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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.