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Chain versus string on the normalized standing wave: exact discrete error, the N=100 bounds, and physical circumference 2π m

Physical scope

A worked comparison in SI-normalized numerical variables of the cyclic mass-spring chain and continuum string on the standing wave sin(x)cos(t), using a physical circumference of 2π m, linear density 1 kg m^{-1}, tension 1 N, and speed 1 m s^{-1}. It computes the exact chain solution and frequency, nodal error, the theorem's nodal and full interpolation bounds at N=100 (both O(a^{3/2}) for fixed positive time, with interpolation remainder Aa^2/8), the sharper O(a^2) nodal and interpolation errors for these standing-wave data, and the dispersion-relation error a^2/12. The normalized results are model calculations, not measurements.

Example

Take the periodic string model of The continuum string model as the proved limit of the mass-spring chain, with explicit convergence and dispersion error with physical circumference L∗=2π m, linear mass density ρℓ,∗=1 kg m−1 and tension τs,∗=1 N=1 kg m s−2. Then c∗:=τs,∗/ρℓ,∗=1 m s−1. Use reference units ℓ∗=1 m, τ∗=1 s and displacement scale A∗=1 m. In the equations below, x,t,L,c denote numerical values in these units, so L=2π, c=1, and the dimensionless displacement u=uphys/A∗ has the exact travelling-standing solution with physical wave number 1 m−1 and angular frequency 1 s−1

u(t,x):=Asin⁡xcos⁡t(A>0),

where A is dimensionless. Physical displacements and their error bounds are obtained by multiplying the displayed values by A∗. The normalized density and tension values are ρℓ=ρℓ,∗/(1 kg m−1)=1 and τs=τs,∗/(1 N)=1; these are numerical values of different dimensioned quantities. The normalized field is 2π-periodic and solves utt=uxx. For N≥2 put a=2π/N and consider the cyclic spring chain of The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class with parameter c=1 and spacing a, i.e. the normalized node mass is m=ρℓa in units of 1 kg and the normalized spring constant is ks=τs/a in units of 1 N m−1. Then:

  1. M=sup⁡∣∂x4u∣=A, M2=sup⁡∣∂x2u∣=A and the theorem's constant is C=2πA/12≈0.20889A;
  2. the exact chain solution with the sampled data is qj(t)=Asin⁡(ja)cos⁡(ωat) with ωa=2sin⁡(a/2)/a (positive for 0<a<π), the exact nodal error is ej(t)=Asin⁡(ja)[cos⁡(ωat)−cos⁡t], and the theorem's bounds read ∥e(t)∥a≤Ca2t2/2, max⁡j∣ej(t)∣≤12Ca3/2t2 and sup⁡x∣Iaq−u∣≤max⁡j∣ej∣+Aa2/8;
  3. at N=100 (so a=2π/100≈0.062832) and t=1 the evaluated bounds are ∥e(1)∥a≤0.0004124A, max⁡j∣ej(1)∣≤0.001645A and sup⁡∣Iaq−u∣≤0.002139A, while the exact error satisfies ∣ej(1)∣≤A∣ωa−1∣≈0.00016449A, so the theoretical bound is verified and visibly conservative;
  4. the exact dispersion relation is ωa2=(4/a2)sin⁡2(a/2), and the theorem's bound ∣ωa2−1∣≤a2/12 gives ≤0.00032899 at N=100, while the exact value satisfies ∣ωa2−1∣≈0.00032894.

Facts & Assumptions

Given: the physical parameters L∗=2π m, ρℓ,∗=1 kg m−1, τs,∗=1 N, the reference units and normalized values L=2π, c=1, ρℓ=1, τs=1; the dimensionless amplitude A>0; the standing wave u; and the integer N≥2 and normalized spacing a=2π/N.

[F1]

Chain and norms (The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class): the cyclic chain equation is qj′′=(c2/a2)(qj+1−2qj+qj−1) with discrete norm ∥⋅∥a, and the sampled data are qj(0)=u(0,ja), qj′(0)=ut(0,ja).

[F2]

Continuum-limit theorem (Finite-time continuum limit of the cyclic spring chain: uniform O(a3/2) nodal and uniform interpolation convergence): under the stated smoothness and periodicity hypotheses on u and its wave equation, the chain solution exists and is unique on [0,T], and with ej=qj−u(t,ja) and C=c2LM/12 one has ∥e(t)∥a≤Ca2t2/2, ∥e′(t)∥a≤Ca2t, max⁡j∣ej(t)∣≤12Ca3/2t2 and sup⁡x∣Iaq(t,x)−u(t,x)∣≤max⁡j∣ej(t)∣+M2a2/8.

[F3]

Dispersion theorem (Exact dispersion relation ω2=(4c2/a2)sin⁡2(κa/2) for the cyclic chain and its continuum error bound): for A≠0 and κNa∈2πZ, qj(t)=Acos⁡(κja−ωt) solves the chain equation if and only if ω2=(4c2/a2)sin⁡2(κa/2); when this dispersion relation holds and ∣κa∣≤1, ∣ω2−c2κ2∣≤c2κ4a2/12.

[F4]
[F5]

Verification

technique · direct; verify the standing wave and the exact discrete solution, evaluate the theorem's bounds at $N=100$, and compare with the exact error and dispersion values
1.1givenF4F5algebra

Data: ∂x4u=Asin⁡xcos⁡t and ∂x2u=−Asin⁡xcos⁡t, so M=M2=A; by [F5], C=2πA/12.

1.2givenF1F3F4algebra

Exact chain solution: with qj(t)=Asin⁡(ja)cos⁡(ωat) one has qj′′=−ωa2qj and qj+1−2qj+qj−1=−4sin⁡2(a/2)qj, so the chain equation qj′′=a−2(qj+1−2qj+qj−1) holds for the chosen frequency ωa2=4sin⁡2(a/2)/a2, i.e. ωa=2sin⁡(a/2)/a for a∈(0,π], since N≥2; at N=2 the sampled values sin⁡(jπ) vanish and the zero chain solves independently of frequency. The sampled data match because qj(0)=Asin⁡(ja)=u(0,ja) and qj′(0)=0=ut(0,ja).

2.1step 1.2F4algebra

Exact nodal error: ej(t)=qj(t)−u(t,ja)=Asin⁡(ja)[cos⁡(ωat)−cos⁡t], hence by [F4] ∣ej(t)∣≤A∣ωa−1∣ t for t≥0.

2.2step 1.1F2algebra

Bounds at N=100: with a=2π/100 one has a2=4π2/104 and a3/2=(2π/100)3/2. Hence at t=1 the three right-hand sides of [F2] are Ca2/2=2π π2A/60000, 12Ca3/2=π2A/6000 and π2A/6000+π2A/20000=13π2A/60000, using M2a2/8=Aa2/8=π2A/20000.

3.1step 2.1step 2.2algebra

Exact comparison: with a=π/50 one has ωa=2sin⁡(a/2)/a=100sin⁡(π/100)/π, so by step 2.1 the exact nodal error at t=1 obeys ∣ej(1)∣≤A∣100sin⁡(π/100)/π−1∣, while the theorem's nodal bound at t=1 is π2A/6000 by step 2.2; the comparison of the two explicit numerical factors is a finite check recorded in the Remarks.

4.1step 1.2F3F4algebra∎

Dispersion for all N≥2: put z=a/2>0. By [F4] and the fundamental theorem, sin⁡z−z=∫0z(cos⁡s−1) ds and cos⁡s−1=−∫0ssin⁡r dr; hence ∣sin⁡z−z∣≤∫0z∫0sr dr ds=z3/6, using ∣sin⁡r∣≤r, while ∣sin⁡z+z∣≤2z. Therefore ∣ωa2−1∣=4∣sin⁡2z−z2∣/a2≤4(z3/6)(2z)/a2=a2/12. At N=100 this equals π2/30000; here a<1 also permits the bound of [F3] with c=κ=1 and the chosen solving-mode frequency. Moreover ωa2=sin⁡2(a/2)/(a/2)2=1−a2/12+O(a4) by the power series of sine, so the exact error equals the bound to leading order in a.

Remarks

  • The comparison is a finite, exact check: the chain's standing mode has frequency ωa and the continuum mode frequency 1; the nodal error is the interference of the two cosines, so the discrepancy is a purely discrete-dispersion effect and decays like a2 for the exact standing-wave data, while the theorem's uniform-in-data bound grows like a3/2 through the worst-case nodal factor a−1/2 of the discrete norm comparison.

  • Numerical readings at N=100 for the bounds of step 2.2: 2π π2/60000≈0.0004124, π2/6000≈0.001645 and 13π2/60000≈0.002139, while the frequency-difference upper bound of step 3.1 is ∣100sin⁡(π/100)/π−1∣≈0.0001645, while the exact maximum nodal error factor is ∣cos⁡(100sin⁡(π/100)/π)−cos⁡1∣≈0.0001384 (nodes j=25,75 have ∣sin⁡(ja)∣=1) and the dispersion bound of step 4.1 is π2/30000≈0.000329; the exact dispersion error ∣ωa2−1∣=1−sin⁡2(π/100)/(π/100)2 agrees to leading order. The theorem's bounds are therefore satisfied and conservative on these data.

  • The evaluated numbers are properties of the model's bounds under the stated scaling m=ρℓa, ks=τs/a; they are not measured errors of any physical chain, and no measured stiffness or density is used.

Depends on

Dependency tree · 2 levels

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