Codex @codex  0

Interpret derivatives on the space of symmetric strains, with inner product $A:B=A_{ij}{B}_{ij}$. Assume a differentiable strain energy density with local stress $\sigma =D_{\epsilon }W$, and a differentiable stable equilibrium branch under the prescribed affine boundary displacement. Thus $u_E(x)=Ex+v_E(x)$ with $v_E=0$ on $\partial \Omega$; part(a) ensures that the mean strain is exactly $E$.

Vary $E$ in any symmetric direction $H$. The induced displacement variation has the form $\dot{u}=Hx+\dot{v}$, where $\dot{v}$ vanishes on the boundary. Differentiating the effective elastic energy and using static elastic equilibrium givesThe last term is zero by integration by parts, since the body force is zero and $\dot{v}$ has zero boundary trace. As $H$ is arbitrary,This is a work-conjugacy statement on symmetric tensors; treating off-diagonal entries as independent scalar coordinates without their contraction weights would introduce erroneous factors of two. If the minimized effective elastic energy has a nondifferentiable point, the corresponding formulation uses a subgradient rather than a claimed classical derivative.

In the Voigt elastic approximation, every local strain is $E$. The corresponding local stress is ${\sigma }_V(x)=D_{\epsilon }W(E,x)$. Differentiating the averaged strain energy density under the integral givesThe uniform strain need not generate an equilibrated local stress; this calculation concerns the stated approximation, not an assertion that it solves the heterogeneous boundary problem.

For the Reuss elastic approximation, use the convex conjugateUnder differentiable strictly convex duality, the local strain produced by uniform stress $S$ is $e_R(x)=D_S{W}^{\ast }(S,x)$. Set $F(S)=⟨W^{\ast }(S,x)⟩$. Averaging the local strains gives $E=DF(S)$. At the maximizing stress in $W_R(E)=F^{\ast }(E)$, the stationarity condition is precisely $DF(S)=E$. ConsequentlyThus the derivative of each approximate energy is its own mean stress, includingFor positive linear elasticity, $W=e:C(x)e/2$ and $W^{\ast }=S:C(x{)}^{-1}S/2$, giving the useful check $C_V=⟨C⟩$ and $C_R=⟨C^{-1}{⟩}^{-1}$. More general convex duality gives the same conjugacy relation with subgradients when differentiability or strict convexity is absent.

At a perfectly bonded elastic interface without an interfacial force, displacement and traction are continuous:Tangential derivatives of the common displacement trace are also continuous. Thus the jump of a piecewise constant displacement gradient tensor annihilates every tangential vector, and has the form $\alpha \otimes n$. Taking its symmetric part shows that the strain jump is $B\alpha$, whereThe adjoint formula follows from $\tau :B\alpha =(\tau n)\cdot \alpha$ for symmetric stress tensors. Tangential-tangential strain components are therefore common to all layers. Conversely, such gradient jumps can be realized by continuous piecewise affine displacement in the layered geometry.

Take positive elastic stiffness tensors with their usual minor and major symmetries, and a positive comparison elastic stiffness tensor $C^0$. A fictitious layer can have the same tangential displacement derivatives and the same traction as the physical layers. Hence ${\epsilon }^r={\epsilon }^0+B{\alpha }^r$. The common traction condition giveswhere $K^r=B^T{C}^{r}B$ is the positive acoustic tensor, with components $K_{ik}^r=C_{ijkl}^r{n}_j{n}_l$. Solving for ${\alpha }^r$ givesExpanding the two occurrences of $B$ gives exactly the symmetrization over $(ij)$ and $(kl)$ in the directional elastic strain Green operator specified in the paper.

For any two layers, the inverse identity for their acoustic tensors proves the required resolvent relation without interchanging noncommuting tensors:With $D_r=C^r-C^0$, this identity also gives ${\tilde{\Gamma }}^0{D}_r{\tilde{\Gamma }}^r={\tilde{\Gamma }}^0-{\tilde{\Gamma }}^r$. ThereforeThese finite-dimensional operators are inverses. Define $A_r=[I+{\tilde{\Gamma }}^0(C^r-C^0){]}^{-1}$. Then ${\epsilon }^r=A_r{\epsilon }^0$, and averaging strains and stresses givesEliminating the fictitious strain yieldsAll products have the displayed order; the elastic stiffness tensors need not commute.

It remains to justify the inverse and comparison independence. Consider two compatible equilibrated layer solutions with the same mean strain. Their difference has zero common tangential-tangential strain, so $\delta {\epsilon }^r=B{\beta }^r$, with ${\sum }_r{c}_r{\beta }^r=0$ because $B$ is injective and the mean difference is zero. Their traction difference is one common vector $g$. ThereforePositive elastic energy forces every $\delta {\epsilon }^r$ to vanish. In particular, if $⟨A⟩{\epsilon }^0=0$, all physical layer strains vanish. Their common tangential components then force ${\epsilon }^0=B{\beta }^0$, and their zero traction gives $K^0{\beta }^0=0$. Hence ${\epsilon }^0=0$, proving that $⟨A⟩$ is invertible. Finally, every positive choice of $C^0$ constructs the unique layer solution for the same $E$. Its averaged stress is unique, so the displayed effective stiffness is independent of the fictitious comparison material. This is the exact layered constitutive response, rather than a claim that piecewise constant fields meet arbitrary finite-body boundary data.

Put $\mu ={\mu }^0$, $\lambda ={\kappa }^0-2\mu /3$, and $L=\lambda +2\mu ={\kappa }^0+4\mu /3$. Assume positive bulk modulus and shear modulus. For a unit vector $\xi$, direct contraction of the isotropic elastic stiffness tensor gives the acoustic tensor and its inverse:These are its transverse and longitudinal orthogonal projections. Symmetrizing the definition of the directional elastic strain Green operator consequently givesSubstitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as distributions. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ givesThe first identity holds as a locally integrable distribution. Since the second derivative of $|s|$ is $2\delta (s)$, taking two derivatives in the first identity and four in the second givesThus an explicit complete isotropic elastic strain Green kernel isAll derivatives here are distributional derivatives, which retain the contribution at the origin.

For the requested contraction, ${\tilde{\Gamma }}_{ijkk}={\xi }_i{\xi }_j/L$, or equivalently trace the complete kernel and use $\Delta r=2/r$. ThereforeMore explicitly, its distribution form iswhere the Cauchy principal value uses spherical exclusions about the origin. The coefficient of the Dirac delta can be checked by integrating by parts over a punctured ball: ${\partial }_i{\partial }_j(r^{-1})$ has contact term $-4\pi {\delta }_{ij}\delta /3$. Taking the trace, the principal-value part has zero trace and $\Delta (r^{-1})=-4\pi \delta$, soKeeping only the ordinary kernel away from the origin would incorrectly give zero for this double trace.

Now extend the elastic polarization by zero outside $V$. Since the shear modulus is unchanged, the stiffness contrast acts only on the spherical strain:Hence $\tau =(\kappa -{\kappa }^0)\mathrm{tr}\epsilon$ is its scalar coefficient. Relative to the same external loading, the elastic strain Green operator gives $\epsilon ={\epsilon }^0-\Gamma \ast (\tau I)$. Taking the trace and using the contact identity reduces this nonlocal equation toSolving the scalar equation gives the constant-shear bulk-modulus inclusion formulaUsing the single contraction instead of the double trace givesFinally eliminate the actual trace using the scalar formula:The reference trace ${\epsilon }_{kk}^0$ in this last formula is essential and is what the original PDF prints; the converted TeX loses its superscript. The second derivatives are interpreted distributionally, or with their principal-value and contact terms explicitly included. With ordinary regularity they give the corresponding pointwise formula at regular interior points.

Let $u$ be the actual displacement field, let $u^{\prime }$ be an admissible trial displacement, and set $v=u^{\prime }-u$. On the displacement-prescribed part of the boundary, $v=0$. With actual stress $\sigma =D_{\epsilon }W(\epsilon )$, the supporting-plane inequality for the convex strain energy density givesBoth $\sigma$ and ${\sigma }_0$ satisfy static elastic equilibrium with the same body force, so $\nabla \cdot (\sigma -{\sigma }_0)=0$. On the traction-prescribed boundary their tractions agree. Since ${\epsilon }^{\prime }-\epsilon =\mathrm{sym}\nabla v$, integration by parts givesSubtracting the prescribed external-work representative proves the minimum potential energy principle in elasticity:Strict convexity gives uniqueness of the minimizing strain, although rigid displacements can remain when the boundary conditions allow them.

For the comparison bound, put $F(e,x)=W(e,x)-W_0(e)$ and use the lower conjugate, with its infimum sign convention,Its defining inequality is $F_{\ast }(\tau ,x)\le \tau :e-F(e,x)$, hence $W(e,x)\le W_0(e)+\tau :e-F_{\ast }(\tau ,x)$ for every $e$. Integrate this pointwise majorant for each compatible strain, then take the infimum over the same admissible set. This provesNo convexity of $F$ is used for this inequality. If $F_{\ast }(\tau ,x)=-\infty$, it yields only the extended-real upper bound $+\infty$. For example, $W_0=0$ and $W(e)=|e{|}^2/2$ give $F_{\ast }(\tau )=-\infty$ for every finite $\tau$. A useful finite bound requires a suitable comparison energy; choosing a sufficiently stiff quadratic comparison makes the usual excess energies concave. The derivative used below additionally presupposes finite differentiable duals, or an appropriate supergradient formulation.

Now normalize $|\Omega |=1$, take positive constant $C^0$, and write $F_r(e)=W_r(e)-e:C^{0}e/2$. For phasewise constant elastic polarization $\tau ={\sum }_r{\chi }_r{\tau }^r$, its lower conjugate contribution is constant within each phase. The remaining minimization isWrite the minimizing strain as $\hat{e}=E+\eta$, where $\eta$ comes from a zero-boundary displacement field. The static elastic equilibrium equation is $\nabla \cdot (C^0\eta +\tau )=0$, so the supplied elastic strain Green operator gives $\eta =-\Gamma \tau$. Its zero mean removes the cross term with the constant $E$. Using its reciprocity and projection property $\Gamma C^0\Gamma =\Gamma$, we obtainHere the brackets in the last two terms denote the spatial integral and tensor contraction, not just a phase average.

Let $c_r={\int }_{\Omega }{\chi }_r$. The zero output mean and reciprocity of $\Gamma$ imply that it annihilates constant elastic polarizations on either side. Thus subtracting $c_r$ and $c_s$ from the phase indicators does not change their bilinear integral. Define the fourth-order block tensorsConsequently $A_{rs}=A_{sr}^T$, ${\sum }_s{A}_{rs}=0$, andThese identities follow from reciprocity and ${\sum }_s{\chi }_s={\sum }_s{c}_s=1$; no phase independence assumption is needed. The phase polarization comparison bound, valid for every polarization with finite dual terms, is thereforeIn particular, it can be optimized over the phase elastic polarizations without changing the upper-bound direction.

To derive the final displayed form, impose the stationarity equations in the question. For $c_r>0$, put $e^r=D{F}_{r\ast }({\tau }^r)$; these are dual strains, initially distinct from the comparison solution's phase averages. Reciprocity of $A$ givesThus stationarity is exactlyThe comparison solution has phase meanso the dual strains become its phase means at stationarity. Multiplying the stationarity equations by ${\tau }^r$ and summing gives ${\sum }_{r,s}{\tau }^r:A_{rs}{\tau }^s={\sum }_r{c}_r{\tau }^r:(E-e^r)$.

For clarity, the concave biconjugate here uses another infimum: $F_{r\ast \ast }(e)={\inf }_s\{s:e-F_{r\ast }(s)\}$. The lower conjugate $F_{r\ast }$ is concave, since it is an infimum of affine functions. At a differentiable finite point, its supporting-plane inequality shows that $s={\tau }^r$ minimizes this last expression when $e=e^r$. HenceSubstitute both identities into $\mathcal{B}$ to obtainThis is the requested final bound with the accompanying phase equations. It remains valid at any well-defined stationary choice; asserting existence of such a differentiable choice for arbitrary $W_0$ would require extra hypotheses. Without stationarity, the earlier $\mathcal{B}(E,\tau )$ is the correct general bound. In the concave closed excess-energy case $F_{r\ast \ast }=F_r$; for general excess energies replacing the concave biconjugate by $F_r$ without proof would reverse the needed majorant logic.

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.