Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 72 / 3

Work in the fixed reference domain ${\Omega }_0$, and write $v_i=\delta x_i$. For the unconstrained hyperelastic material considered here, the material-first nominal stress is $N_{\alpha i}=W_{,A_{i\alpha }}$. The increment of the deformation gradient and the incremental elastic moduli areAll coefficients are evaluated at the base deformation. The Hessian symmetry is $c_{\alpha i\beta j}=c_{\beta j\alpha i}$; minor interchange symmetry is unnecessary. Linearizing the momentum equation givesThe reference mass density and reference outward unit normal vector are fixed, so the incremental boundary conditions areIn particular, no variation of a current surface normal belongs in this material-coordinate traction condition. If a material constraint were imposed, one would additionally linearize its constraint reaction in hyperelastic stress and its admissibility equation; the present moduli formula is the unconstrained one.

For a uniform material and an affine static base deformation, the deformation gradient, incremental elastic moduli and reference mass density are constant. With zero incremental body force, the equation reduces to $c_{\alpha i\beta j}{v}_{j,\beta \alpha }={\rho }_0{\ddot{v}}_i$. Let $\nu$ be a unit material propagation direction, $m$ a constant polarization and $s=t-{\nu }_{\alpha }{\xi }_{\alpha }/c$. Direct differentiation of the plane wave givesThus every twice differentiable profile $f$ satisfies the incremental equation ifThis is the acoustic tensor eigenvalue problem. The acoustic tensor is a real symmetric matrix: interchange the index pairs in the incremental elastic moduli and then interchange the dummy material indices to obtain $Q_{ji}=Q_{ij}$. Its eigenvectors give the displacement polarizations. A positive eigenvalue gives a real nonzero speed $c$; arbitrary-profile propagation imposes precisely this algebraic condition. Boundary conditions determine which combinations of these local plane waves are allowed in a finite body.

For a static base state under dead loading, put $v_i=u_i(\xi )e^{i\omega t}$ and take zero incremental body force and traction. The admissible squared frequencies are the eigenvalues of the weighted boundary value problemMultiply by $\overline{u_i}$ and integrate over the reference body. The boundary term in integration by parts vanishes, giving the elastic normal-mode energy identityFor a nonzero mode and positive mass density, the denominator is real and positive. The numerator is real by major symmetry of the incremental elastic moduli. Explicitly, if $u=p+iq$ with real $p,q$, the two imaginary cross terms cancel, leaving the sum of the quadratic energies of $\nabla p$ and $\nabla q$. Consequently every admissible ${\omega }^2$ is real. A negative value, when allowed by indefinite moduli, corresponds to exponential growth or decay rather than an oscillation with real frequency.

If the moduli quadratic form is strictly positive on every nonzero real matrix, that same decomposition shows that the numerator is positive whenever $\nabla u$ is nonzero. Thus ${\omega }^2>0$ for every nontranslation mode. Under the usual assumptions of a bounded connected body, regular coefficients and uniform positivity, the Rayleigh quotient defines a self-adjoint operator with discrete positive squared frequencies on the mass-weighted mean-zero displacement space. Its eigenfunctions give the nontrivial small harmonic oscillations.

There is an unavoidable zero-frequency qualification for the pure traction problem as printed: $u_i=b_i$, for any nonzero constant vector $b$, has zero gradient, zero incremental nominal stress and ${\omega }^2=0$. These three constant translations obey all the homogeneous incremental boundary conditions. Under the strict quadratic-form hypothesis they are the entire zero eigenspace on a connected body. Therefore the unrestricted conclusion isStrict positivity for every nonzero displacement requires excluding these modes, for example by imposing ${\int }_{{\Omega }_0}{\rho }_{0}udV=0$, or by adding a displacement constraint. If harmonic perturbations are defined to have nonzero frequency from the outset, the requested positivity holds for all those perturbations.

The fixed-displacement boundary value problem now has $u=0$ on the whole boundary. The elastic normal-mode energy identity still holds, because its boundary term contains $\overline{u}$. DefineFor real constant $K$, $B$ has major symmetry: both Levi-Civita symbols change sign upon exchanging the two index pairs. Its quadratic density is a quadratic minor null Lagrangian. Indeed,The boundary integral vanishes by the Dirichlet boundary condition, and the volume integral vanishes because mixed derivatives are symmetric in $\alpha ,\beta$ whereas ${ϵ}_{\alpha \beta \gamma }$ is antisymmetric. Constancy of $K$ is essential: otherwise its derivatives would produce extra terms. This calculation initially applies to smooth fields and extends to fields with zero Sobolev boundary trace by density.

It follows that the Rayleigh quotient numerator can be replaced by the quadratic form with $H$:The assumed positivity of $H$ on all nonzero real matrices also gives positivity on nonzero complex matrices by splitting real and imaginary parts. A nonzero $u$ satisfying $u=0$ on the boundary cannot have identically zero gradient. Hence the right side is strictly positive, andThis works even when the original moduli quadratic form is not pointwise positive on all matrices: its integrated energy on fixed-boundary gradients agrees with a positive density after adding the quadratic minor null Lagrangian.

At a fixed material point, freeze the incremental elastic moduli at their local values and fix a unit propagation direction $\nu$. For an arbitrary nonzero polarization $m$, the rank-one matrix $a_{i\alpha }=m_i{\nu }_{\alpha }$ annihilates the added quadratic minor null Lagrangian, sinceThe assumed positivity therefore givesThus the acoustic tensor is positive-definite; this is strong ellipticity in elasticity. Major symmetry also makes it a real symmetric matrix. By the spectral theorem for real symmetric matrices, it has three real positive eigenvalues $q_1,q_2,q_3$ and an orthonormal basis of real polarizations $m^{(1)},m^{(2)},m^{(3)}$. ThereforeEach polarization gives a real plane wave profile through the equation of part (a), in either sign of the propagation speed. Repeated eigenvalues still admit three independent polarizations; the claim is not that the three speeds must be distinct. For an inhomogeneous base deformation these are local, frozen-coefficient wave modes, rather than exact plane waves throughout the entire body.