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Exact dispersion relation ω2=(4c2/a2)sin⁡2(κa/2) for the cyclic chain and its continuum error bound

Statement

Let N≥2, a>0, c>0, let κ,ω∈R satisfy κNa∈2πZ, and let A∈R∖{0}. Consider the travelling cosine ansatz for the cyclic chain of The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class:

qj(t):=Acos⁡(κja−ωt)(j∈Z/N).

(a) The curve q:R→RN solves qj′′=c2a2(qj+1−2qj+qj−1) for all t∈R and all j if and only if

ω2=4c2a2sin⁡2 ⁣(κa2).(DR)

(b) If (DR) holds and ∣κa∣≤1, then

∣ω2−c2κ2∣≤c2κ4a212,

so the continuum dispersion ω2=c2κ2 is recovered with an explicit error bound of second order in the spacing a and fourth order in the wave number κ.

Facts & Assumptions

Given: N,a,c,κ,ω,A with A≠0 and the ansatz qj(t)=Acos⁡(κja−ωt).

[F1]

Trigonometric addition formulas, 1−cos⁡u=2sin⁡2(u/2) and the bound ∣cos⁡r∣≤1 (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Sine and cosine defined by their real power series). The additional bound ∣sin⁡r∣≤∣r∣ follows locally: by the derivative identity [F2] and integral evaluation [F3], sin⁡r=∫0rcos⁡s ds, so ∣sin⁡r∣≤∫min⁡(0,r)max⁡(0,r)∣cos⁡s∣ ds≤∣r∣.

Proof

technique · direct; evaluate both sides of the ODE on the ansatz and bound the trigonometric remainder
1.1givenF1F2algebra

Since κNa∈2πZ, qj+N(t)=qj(t) for every integer j and every t by cosine periodicity, so the ansatz is well-defined on the cyclic sites and its shifted values agree across the seam. Put θj:=κja−ωt. By [F2], qj′′(t)=−ω2Acos⁡θj. By [F1], cos⁡(θj+κa)+cos⁡(θj−κa)=2cos⁡θjcos⁡(κa), so qj+1−2qj+qj−1=Acos⁡θj(2cos⁡(κa)−2)=−2Acos⁡θj(1−cos⁡(κa))=−4Acos⁡θjsin⁡2 ⁣(κa2).

1.2givenF1F3

For the error bound, first prove ∣sin⁡z−z∣≤∣z∣3/6 for all z. Write sin⁡z−z=∫0z(cos⁡s−1) ds and cos⁡s−1=−∫0ssin⁡r dr; by [F3] and ∣sin⁡r∣≤∣r∣, ∣cos⁡s−1∣≤s2/2; hence ∣sin⁡z−z∣≤∫0∣z∣s2/2 ds=∣z∣3/6.

2.1step 1.1givenalgebra

The ODE holds for all t,j if and only if −ω2Acos⁡θj=−4c2a2Acos⁡θjsin⁡2 ⁣(κa2) for all t,j. Since A≠0 and cos⁡θj=1 for θj=0 (a value attained for appropriate t, for instance t=κja/ω when ω≠0, and the identity is also immediate at t=0, j≡0), the coefficients must agree; conversely (DR) makes the two sides equal identically. If ω=0 the same comparison at t=0 and j=0 gives the same conclusion. This proves (a).

3.1step 1.2step 2.1F1algebra∎

Put z:=κa/2 and assume ∣z∣≤1/2 (equivalently ∣κa∣≤1). Then ∣4sin⁡2z−(2z)2∣=4 ∣sin⁡z−z∣ ∣sin⁡z+z∣≤4⋅∣z∣36⋅2∣z∣=4∣z∣43, using ∣sin⁡z∣≤∣z∣ in ∣sin⁡z+z∣≤2∣z∣. Substituting ∣z∣=∣κa∣/2 gives ∣4sin⁡2(κa/2)−(κa)2∣≤(κa)4/12. Multiplying by c2/a2 and using (DR) yields the claimed bound ∣ω2−c2κ2∣≤c2κ4a2/12.

Remarks

  • The exact relation (DR) is bounded between 0 and 4c2/a2; the long-wavelength branch has ω2→c2κ2 as κa→0, and the bound of (b) makes that limit quantitative. The bound is stated for ∣κa∣≤1 only; no claim is made for short waves.

  • The ansatz is a real travelling wave of constant amplitude; it is not asserted that every chain solution is a superposition of such waves, and no stability or energy estimate for the ansatz is claimed here. The continuum comparison of this item is strictly local in wave number.

Remark

The scalar remainder calculation in steps 1.2 and 3.1 proves ∣sin⁡z−z∣≤∣z∣3/6 and ∣4sin⁡2z−4z2∣≤4∣z∣4/3 for every real z: those inequalities use only ∣sin⁡z∣≤∣z∣, not the additional long-wave restriction of clause (b). They can therefore be used directly to check a specified compatible mode outside that restricted clause.

Depends on

Dependency tree · 2 levels

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