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Post-receptoral processing compares cone signals, organizes those comparisons across space and time, and uses them to represent surfaces and objects. Three receptor classes do not imply three neurons labelled with three color names. A complete answer must distinguish the information available at the receptors from its neural transformation and its perceptual interpretation.

The principle of univariance supplies the starting constraint. For one cone class, an absorption-dependent response does not identify wavelength independently of intensity. Comparing differently sensitive classes makes chromatic information available. Relative to an adapted background, cone contrasts can be written $c_j=(Q_j-Q_{j0})/Q_{j0}$, where $Q_{j0}>0$ is the background catch. This representation separates a change relative to the prevailing input from the raw number of absorbed photons. It is a useful coordinate system, not a claim that every neuron implements this exact formula.

Retinal circuitry immediately transforms the signals. Light hyperpolarizes cone cells and changes their transmitter release. ON and OFF bipolar pathways respond differently to changes in cone output, while horizontal and amacrine circuitry contribute spatial and temporal interactions. Retinal ganglion cells transmit spike trains rather than simply forwarding a cone's voltage. Their center–surround organization compares regions, so chromatic and spatial structure are already intertwined. Near the central retina, the midget pathway can oppose an L- or M-dominated center to a surround containing other cone input. The same circuitry can carry information useful for fine spatial structure and for chromatic contrast; these are not necessarily separate wires.

A distinct S-cone route illustrates the anatomical specificity. Dennis M. Dacey and colleagues combined physiological characterization with identification of ganglion-cell morphology in primate retina, finding that the S-ON opponent response belonged to a small bistratified ganglion-cell type rather than simply another midget cell.^[ref] This supports parallel circuits with different cone combinations. Calling the pathway “blue-on” is shorthand for its stimulus response, not proof that the cell itself produces a unique blue experience.

At a simplified computational level, an achromatic-like combination and two chromatic combinations can be represented asThis invertible three-channel opponent transform reorganizes three coordinates rather than creating additional independent spectral information. Their matrix has determinant $-2$, so the transform is invertible: $c_L=(y_0+y_1)/2$, $c_M=(y_0-y_1)/2$ and $c_S=y_2+y_0/2$. Actual weights depend on cell, location, adaptation and stimulus conditions. This algebra is an illustrative opponent transform, not a fitted retinal wiring diagram or a complete perceptual color space.

The lateral geniculate nucleus preserves several parallel retinal pathways while also participating in cortical interactions. A. M. Derrington and colleagues studied macaque geniculate cells with gratings whose luminance and chromatic modulation could be varied independently on a calibrated color television display. Finding directions that changed cone stimulation without driving a cell constrained its linear cone weights.^[ref] For a nonzero linear weight vector, the zero-response contrasts obey $r={\mathbf{w}}^{\mathsf{T}}\mathbf{c}=0$, defining a chromatic null plane. A null response therefore tests a relation between inputs, rather than merely assigning the cell a preferred color name. The resulting opponent directions need not coincide with human unique hues.

Cortical processing adds more complex spatial comparisons. Bevil R. Conway mapped cone inputs to color-responsive neurons in alert macaque V1 using cone-isolating spots and eye-position-corrected reverse correlation. Some cells combined chromatic opponency with opposite tuning in adjacent spatial subregions.^[ref] Such double-opponent cells can respond strongly to a chromatic boundary while responding less to a spatially uniform change. This offers a mechanism useful for separating surface differences from shared illumination changes. It does not establish that one V1 population alone solves color constancy under arbitrary scenes.

Further processing must relate color to form, illumination, material and the current task. The same cone contrasts can be interpreted differently in a shadow, highlight or translucent object. Surface color judgments, discrimination thresholds and color naming therefore constrain different computations. Feedback and attention can influence which information is selected, while learning can alter how a signal is interpreted or normalized. A lesion or imaging response can constrain contributions to those tasks, but an area responsive to color is not automatically an isolated “color center.”

The strongest picture is consequently a hierarchy of interacting computations: cone-specific signals, opponent and spatial filtering, combinations tuned to richer image structure, and object-level judgments. Trichromatic sampling limits spectral information; later processing determines how that information supports boundaries, stable surface appearance and behavior.

Factor $A+H=A(I+A^{-1}H)$. The determinant expansion near the identity isthe linear terms in the permutation formula come from choosing one diagonal increment and all remaining identity entries. ThereforeThe required Fréchet derivative is the linear functionalThe quadratic remainder proves full differentiability, rather than only the existence of each directional derivative. This is the derivative of the determinant.

Now use $(ct,x{)}^T$ and ${\eta }_2=\mathrm{diag}(1,-1)$. WritePreservation of the Minkowski metric says that the first column has squared norm $1$, the second has squared norm $-1$, and the columns are orthogonal for the Minkowski inner product:Orthochronousness gives $a\ge 1$. From the orthogonality equation, $b=de/a$; substituting in the second norm identity gives $e^2=a^2$. Hence $(b,e)=\epsilon (d,a)$ with $\epsilon =\pm 1$. The determinant is then $\epsilon (a^2-d^2)=\epsilon$, so properness forces $\epsilon =1$. Thus $b=d$ and $e=a$.

Set $\beta =d/a$. The first norm identity implies $|\beta |<1$ and $a=(1-{\beta }^2{)}^{-1/2}$. With $u=c\beta$, this givesConversely, the displayed matrix has ${\gamma }^2-(\gamma u/c{)}^2=1$, equal off-diagonal entries, determinant one, and positive time-time entry. Direct multiplication therefore gives $L^T{\eta }_{2}L={\eta }_2$, proving both directions. The plus signs are the Lorentz boost convention used in the paper; reversing the boost direction replaces $u$ by $-u$.

For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

Take $\Omega ={\mathbb{R}}^2$ andIts Jacobian matrix and determinant areThus its total derivative is invertible everywhere, and the inverse function theorem makes it locally invertible. But $F(s,t+2\pi )=F(s,t)$, so it is not injective and has no global inverse on its image. The domain is open and connected as required.

For $r>s$, $a>0$. When $0<a\le 1/4$, the roots of ${\mu }^2-\mu +a$ are real and lie strictly between zero and one. When $a>1/4$, they are a complex-conjugate pair of modulus $\sqrt{a}$. Thus the positive steady state is linearly asymptotically stable exactly when $0<a<1$, and unstable when $a>1$. Since $a=2s-2s^2/r$ increases strictly with $r$, the crossing occurs atAt the crossing,Every nonzero real linearized perturbation has period six. This establishes the intended linear stability threshold and linearized period six.

The nonlinear period-six claim in the printed paper does not follow, and is false as a local bifurcating-branch claim for this map. One can check this directly rather than infer a nonlinear period from the eigenvalues. At $r=r_c$ normalize $N_t=N^{\ast }(1+u)$, $N_{t-1}=N^{\ast }(1+v)$. The exact planar map isIts cubic Taylor expansion isHere is a concrete check of the sixfold return. Put $z=(u,v{)}^T$, $Q=u^2-uv+v^2$, andComposing the displayed cubic expansion six times cancels all quadratic terms and givesIn particularThe quartic coefficient is strictly negative. Hence, sufficiently near the equilibrium, every nonzero point has smaller $Q$ after six iterations: the critical nonlinear map has no nearby nontrivial period-six orbit, and the equilibrium itself is locally attracting despite its neutral linearization.

One can also rule out a branch of small period-six orbits on either side of the crossing. Normalize about the positive equilibrium at each nearby $r$, and put $\eta =a-1$. The derivative of the sixth power of $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ at $a=1$ is $3I+K$. Smooth dependence on $r$ therefore givesThe leading matrix has determinant $(3\eta +\alpha Q{)}^2+3(\eta +\beta Q{)}^2$. Both terms could vanish with $Q>0$ only if $\alpha =3\beta$, butThus its norm on nonzero $z$ is bounded below by a positive constant times $(|\eta |+\Vert z{\Vert }^2)\Vert z\Vert$, and the smaller remainder cannot cancel it. There is no nearby nonzero solution of $F_{\eta }^6(z)=z$. The eigenvalues satisfy the nonresonance conditions through order four, their moduli cross outward with increasing $r$, and the cubic radial coefficient is negative. The standard Neimark–Sacker bifurcation theorem therefore gives a supercritical invariant closed curve. Its rotation is near the linear resonance $1/6$; it is not a local branch of exact period-six orbits.

Finally, the discrete population map does not preserve nonnegativity for arbitrary large lagged populations: $r/(1+b{N}_{t-1}^2)-\delta$ can be negative. The positive-equilibrium stability conclusions above are local and do not assume a global positivity property that the formula lacks.

The oscillator flow isIts determinant is ${\cos }^2\omega t+{\sin }^2\omega t=1$, explicitly preserving area. The orbits are energy ellipses; stretching one coordinate during transport is compensated by the conjugate direction.

Take a reference point immediately before the focusing quadrupole and let matrices act on $(x,x^{\prime })$ from the right. With $q=1/f$, define $Q_F=\left(\begin{array}{cc}1& 0\\ -q& 1\end{array}\right)$, $Q_D=\left(\begin{array}{cc}1& 0\\ q& 1\end{array}\right)$ and $D(L)=\left(\begin{array}{cc}1& L\\ 0& 1\end{array}\right)$. One cell isIts determinant is one. A different starting point gives a conjugate matrix with the same trace and stability; the orthogonal transverse plane replaces $q$ by $-q$.

Substitution of the exponential ansatz gives the homogeneous amplitude systemA nonzero mode requires zero determinant. Thus $\boxed{({\omega }_1^2-{\omega }^2)({\omega }_2^2-{\omega }^2)=4{\Omega }^2{\omega }^2}$ with the stated two effective stiffnesses. The off-diagonal product is positive, which determines the sign on the right.

TODO: user-deferred biological kinetic mechanism and its biochemical pattern-forming applications. The remainder treats generic two-field reaction-diffusion physics only.

Let $\mathbf{u}=(u,v)$ obey ${\partial }_t\mathbf{u}=\mathbf{f}(\mathbf{u})+\mathrm{diag}(D_u,D_v){\nabla }^2\mathbf{u}$. At a uniform steady state, write its reaction Jacobian as $J=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Without diffusion stability requires $a+d<0$ and $\Delta =ad-bc>0$. A Fourier perturbation evolves under $J-q^2\mathrm{diag}(D_u,D_v)$, whose determinant isIts trace becomes more negative with increasing $q$. A stationary diffusion-driven instability therefore requireswith an allowed domain mode in the resulting negative-determinant band. Equal diffusivities cannot satisfy this while $a+d<0$. The physical mechanism is that transport can destabilize coupled fields even when the local kinetics relaxes a spatially uniform perturbation. Faster diffusion of an inhibitory response is one possible realization.

The selected wave numbers depend on growth rates and boundaries; nonlinear saturation determines the final stripes, spots or other states. Linear instability alone neither supplies a biochemical mechanism nor uniquely predicts a final finite-amplitude pattern.

We first derive the stress and energy facts used in both parts. The Cauchy traction $s=Tn$ is force per unit area in the deformed configuration. The first Piola-Kirchhoff traction $s_R=T_{R}N$ is force per unit reference area. Their force vectors are measured in the same spatial coordinates, but the area measures and the normals used to specify the surface are different.

Parametrize a reference surface by $x=x(u,v)$. Its oriented area vector is $(x_u\times x_v)dudv$. Under deformation, the two tangents become $F{x}_u,F{x}_v$. The cross-product identitytherefore proves Nanson's formula $nda=J{F}^{-T}NdA$. Equality of the physical force on corresponding surface patches now givesSince this holds for every reference normal,The first Piola-Kirchhoff stress tensor is generally not symmetric, whereas the Cauchy stress tensor is symmetric in the absence of couple stresses.

For the stated energy on orientation-preserving gradients, differentiation givesA stress-free state satisfies $F{F}^T=J^{-1}I$. Taking determinants yields $J^2=J^{-3}$, so $J=1$ and $F{F}^T=I$: precisely $F\in SO(3)$. These states are also the global energy minimizers. If $v_i>0$ are the singular values, the arithmetic-geometric mean inequality gives $|F{|}^2\ge 3J^{2/3}$, with equality only when the singular values are equal. The function $3J^{2/3}/2+J^{-1}$ decreases up to $J=1$ and increases thereafter, with minimum $5/2$. Thus $W\ge 5\mu /2$, with equality precisely at rotations. A constant shift of the energy can set this minimum to zero without changing the stress.

For the given simple shear,For a unit normal $N$, $|T_{R}N{|}^2={\mu }^2(N_1^2+N_2^2)={\mu }^2(1-N_3^2)$. Hence every reference plane whose normal lies in the $e_1,e_2$ plane is maximizing:There is a family of maximizing planes, rather than one distinguished normal.

Put $n=x/R$ and $q=r(R)/R$. Differentiating $y=qx$ uses $\nabla R=n$ and $\nabla q=(r^{\prime }-q)n/R$. HenceIts radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $q$. They are positive by hypothesis, so they are also the principal stretches:For an isotropic stored energy the nominal stress has the same radial/tangential projectors,where symmetry of $\Phi$ gives equal tangential derivatives. Since $\mathrm{Div}(n\otimes n)=2n/R$ and the scalar coefficients depend only on $R$,Thus equilibrium is exactly $\boxed{d{\Phi }_1/dR=2({\Phi }_2-{\Phi }_1)/R}$.

For the particular energy define $\delta =r^{\prime }{q}^2=\det F>0$. Its two derivatives areAlso $q^{\prime }=(r^{\prime }-q)/R$. Differentiating the first formula and substituting the equilibrium equation givesThe last terms cancel. Strict positivity of $h^{\prime \prime }$ and $q$ yields $\boxed{{\delta }^{\prime }=0}$, so $r^2{r}^{\prime }=D{R}^2$ for one positive constant $D$. Integration gives $r^3=D{R}^3+C$.

The displacement data mean $r(1)={\lambda }_1$ and $r(2)=2{\lambda }_2$, not $r(2)={\lambda }_2$. Solving the two scalar equations givesThe given inequality makes $D>0$. Since $r^3$ increases from ${\lambda }_1^3>0$ to $8{\lambda }_2^3$, it is positive on the whole shell, and $r^{\prime }=D{R}^2/r^2>0$. Thus the formula is smooth and admissible and satisfies equilibrium. Conversely every admissible radial equilibrium had to have this constant determinant and these two integration constants, proving uniqueness. This is a constant-Jacobian radial equilibrium for a stretch-sum energy.

On each half-space the deformation is affine. Continuity on the separating plane makes the two affine constants equal and gives $(A-B)t=0$ for every tangent vector $t\perp N$. Decomposing any $x$ into its tangential part and $(x\cdot N)N$ givesThis is the Hadamard jump condition. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the matrix determinant lemma givesBoth determinants are positive, hence $\boxed{A^{-1}a\cdot N<1}$.

Write $a_1={\eta }_1$, $a_2={\eta }_2$, $S=a_1^2+a_2^2$ and $\Delta =a_2^2-a_1^2\ne 0$. The two proposed angles have $c=2a_1{a}_2/S$ and $s=\kappa \Delta /S$, and $c^2+s^2=1$. Therefore the displayed rotations lie in $SO(3)$. Direct multiplication givesThis is a rank-one connection for each sign. With a unit normal one may choose $N=(1,\kappa ,0)/\sqrt{2}$ and jump vector $a=\sqrt{2}\Delta (-a_2,\kappa a_1,0)/S$. The two rotations are distinct because $\Delta \ne 0$; the permitted at-most-two result therefore proves these are all solutions. Flipping both $a$ and $N$ changes no connection.

It remains to test the material plane. Because $y(0)=0$, the affine constants vanish. Let $t=(-p_2,p_1,0)$, so the plane is spanned by $t$ and $e_3$. Both deformation gradients send $e_3$ to $a_1{e}_3$. If the plane is not the twin plane, it cuts the two half-spaces; its two image halves are coplanar exactly when the in-plane vectors $Bt$ and $At$ are parallel. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa )$ in their two-dimensional sections. ThenIts prefactor is nonzero, so coplanarity requires $p_1=\kappa p_2$, precisely $p\parallel (1,\kappa ,0)\parallel N$. Conversely, on that twin plane $(A-B)x=0$, so the two affine maps agree and its image is a plane. Thus $\boxed{\Pi \text{ remains plane}⟺p\parallel N}$. All other allowed planes acquire a kink along their intersection with the twin. This is the planarity test across a cubic-to-tetragonal twin.

The matrix exponential is the absolutely convergent series $e^{tA}={\sum }_{n=0}^{\infty }{t}^n{A}^n/n!$, with derivative $A{e}^{tA}$ and inverse $e^{-tA}$. If $K^T=-K$, then $(e^{tK}{)}^T{e}^{tK}=e^{-tK}{e}^{tK}=I$. Its determinant is continuous in $t$, is never zero, and starts at one, so it is one throughout: $e^{tK}\in SO(3)$. Conversely, differentiating $(e^{tK}{)}^T{e}^{tK}=I$ at zero gives $K^T+K=0$. ThusThe quantifier matters: membership at one isolated time alone would not imply skewness.

Write $P(F)=DW(F)$ and $J=\det F>0$. For any skew matrix $K$, differentiate frame indifference along $e^{tK}F$:A matrix orthogonal to every skew matrix is symmetric. Therefore $P(F)F^T$ and the Cauchy stress tensor $T=J^{-1}P(F)F^T$ are symmetric.

Let $\mathcal{C}=D^{2}W(I)$ and assume $P(I)=0$. Differentiating symmetry of $P(F)F^T$ at $I$ in direction $H$ shows $\mathcal{C}[H]$ is symmetric for every $H$. Hence $c_{ijkl}=c_{jikl}$. As a Hessian, $\mathcal{C}$ also has pair-interchange symmetry $c_{ijkl}=c_{klij}$. Combining the two gives $c_{ijkl}=c_{ijlk}$. Equivalently $\mathcal{C}$ annihilates infinitesimal rotations. This proves both requested minor symmetries and shows exactly where the stress-free assumption enters. It is stress-free linearization of an objective energy.

For the particular energy, ${\sum }_i{v}_i^2=\mathrm{tr}(F^{T}F)$ and ${\prod }_i{v}_i=J$, soAt $I$ this is $[4\alpha +h^{\prime }(1)]I$. Therefore the reference is stress-free exactly when $\boxed{h^{\prime }(1)=-4\alpha }$. To differentiate again, use $\mathrm{cof}F=J{F}^{-T}$. The determinant and inverse derivatives at $I$ givewhich also proves the supplied hint. It follows thatFor a symmetric $e$ this becomesNo additional positivity of the bulk modulus follows from the stated smoothness of $h$ alone; this calculation identifies the moduli rather than imposing extra material assumptions.

The material symmetry group consists of invertible admissible reference transformations $H$ for which $W(FH)=W(F)$ for every admissible $F$. Its rotational part lies in $SO(3)$; a determinant-one reference convention restricts the full group to the corresponding special linear group. This right action is distinct from the spatial left action $W(QF)=W(F)$ of material frame indifference.

The proper cube-rotation group has $\boxed{24}$ elements. Choose the image of one outward face normal in six ways, then the image of an adjacent perpendicular normal in four ways; orientation fixes the third direction. No reflections are included. For an isotropic material every proper rotation belongs to the material symmetry group.

Combine material isotropy with the physical frame-indifference condition. The singular value decomposition may be written $F=R\mathrm{diag}(v_1,v_2,v_3)Q$ with $R,Q\in SO(3)$ because $\det F>0$. Left and right rotational invariance remove $R,Q$. Signed permutation rotations permit any permutation of the three diagonal stretches, soRight invariance alone, without the usual material frame indifference assumption, would not suffice to remove spatial orientation.

The strict Baker-Ericksen inequalities say that larger distinct principal stretches carry larger principal Cauchy stresses:A natural state of a hyperelastic material has zero stress. If two stretches differed, their equal zero stresses would contradict the strict inequality. Thus all three equal some $v>0$, and polar decomposition gives $\boxed{F=vQ}$, $Q\in SO(3)$. A weak non-strict inequality alone would not prove this conclusion, so its strict convention is explicit here.

For the radial deformation put $f(R)=r(R)/R$, $R=|\mathbf{x}|$, $\mathbf{\theta }=\mathbf{x}/R$. Since ${\partial }_{j}R={\theta }_j$,This symmetric tensor has radial eigenvalue $r^{\prime }$ and the repeated tangential eigenvalue $r/R$. Both are positive in the open annulus under the stated monotonicity/nonnegativity conditions. Therefore these are its principal stretches. The two cases below establish radial zero-energy rigidity by different arguments.

A homogeneous hyperelastic material has a stored energy per reference volume $W(F)$ depending on the local deformation gradient but not explicitly on the material point. Its first Piola-Kirchhoff response isA superposed rigid rotation changes the observer's description without changing the energy of the deformation. Material frame indifference is thereforeThis left-rotation invariance is distinct from a material isotropy assumption on right rotations. Differentiate it in an arbitrary matrix direction $H$:The Frobenius inner product then gives $Q^T\hat{T}(QF)=\hat{T}(F)$, proving $\hat{T}(QF)=Q\hat{T}(F)$.

For the converse, that stress covariance makes the derivative in $F$ of $W(QF)-W(F)$ zero. The positive-determinant matrices form a connected set, as also follows from polar decomposition, so this difference is a constant $c(Q)$ independent of $F$. Composition gives $c(Q_1{Q}_2)=c(Q_1)+c(Q_2)$. For fixed $F$, continuity of $W$ and compactness of $\mathrm{SO}(3)$ make $c$ bounded. But $c(Q^n)=nc(Q)$ for every positive integer, forcing $c(Q)=0$. Thus stress covariance and frame indifference of the energy are equivalent. Compactness removes the possible additive energy constant in the converse.

The determinant condition is $r^{\prime }(r/R{)}^2=1$, or $r^2{r}^{\prime }=R^2$. Integration gives $r^3=R^3+C$ and the outer boundary value fixes $C=t^3-1$. ThusFor $t>1$, $r(R)$ tends to the positive cavity radius $(t^3-1{)}^{1/3}$ as $R\downarrow 0$. Its tangential stretches grow like that radius divided by $R$, so $|F|\to \infty$. The deformation is singular at the origin and is interpreted as a finite-energy Sobolev deformation, with a point value irrelevant to its energy.

Since $r^{\prime }=R^2/r^2$ and the determinant penalty is zero almost everywhere,For fixed $t>1$, the first integrand is $O(R^6)$ near zero and the second is bounded. Thus the energy is finite despite the unbounded gradient. In particular the gradient is square integrable, as required for this competitor.

For large $t$, $r(R)/t\to 1$ uniformly on $[0,1]$. The first integral is $O(t^{-4})$, while ${\int }_0^{1}r(R{)}^{2}dR\sim t^2$. ConsequentlyThe spherical volume factor $R^{2}dR$ is what makes the $R^{-1}$ tangential-gradient singularity integrable in three dimensions.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

In the classical smooth setting, a deformation in continuum mechanics is a sufficiently regular, injective map $y:\Omega \to {\mathbb{R}}^3$ with $\det Dy>0$. Injectivity excludes material interpenetration, and a positive determinant preserves orientation and local volume. We use $C^1$ regularity for these kinematics and additional differentiability where stress-divergence identities are applied.

Cauchy's stress theorem states that the contact force on an oriented surface element at a current point depends on the point and its unit normal, rather than on the shape of the chosen subbody. Thus the contact force on the boundary of a current subbody $v$ has the form ${\int }_{\partial v}t(z,n)da$. Static force balance, for a body force $b$ per unit current volume, isfor every subbody. A shrinking pillbox gives $t(z,-n)=-t(z,n)$; a shrinking tetrahedron then gives linear dependence on the normal, $t(z,n)=S(z)n$. Applying the divergence theorem and using arbitrary subbodies givesIf body force is specified per unit mass instead, replace $b$ by $\rho b$. Symmetry of $S$ follows additionally from conservation of angular momentum in the absence of body couples; it is not a consequence of force balance alone.

Taking traces of the constitutive equation givesThe assumed finite limits of stress and strain make the right side bounded. Therefore $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$, and the limiting strain satisfies $\mathrm{tr}e=0$. The permitted interchange of derivatives and limits gives $\mathrm{tr}e=\mathrm{div}u$. DefineThen the limit stress is $S=-pI+2\mu e$. Passing to the limit in force balance and using $2\mathrm{Div}e=\Delta u+\nabla \mathrm{div}u$ givesThe pressure acts as the Lagrange multiplier enforcing the volume constraint. The differential equations determine it up to an additive constant, while a specified absolute limit stress can fix that constant.

Nonlinear elastic incompressibility is $\det Dy=1$. Expanding the determinant for $Dy=I+Du$ gives $\det (I+Du)=1+\mathrm{div}u+O(|Du{|}^2)$, so its first-order form is precisely $\mathrm{div}u=0$. The volumetric energy term $\lambda (\mathrm{div}u{)}^2/2$ becomes increasingly expensive as $\lambda$ grows, while the fixed positive $\mu$ leaves shear strain possible. Thus the infinite-bulk-stiffness limit is the incompressible limit of isotropic linear elasticity, with the same equilibrium operator as the steady Stokes equations.