= Solution
For uniform <density> $\rho>0$ and stiffness, linear momentum balance without <body force> is $\rho\ddot u_i=\partial_j\sigma_{ij}$. The minor elastic symmetries make the two symmetric-strain terms equal, so $\partial_j\sigma_{ij}=c_{ijkl}u_{k,lj}$. Write $s=ct-\hat p_jx_j$ and $u_i=a_if(s)$. Then $\ddot u_i=c^2a_if''$ and $u_{k,lj}=a_k\hat p_l\hat p_jf''$. Thus a nontrivial plane-wave profile satisfies
$$
\rho c^2a_i=c_{ijkl}\hat p_j\hat p_la_k.
$$
The density-normalized <elastic acoustic tensor> is consequently
$$
\boxed{B_{ij}=\frac1\rho c_{ikjl}\hat p_k\hat p_l,\qquad B_{ij}a_j=c^2a_i.}
$$
The alternative dummy-index form $B_{ik}=\rho^{-1}c_{ijkl}\hat p_j\hat p_l$ is identical. In ordinary conservative <elasticity>, major symmetry $c_{ikjl}=c_{jlik}$ gives $B_{ji}=B_{ij}$ after interchanging the dummy indices. This symmetry comes from a strain-energy <density>; it is an assumption of the ordinary elastic tensor, not a consequence of homogeneity alone.
For any such eigenpair the profile $f$ is arbitrary, since both sides carry the same $f''$. This nondispersive result uses linear, local, time-independent <elasticity> without attenuation or spatial dispersion. A profile with $f''=0$ is a trivial uniform or affine <displacement> field and cannot be used by itself to infer the wave-speed <eigenvalue> condition.
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