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.

Let the boundary split into a displacement part ${\Gamma }_D$ and a traction part ${\Gamma }_N$. The linearized stored elastic energy functional isIts corresponding total potential subtracts the work of fixed body and surface loads:The problem in the calculus of variations is to minimize $\Pi$ among admissible displacements with $u=u_D$ on ${\Gamma }_D$. For a stable quadratic elastic energy this is the usual energy minimization formulation. Its stationarity condition suffices to derive equilibrium.

Vary by $u+sv$, with $v=0$ on ${\Gamma }_D$. Major symmetry combines the two differentiated quadratic terms, givingIntegration by parts, first with test functions compactly supported in the interior, gives $-\mathrm{Div}{S}^{\mathrm{lin}}=f$. Then arbitrary traces on ${\Gamma }_N$ give the natural traction condition. The complete boundary value problem isFor a homogeneous hyperelastic material the elasticity tensor is constant. In isotropic linear elasticity the constitutive equation is $S^{\mathrm{lin}}=\lambda \mathrm{div}uI+\mu (Du+D{u}^T)$, giving the familiar Navier-Cauchy equation $-\mu \Delta u-(\lambda +\mu )\nabla \mathrm{div}u=f$.

The displacement field is $u(x)=y(x)-x$, and its gradient is $Du=Dy-I$. Linear elasticity expands about an undeformed stress-free reference state, taking $y=x+ϵv$ with $|Du|\ll 1$ and external loads of the same small order. A rigid translation need not itself be small; it is the deformation gradient that controls the infinitesimal strain tensor.

For $W\in C^2$ with $DW(I)=0$, a Taylor series givesThe constant $W(I)$ can be set to zero. The tensor $\mathbb{A}$ is the elasticity tensor. Commutation of continuous mixed partial derivatives proves its major symmetry:Material frame indifference at a stress-free reference also makes the quadratic energy depend only on the infinitesimal strain tensor $e=(H+H^T)/2$, leading to the usual minor symmetries after spatial and reference indices are identified.

In this scaling, an external force of order $ϵ$ may be evaluated on the reference geometry: its change due to the order-$ϵ$ displacement contributes only at order ${ϵ}^2$. Thus the leading forces are fixed dead loads. Likewise $F=I+O(ϵ)$, $J=1+O(ϵ)$ and $S=O(ϵ)$, so the Piola transformation gives $T=JS{F}^{-T}=S+O({ϵ}^2)$. Hence Cauchy and first Piola stresses agree to linear order. This reasoning uses the usual small-load, stress-free linearization; an order-one prestress or leading follower-load effect requires a different expansion.

The Green-Lagrange strain tensor has eigenvalues $({\lambda }_j^2-1)/2$. Its Frobenius norm and trace are invariant under the orthogonal principal-axis change, soFor the homogeneous dilation $F=\rho I$, put $K=\mu +3\lambda /2$. Then $E=({\rho }^2-1)I/2$. Differentiating the energy through $E$ gives the second Piola-Kirchhoff stress tensor $2\mu E+\lambda \mathrm{tr}(E)I=K({\rho }^2-1)I$, and multiplication by $F$ gives the first Piola-Kirchhoff stress tensor. ThusThe normals of the reference and deformed spheres point in the same radial direction. As $\rho \downarrow 0$, the reference traction magnitude $|TN|$ tends to zero, while the current traction magnitude $|Sn|$ tends to infinity, provided $K>0$, as for physically stable Lamé parameters. The negative signs of the stress coefficients for $0<\rho <1$ indicate compression.

There is no contradiction: the current area is ${\rho }^2$ times the reference area, so the same force divided by the vanishing current area has the larger stress. Indeed $TN={\rho }^{2}Sn$. The limiting map at $\rho =0$ is not itself an admissible deformation in continuum mechanics; the statement concerns the approach through positive $\rho$. A degenerate choice $K=0$ would not have the asserted growth without bound.

For $R=|x|>0$, $DR=\theta$ and $D\theta =(I-\theta \otimes \theta )/R$. Differentiating the spherically symmetric deformation givesFor an orientation-preserving spherically symmetric deformation with positive radius, $r>0$ and $r^{\prime }>0$. The radial direction and its perpendicular plane are eigenspaces of this positive symmetric matrix, soAt a positive diagonal matrix $D$, every diagonal rotation with two minus signs commutes with $D$. The supplied equivariance relation therefore forces $\hat{T}(D)$ to be a diagonal matrix: each off-diagonal entry changes sign under one such rotation while $D$ stays fixed. Varying a positive diagonal entry shows that the diagonal entries of $\hat{T}(D)$ are ${\Phi }_{,j}$. Permutation symmetry of $\Phi$ implies ${\Phi }_{,2}={\Phi }_{,3}$ when ${\lambda }_2={\lambda }_3$.

Choose a rotation whose first column is $\theta$. Applying equivariance to $F=QD{Q}^T$ now giveswith the derivatives evaluated at $(r^{\prime },r/R,r/R)$. Thus the stress has one radial and one repeated tangential principal value, independently of the choice of tangent basis.

The surface element relation is Nanson's formula. The force on a corresponding pair of current and reference surface elements must agree:Therefore the first Piola-Kirchhoff stress tensor isIt measures force per reference area, with a current force direction and a reference normal direction.

In components, differentiate $T_{iA}=S_{ij}(y(x))(\mathrm{cof}F{)}_{jA}$. The Piola identity says ${\partial }_A(\mathrm{cof}Dy{)}_{jA}=0$. One proof writes the cofactor matrix as $\frac{1}{2}{\epsilon }_{jkl}{\epsilon }_{ABC}{y}_{k,B}{y}_{l,C}$; differentiating in $A$ produces symmetric second derivatives contracted against antisymmetric epsilon symbols, hence zero. The chain rule consequently givessince $F_{kA}(\mathrm{cof}F{)}_{jA}=J{\delta }_{kj}$. Thus, with the coordinate composition made explicit,The distinction between reference divergence and current divergence is essential to this identity.

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.

Retain the material frame indifference assumption used in part (c), in addition to material isotropy. Put $p=u_{,1}$, $q=u_{,2}$ and $s=p^2+q^2$. For this antiplane shear,The invariant identities give $I_B=I{I}_B=3+s$ and $II{I}_B=1$. Hence the restriction of the strain energy density isAntiplane energy does not determine the full stress: the restriction gives one scalar energy along antiplane gradients. It does not determine the full stress derivative with respect to every entry of $F$.

Indeed, using the isotropic hyperelastic stress invariant representation and differentiating the energy, with $h_j$ denoting the corresponding partial derivatives, givesUsing $S=J^{-1}T{F}^T$, the general objective isotropic Cauchy stress tensor isAll derivatives are evaluated at the current invariant triple. The extra terms generally remain nonzero on antiplane gradients, so the printed assertion that the full stress must be $\beta B$ does not follow.

A stress-free, objective isotropic counterexample is the compressible neo-Hookean solid $W(F)=\frac{\mu }{2}(|F{|}^2-3)-\mu \log J$, with $\mu >0$. Its derivative is $T=\mu F-\mu F^{-T}$, so on antiplane gradientsTake $u(x_1,x_2)=x_1$. Then $B_{22}=1$ but $S_{22}=0$, whereas $B_{13}=1$ and $S_{13}=\mu$. The diagonal entry would force $\beta =0$, contradicting the shear entry. Thus this is not a matter of choosing the right scalar coefficient.

For the additional one-invariant constitutive assumption $W(F)=\hat{w}(I_B)$, the stronger formula does hold, withThis is a valid special case, rather than a consequence of material isotropy alone. For the general restricted energy, the shear entries are still fixed: $S_{31}=2w^{\prime }(s)u_{,1}$ and $S_{32}=2w^{\prime }(s)u_{,2}$, since $w^{\prime }=h_1+h_2$ along $(3+s,3+s,1)$. The full normal stress needs the constitutive information that the printed claim omits. If material isotropy alone is interpreted without material frame indifference, even the first reduction to $|Du{|}^2$ needs that additional material frame indifference assumption.

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.

Define the invariants by the characteristic polynomialBecause $B=F{F}^T$ has positive eigenvalues ${\lambda }_j^2$, its invariants are the elementary symmetric functions in those eigenvalues:The Frobenius norm gives $|F{|}^2=\mathrm{tr}(F^{T}F)={\sum }_j{\lambda }_j^2$. Write $J=\det F>0$. Since $\mathrm{cof}F=J{F}^{-T}$, its singular values are $J/{\lambda }_j$, soFinally $\det B=\det F\det F^T=J^2$. ThereforeThese identities also give $I_B=\mathrm{tr}B$ and $I{I}_B=((\mathrm{tr}B{)}^2-\mathrm{tr}(B^2))/2$.

A homogeneous hyperelastic material has a strain energy density per unit reference volume $W(F)$ depending on the deformation gradient, with no explicit dependence on the reference configuration. For orientation-preserving gradients $F\in G{L}^{+}(3)$, its first Piola-Kirchhoff stress tensor is the derivative:Equivalently the infinitesimal stored-energy change is $DW(F)[H]=T:H$. Reference and spatial indices are distinguished here by upper- and lowercase index letters.

The material symmetry group may be written $G_W=\{H\in G{L}^{+}(3):W(FH)=W(F)\text{ for every }F\in G{L}^{+}(3)\}$. A volume-preserving relabelling convention restricts the candidate $H$ to $SL(3)$; the material isotropy criterion is the same. The material has material isotropy when $SO(3)\subseteq G_W$, meaning $W(FQ)=W(F)$ for every proper rotation $Q$. If only rotational symmetries are included in the definition of the group, material isotropy means that group equals $SO(3)$. This right-rotation invariance is distinct from material frame indifference, the left-rotation invariance $W(QF)=W(F)$.

The polar decomposition in continuum mechanics gives the unique factorizationswhere $U,V$ are positive-definite symmetric matrices. Their common positive eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$ are the principal stretches, equivalently the singular values of $F$. They describe the local stretch magnitudes after separating out a rigid rotation.

For a differentiable crack path, the local quasistatic Griffith laws with constant positive toughness areThey express irreversibility, local stability, and energy balance during growth. A strictly subcritical release rate forces the crack to stop.

Fix a crack parameter $s$ and let $u(s,t)$ denote the equilibrium displacement with the prescribed time-dependent loads. The scalar equilibrium problem is linear and has a unique solution for its fixed Dirichlet data. Both body force and boundary displacement scale by $t$, soThe minimized potential energy is quadratic under this scaling: the strain term scales by $t^2$ and so does the load term $-f(t)u(s,t)$. Hence $E(s,t)=t^{2}E(s,1)$ andEquivalently the crack-tip coefficient scales by $t$ and its squared energy-release formula gives the same result.

Put $s_0=\ell (0)$. Finiteness and continuity of $G(s,1)$ make it bounded by some finite $M\ge 0$ on a neighborhood of $s_0$. Continuity of $\ell$ supplies a positive interval on which $\ell (t)$ stays in that neighborhood. Shorten it to $[0,t_0]$ so that $t_0^{2}M<G_c$. Then $G(\ell (t),t)\le t_0^{2}M<G_c$, and complementarity forces $\dot{\ell }(t)=0$ throughout. ThereforeNo differentiating an unknown crack path or assuming its length already constant is needed; its continuity first gives the local release-rate bound.

Use $-\pi <{\theta }_s<\pi$, with the crack along the negative radial axis. In the coefficient convention used here, the leading crack-tip expansion isThe square-root term is harmonic and has zero angular derivative on both crack faces. Regular contributions may include a constant and bounded-gradient terms under smooth loading.

If $E(s)$ is the minimized elastic-plus-load potential, the energy release rate is $\boxed{G(s)=-dE(s)/ds}$. With this particular square-root coefficient normalization,For example, the small-circle energy flux is $\int [\mathcal{W}{n}_1-\mu u_r{u}_{x_1}]R_{s}d{\theta }_s$. The singular term gives $\mathcal{W}=\mu k_s^2/(8R_s)$, $u_r=k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$ and $u_{x_1}=-k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$. Integration over $(-\pi ,\pi )$ produces the stated factor. Other stress-intensity normalizations change that factor.

For toughness $G_c>0$, local stability against infinitesimal extension means $\boxed{G(s)\le G_c}$, since the derivative of $E(s)+G_c(s+1)$ is nonnegative. Global energetic stability against every extension is the stronger condition $E(s)\le E(r)+G_c(r-s)$ for all $r\ge s$.

In full isotropic linearized elasticity,For the antiplane field this becomesAhead of the tip, ${\theta }_s=0$, the singular gradient is $(0,k_s/(2\sqrt{R_s}))$. ThusThis is stronger than the printed remainder $o(R_s^{-1})$. If $k_s$ denotes a nonnegative intensity, it has the printed leading coefficient; for a signed singular coefficient its absolute value is essential because the left side is a norm.

An antiplane displacement has the form $v(x_1,x_2,x_3)=(0,0,u(x_1,x_2))$, independent of the out-of-plane coordinate. Only $e_{13}=e_{31}=u_{x_1}/2$ and $e_{23}=e_{32}=u_{x_2}/2$ are nonzero, so its strain energy per unit out-of-plane thickness is $\mu |\nabla u{|}^2/2$.

The equilibrium minimizes total potential energyHere $f$ is the out-of-plane body-force component; the two crack faces have independent traces. For a variation $\eta$ with zero outer-boundary trace, stationarity givesIntegrating by parts first for interior test functions and then for variations free on either crack face yieldsThese are the weak equilibrium equation, prescribed displacement, and zero traction. The tip behavior is selected by finite energy; it is not an additional prescribed tip value.