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Two-mode fixed-end string: exact solution, d'Alembert cross-check and the O(N^{-4}) tail bound at N=1

Physical scope

A worked fixed-end string problem of the adopted model: an initial displacement with two sine modes and zero initial velocity, its exact two-mode solution, agreement with the d'Alembert formula, mode frequencies and a finite check of the O(N^{-4}) tail bound at N=1. The example is an exact conditional computation within the linear string model and reports no measurement.

Example

Let L>0 be a length, c>0 a speed, A∗>0 a displacement scale with [A∗]=ℓ, and β∈R∖{0} dimensionless. Let [x]=ℓ and [t]=τ, and consider the fixed-end string model of The string model: d'Alembert solutions, fixed-end modes and uniqueness for the adopted one-dimensional continuum with initial displacement and velocity

F(x):=A∗[sin⁡(πxL)+βsin⁡(3πxL)],G(x):=0.

Then F is odd and 2L-periodic of class C∞ and G is odd and 2L-periodic of class C4, with M4=sup⁡∣G(4)∣=0 and M5=sup⁡∣F(5)∣≤A∗(π/L)5(1+243∣β∣)=:M‾5 (Sine coefficients, sine series and the odd periodic extension). Put ωn:=nπc/L and

u(t,x):=A∗[cos⁡(ω1t)sin⁡(πxL)+βcos⁡(ω3t)sin⁡(3πxL)].

Then:

  1. the sine coefficients of F are b1(F)=A∗, b3(F)=βA∗ and bn(F)=0 otherwise, and the coefficients of G all vanish; hence the series solution of The classical C2 fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound is exactly the displayed finite sum u;
  2. u is C∞ on R2, satisfies utt=c2uxx, vanishes at x=0 and x=L for every t, and has u(0,⋅)=F, ut(0,⋅)=0; by the uniqueness clause of the theorem it is the unique C2 fixed-end solution with these data;
  3. the d'Alembert formula of d'Alembert's formula solves the wave equation, is the unique C2 solution, and the fixed-end sine modes solve it exactly with data (F,0) gives the same field: 12[F(x−ct)+F(x+ct)]=u(t,x) by the product-to-sum identity (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine);
  4. the tail bound at N=1 is verified: with S1u=A∗cos⁡(ω1t)sin⁡(πx/L) one has ∣u−S1u∣=A∗∣β∣∣cos⁡(ω3t)∣∣sin⁡(3πx/L)∣≤A∗∣β∣, while the theorem gives ∣u−SNu∣≤C(N+1)−4 with C≤C‾:=4A∗(1+243∣β∣), so the explicit upper estimate is A∗(1+243∣β∣)/4 at N=1, which also dominates the direct bound A∗∣β∣ for every β;
  5. the mode frequencies satisfy [ωn]=τ−1 under [c]=ℓτ−1 and [L]=ℓ (Continuum body data, the fixed inertial frame and the dimensions of the fields).

Facts & Assumptions

Given: the string data L>0, c>0, A∗>0 with [A∗]=ℓ; the dimensionless parameter β≠0; the functions F,G and the field u.

[F1]

Fixed-end theorem (The classical C2 fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound, Sine coefficients, sine series and the odd periodic extension): for odd 2L-periodic C5 displacement data F and odd 2L-periodic C4 velocity data G, the sine series with coefficients an=bn(F), dn=bn(G)/ωn solves the fixed-end problem, is unique among C2 solutions, and satisfies the tail bound ∣u−SNu∣≤C(N+1)−4 with C=2(2L5M5/π5+2L5M4/(cπ5)).

[F2]

d'Alembert formula (d'Alembert's formula solves the wave equation, is the unique C2 solution, and the fixed-end sine modes solve it exactly): for F∈C2(R) and G≡0 the field 12[F(x−ct)+F(x+ct)] is C2 and solves utt=c2uxx with u(0,⋅)=F,ut(0,⋅)=0; uniqueness among C2 fields on R2 holds.

[F3]

Trigonometry and derivatives (The addition formulas for sine and cosine, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, Ck Euclidean maps and diffeomorphisms): F and G have the stated smoothness and parity; sin⁡acos⁡b=12[sin⁡(a+b)+sin⁡(a−b)]; ω3=3ω1; and ∣F(5)∣=A∗∣(π/L)5cos⁡(πx/L)+243β(π/L)5cos⁡(3πx/L)∣≤A∗(π/L)5(1+243∣β∣).

[F4]

String model (The string model: d'Alembert solutions, fixed-end modes and uniqueness for the adopted one-dimensional continuum): the model equation is utt=c2uxx for the transverse displacement with the stated idealizations and dimension conventions (Continuum body data, the fixed inertial frame and the dimensions of the fields).

Verification

technique · direct; identify the finite sine expansion of the two-mode data, verify the equation and data mode by mode, cross-check with the d'Alembert formula and evaluate the tail bound at $N=1$
1.1givenF1algebra

Coefficients: by the orthogonality of the sine family, bn(F)=2L∫0LF(x)sin⁡(nπx/L) dx equals A∗ for n=1, βA∗ for n=3 and 0 otherwise (Sine coefficients, sine series and the odd periodic extension); likewise bn(G)=0 for all n. Hence the series solution of [F1] is ∑n[bn(F)cos⁡(ωnt)+0]sin⁡(nπx/L)=u(t,x), a finite sum.

2.1step 1.1F3F4algebra

Equation and data: the model equation is the one of [F4], and differentiating the two terms twice in t and x gives utt=−∑n∈{1,3}ωn2[…] and c2uxx=−c2∑n(nπ/L)2[…], which agree because ωn=cnπ/L; at t=0 the cosines equal 1 so u(0,⋅)=F, and ut(t,x)=−∑nωnsin⁡(ωnt)[…] vanishes at t=0, giving ut(0,⋅)=0. Boundary conditions hold since each sin⁡(nπx/L) vanishes at x=0,L.

3.1step 2.1F1

Uniqueness: by [F1] the solution is unique among C2 fields on [0,T]×[0,L] with these boundary and initial data; the displayed u is therefore the model's solution.

3.2step 2.1F2F3algebra

d'Alembert cross-check: by the product-to-sum identity of [F3], 12[F(x−ct)+F(x+ct)]=A∗2[sin⁡(π(x−ct)/L)+sin⁡(π(x+ct)/L)]+βA∗2[sin⁡(3π(x−ct)/L)+sin⁡(3π(x+ct)/L)]=A∗cos⁡(ω1t)sin⁡(πx/L)+βA∗cos⁡(ω3t)sin⁡(3πx/L)=u(t,x), so the two formulas agree for these data.

4.1step 2.1F1F3algebra∎

Tail bound at N=1: S1u=A∗cos⁡(ω1t)sin⁡(πx/L), so ∣u−S1u∣=A∗∣β∣∣cos⁡(ω3t)∣∣sin⁡(3πx/L)∣≤A∗∣β∣. Differentiating F five times by [F3] gives F(5)(x)=A∗(π/L)5(cos⁡(πx/L)+243βcos⁡(3πx/L)), hence M5≤M‾5=A∗(π/L)5(1+243∣β∣), while M4=0. Thus the constant in [F1] satisfies C=4L5M5/π5≤C‾:=4A∗(1+243∣β∣); applying [F1] yields ∣u−S1u∣≤C2−4≤C‾2−4=A∗(1+243∣β∣)/4. This explicit upper estimate also dominates the direct bound A∗∣β∣, since 1+243∣β∣≥4∣β∣, and preserves the theorem’s O(N−4) estimate with C≤C‾ for every N≥1. Equality between M5 and M‾5 is not required.

Remarks

  • The two-mode data make every series finite, so the check isolates the theorem's content — the mode frequencies, the boundary and initial conditions, uniqueness, the d'Alembert correspondence and the tail constant — without requiring convergence arguments beyond the supplier theorem.

  • Modes with larger n oscillate faster and the displayed bound is deliberately conservative at small N; the O(N−4) decay becomes informative for large mode numbers and smooth data, as the theorem states. No measured frequency, tension or density appears.

Depends on

Dependency tree · 2 levels

Sources