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A compressional wave is a longitudinal wave whose propagation produces changes of volume or density. In an isotropic elastic solid its displacement is parallel to the wavevector; substituting that polarization into the dynamic Navier-Cauchy equation gives the displayed speed. In a fluid without static shear stiffness the analogous speed is $\sqrt{K/\rho }$, with the appropriate compressional bulk modulus. Reflection at a rigid elastic boundary can convert part of its energy into a shear wave.

For the one-dimensional reduction of isotropic linear elasticity, nondegeneracy means that the effective longitudinal modulus $\lambda +2\mu$ is nonzero. The static Navier-Cauchy equation then forces the displacement to be affine.

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.

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.

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.

For antiplane shear, the displacement field is $\mathbf{u}=(0,0,w(x,y,t))$, independent of $z$. Its divergence is zero. Consequently the only possibly nonzero components of the stress tensor are the shear stressesThe $x$ and $y$ components of the Navier-Cauchy equation vanish identically. Its $z$ component gives $\rho w_{tt}={\partial }_x(\mu w_x)+{\partial }_y(\mu w_y)$. Each material has constant shear modulus and mass density, so the required wave equation and wave speeds are

At a perfectly bonded elastic interface, both the displacement field and the traction are continuous. Using the same normal ${\mathbf{e}}_y$ on both sides, the only nonzero traction component is ${\tau }_{zy}$. Thus the interface boundary conditions areContinuity of “stress” here means continuity of traction; it does not require equality of the full stress tensors. The rigid plate supplies the remaining boundary condition $w_1(x,0,t)=0$.

In the one-dimensional limit, $v=0$ and $u=u(x)$. The first component of the Navier-Cauchy equation reduces toFor a nondegenerate one-dimensional elastic material, $\lambda +2\mu \ne 0$, so $u^{\prime \prime }=0$ and $u(x)=Ax+B$. The boundary conditions $u(0)=0$ and $u(L)=\Delta$ give $B=0$ and $A=\Delta /L$. Therefore the uniform extension of a one-dimensional elastic body is

Let the displayed integrand be $F$. Since it has no explicit dependence on $u$ or $v$, the Euler-Lagrange equations for two fields areIts derivatives areThusUsing the Laplacian, gradient, and divergence, these combine into the static Navier-Cauchy equation

Let the boundary split into a displacement part ${\Gamma }_D$ and a traction part ${\Gamma }_N$. The linearized stored elastic energy functional isIts corresponding total potential subtracts the work of fixed body and surface loads:The problem in the calculus of variations is to minimize $\Pi$ among admissible displacements with $u=u_D$ on ${\Gamma }_D$. For a stable quadratic elastic energy this is the usual energy minimization formulation. Its stationarity condition suffices to derive equilibrium.

Vary by $u+sv$, with $v=0$ on ${\Gamma }_D$. Major symmetry combines the two differentiated quadratic terms, givingIntegration by parts, first with test functions compactly supported in the interior, gives $-\mathrm{Div}{S}^{\mathrm{lin}}=f$. Then arbitrary traces on ${\Gamma }_N$ give the natural traction condition. The complete boundary value problem isFor a homogeneous hyperelastic material the elasticity tensor is constant. In isotropic linear elasticity the constitutive equation is $S^{\mathrm{lin}}=\lambda \mathrm{div}uI+\mu (Du+D{u}^T)$, giving the familiar Navier-Cauchy equation $-\mu \Delta u-(\lambda +\mu )\nabla \mathrm{div}u=f$.

In a homogeneous isotropic elastic half-space $Y>0$, a Rayleigh wave with factor $e^{ik(X-ct)}$ has two depth factors $e^{-b_{L}Y}$ and $e^{-b_{T}Y}$. Substitution into the Navier-Cauchy equation factors the amplitude determinant into longitudinal and transverse factors, giving the displayed exponents. Strict exponential decay requires $0<c<\min (c_L,c_T)$. The longitudinal constituent has $(A,B)\propto (ik,-b_L)$; the transverse constituent has $ikA-b_{T}B=0$. The free-surface traction conditions select a combination of both constituents.

A shear wave in an isotropic elastic material has displacement perpendicular to its wavevector, with zero divergence for a plane wave. The Navier-Cauchy equation with inertia then reduces to $\rho {\mathbf{u}}_{tt}=\mu \Delta \mathbf{u}$, giving the displayed wave speed. At a boundary it can be generated by mode conversion from a compressional wave, even when the incident displacement is purely longitudinal.