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Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds

Statement

Let ρ0>0 and Lam'e parameters λ,μ∈R with μ>0 and λ+2μ3>0, let a,k∈R3 and ω∈R∖{0}, and define the plane wave

ξ:R×R3→R3,ξ(t,x):=acos⁡(⟨k,x⟩2−ωt),

with the Euclidean inner product ⟨⋅,⋅⟩2 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn). For each fixed t, the spatial displacement map is ξt(x):=ξ(t,x). Define the strain and stress fields using the spatial derivative at fixed time:

e(t,x):=12(Dxξ(t,x)+(Dxξ(t,x))T),σ(t,x):=λ(tr⁡e(t,x))I3+2μe(t,x).

These are the pointwise constitutive formulas of the isotropic linear elastic model (The small-strain isotropic linear elastic constitutive law and its strain energy); in particular Dxξ is the total derivative of x↦ξt(x), not the derivative of the full spacetime map.

Then the componentwise identity

ρ0 ∂t2ξi=∑j=13∂jσij(i=1,2,3)

holds on all of R×R3 if and only if

ρ0ω2a=μ∣k∣22 a+(λ+μ) ⟨k,a⟩2 k.(D)

When (D) holds and a,k are nonzero, define the phase speed by cph:=∣ω∣/∣k∣2. If a=αk for some α≠0 (longitudinal polarization), then ρ0ω2=(λ+2μ)∣k∣22 and cph2=(λ+2μ)/ρ0. If ⟨k,a⟩2=0 (transverse polarization), then ρ0ω2=μ∣k∣22 and cph2=μ/ρ0. The constitutive positivity hypotheses imply λ+2μ=(λ+2μ/3)+4μ/3>0 and μ>0, so both branch phase speeds are real and positive. (When k=0, (D) has no solution with a≠0 and ω≠0.)

Facts & Assumptions

Given: The parameters ρ0,λ,μ, the vectors a,k, the frequency ω≠0, and the field ξ as constructed.

Proof

technique · direct computation; compare the coefficients of the single scalar factor $\cos(\langle k,x\rangle_2-\omega t)$
1.1givenF1algebra

Put θ(t,x):=⟨k,x⟩2−ωt. Then ∂tθ=−ω and ∂jθ=kj, so by [F1] and the chain rule ∂tξi=aiωsin⁡θ,∂jξi=−aikjsin⁡θ,∂t2ξi=−aiω2cos⁡θ.

2.1step 1.1F2algebra

From step 1.1, eij=12(∂jξi+∂iξj)=−12(aikj+ajki)sin⁡θ and tr⁡e=−⟨k,a⟩2sin⁡θ. Hence σij=−sin⁡θ(λ⟨k,a⟩2 δij+μ(aikj+ajki)).

3.1step 2.1F1algebra

Differentiate step 2.1 and use [F1]: ∑j∂jσij=−cos⁡θ∑jkj(λ⟨k,a⟩2δij+μ(aikj+ajki))=−cos⁡θ(μ∣k∣22ai+(λ+μ)⟨k,a⟩2ki), where ∑jkjδij=ki, ∑jkj2=∣k∣22 and ∑jkjaj=⟨k,a⟩2.

4.1step 1.1step 3.1algebra

The identity ρ0∂t2ξi=∑j∂jσij for all (t,x) reads, by steps 1.1 and 3.1, cos⁡θ(ρ0ω2ai−μ∣k∣22ai−(λ+μ)⟨k,a⟩2ki)=0 for all (t,x). Choosing t=0, x=0 gives cos⁡θ=1, so the bracket vanishes; conversely vanishing of the bracket makes the identity hold everywhere. This is exactly (D).

5.1step 4.1givenalgebra∎

Assume (D) and a,k≠0. If a=αk with α≠0, then (D) reads ρ0ω2αk=μ∣k∣22αk+(λ+μ)α∣k∣22k=α∣k∣22(λ+2μ)k, so ρ0ω2=(λ+2μ)∣k∣22. If ⟨k,a⟩2=0, then (D) reads ρ0ω2a=μ∣k∣22a, so ρ0ω2=μ∣k∣22. Dividing by ρ0∣k∣22 gives the phase-speed formulas. Positivity follows from λ+2μ=(λ+2μ/3)+4μ/3>0 and μ>0.

Remarks

  • The theorem verifies exact solutions of the adopted constant-coefficient PDE; it does not assert that every solution of the elastic model is a plane wave, that a wave packet propagates unchanged, or that any material realises the model. General initial-value problems for the elastic system are not solved here.

  • The identity (D) is the dispersion relation of the plane-wave ansatz. The two eigenvalue branches computed in the last step are the longitudinal and transverse speeds; they require λ+2μ>0 and μ>0 respectively, and no claim is made about λ alone.

Depends on

Dependency tree · 2 levels

Sources