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For $\Delta t=\Delta x/c$,Equivalently $\boxed{y(x,t+\Delta t)=y(x-c\Delta t,t)}$. Thus every value of the displacement is translated by the same distance $c\Delta t$: the entire travelling wave profile keeps its shape. The wave speed is independent of the profile, as expected for this nondispersive wave equation.

Take an ideal flexible mechanical string with constant tension $T>0$ and uniform linear mass density $\mu >0$. Let $x$ measure distance along its straight equilibrium position and let $y(x,t)$ be its small transverse displacement. Neglect damping, bending stiffness and external distributed forces, and assume $|y_x|\ll 1$.

A segment of reference length $\Delta x$ has mass $\mu \Delta x$. The endpoint tension forces have transverse resultantsince $\sin \theta \simeq \tan \theta =y_x$ for small slopes. Newton's second law, divided by $\Delta x$ and taken to the limit, givesHere $t$ is time, the subscripts denote partial derivatives, and $c$ is the wave speed. The transverse acceleration is $y_{tt}$; $y_{xx}$ describes the small-slope curvature responsible for the net transverse force.

Put the fixed end at $x=0$, with the mechanical string occupying $x\le 0$. Let $H$ be the incident peak amplitude, set $u=x/L$ and $\tau =ct/L$, and take $t=0$ when the incident front first reaches the end. The symmetric triangular pulse in the source has initial profile $HT(u)$, whereThe fixed-end reflection of a triangular pulse follows from the Dirichlet boundary condition $y(0,t)=0$ and the superposition principle, giving the reflected displacementThe minus sign is the inversion on reflection. At $\tau =1/4$, the resulting profile is zero outside $[-3/4,0]$ and consists of the straight segments throughThe segment beside the wall is steeper because incident and inverted reflected pulses overlap there.

At $\tau =3/4$, the reflected pulse dominates the overlap. The total displacement is zero outside $[-3/4,0]$ and consists of straight segments throughThis is the inverted version of the quarter-reflected profile. The entire negative pulse has not yet separated from the incident pulse.

A normal mode of a stable undamped linear coupled oscillator has a single angular frequency and a fixed ratio of component displacements. All components share a sinusoidal time factor, with relative oscillation phases of zero or $\pi$ for real mode shapes. A general free motion is a superposition of the normal modes. For this system the symmetric stiffness matrix gives real eigenvalues and orthogonal eigenvectors; positive spring constants make the squared angular frequencies positive.

Write $m=m_1=m_2>0$ and $\kappa =k_{12}$. Assume ideal massless linear springs, negligible damping and positive $k_1,k_2,\kappa$. By Hooke's law and Newton's second law,The coupling spring extends by $x_2-x_1$, giving equal and opposite coupling forces.

For a normal mode, take $\mathbf{x}=\mathrm{Re}(\mathbf{a}{e}^{i\omega t})$. Nonzero eigenvectors requireSolving this quadratic equation in $m{\omega }^2$ givesA corresponding mode shape has ratioand the general real solution is $\mathbf{x}=C_{-}{\mathbf{a}}_{-}\cos ({\omega }_{-}t+{\varphi }_{-})+C_{+}{\mathbf{a}}_{+}\cos ({\omega }_{+}t+{\varphi }_{+})$. The arbitrary amplitudes and oscillation phases account for four independent initial data.

Weak coupling alone does not guarantee large beats. A useful representative sketch assumes nearly equal uncoupled angular frequencies. In the resonant case $k_1=k_2=k$ with $\kappa \ll k$, the mode shapes are $(1,1)$ and $(1,-1)$, andFor $x_1(0)=A$, $x_2(0)={\dot{x}}_1(0)={\dot{x}}_2(0)=0$, both normal modes have equal coefficients. With $\bar{\omega }=({\omega }_{+}+{\omega }_{-})/2$ and $\Delta \omega ={\omega }_{+}-{\omega }_{-}$,These rapid oscillations have slowly varying complementary wave envelopes. The initial displacement shifts almost entirely to the other mass at $t=\pi /\Delta \omega$, and returns after $2\pi /\Delta \omega$. Arbitrary initial data give a different superposition; if $|k_1-k_2|\gg \kappa$, the normal modes are localized mainly on individual masses, and strong exchange is suppressed.

Holding one mass fixed leaves the coupling spring attached and stretched by the other mass. Therefore the two constrained angular frequencies areThese differ from the angular frequencies obtained by physically removing the coupling spring. Ordinary frequencies in hertz are each corresponding $\omega /(2\pi )$.

Finally, for $k_1>k_2$ and $\kappa >0$, put $D=\sqrt{(k_1-k_2{)}^2+4{\kappa }^2}>k_1-k_2$. ThenSince ${\omega }_1>{\omega }_2$, taking positive square roots provesAt $\kappa =0$, the outer inequalities become equalities.

First solve the supplied equations as a linear system, with ${\omega }_0^2=g/l$ and $\alpha >0$. A normal mode $\mathbf{x}=\mathbf{a}{e}^{i\omega t}$ obeysIts characteristic polynomial is ${\lambda }^2-(1+3\alpha )\lambda +\alpha (1+\alpha )$. HenceBoth are positive, since the trace and determinant of $D$ are positive.

There is a physical discrepancy in the source. The horizontal-coordinate mass matrix of an unequal-mass double pendulum follows from the stated horizontal displacements, which giveThe Euler-Lagrange equations reproduce the supplied upper equation, but give ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, without its extra $\alpha$. The true normal modes of a double pendulum consequently have ${\omega }_{\pm }^2=(g/l)[1+\alpha \pm \sqrt{\alpha (1+\alpha )}]$. The boxed answer and the remaining parts below concern the explicitly supplied linear system. The two models agree at $\alpha =1$, but not for a general mass ratio.

At $\alpha =1$, the supplied and physical systems agree. Their squared angular frequencies are $(g/l)(2\pm \sqrt{2})$. The upper row gives $a_2/a_1=3-\lambda$, soIn the slower normal mode, both masses move horizontally in phase and the lower has the larger amplitude. In the faster normal mode, the lower mass moves in the opposite direction, with magnitude $(\sqrt{2}-1)|a_1|$. These are horizontal displacement ratios. The rod angles instead satisfy ${\theta }_2/{\theta }_1=\pm \sqrt{2}$, because $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$; confusing these ratios changes the sketch.

For the supplied equations, ${\lambda }_{+}\to 1$ and $a_2/a_1=\alpha /(\alpha -{\lambda }_{+})\sim -\alpha$. Thus the higher-frequency normal mode becomes an upper-mass oscillation at $\sqrt{g/l}$ with a vanishing, opposite-sign lower horizontal displacement. At $\alpha =0$ the supplied lower equation is ${\ddot{x}}_2=0$ and its other mode has zero frequency.

This limiting picture belongs to the given matrix. For the physical unequal-mass double pendulum, both normal mode angular frequencies tend to $\sqrt{g/l}$ as the lower mass tends to zero; at exactly zero mass its independent equation needs a limiting test-body interpretation. The source's extra $\alpha$ changes precisely this conclusion.

Expand the initial velocity in the Fourier sine basis, $V(x)={\sum }_{n\ge 1}{V}_n\sin (n\pi x/L)$, where $V_n=(2/L){\int }_0^{L}V(x)\sin (n\pi x/L)dx$. Zero initial displacement removes the cosine-in-time coefficients. ThusThe trigonometric identity $2\sin a\sin b=\cos (a-b)-\cos (a+b)$ rewrites each summand asIt is therefore a superposition of travelling waves of speeds $+c$ and $-c$, whose interference enforces the fixed endpoints. For general square-integrable initial velocity, the series is interpreted in the standard finite-energy sense.

The Lighthill elongated-body theory is a reactive, nearly inviscid flow description of a long slender swimmer, appropriate when the longitudinal Reynolds number is large but its lateral displacement and slope are small. To leading order, crossflow past each section is the two-dimensional potential flow generated by lateral motion of that section. Longitudinal variation is slow on the cross-sectional scale, and axial speed $U$ is prescribed at this order. With $x$ increasing from nose to tail, fluid passes sections in increasing $x$, and the linearized relative lateral velocity isThe section's added mass per unit length $m(x)$ is defined by its crossflow kinetic energy $\frac{1}{2}m(x)w^2$. Equivalently, if $\Phi$ is the velocity potential for unit lateral section motion, then $m=\rho \int |{\nabla }_{\perp }\Phi {|}^{2}dA$ over the exterior crossflow plane. It has units of mass per length and depends on cross-sectional shape and density. For a circular section of radius $a$, $\Phi =-a^2\cos \phi /r$ gives $m=\rho \pi a^2$. It is not the body's mass per unit length.

The transverse fluid impulse attached to a section is $m(x)w$. A control segment has both changing impulse and an impulse flux $Umw$ through its ends. Taking their sum per unit length gives the reaction on the fluid:The force on the fish is $-F_z$. In particular, $m$ is inside the convective derivative; dropping its spatial derivative or the endpoint fluxes is not justified merely because the fish is slender.

The recoil correction in elongated-body theory adds unknown whole-body translation and yaw to a prescribed active bending shape:They are determined by total lateral linear momentum and yaw angular momentum balance, rather than by imposing that the deforming swimmer remains on its original axis. Body inertia and hydrodynamic added inertia have the same sign. If the translation variable is required to be the literal centre of mass, the active deformation must first have its mass-weighted mean removed: the reference displacement $Z_r$ is otherwise not the centre of mass displacement during the stroke. The corresponding deformation contribution to angular momentum must also be retained when it is significant.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.

The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.

Linearizing the vertical velocity requires retaining advection of ${\bar{\psi }}_z=-\Lambda y$ by $v^{\prime }={\psi }_x^{\prime }$. ThereforeThe impermeable boundary at $z=0$ imposesThis is conservation of boundary buoyancy, not merely a condition that ${\psi }_z^{\prime }$ vanish.

For zero interior potential vorticity, write ${\psi }^{\prime }=\mathrm{Re}\{\varphi (z,t)e^{i(kx+ly)}\}$. With $K_h=\sqrt{k^2+l^2}>0$, stratified quasi-geostrophic inversion givesThe decaying solution isNo time dependence has been assumed to obtain this vertical structure. Applying the boundary condition gives $-\mu \dot{A}-i\Lambda kA=0$, henceThus the Eady edge wave travels eastward with horizontal phase velocity $\omega /k=\Lambda /\mu$ for $k\ne 0$.

The propagation mechanism can be made explicit using boundary parcels. Thermal wind gives the background buoyancy gradient ${\bar{b}}_y=f{\bar{\psi }}_{zy}=-f\Lambda$. A small meridional parcel displacement $d$ produces boundary buoyancy perturbation $b^{\prime }=-d{\bar{b}}_y=f\Lambda d$. Meanwhile the decaying potential-vorticity inversion gives $b^{\prime }=f{\psi }_z^{\prime }=-f\mu {\psi }^{\prime }$ at the boundary. Consequently ${\psi }^{\prime }=-\Lambda d/\mu$ there, and the parcel equation $d_t=v^{\prime }={\psi }_x^{\prime }$ becomesA displaced buoyancy contour therefore induces a velocity field that advances that displacement eastward. No interior potential-vorticity gradient is needed because the boundary buoyancy supplies the invertibility data.

Vortex stretching is what sets the induced flow's vertical reach. In the interior, the zero-potential vorticity condition requiresBoundary isopycnal displacement decreases with height as $e^{-\mu z}$, so fluid columns are stretched or compressed over depth ${\mu }^{-1}=f/(N{K}_h)$. Conservation of potential vorticity makes this stretching balance the relative vorticity. The resulting inversion factor $b^{\prime }=-f\mu {\psi }^{\prime }$ gives the propagation speed: stronger density stratification reduces penetration and speed, while larger $f$ increases them. Zero interior potential vorticity therefore does not imply zero interior flow.

For imposed boundary vertical velocity, retain the same decaying structure and writeThe boundary relation becomes the forced Eady edge wave equationThe integrating factor $e^{i\omega t}$ gives, for ${\omega }_0\ne \omega$,The arbitrary homogeneous coefficient $B$ sets the initial disturbance. With $A(0)=0$, this is $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$.

At resonance, ${\omega }_0=\omega$, integration instead yieldsAn initially undisturbed fluid has ${\psi }^{\prime }=Ct{e}^{-\mu z}\cos (kx+ly-\omega t)$. The imposed pumping reinforces the same boundary normal mode every cycle, with no dissipative loss, so the linear amplitude grows secularly. The divergence describes resonance of the undamped linear model; sufficiently large wave amplitude invalidates the assumed small-disturbance approximation.

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.

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.

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.

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 ${\sum }_{r,s}{\tau }^r:A_{rs}{\tau }^s={\sum }_r{c}_r{\tau }^r:(E-e^r)$.