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Take the trace of the constitutive relation at zero stress. It gives $0=(2\mu +3\lambda )e_{kk}-3c\alpha (T-T_0)$. With the observed stress-free volume strain this requiresThis is the bulk modulus. A stable solid resists uniform compression, so $\boxed{K>0}$; equivalently the elastic energy of a nonzero pure dilation is positive. The thermal strain is isotropic, $e_{ij}^{\mathrm{th}}=\alpha (T-T_0){\delta }_{ij}/3$, because $\alpha$ here is the volumetric, not linear, thermal expansion coefficient.

Let $\beta =K\alpha$. Using $e_{ij}=(u_{i,j}+u_{j,i})/2$, stress divergence isZero body force and equilibrium therefore giveFor an applied body force per unit volume $b$, equilibrium is $\nabla \cdot \tau +b=0$, so its right-hand side becomes $\beta \nabla T-b$. Consequentlycancels the effective thermal forcing in the displacement equation. Per unit mass the force is $\beta \nabla T/\rho$. This sign follows from the divergence of the thermal stress $-\beta (T-T_0)I$, which is $-\beta \nabla T$. Cancelling that divergence is distinct from removing the stress tensor itself: boundary tractions and thermal compatibility still matter.

For spatially uniform heating with the entire boundary fixed, $u=0$ is the unique equilibrium displacement in a stable elastic solid, andThe stress is compressive hydrostatic stress for a temperature rise. The fixed boundary supplies the tractions that prevent thermal expansion. If $T$ varies spatially and no compensating body force is imposed, fixing the boundary alone does not generally force the interior displacement to vanish: one must solve the displayed forced Navier-Cauchy equation with $u=0$ on the boundary, then recover the stress. The simple uniform hydrostatic expression should not be asserted for arbitrary temperature fields. This is uniform thermal stress in a fully constrained solid.

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.

Multiply equilibrium by the displacement and integrate. Integration by parts and the traction condition giveThe stress is symmetric, so it contracts only with the symmetric part of $\nabla \mathbf{u}$, namely $e$. Hooke's law then gives $T:e=\lambda (\mathrm{tr}e{)}^2+2\mu e:e=2W$. HenceThe identity is the linear-elastic work balance: twice the stored quadratic energy equals the body-force and boundary-traction work paired with the final displacement.

Apply energy uniqueness for a linear elastic traction problem. Subtract two solutions with the same loading, writing their displacement difference as $\mathbf{v}$. Linearity makes its body force and boundary traction zero. The preceding energy identity therefore yields ${\int }_{B}2W(e(\mathbf{v}))dV=0$. Positive definiteness and continuity imply $e(\mathbf{v})=0$ everywhere. By the permitted zero-strain characterization,Thus the solution is unique modulo infinitesimal rigid translations and rotations, on each connected body. Conversely adding such a displacement changes neither strain nor stress, so these are exactly the unconstrained modes. This is conditional uniqueness; existence of a pure-traction solution additionally requires balanced resultant force and torque.

Differentiating the displacement givesThe isotropic stress is diagonal, withThe two transverse traction-free conditions set the latter entries to zero. Hence $\lambda =2\nu (\lambda +\mu )$ andSubstitution shows $E=2\mu (1+\nu )$; solving for the two Lamé parameters givesTo determine the allowed interval without assuming signs not given, put $K=\lambda +2\mu /3>0$. Then $\nu =(3K-2\mu )/[2(3K+\mu )]$, soTherefore $\boxed{-1<\nu <1/2}$. The endpoints would require a degenerate elastic modulus, so they are excluded by the strict material inequalities.

The perturbation preserves the prescribed displacement because $\mathbf{w}=0$ on the boundary. Strain varies as $e_{ij}(\mathbf{u}+\alpha \mathbf{w})=e_{ij}(\mathbf{u})+\alpha (w_{i,j}+w_{j,i})/2$. Taylor expansion of the strain energy density givesSymmetry of the stress makes its contraction with the two symmetric terms equal, henceA minimizer must have zero first variation for both signs of $\alpha$. Integrating by parts and using the zero boundary variation yields ${\int }_B(-{\partial }_j{\tau }_{ij}-\rho g_i)w_{i}dV=0$. The fundamental lemma of the calculus of variations, applied to arbitrary compactly supported variations, givesThis is the static momentum equation. Fixed-displacement data remove the surface variation; prescribed tractions would instead appear through boundary work in the potential energy.

Multiplication of the preceding displacement formula by $-ik$ givesFor the supplied kernels, $\widehat{f_{1,x}}=-ik{e}^{-qy}$ and ${\hat{f}}_2=i(k/q)e^{-qy}$. The convolution theorem therefore givesSince $f_{1,x}(s,y)=-2ys/[\pi (s^2+y^2{)}^2]$ and $f_2(s,y)=s/[\pi (s^2+y^2)]$, combine the two fractions and divide by two:The sign is tied to the positive forward Fourier phase used here. This is a displacement derivative, so an additive displacement reference drops out. At $y>0$ the kernel is regular near $s=0$; the usual decay/integrability of the applied load controls the tails. A boundary-limit singular integral, if wanted at $y=0$, would require its own principal-value interpretation.

The source uses the forward Fourier phase $e^{ikx}$, so $\widehat{{\partial }_{x}A}=-ik\hat{A}$. For a nonzero wavenumber put $q=|k|$. Biharmonicity becomes $({\partial }_y^2-q^2{)}^2\hat{A}=0$. The decaying solutions are $(C+Dy)e^{-qy}$. The normal-stress condition gives $-k^2\hat{A}(k,0)=-\hat{P}(k)$, hence $C=\hat{P}/k^2$. The shear condition gives ${\hat{A}}_y(k,0)=0$, so $D=qC$. ConsequentlyIts second derivatives are $\widehat{A_{xx}}=-\hat{P}(1+qy)e^{-qy}$ and $\widehat{A_{yy}}=\hat{P}(qy-1)e^{-qy}$. The given strain/stress inversion therefore yieldsIntegrating in $y$ with the stated decaying-mode condition givesThis is the Fourier solution of a loaded elastic half-space.

The zero wavenumber needs separate treatment: these displayed divisions by $k^2$ and $|k|$ are not formulas at $k=0$. In particular a nonzero resultant $\hat{P}(0)=\int Pdx$ is incompatible with an ordinary decaying zero-mode displacement of the stipulated kind; two-dimensional net loading produces a logarithmic displacement reference issue. Balanced loads or an appropriate distributional/reference-displacement formulation are needed for a global transform solution. The nonzero modes and the displacement-gradient kernel below remain the correct formal calculation; the singular zero mode must not be silently assigned a finite value.

The displacement follows a material point from its reference position to its current position:The deformation gradient is the Jacobian matrix $F=\partial \mathbf{x}/\partial \mathbf{X}=I+{\nabla }_X\mathbf{u}$, and the volume Jacobian isFor an orientation-preserving physical deformation, $J>0$ measures the local volume expansion factor. Incompressibility requires $J=1$; conservation of mass gives $\rho (\mathbf{x},t)J={\rho }_0(\mathbf{X})$. For small displacement gradients, $J=1+{\nabla }_X\cdot \mathbf{u}+O(|{\nabla }_X\mathbf{u}{|}^2)$. If a spatial representation is desired, the same displacement is $\mathbf{u}(\mathbf{x},t)=\mathbf{x}-\mathbf{X}(\mathbf{x},t)$.

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.

Apply the wave ansatz to the oscillatory displacement about a static body-force balance, or equivalently to the zero-body-force homogeneous equation. The parallel and perpendicular polarizations have speedsThe first is longitudinal and the second transverse. A wave family is dispersive when its phase speed $\omega /|\mathbf{k}|$ varies with wavenumber. In each of these homogeneous bulk families, $\omega =c|\mathbf{k}|$, so both are nondispersive and group and phase speeds coincide.

The positivity assumptions imply $\lambda >-2\mu /3$, hence $\lambda +\mu >\mu /3>0$. Therefore $c_p^2-c_s^2=(\lambda +\mu )/\rho >0$: longitudinal waves travel faster.

Substitution of $w_i=f_i(y)e^{i(kx-\omega t)}$ givesThe oscillatory seam solution requires $m^2={\omega }^2/c_{s1}^2-k^2$, givingFor an exponentially decaying exterior, $f_2^{\prime \prime }={\ell }^2{f}_2$ with $\ell >0$, so the same differential equation instead requiresThe printed exterior plus sign is inconsistent with its decaying ansatz and must be a sign error. The guided mode has $c_{s1}<c_L<c_{s2}$, so a slower shear-wave seam than its cladding is necessary.

For an even seam mode $f_1=A\cos (my)$, upper-interface displacement and traction continuity giveThe lower interface gives the same equations when $B_2=A_2$. Eliminating the amplitudes yieldsFor an odd seam mode $f_1=B\sin (my)$, the equations areWith $B_2=-A_2$ the lower condition agrees, and elimination givesThese interface conditions together with the two corrected dispersion identities determine the guided-wave branches.

In the formal stiff-cladding limit with a nonvanishing exterior decay rate, the interface displacement tends to zero. The even seam mode then has $\cos (mh)=0$, whereas the odd mode has $\sin (mh)=0$. The positive wavenumbers areThe zero odd index would give an identically zero mode and is excluded. The smallest even transverse wavenumber is $\pi /(2h)$, so its lowest branch satisfiesThis is the symmetric-mode cutoff: the actual limiting branches have the discrete listed values of $m$. If the exterior shear speed remains finite, the separate decay requirement ${\omega }^2<c_{s2}^2{k}^2$ still applies; the cutoff alone does not replace that guided-wave condition.

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.

Choose $y$ northward and $z$ upward in the Northern Hemisphere. In a locally linear Boussinesq approximation, write the background buoyancy as $\bar{b}=b_{\ast }+By+N^{2}z$, with horizontal buoyancy gradient $B<0$ and positive squared buoyancy frequency $N^2$. The stable density stratification resists a purely vertical displacement, but the isopycnals slope upward poleward:The geopotential surfaces are horizontal in this local approximation. Their inclination differs from that of the isopycnals, which is the relevant baroclinicity.

Exchange two equal small fluid elements, each of volume $V$, between positions separated by $(\delta y,\delta z)$. The adiabatic parcel exchange has no thermal diffusion, so each parcel retains its buoyancy. The background buoyancy difference is $\Delta \bar{b}=B\delta y+N^2\delta z$. Because the variable-density contribution to gravitational potential energy is $-{\rho }_{0}Vbz$, the change in the pair's gravitational potential energy isFor a poleward-rising exchange, set $\delta z=s\delta y$ with $s>0$. ThenThe trajectory lies between a geopotential surface and an isopycnal. The poleward-rising parcel starts more buoyant than the parcel it replaces, and the equatorward-sinking parcel starts less buoyant: the buoyancy does positive work on both. This releases available potential energy into kinetic energy, allowing eddies to amplify if the full rotating dynamics supports the displacement and dissipation does not remove the gain.

A trajectory steeper than the isopycnals, $s>s_{\rho }$, instead costs gravitational potential energy; an equatorward-rising trajectory also costs it. Motion exactly along a geopotential surface or an isopycnal gives zero change in this ideal exchange. At fixed $\delta y$, the largest release occurs at $s=s_{\rho }/2$, since $N^{2}s(s-s_{\rho })$ is minimized there.

This is an energetic possibility argument for baroclinic instability, not a sufficient normal mode instability criterion. In particular, unrestricted parcel motion must not be assumed in a rotating fluid: conservation of angular momentum can impose an additional restoring effect. Symmetric instability has a separate Ertel potential vorticity criterion; ordinary Eady instability can occur without violating that criterion.

The scalar product $\mathbf{r}\cdot \hat{\mathbf{n}}=d$ fixes the component of the position vector along the given unit normal vector. Thus the locus is a plane normal to $\hat{\mathbf{n}}$, through $d\hat{\mathbf{n}}$. Its signed distance from the origin along that normal is $d$, and its perpendicular distance from the origin is $|d|$. Any displacement lying within the plane is orthogonal to $\hat{\mathbf{n}}$.

Measure the car's vertical displacement $y$ from its static equilibrium, so gravity's constant contribution is already balanced. Both the spring and the viscous damper act on motion relative to the wheel. The base-excited oscillator therefore satisfiesFor dimensional clarity write the road profile as $B\sin (qx)$; the printed notation sets the implicit wave number $q=1$ in the chosen length units. Constant forward speed gives forcing angular frequency $\omega =qv$. With complex amplitudes $Y,Y_0$, the steady response isSet $q=1$ for the expression in the question's notation. Assuming $M,k>0$ and $\gamma >0$, the transient response decays, leaving this steady-state response.

For a useful sketch define ${\omega }_n=\sqrt{k/M}$, $r=\omega /{\omega }_n$ and damping ratio $\zeta =\gamma /(2\sqrt{kM})$. The displacement transmissibility isIt starts at $1$, rises to a finite maximum for finite positive damping, crosses $1$ at $r=\sqrt{2}$, then tends to zero. Differentiation with respect to $r^2$ gives the maximum at $r^2=2/[1+\sqrt{1+8{\zeta }^2}]$.

Very small damping permits large resonant oscillations near ${\omega }_n$, producing discomfort and potentially large suspension travel. At exactly zero damping the assumed decay of transients fails, and resonant forcing produces secular growth. Very large damping makes the car follow the bumps over a wide range of speeds, sacrificing vibration isolation: at each fixed forcing frequency, $Y/Y_0\to 1$ as $\gamma \to \infty$. For any fixed finite damping the response still eventually decreases at sufficiently high frequency. These two limits should not be interchanged.

With the same positive displacement convention for both masses, the first eigenvector equation is $(k+k_2-m{\Omega }^2)a_1-k_2{a}_2=0$. At the symmetric frequency it reduces to $k_2(a_1-a_2)=0$, while at the antisymmetric frequency it becomes $-k_2(a_1+a_2)=0$. HenceIn both normal modes the displacement magnitudes are equal. For the first, the masses move in phase and their separation stays unchanged; for the second they move in opposite phase, alternately approaching and separating. A negative signed amplitude ratio records a phase difference of $\pi$, not a negative displacement magnitude. The left panel of the later figure plots these signed normal mode shapes with the fixed wall values included.

Let $x_1,x_2$ be displacements from equilibrium, with positive directions chosen alike. Hooke's law and Newton's second law giveFor a normal mode $x_j=a_j\cos \Omega t$, the determinant condition isSolving the quadratic for ${\Omega }^2$ givesThe angular frequencies are the positive square roots. Positive masses and spring constants make the stiffness quadratic form $k_1{x}_1^2+k_2(x_2-x_1{)}^2+k_3{x}_2^2$ positive definite, ensuring both squared frequencies are positive. Distinct spring constants do not require equal masses or any symmetry approximation.

Both masses in the source's absorber diagram are free: there is no additional spring attaching $M$ to a wall. Let their displacements be $X$ and $x$. The equations areFor a periodic particular response $X=A\cos \omega t$, $x=B\cos \omega t$, the amplitude equations are $(k-M{\omega }^2)A-kB=F_0$ and $-kA+(k-m{\omega }^2)B=0$. Their solution isAt $\boxed{k=m{\omega }^2}$, $A=0$ and $B=-F_0/k$: the spring force on the larger mass cancels the applied force at every instant. This is antiresonance, a frequency-selective cancellation, rather than dissipation of energy.

There is a nearby undamped resonance at $k_{\mathrm{res}}=Mm{\omega }^2/(M+m)$. For $m\ll M$ it lies close to the cancellation setting. More quantitatively, write $k=m{\omega }^2(1+ϵ)$. ThenFor $|ϵ|\ll m/(M+m)$, the amplitude relative to the unconnected mass's amplitude $F_0/(M{\omega }^2)$ is approximately $(M/m)|ϵ|$. A light absorber therefore needs very accurate tuning and loses suppression when the driving frequency changes. Its own travel $|B|=F_0/(m{\omega }^2)$ can also become large, limiting linearity and available stroke.

Without damping, resonance growth and persistent free oscillations remain; special initial conditions select the displayed periodic particular response. The free pair also has rigid-translation motion. These limitations prevent interpreting the cancellation as broadband damping or as a unique attracting late-time state.