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Divide the Navier-Cauchy equation by mass density. Since $\mu /\rho =c_T^2$ and $(\lambda +\mu )/\rho =c_L^2-c_T^2$, it becomesEquivalently, using the vector calculus identity $\Delta u=\nabla \mathrm{div}u-\nabla \times (\nabla \times u)$,The divergence and curl consequently propagate at $c_L$ and $c_T$, respectively.

Resolve each amplitude by the orthogonal projectionsThe acoustic tensor acts by $\lambda +2\mu$ on the longitudinal line and by $\mu$ on its perpendicular plane. ThusThe longitudinal component is a longitudinal wave, while the two independent transverse polarizations are transverse waves. Each projection obeys its own wave equation, and superposition adds the corresponding solutions.

A general planar disturbance is therefore a sum of the two types. For a single sinusoid with a common speed, both projections cannot be nonzero unless the two speeds coincide: projecting the eigenvalue equation gives $(\rho c^2-\lambda -2\mu )a_L=0$ and $(\rho c^2-\mu )a_T=0$. This distinguishes superposition of waves with different speeds from a mixed polarization traveling rigidly at one speed.

Substituting the sinusoidal plane wave into the Navier-Cauchy equation gives $\Delta u=-u$, $u_{tt}=-c^{2}u$, and $\nabla \mathrm{div}u=-k(k\cdot a)\sin (k\cdot X-ct)$ because $|k|=1$. Equality of the amplitudes is thereforeThis is an eigenvalue problem for the acoustic tensor $\mu I+(\lambda +\mu )k\otimes k$. Choosing a nonzero amplitude parallel to $k$ gives $\rho c^2=\lambda +2\mu$; choosing any nonzero amplitude perpendicular to $k$ gives $\rho c^2=\mu$. For positive mass density and stable Lamé parameters, both squared speeds are positive. These choices construct nonzero solutions explicitly, proving existence.

Integrate the radial equilibrium equation across the thickness, retaining the integral rather than evaluating its antiderivative. The traction conditions give $t_r(b)-t_r(a)=P$. Hence $a$ must satisfyAll endpoints and coefficients here depend only on $a$ and the specified material and reference geometry. This is the required implicit pressure–radius relation for inflation and extension of an incompressible tube with unit axial stretch. In particular $a=A$ gives zero integrand and $P=0$. Once an admissible root is selected, the radial stress follows by integrating from the outer face, $t_r(r)=-{\int }_r^b(d{t}_r/ds)ds$, and the remaining stress components follow from the constitutive equation.

The radial equilibrium equation gives $d{t}_r/dr=(t_{\theta }-t_r)/r$. Set $C=a^2-A^2$, so that $R^2=r^2-C$. Substitution of the neo-Hookean solid stress difference eliminates both $q$ and $t_{\theta }$:This is a first-order ordinary differential equation for the radial stress, parametrized by the unknown inner radius. For inflation $C>0$, its right side is positive, so $t_r$ increases from the negative inner pressure to zero at the outer surface. The denominator stays positive because $r^2-C=R^2\ge A^2>0$.

The multiplier enforcing elastic incompressibility contributes an arbitrary hydrostatic stress. Denote it by $q(r)$ to distinguish it from the applied pressure $P$. For the neo-Hookean solid, the Cauchy stress tensor is $\tau =-qI+\mu F{F}^T$. Thus its principal components areThe axial component is $t_z=-q+\mu$, despite the unit axial stretch. The quantity $q$ is a Lagrange multiplier determined together with equilibrium and the traction conditions; it is not obtained by differentiating the reduced stretch energy while ignoring its constraint.

In particular the stress difference eliminates this multiplier:

Use orthonormal cylindrical bases in the reference configuration and current configuration. The deformation gradient maps a radial increment $dR$ to $r^{\prime }dR$, an azimuthal arc $Rd\Theta$ to $rd\Theta$, and an axial increment $dZ$ to itself. ThereforeDifferentiating the radius relation gives $r^{\prime }=R/r$. Its three principal stretches areTheir product is one, as required by elastic incompressibility. The symbols called “strains” in the source are these stretch ratios; engineering strain would instead subtract one. The axial stretch is one because this is plane strain.

For the axisymmetric deformation map, take $\theta =\Theta$, $z=Z$, and $r=r(R)$. A reference cylindrical shell of unit axial length has volume $2\pi RdR$, and the corresponding current shell has volume $2\pi rdr$. Elastic incompressibility therefore gives $rdr=RdR$.

Integrating from the inner surface, where $r(A)=a$, gives $r^2-a^2=R^2-A^2$. The positive radius is consequentlyThe outer surface is not an additional independent unknown: $\boxed{b=\sqrt{B^2+a^2-A^2}}$, and hence $h=b-a$.

Write $B=A+H$ for the reference outer radius and $b=a+h$ for the current outer radius. The inner material normal is $-e_r$. A fluid pressure $P$ applies traction $-Pn=P{e}_r$ there, so the radial Cauchy stress tensor component must be negative:The second condition uses zero external gauge pressure. Both faces also have zero shear traction; the axisymmetric diagonal stress field satisfies those conditions automatically. The axial constraint is plane strain, so supports may supply an axial reaction; it does not imply zero axial stress.

In the current configuration, balance of linear momentum is the Cauchy momentum equationHere $\tau$ is the Cauchy stress tensor, $\rho$ is current mass density, $b$ is body force per unit mass, and $v(x,t)$ is the velocity of the material particle currently at $x$. The expression in parentheses is the material acceleration. The divergence is taken in current spatial coordinates, with $(\mathrm{div}\tau {)}_i={\partial }_j{\tau }_{ij}$.

Equivalently the right side is $\rho {\partial }_t^{2}x(X,t)$ evaluated at the particle, because this is the same material acceleration in Lagrangian coordinates. For the subsequent static inflation with negligible body force, the equation reduces to $\mathrm{div}\tau =0$.

Choose the outward unit normal $n$ to the boundary of the chosen current material subbody. The traction $t(x,n)$ is the contact force exerted on that subbody per unit current area at the boundary point. Reversing the subbody side reverses $n$ and $t$.

By Cauchy's stress theorem, the dependence on $n$ is linear. With the usual column-vector convention the Cauchy stress tensor satisfiesThe same point has different traction vectors on differently oriented planes, but a single Cauchy stress tensor encodes all of them. In the absence of body couples, conservation of angular momentum makes $\tau$ symmetric.

The deformation gradient is the derivative of the deformation map with respect to the reference configuration coordinates, at fixed time:Its determinant $J=\det F$ is the local ratio of current to reference volume: $dv=JdV$ for an orientation-preserving deformation in continuum mechanics. Thus elastic incompressibility is the pointwise constraint $\boxed{J=1}$, rather than a requirement that each individual principal stretch be one.

Use the same first variation on the half-plane, with no restrictions on the edge traces of $\eta$. Write $M=w_{xx}+\nu w_{yy}$. Two $x$ integrations of the $M{\eta }_{xx}$ term produce the boundary term $-M{\eta }_x+M_x\eta$ at $x=0$. The mixed term, after one integration in $x$ and one along the edge in $y$, contributes $2(1-\nu )w_{xyy}\eta$. All other boundary terms vanish at infinity. HenceThe bulk equation makes its area term zero. Stationarity therefore gives exactlyThe two edge traces can be prescribed independently: use $\eta (x,y)=\chi (x)[f(y)+xg(y)]$ with $\chi =1$ near zero and compactly supported smooth $f,g$. The fundamental lemma on the edge then givesThey are zero bending moment and zero effective transverse shear at a free straight edge. The second includes the edge derivative of the twisting moment; omitting the mixed contribution would give the wrong coefficient.

Contract the tensors, counting both off-diagonal shear terms:Since ${\int }_{-h/2}^{h/2}{z}^{2}dz=h^3/12$, integration givesFor the general midplane deflection, expand the energy density as $w_{xx}^2+2\nu w_{xx}{w}_{yy}+w_{yy}^2+2(1-\nu )w_{xy}^2$. A variation $w+\epsilon \eta$ givesFor clamped variations, $\eta ={\partial }_n\eta =0$ on the boundary, so integrating twice transfers all derivatives to $w$ without a boundary contribution. The bulk coefficient isStationarity for all compactly supported variations and the fundamental lemma therefore give

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.

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.

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.

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.

Assume $\mathbf{k}\ne 0$. Dotting the decomposition with $\mathbf{k}$ givesCrossing it on the left by $\mathbf{k}$ and using $\mathbf{B}\cdot \mathbf{k}=0$ gives $\mathbf{k}\times \mathbf{a}=|\mathbf{k}{|}^2\mathbf{B}$. ThusIndeed $(\mathbf{k}\times \mathbf{a})\times \mathbf{k}/|\mathbf{k}{|}^2=\mathbf{a}-A\mathbf{k}$, verifying the decomposition and the imposed transverse condition.

The traction on the oriented solid element isWith constant Lamé coefficients and derivatives in the reference variables,Substitution into balance of linear momentum givesTo check the vector identity componentwise, contract the two alternating tensors:Rearranging proves ${\nabla }^2\mathbf{u}=\nabla (\nabla \cdot \mathbf{u})-\nabla \times (\nabla \times \mathbf{u})$. Therefore