A longitudinal plane wave in the linear elastic branch: stress, equation of motion and strain energy computed

Adopt the isotropic linear elastic branch of Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy with reference density ${\rho }_0>0$ of dimension $\mu {\mathrm{\ell }}^{-3}$ and moduli $\lambda ,{\mu }_s$ of dimension $\mu {\mathrm{\ell }}^{-1}{\tau }^{-2}$, with ${\mu }_s>0$ and $\lambda +\frac{2{\mu }_s}{3}>0$ (Continuum body data, the fixed inertial frame and the dimensions of the fields). Use $I=\mathbb{R}$, $O={\mathbb{R}}^3$ and let $x_1$ be the reference position in metres and $t$ time in seconds. The amplitude $A\mathrm{\ne }0$ has dimension $\mathrm{\ell }$, and the wave number $\kappa \mathrm{\ne }0$ has dimension ${\mathrm{\ell }}^{-1}$. Define the positive longitudinal speed and angular frequency by $c_L:=\sqrt{(\lambda +2{\mu }_s)\mathrm{/}{\rho }_0}$ and $\omega :=\kappa c_L$; then $[c_L]=\mathrm{\ell }{\tau }^{-1}$, $[\omega ]={\tau }^{-1}$ and the phase is dimensionless. Let

$\xi (t,x):=A\cos (\kappa x_1-\omega t)e_1,$

with $A\mathrm{\ne }0$, $\kappa \mathrm{\ne }0$ and $e_1$ the first standard basis vector: a longitudinal plane wave travelling along $e_1$. This is a solution of the linear reference-field model for every declared $A,\kappa$; physical small-strain use requires small $\mathrm{\mid }A\kappa \mathrm{\mid }$. No finite-motion diffeomorphism is inferred: its candidate placement has determinant $1-A\kappa \sin (\kappa x_1-\omega t)$, which need not be positive for these unrestricted algebraic data. Then:

1. the linear strain is $e_{11}(t,x)=-A\kappa \sin (\kappa x_1-\omega t)$ with all other entries zero, so $\mathrm{tr}e=e_{11}$ and $e_0=e-\frac{1}{3}(\mathrm{tr}e)I_3$ has Frobenius squared norm $\frac{2}{3}{e}_{11}^2$;
2. the stress matrix is $\sigma (t,x)=(\lambda +2{\mu }_s)e_{11}(t,x)e_1{e}_1^{\mathsf{T}}+\lambda e_{11}(t,x)(e_2{e}_2^{\mathsf{T}}+e_3{e}_3^{\mathsf{T}});$
3. the displacement equation (E) of The displacement equation of the linear elastic branch: rho_0 partial_t^2 xi = rho_0 b + mu_s Laplacian xi + (lambda + mu_s) grad div xi with $b\equiv 0$ holds exactly: ${\rho }_0{\mathrm{\partial }}_t^2{\xi }_1=(\lambda +2{\mu }_s){\mathrm{\partial }}_1^2{\xi }_1$;
4. the strain-energy density, of dimension $\mu {\mathrm{\ell }}^{-1}{\tau }^{-2}$, is $W(e)=(\frac{\lambda }{2}+{\mu }_s)e_{11}^2={\mu }_s\frac{2}{3}{e}_{11}^2+(\frac{\lambda }{2}+\frac{{\mu }_s}{3})e_{11}^2\ge 0$, with equality only where $e_{11}=0$; both coefficients in the decomposition are positive under the branch's hypotheses, and the squared speed $c_L^2=(\lambda +2{\mu }_s)\mathrm{/}{\rho }_0$ is positive with $[c_L]=\mathrm{\ell }{\tau }^{-1}$.

• The computation exhibits the branch's structure concretely: the wave is longitudinal (the displacement is parallel to the propagation direction) and its speed is the longitudinal speed of Plane elastic waves in the adopted model: the dispersion condition and the longitudinal and transverse speeds; the transverse branch, with speed $\sqrt{{\mu }_s\mathrm{/}{\rho }_0}$, is not recomputed here.

• The energy density computed is the model's strain energy of the branch (The small-strain isotropic linear elastic constitutive law and its strain energy); it is a nonnegative function of the strain and is zero exactly when the linear strain vanishes, under the branch's positivity hypotheses. Rigid translations and infinitesimal rigid rotations both have zero linear strain; for this particular longitudinal wave, zero energy is exactly $e_{11}=0$. No measured modulus or speed appears.