formulation, the piecewise-linear hat functions, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.^[ref]

Take the uniform finite element mesh $x_j=jh$, $j=0,\dots ,n$, with $h=1/n$. The continuous $P_1$ trial space isHere $P_1$ means polynomials of degree at most one on each interval. This is a conforming finite element space in $V$, with $n-1$ interior piecewise-linear hat functions ${\varphi }_i$ as a basis. Their endpoint values enforce the Dirichlet boundary conditions automatically. Set $u_h={\sum }_{i=1}^{n-1}{c}_i{\varphi }_i$ and test the weak formulation against each ${\varphi }_i$:Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared interval, giving $-1/h$. All other derivative products vanish. The stiffness matrix is the tridiagonal matrixThe load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The piecewise-linear hat function article explains how the same entries arise by adding local two-node contributions.

With four mesh intervals, $h=1/4$, there are three unknowns:For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The coefficients are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, linear interpolation gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.

In this model the Galerkin method solution equals the nodal approximation $I_{h}u$ obtained by linear interpolation on any finite element mesh. Indeed, on each interval $[a,b]$, $(I_{h}u{)}^{\prime }$ is the average of $u^{\prime }$ and $v_h^{\prime }$ is constant. Therefore ${\int }_a^b(u^{\prime }-(I_{h}u{)}^{\prime })v_h^{\prime }dx=0$. Summing proves Galerkin orthogonality for $I_{h}u$, and uniqueness gives $u_h=I_{h}u$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for finite element methods with other operators or integration rules.

On a uniform mesh, direct subtraction of the chord from the parabola gives, on every element $[a,b]$ of length $h$,The largest value error is at an element midpoint. Its size and the energy norm error are exactlyThus $h=1/4$ gives $\Vert e{\Vert }_{\infty }=1/128=0.0078125$ and $\Vert e{\Vert }_a=1/(4\sqrt{12})\approx 0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx 0.0360844$. Halving the mesh spacing quarters the maximum value error but only halves the energy norm error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.

Material isotropy makes the elastic response independent of a rotation of reference material directions: $\sigma (FR)=\sigma (F)$ for proper rotations $R$. Combined with material frame indifference, it yields coaxiality of isotropic elastic stress.

A hyperelastic material derives its elastic stress from a strain energy density. In reference-volume units its first Piola-Kirchhoff stress tensor is the derivative of that density with respect to the deformation gradient.

A unit reference fibre direction $m_0$ retains its length under this constraint on the deformation gradient. Its reaction contributes an undetermined tension along the current fibre direction, independently of the pressure reaction enforcing incompressibility.