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A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

An elastic deformation is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous stress is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the displacement field $u_i$ gives the infinitesimal strain tensor $e_{ij}=(u_{i,j}+u_{j,i})/2$. The Cauchy stress tensor is defined by traction $t_i={\sigma }_{ij}{n}_j$ on a plane of unit normal $\mathbf{n}$; absence of body couples makes ${\sigma }_{ij}={\sigma }_{ji}$.

Let $\mathcal{E}(e)$ be the recoverable elastic energy per unit mass. The local mechanical power per unit volume isbecause the symmetric Cauchy stress tensor annihilates the antisymmetric part of the velocity gradient. Conservation of energy during the deformation gives $\rho \dot{\mathcal{E}}={\sigma }_{ij}{\dot{e}}_{ij}$. Since $\dot{\mathcal{E}}=(\partial \mathcal{E}/\partial e_{ij}){\dot{e}}_{ij}$ for arbitrary symmetric strain rates,Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.

Take a stress-free reference state of mass density ${\rho }_0$. The Taylor series of the elastic energy starts quadratically:Using $\rho ={\rho }_0+O(|e|)$ gives ${\sigma }_{ij}=c_{ijkl}{e}_{kl}+O(|e{|}^2)$, the relation of linear elasticity. If the reference state has initial stress, the first-order expansion also retains that constant stress. Symmetry of stress and strain gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of elastic energy gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the elastic stiffness tensor is a symmetric map on a six-dimensional strain space: there are $6\cdot 7/2=21$ independent coefficients.

For twofold elastic symmetry, choose the symmetry axis as $x_3$. Rotation through $\pi$ is $Q=\mathrm{diag}(-1,-1,1)$, and transforms the infinitesimal strain tensor into $Qe{Q}^T$. Components $e_{11},e_{22},e_{33},e_{12}$ are even and $e_{23},e_{13}$ are odd. Invariance of elastic energy excludes every coupling between these parity classes. In the engineering strain ordering $(e_{11},e_{22},e_{33},2e_{23},2e_{13},2e_{12})$, with corresponding stress ordering $({\sigma }_{11},{\sigma }_{22},{\sigma }_{33},{\sigma }_{23},{\sigma }_{13},{\sigma }_{12})$, the symmetric stiffness matrix has the formIts symmetric even block has ten independent entries and its symmetric odd block has three. Hence a twofold axis leaves $13$ independent coefficients. No invariance under other rotation angles has been assumed.

For a homogeneous lossless elastic solid, insert the plane wave $u_i=\mathrm{Re}\{U{a}_i{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}\}$ into the equations of linear elasticity. Put $\mathbf{k}=\kappa \mathbf{n}$, $|\mathbf{n}|=1$, and $v=\omega /\kappa$. The resulting eigenvalue problem isThe minor and major symmetries make this acoustic tensor a real symmetric matrix. For a stable material it is positive-definite, so its three positive eigenvalues give three real positive phase velocities and therefore three slownesses $s_a=1/v_a$. Coincident eigenvalues are possible, so there need not be three distinct speeds. For distinct eigenvalues the displacement polarization are orthogonal; in an anisotropic solid they are generally quasi-longitudinal and quasi-transverse rather than exactly longitudinal and transverse. Stability, or at least strong ellipticity in elasticity, is required for the assertion about positive real speeds.

The elastic slowness surface is the locus of the three vectors $\mathbf{p}=\mathbf{n}/v_a(\mathbf{n})=\mathbf{k}/\omega$. Its equation isFor each wavefront normal there are three radial sheets, which may meet at degeneracies. To obtain the energy flux directly, multiply the elastic equation by ${\dot{u}}_i$. Forthe symmetry of stress gives ${\partial }_{t}W={\partial }_j({\sigma }_{ij}{\dot{u}}_i)$. Thus the instantaneous elastic-wave energy flux is $F_j=-{\sigma }_{ij}{\dot{u}}_i$. With a real branch polarization $\mathbf{a}$, harmonic averaging givesNow differentiate $c_{ijkl}{p}_j{p}_l{a}_k=\rho a_i$ along a regular slowness surface sheet and contract with $a_i$. The terms containing $d\mathbf{a}$ cancel by symmetry and the original eigenvalue equation, leavingEvery tangent $d\mathbf{p}$ is consequently perpendicular to $⟨\mathbf{F}⟩$: the energy flux is normal to the slowness surface. At a degenerate point this statement is made on the limiting branches, rather than assuming a unique normal.

For completeness, equality of the mean kinetic and strain energies gives $⟨W⟩=\rho {\omega }^2|U{|}^2|\mathbf{a}{|}^2/2$. Differentiating the acoustic tensor equation in the wavevector variables similarly givesThus the group velocity is the elastic energy velocity. The homogeneous elastic dispersion relation is degree one in $\mathbf{k}$, so differentiating $\omega (\lambda \mathbf{k})=\lambda \omega (\mathbf{k})$ at $\lambda =1$ gives $\mathbf{p}\cdot \mathbf{g}=1$.

The elastic wave surface is the unit-time locus of these group velocities. From an impulsive point source, the propagating wavefronts at time $t$ are its sheets scaled to $\mathbf{x}=t\mathbf{g}$. A particular source may fail to excite a mode in some directions, and its amplitudes require the source polarization and geometrical spreading; the wave surface determines the arrival geometry. It does not specify all near-field motion between the radiative fronts. Geometrically it is the envelope of the planes $\mathbf{p}\cdot \mathbf{x}=1$: tangent variation gives $d\mathbf{p}\cdot \mathbf{x}=0$, so $\mathbf{x}$ is the appropriately normalized normal to the slowness surface.

In a symmetry-plane cross-section write $\mathbf{p}=r(\theta )\mathbf{n}$, $\mathbf{n}=(\cos \theta ,\sin \theta )$ and $\mathbf{t}=(-\sin \theta ,\cos \theta )$. The two envelope conditions giveThe numerator is the signed curvature numerator of the slowness surface cross-section. At a simple inflection point, $h=v+v^{\prime \prime }=0$ and $h^{\prime }\ne 0$. Then ${\mathbf{g}}^{\prime \prime }=h^{\prime }\mathbf{t}$ and ${\mathbf{g}}^{\prime \prime \prime }=h^{\prime \prime }\mathbf{t}+2h^{\prime }{\mathbf{t}}^{\prime }$ are linearly independent. These are precisely the local conditions for an ordinary cusp. The slowness curvature and wave-surface cusps relation explains why a smooth nonconvex slowness surface can generate a cusped wavefront and multiple ray arrivals.

The example below uses a positive orthotropic stiffness in normalized units: $\rho =1$, $C_{11}=C_{22}=C_{33}=4$, $C_{44}=C_{55}=1.2$, $C_{66}=1$, and all normal off-diagonal stiffnesses zero. In the symmetry plane $n_3=0$, its two in-plane acoustic tensor eigenvalues are $v_{\pm }^2=(5\pm \sqrt{5+4\cos 4\theta })/2$, and its out-of-plane shear speed is $\sqrt{1.2}$. The in-plane shear sheet is nonconvex and produces the marked cusps; the construction uses an actual stable elastic tensor.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.

Use fixed spatial coordinates in a stationary heterogeneous Earth and define the operatorHere the elastic stiffness tensor has its minor and major symmetries, and the physical body force per unit volume is $q_i=\rho f_i$. The retarded elastodynamic Green tensor $G_i^j(\mathbf{x},\mathbf{\xi },t)$ is the displacement in direction $i$ due to a unit physical force impulse in direction $j$ at $\mathbf{\xi }$:It satisfies the actual Earth's homogeneous boundary conditions, including zero traction on the free surface and continuity of displacement and traction at bonded internal interfaces. Integration through $t=0$ gives $\rho (\mathbf{x}){\partial }_t{G}_i^j(0^{+})={\delta }_i^j\delta (\mathbf{x}-\mathbf{\xi })$, with $G(0^{+})=0$. No spherical symmetry is required.

For zero initial conditions, form the convolutionApplying $\mathcal{L}$ to this expression uses the defining Dirac delta function to give $\rho (\mathbf{x})f_i(\mathbf{x},t)$. Causality gives zero initial conditions, and each Green column supplies the required boundary conditions. The positive-energy uniqueness argument then identifies this expression with the physical response. The source factor is $\rho (\mathbf{\xi })$ because $f$ is force per unit mass, whereas the Green tensor was normalized to unit force.

A force dipole is a limiting pair of opposite point forces separated by a small vector. Put $+F_j$ at $\mathbf{\xi }+\epsilon \mathbf{d}/2$ and $-F_j$ at $\mathbf{\xi }-\epsilon \mathbf{d}/2$, and hold $D_{jk}=\epsilon F_j{d}_k$ fixed as $\epsilon \to 0$. Expansion of the two Dirac delta functions givesIts resultant force is zero, and integration by parts gives its radiated displacement as ${\int }_0^t{D}_{jk}(\tau ){\partial }_{{\xi }_k}{G}_i^j(\mathbf{x},\mathbf{\xi },t-\tau )d\tau$. Thus the source derivative of the elastodynamic Green tensor represents a dipole with force direction $j$ and separation direction $k$.

For the surface source choose a normal $\mathbf{n}$ from the minus side to the plus side and defineThe same normal is used in both tractions. If the elastic coefficients are continuous through the surface, the distributional gradient of a piecewise smooth displacement is ${\partial }_j{u}_i=({\partial }_j{u}_i{)}_{\mathrm{reg}}+b_i{n}_j{\delta }_{\Sigma }$. Its stress therefore has a singular part $c_{ijkl}{b}_k{n}_l{\delta }_{\Sigma }$. Taking the divergence includes the regular stress jump, so a field satisfying the source-free equations on both sides has equivalent sourceIn particular, the signs follow from the chosen jump convention, not from an unspecified orientation. The elastodynamic surface-jump representation is consequentlywith the Green tensor evaluated at $(\mathbf{x},\mathbf{\xi },t-\tau )$. The first term is a surface distribution of forces, and the second is a surface distribution of dipoles.

One can derive the same formula without multiplying discontinuous material coefficients by a surface delta. Apply integration by parts on the two sides of $\Sigma$ to the field $u$ and the reciprocal Green field $v$. Major symmetry cancels the volume strain terms; the spatial identity iswhen the temporal terms are handled by the retarded time convolution. More explicitly, take $v_i(\mathbf{\xi },\tau )=G_i^a(\mathbf{\xi },\mathbf{x},t-\tau )$, extend the zero-initial-data solution causally to $\tau <0$, and integrate over all source times. Integration by parts cancels the two temporal terms, with zero endpoint terms; the causal supports subsequently reduce the representation to $0\le \tau \le t$. The Green impulse contributes $-u_a(\mathbf{x},t)$ on the left. The two outward face normals are $\mathbf{n}$ and $-\mathbf{n}$, so their sum on the right is $-[u_i]t_i(v)+v_i[t_i]$. Moving it to the other side gives $[u_i]t_i(v)-v_i[t_i]$. The same identity applied to two force Green fields with interchanged source and receiver proves the reciprocal relation $G_i^j(\mathbf{x},\mathbf{\xi },t)=G_j^i(\mathbf{\xi },\mathbf{x},t)$; hence the displayed surface formula follows. On a bonded material interface the Green displacement and common Green traction are continuous: $[u_i]t_i(v)$ is then the unambiguous version of the dipole term, evaluated with the matching one-sided coefficients. External free-surface terms vanish because both tractions vanish.

For a fault in isotropic material, the Lamé parameters giveTangential slip obeys $\mathbf{b}\cdot \mathbf{n}=0$, and ordinary fault traction is continuous across the two faces. Therefore the source is a double-couple fault source withFor a uniform patch of area $A$ with $\mathbf{b}=b{\mathbf{e}}_1$ and $\mathbf{n}={\mathbf{e}}_3$, its integrated seismic moment tensor isThe $13$ dipole is a force in the slip direction separated in the normal direction; the $31$ dipole is a normal force separated in the slip direction. Their equal and opposite torques cancel. The symmetric trace-free seismic moment tensor has eigenvalues $M_0,-M_0,0$, with zero resultant force and torque. A temporal stress drop on a fault is not a jump of traction between its two simultaneously matching faces; it does not add a single-force layer to this tangential-slip representation.

For antiplane shear, the only nonzero infinitesimal strains are $e_{13}=e_{31}=w_{,1}/2$ and $e_{23}=e_{32}=w_{,2}/2$. Their trace is zero. The isotropic constitutive law ${\sigma }_{ij}={\lambda }_L{e}_{kk}{\delta }_{ij}+2\mu e_{ij}$ consequently giveswith every other component zero. The first two equations of motion vanish because the fields are independent of $x_3$. The third is $\rho w_{,tt}={\sigma }_{31,1}+{\sigma }_{32,2}$, henceThe PDF omits the final equals-zero; this is the equation supplied by momentum balance.

For a steadily moving field put $x=x_1-Vt$ and $\beta =\sqrt{1-V^2/c^2}$. In the subsonic case $|V|<c$, the equation becomes ${\beta }^2{w}_{xx}+w_{x_2{x}_2}=0$. With $y=\beta x_2$ it reduces to $w_{xx}+w_{yy}=0$. Thus $w$ is a harmonic function and, on a simply connected region, has a harmonic conjugate; there is a holomorphic function $H(z)$ such thatThis is steadily moving antiplane shear. The representation is local on more general domains unless the conjugate's periods vanish. The subsonic condition is essential: at or above the shear-wave speed the coordinate rescaling is degenerate or the equation is hyperbolic rather than this elliptic harmonic problem.

Let $a_c\ll R$ be the contact radius of the adhering neutrophil. Indentation is of order $a_c^2/R$, so the elastic strain is of order $a_c/R$ within a region of volume $a_c^3$. The elastic energy is consequently $G{a}_c^5/R^2$, while adhesion lowers the energy by order $J{a}_c^2$. Balancing these terms givesThis estimate requires weak adhesion, $J\ll GR$.

For an adhesive-cell rolling threshold, assume peeling the rear bonds dissipates work of order $J$ per unit contact area. Rolling through angle $d\theta$ sweeps contact area of order $a_{c}Rd\theta$, giving resisting moment $M_r\sim J{a}_{c}R$. A surrounding fluid of dynamic viscosity $\eta$ supplies a moment of order $\eta \dot{\gamma }{R}^3$. Equating moments yieldsA dimensionless numerical coefficient depends on the contact and near-wall Stokes flow. Fluid viscosity is indispensable to a rate estimate, even though the contact model alone specifies only $R,G,J$.

A perfectly reversible elastic adhesive contact does not lose net energy when an unchanged contact footprint translates. It therefore does not, by itself, justify a finite rolling threshold. The above estimate assumes adhesive hysteresis or irreversible bond peeling. If equilibrium adhesion is $J$ but the work lost during peeling and rebinding is $\Delta J$, retain $a_c\sim (J{R}^2/G{)}^{1/3}$ and replace $J$ by $\Delta J$ only in the resisting moment: ${\dot{\gamma }}_c\sim \Delta J{a}_c/(\eta R^2)$.

The printed diagram and the word “parallel” describe a Kelvin-Voigt model, not a series Maxwell fluid. In the depicted model both elements have displacement $x$ and their forces add:For $x=a{e}^{i\omega t}$ and the convention $G(\omega )=F_{\mathrm{amplitude}}/x_{\mathrm{amplitude}}$,The storage modulus and loss modulus here are element stiffnesses with units $\mathrm{N}{\mathrm{m}}^{-1}$; conversion to stress/strain moduli requires a geometric factor. If “response” instead denotes displacement per force, it is the compliance $1/(k+i\omega \mu )$. Specifying which ratio is used removes that convention ambiguity. The mean dissipated power is $\mu {\omega }^2|a{|}^2/2$, confirming the positive loss sign for $e^{i\omega t}$.

For completeness, the actual series Maxwell fluid analogue has a common force and additive spring/dashpot extensions. Thus $\dot{x}=\dot{F}/k+F/\mu$. With $\tau =\mu /k$,These different functions cannot both describe the printed parallel arrangement. The distinction also changes the elastic beam's response to a constant load.

For the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

If two harmonic solutions have the same body force and prescribed boundary traction, their difference has no body force and is traction-free. A nonzero difference would therefore be a free eigenfunction at that frequency. For any frequency outside the countable exceptional set found above, the difference must vanish, so the forced boundary-value solution, when it exists, is unique. At an eigenfrequency, a free eigenfunction can be added to any existing solution, and a resonant forcing may fail to admit a steady harmonic response. This distinguishes uniqueness from existence.

For the full time-dependent problem, let $\mathbf{w}$ be the difference between two fields with identical forces, tractions and initial displacement field and velocity. It obeys $\rho {\mathbf{w}}_{tt}=\nabla \cdot \mathbf{\sigma }(\mathbf{w})$, zero boundary traction, and zero initial data. Its energy uniqueness for traction-driven elasticity follows by direct integration:Multiply the equation by ${\mathbf{w}}_t$ and integrate by parts. Major symmetry turns the strain term into the derivative of elastic energy, giving $E^{\prime }(t)={\int }_{\partial \mathcal{D}}{\mathbf{w}}_t\cdot \mathbf{t}(\mathbf{w})dS=0$. Stable elasticity makes $E\ge 0$, and $E(0)=0$ gives ${\mathbf{w}}_t=0$. Thus $\mathbf{w}$ equals its initial value zero. Even rigid zero-strain motions cannot survive these initial conditions.

The PDF correctly specifies zero initial displacement field and zero initial velocity; the TeX duplicates displacement field in the second condition. The complete causal displacement field is unique, including when its Fourier representation contains resonant modal contributions fixed by the initial data.

In infinitesimal elasticity and plasticity in continuum mechanics, decompose the symmetric strain as $e=e^e+e^p$. For incompressible isotropic elasticity, ${\sigma }^{\prime }=2G{e}^e$, with $G$ the shear modulus and both strains traceless. The associated flow rule for the Von Mises yield criterion has ${\dot{e}}^p=\Lambda {\sigma }^{\prime }$ after absorbing a positive factor into $\Lambda \ge 0$. Plane strain has $e_{33}=0$, soIf ${\sigma }_{33}^{\prime }=0$ initially, it remains zero by this homogeneous scalar differential equation. Then $e_{33}^e=0$ and ${\dot{e}}_{33}^p=0$, so $e_{33}^p=0$ is consistent with zero initial plastic component. This proves consistency, not a claim that arbitrary prestressed plane-strain histories must have zero separate components. The out-of-plane shear components can consistently be zero by the same argument.

Write $p=-\mathrm{tr}\sigma /3$, so ${\sigma }_{33}=-p$. The proposed in-plane parameterization hasIts squared contraction is $2k^2({\sin }^{2}2\varphi +{\cos }^{2}2\varphi )=2k^2$, proving the yield condition identically. Its principal deviatoric stresses are $k,-k,0$. Define the two slip line families by tangent unit vectorsThus $\alpha$ lines have tangent angle $\varphi$ and $\beta$ lines have tangent angle $\varphi +\pi /2$. Their normal extensions from the associated plastic strain rate vanish, while their shear stresses have maximum magnitude $k$; they are not the principal stress directions.

In the absence of body force, differentiate the stress parameterization in the two equilibrium equations. One obtainsThe symmetric matrix multiplying $2k\nabla \varphi$ has eigenvectors $a,b$ with eigenvalues $+1,-1$. Taking its contraction with these vectors provesHence the Hencky stress relations areConstants can differ between lines, and these signs depend on the specified negative shear-stress convention.

Place the notch tip at the origin, with the opening towards negative $x_1$ and tensile loading in the transverse direction. For a usual reentrant notch, $0\le \gamma \le \pi /2$. The upper face has tangent angle ${\vartheta }_f=\pi -\gamma$. Choose its tangent $t=(\cos {\vartheta }_f,\sin {\vartheta }_f)$ directed away from the tip and outward material normal $n=(-\sin {\vartheta }_f,\cos {\vartheta }_f)$ pointing into the notch. Direct rotation of the stress gives the normal and tangential traction componentsTake the notch faces traction-free, as intended by the notch problem. Then the cosine vanishes. Select the tensile branch $\sin (2\varphi -2{\vartheta }_f)=1$, for which the face-tangential normal stress is $2k$. Thus the upper face has $\boxed{p=-k,{\varphi }_f=\pi /4-\gamma (\mathrm{mod}\pi )}$. Choosing the opposite branch would describe compression rather than the tensile field.

Construct the centred slip-line fan at a traction-free V-notch to connect this surface state to the forward symmetry axis. Near the upper face there is a constant-stress region. It joins a fan bounded by raysInside the fan choose $\beta$ lines radial and $\alpha$ lines circular about the tip, so their angle parameter is $\varphi =\vartheta -\pi /2$. Along a circular $\alpha$ line the first invariant givesThe second invariant is automatically constant along each radial $\beta$ line because $p,\varphi$ are independent of radius. Thus this is an actual equilibrium fan, not just an assumed transport of a surface value. At ${\vartheta }_2$ it matches the free-face constant state. At ${\vartheta }_1$ it matches a forward constant-stress region withReflect the upper field across the symmetry axis to obtain the lower field. The forward region has zero shear, respects this symmetry, and contains the positive $x_1$ axis. There $\sin 2\varphi =-1$, and henceThis value holds in the fully plastic notch-tip region described by the field. It is not a uniform stress assertion for the entire axis of an arbitrary finite specimen, including any elastic or far-field region.

Take $h$ positive downwards and write $B$ for the filament bending modulus. Under the material-modulus interpretation of the printed $A$ as Young's modulus, the circular cross-section gives the second moment of area and bending modulusIndeed a small curvature $\kappa$ gives axial strain $-y\kappa$, so integrating $A{y}^2{\kappa }^2/2$ over the section gives bending energy $B{\kappa }^2/2$ per unit length. Keeping this distinction is essential to the powers of radius in the answer.