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When only the bulk modulus differs from an isotropic matrix, the elastic polarization is spherical: $\tau =(\kappa -{\kappa }^0)\mathrm{tr}(\epsilon )I$. The double trace of the isotropic elastic strain Green kernel makes the trace equation local, giving the displayed formula. Individual strain components still depend on a nonlocal potential of the spherical polarization, so the complete infinitesimal strain tensor is not generally a local multiple of its reference value.

The deviatoric strain is the trace-free part of the infinitesimal strain tensor. It removes the isotropic volume change. The decomposition is orthogonal for the Frobenius inner product, because a trace-free matrix has zero Frobenius inner product with $I$. ConsequentlyFor $\epsilon =\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$, its trace is $\mathrm{div}v$.

A tensor contraction over two paired indices. For real second-order tensors, it gives the displayed Frobenius inner product. For a fourth-order elasticity tensor and a second-order infinitesimal strain tensor, $(C:ϵ{)}_{ij}={\sum }_{kl}{C}_{ijkl}{ϵ}_{kl}$ remains a second-order tensor; double contraction need not always give a scalar.

An elastic deformation is recoverable when the deforming loads are removed. In a local elastic constitutive model the stress depends on the present deformation. When this response also derives from a stored elastic energy, the material is a hyperelastic material. In the small-deformation approximation, the relevant kinematic variable is the infinitesimal strain tensor.

For a homogeneous comparison elastic stiffness tensor $C^0$, the elastic strain Green operator maps an elastic polarization to minus the induced infinitesimal strain tensor. In a bounded domain the induced displacement field has zero boundary data; in an infinite medium it is the decaying perturbation. The defining static elastic equilibrium equation is $\nabla \cdot (C^0\mathrm{sym}\nabla u+\tau )=0$. The bounded-domain operator is self-adjoint, annihilates constant elastic polarizations, has zero mean output, and satisfies $\Gamma C^0\Gamma =\Gamma$.

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.

The infinitesimal strain tensor is the symmetric part of a displacement gradient tensor. It is the linear approximation to a finite strain measure for small displacement gradients. An infinitesimal rigid rotation contributes no strain, so the elastic energy depends on the symmetric part rather than the entire gradient.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.

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.

The infinitesimal strain is $e_{ij}=(u_{i,j}+u_{j,i})/2$. A constant translation has zero gradient. For the remaining term, $u_i={ϵ}_{ikl}{\omega }_k{x}_l$ has $u_{i,j}={ϵ}_{ikj}{\omega }_k$, a skew-symmetric matrix. ThereforePhysically $\mathbf{a}$ is a translation and $\mathbf{\omega }\times \mathbf{x}$ is an infinitesimal rigid rotation, with axis along $\mathbf{\omega }$ and rotation angle $|\mathbf{\omega }|$. It is the linearized rotation displacement, not the exact displacement of a general finite rotation. This distinction matters because the infinitesimal strain omits quadratic displacement-gradient terms.

The infinitesimal strain tensor is the symmetric part of the displacement gradient tensor:Its diagonal entries are infinitesimal normal extensions; twice an off-diagonal entry is the corresponding engineering shear strain. An infinitesimal rigid rotation has an antisymmetric displacement gradient tensor and therefore contributes no linearized strain.

A Cartesian Cartesian second-rank tensor has components transforming under a constant orthogonal change of axes asHere ${\mathbf{X}}^{\prime }=P\mathbf{X}$ and ${\mathbf{u}}^{\prime }=P\mathbf{u}$. By the chain rule and orthogonality,Thus the displacement gradient is a Cartesian second-rank tensor. Its symmetric part transforms in the same way:Hence the infinitesimal strain tensor is also second-rank. This is a passive coordinate rotation, under which both reference and spatial coordinates rotate; it does not assert finite-deformation frame indifference of the linearized strain formula.

Differentiate the displacement components. The shear strains involving $z$ vanish because $u_z=-w_x$ and $v_z=-w_y$. The full infinitesimal strain tensor isIn particular $\mathrm{tr}\mathcal{E}=(2\nu -1)(a+c)z/(1-\nu )$. Substituting the Lamé coefficients into the constitutive law givesFor example ${\tau }_{zz}=\lambda \mathrm{tr}\mathcal{E}+2\mu e_{zz}=0$: the thickness-direction strain supplies exactly the plane-stress Poisson correction. Also ${\tau }_{xy}=-Ebz/(1+\nu )$, confirming the off-diagonal factor.

Write the in-plane displacement field as $(u,v)$. The infinitesimal strain tensor gives $2e_{XY}=u_Y+v_X$, $e_{YY}=v_Y$, and $\mathrm{tr}e=u_X+v_Y$. The two nontrivial zero-traction conditions becomeEquivalently for the complex amplitudes they are $U^{\prime }(0)+ikV(0)=0$ and $ik\lambda U(0)+(\lambda +2\mu )V^{\prime }(0)=0$. Because the constitutive equation is real and linear, vanishing of these complex amplitudes enforces the conditions at every position and time.

The infinitesimal strain tensor $\epsilon (v)=\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$ and the divergence are linear in $v$. Consequently the given $a$ is a symmetric bilinear form, andFor a matrix-valued function, $\Vert \cdot {\Vert }_0$ here uses the Frobenius inner product pointwise and then the L2 norm. All expressions are finite for $v\in H^1(\Omega ;{\mathbb{R}}^3)$.

Expand the quadratic term at $\theta v+(1-\theta )w$. The linearity of $\ell$ cancels its contribution to the difference, givingFor $0\le \theta \le 1$, this is nonnegative. Thereforeso $J$ is a convex functional on the whole Sobolev space. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the zero-boundary Sobolev space will instead use the Korn inequality in part (b).

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

Take the difference $w$ of two solutions with the same body force, initial displacement and initial velocity. It satisfies the homogeneous linear elasticity equations. With $e=e(w)$ the infinitesimal strain tensor, define the bulk energyThe usual stability assumptions make the elastic quadratic form nonnegative and $\rho >0$. Multiplying momentum balance by $w_t$, using integration by parts and using symmetry of the stress tensor givesAll interior terms cancel: the strain term is exactly the time derivative of elastic energy.

For the restoring boundary law, $t_w=kw$. The prescribed support displacement cancels in the difference. Since $k$ is a time-independent symmetric matrix,Hence the elastic energy uniqueness with restoring boundary springs identity isThe boundary contribution is nonnegative because $k$ is negative semidefinite. The initial difference and its velocity vanish, so the conserved energy is zero. Its kinetic term implies $w_t=0$ everywhere; the zero initial displacement then gives $w=0$. This proves uniqueness without needing to exclude rigid displacements separately.

Physically, $-k$ is a nonnegative boundary spring-stiffness matrix, and the prescribed vector $U$ is the moving support position. The traction restores the displacement towards that support; null directions of $k$ have no spring force. The interpretation and conserved-energy proof use a fixed spring matrix, as indicated by the given boundary law.

On the mixed part $S_1$, write $t=\sigma n$. The condition isThus the tangential traction and normal displacement are prescribed. The normal traction is a constraint reaction; tangential displacement is free. Only the tangential component of $F$ matters. On $S_2$, the entire displacement is prescribed.

For the difference of two such solutions, $w\cdot n=0$ and $(I-n{n}^{\mathsf{T}})t_w=0$ on $S_1$. Consequently $w_t$ is tangential and $t_w$ normal, so $w_t\cdot t_w=0$. On $S_2$, $w_t=0$. The boundary power vanishes everywhere, giving $d{E}_{\mathrm{bulk}}/dt=0$. The same zero-initial-energy argument gives $w=0$. Therefore the mixed boundary problem also has at most one solution.

Use the local momentum equation in the current volume:Symmetry of the Cauchy stress tensor and the product rule giveApply the divergence theorem and use $t_j={\sigma }_{jk}{n}_k$. The mean stress and strain boundary identities follow:For infinitesimal strain tensor, $e_{ij}=(u_{i,j}+u_{j,i})/2$, and a second application of the divergence theorem gives

To first order, evaluate self-gravity in the undeformed uniform sphere. The shell theorem says that only the enclosed mass $M(r)=4\pi \rho r^3/3$ contributes to the inward field. ConsequentlyThe field vanishes at the centre and increases linearly to its surface value.

The requested radius change concerns the final traction-free static elastic equilibrium, not the acceleration immediately after gravity is switched on. In equilibrium the boundary traction and acceleration vanish. Spherical symmetry givesThe homogeneous linear elasticity law commutes with volume averaging:Thus $\overline{e}=e_{0}I$ with $e_0=-\rho ga/(15\kappa )$. This conclusion is about the mean strain, not a claim that the local strain is uniform.

Let the boundary displacement be $u(a)=u_{a}n$. Since ${\int }_S{n}_i{n}_{j}dS=(4\pi a^2/3){\delta }_{ij}$, the boundary identity gives $\overline{e}=(u_a/a)I$. Therefore the equilibrium contraction of a self-gravitating elastic sphere isThe shear modulus cancels because the mean deformation is purely volumetric.

The full radial field provides a useful check without assuming uniform deformation. For $u=f(r)e_r$, momentum balance isRegularity at the centre gives $f=Ar+B{r}^3$, withfrom ${\sigma }_{rr}(a)=0$. Substitution reproduces $f(a)=-\rho g{a}^2/(15\kappa )$. During the transient, the acceleration term in the mean-stress identity remains; gravitational feedback from the small density and shape changes is higher order in the equilibrium calculation.

Write $⟨q⟩=|\Omega {|}^{-1}{\int }_{\Omega }qdx$, $E=⟨\epsilon ⟩$ and $S=⟨\sigma ⟩$. Use the infinitesimal strain tensor, symmetric stress, zero body force, and the outward unit normal $n$. The divergence theorem applied separately to the two displacement derivatives givesFor static elastic equilibrium, ${\sigma }_{ik,k}=0$, and therefore ${\partial }_k({\sigma }_{ik}{x}_j)={\sigma }_{ij}$. With traction $t_i={\sigma }_{ik}{n}_k$, the divergence theorem now givesThe right-hand side is automatically symmetric because the volume stress tensor is symmetric; it can also be written as the integral of $(t_i{x}_j+t_j{x}_i)/2$.

To prove the work identity, put $v=u-Ex$. The constant symmetric tensor $S$ is equilibrated, so integration by parts givesHere the two cross terms each equal $|\Omega |S:E$, and the last constant term restores one of them. Division by the volume proves the required identity. In particular, affine boundary displacement makes $v=0$ on the boundary and yields the Hill-Mandel energy identity. The zero-body force hypothesis matters: otherwise the first integration by parts has an additional volume-force term.