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Derivation

Fourier Decomposition into Normal Modes

D-067 Home PU-103 Threads waves · symmetry Depends on Standing Waves and Boundary Quantization, function-space-orthogonality
Statement

For a uniform string of length L fixed at both ends, the normal modes φn(x) = sin(nπx/L), n = 1, 2, 3, … form a complete orthogonal set on [0, L]. Any admissible transverse configuration y(x) obeying the boundary conditions expands uniquely as y(x) = Σn bn sin(nπx/L), and each modal amplitude is recovered by projection, bn = (2/L) ∫0L y(x) sin(nπx/L) dx.

Why it matters

The wave equation is linear, so its general motion is a superposition of independently oscillating normal modes. Decomposition converts an intractable boundary-value problem for an arbitrary initial shape into a countable list of decoupled harmonic oscillators, one per mode, each evolving as cos(ωnt) with ωn = nπc/L.

The projection formula is the physical content of orthogonality: it says the modes do not "leak" into one another, so the energy and amplitude carried by each harmonic can be read off the initial data alone. This same machinery — expand over eigenfunctions, extract coefficients by inner product — recurs in quantum mechanics, optics, and signal analysis.

Assumptions
Fixed ends (Dirichlet boundary conditions).If the ends are free or loaded, the mode set becomes cosines or a mixed set with shifted eigenvalues; the sine basis no longer spans the configurations.
Uniform linear density and tension.Non-uniform μ(x) or T(x) gives a Sturm–Liouville problem whose eigenfunctions are not sines, though they remain orthogonal under a weighted inner product.
Linear (small-amplitude) transverse displacement.Large amplitude introduces nonlinearity and mode coupling, so superposition fails and coefficients cease to be constant.
The configuration lies in L2[0, L] and completeness holds.If y(x) is not square-integrable, the projection integral may diverge; if the eigenset were incomplete, some configurations could not be represented at all and the equality would be false.
Derivation
1
y(x) = Σn=1∞ bn sin(nπx/L)
Posit the expansion over the normal modes; legitimate because the fixed-end modes are a complete orthogonal basis of L2[0, L] (completeness theorem for the Dirichlet Laplacian). B
2
∫0L sin(nπx/L) sin(mπx/L) dx = (L/2) δnm
Orthogonality of the modes on [0, L] (prior result function-space-orthogonality); the self-adjoint operator −d2/dx2 with Dirichlet conditions has orthogonal eigenfunctions. A
3
∫0L y(x) sin(mπx/L) dx = ∫0L [ Σn bn sin(nπx/L) ] sin(mπx/L) dx
Project onto mode m: multiply both sides by sin(mπx/L) and integrate over the string. Both sides are equal as functions, so their inner products with any fixed mode agree. A
4
= Σn bn ∫0L sin(nπx/L) sin(mπx/L) dx
Interchange sum and integral. Legal because the series converges in L2 (and uniformly for continuous, piecewise-smooth y), so the integral of the sum equals the sum of the integrals (dominated/uniform convergence). C
5
= Σn bn (L/2) δnm = bm (L/2)
Apply orthogonality from Step 2; the Kronecker delta collapses the infinite sum to the single n = m term. A
6
bm = (2/L) ∫0L y(x) sin(mπx/L) dx
Divide by L/2 and relabel m → n. Symbols rearranged; no numbers introduced. A
Result
y(x) = Σn=1∞ bn sin(nπx/L), bn = (2/L) ∫0L y(x) sin(nπx/L) dx

Reading. The string's shape is a weighted sum of standing modes, and the weight of each mode is its overlap with the shape, measured by the inner product and normalised by the mode's own norm L/2. A mode "hears" only the part of y(x) that has its spatial rhythm; everything orthogonal to it is invisible to the projection.

Units check. With [y] = m and the sine dimensionless, ∫ y sin dx has units m·m = m2. Multiplying by 2/L (units m−1) gives bn in metres — correct, since bn is an amplitude.

Limiting cases
  • Single-mode initial shape. If y(x) = A sin(kπx/L), orthogonality gives bn = A δnk — the projection returns only the mode present.
  • Smooth shape. For an infinitely differentiable y vanishing at both ends, coefficients decay faster than any power of 1/n; the series converges rapidly and truncation is safe.
  • Kinked shape (plucked string). A corner makes bn ∼ 1/n2; a jump in y itself would give 1/n and Gibbs overshoot near the discontinuity.
  • Long-wavelength dominance. Because higher modes cost more curvature energy, physical excitations usually have |b1| largest, with rapidly falling overtones.
Breaks when
  • Non-Dirichlet or moving boundaries. Free ends, an attached mass, or a time-varying length break the fixed sine eigenset; the correct basis is different, and the 2/L normalisation no longer holds.
  • Nonlinear or large-amplitude motion. When tension varies with amplitude, modes exchange energy; the coefficients bn become time-dependent and superposition is invalid.
  • Non-L2 data. A configuration with a non-integrable singularity has a divergent projection integral, so no finite bn exists.
  • Dispersive or stiff strings. Bending stiffness adds a ∂4y/∂x4 term; the spatial modes shift and the pure-sine expansion no longer diagonalises the dynamics.
Failure modes
  • Dropping the 2/L. Writing bn = ∫ y sin dx forgets to divide by the mode norm L/2, giving amplitudes too large by L/2.
  • Full-period normalisation. Using 1/L (as for a 2L-period Fourier series) instead of 2/L; the sine series on [0, L] is the odd half-range extension and carries the factor 2.
  • Integrating a cosine or a shifted sine. Projecting with the wrong basis function (free-end cosines) against fixed-end data yields spurious nonzero overlaps.
  • Assuming all modes contribute. For a symmetric pluck at the centre, even modes have a node there and vanish identically; students often "find" small nonzero even coefficients from arithmetic slips.
  • Interchanging sum and integral without justification. Treating a non-convergent or non-uniformly-convergent series as if term-by-term integration were automatic.
Discussion

The decomposition is an eigenfunction expansion of a self-adjoint operator. The spatial operator −d2/dx2 acting on functions that vanish at x = 0, L is Hermitian under the inner product ⟨f, g⟩ = ∫0L f g dx. Hermiticity forces distinct eigenvalues (nπ/L)2 to have orthogonal eigenfunctions, and Sturm–Liouville theory guarantees the set is complete. Projection is then nothing more than resolving a vector along an orthonormal basis — the string configuration is a vector in an infinite-dimensional function space, and bn are its components.

Inserting the time dependence, each coefficient becomes bn cos(ωnt) + (an/ωn) sin(ωnt) with ωn = nπc/L and c = √(T/μ). The full solution y(x, t) is thus fixed entirely by the initial shape (setting the bn) and initial velocity (setting the an). Because the modes are orthogonal, the total energy is a sum of independent modal energies with no cross terms — Parseval's theorem, ∫0L y2 dx = (L/2) Σn bn2, is energy bookkeeping made exact.

The rate at which bn decays with n encodes the smoothness of the shape: this is a spectral-analysis principle. A kink (discontinuous slope) gives 1/n2; a genuine jump gives 1/n and audible bright harmonics. This is why a string plucked with a hard plectrum sounds brighter than one bowed smoothly — the excitation's spatial sharpness directly sets the overtone spectrum.

Rigorously, convergence is subtle: the sine series converges to y in the L2 norm for any square-integrable configuration, but pointwise convergence requires Dirichlet-type conditions (piecewise monotone, bounded variation), and at a jump the series converges to the midpoint value with Gibbs overshoot of about 9% that does not vanish as more terms are added. The interchange in Step 4 is licensed by L2 completeness for the coefficient extraction even when pointwise reconstruction misbehaves — a distinction that matters when the "shape" is only defined almost everywhere.

Common misconceptions. The expansion does not require y(x) to be periodic or even continuous — only square-integrable and consistent with the boundary conditions; periodicity is a property of the odd extension, not a prerequisite. And orthogonality is a statement about spatial overlap at a fixed instant, not about time: modes with commensurate frequencies still stay spatially orthogonal.

Worked examples
1
Centre-plucked string. y(x) = 2hx/L (0 ≤ x ≤ L/2), 2h(L−x)/L (L/2 ≤ x ≤ L)
Triangular initial shape, height h at the midpoint. Symbols first, then numbers.
2
bn = (2/L) ∫0L y(x) sin(nπx/L) dx = (8h / n2π2) sin(nπ/2)
Evaluate the projection integral by parts on each half; sin(nπ/2) vanishes for even n, so only odd modes survive.
3
h = 5.0 mm = 5.0×10−3 m, L = 0.65 m
Guitar-string values. Compute the leading odd coefficients.
4
b1 = 8(5.0×10−3)/π2 = 4.05×10−3 m, b3 = −8(5.0×10−3)/(9π2) = −4.50×10−4 m, b5 = 8(5.0×10−3)/(25π2) = 1.62×10−4 m
Note L cancels; amplitudes depend only on h. The 1/n2 decay reflects the corner at the pluck point.
b1 = 4.05 mm, b3 = −0.450 mm, b5 = 0.162 mm; even modes zero

Reading. The fundamental holds ~90% of the amplitude; overtones fall as 1/n2 and alternate in sign. Even modes have a node at the centre, so the symmetric pluck cannot excite them.

Units check. Every bn is a pure multiple of h, so all are in metres — correct for amplitudes.

1
Parabolic release. y(x) = (4A/L2) x(L − x), peak A at x = L/2
A smooth, corner-free initial displacement — contrast with the plucked case.
2
∫0L x(L−x) sin(nπx/L) dx = (2L3/n3π3) [1 − (−1)n]
Twice integrate by parts; the bracket is 0 for even n and 2 for odd n.
3
bn = (2/L)(4A/L2)(2L3/n3π3)[1−(−1)n] = 32A/(n3π3) (odd n)
Collect factors; L cancels again. Coefficients now decay as 1/n3 — faster, because the shape has no kink.
4
A = 3.0 mm: b1 = 32(3.0×10−3)/π3 = 3.10×10−3 m, b3 = 32(3.0×10−3)/(27π3) = 1.15×10−4 m
Insert numbers. The fundamental slightly exceeds A because the parabola is "fuller" than a half-sine.
b1 = 3.10 mm, b3 = 0.115 mm, b5 = 0.0248 mm; even modes zero

Reading. The 1/n3 decay makes the third harmonic ~27× weaker than in the plucked case relative to the fundamental — smoother shape, purer tone.

Units check. Each bn is a dimensionless number times A (metres), so all amplitudes are in metres.

Problems
  1. Show that a string in the pure shape y(x) = 0.4 sin(3πx/L) mm has b3 = 0.4 mm and all other coefficients zero.
    Solution Project: bn = (2/L)∫0L 0.4 sin(3πx/L) sin(nπx/L) dx. By orthogonality the integral is (L/2)δn3, so bn = 0.4 δn3 mm. Only b3 = 0.4 mm survives; every other mode is orthogonal to the shape.
  2. A string of length L = 1.20 m is plucked at its centre to height h = 8.0 mm. Find b1 and b2.
    Solution Use bn = (8h/n2π2) sin(nπ/2). For n = 1: sin(π/2) = 1, b1 = 8(8.0×10−3)/π2 = 6.48×10−3 m = 6.48 mm. For n = 2: sin(π) = 0, so b2 = 0. The even mode has a node at the centre and cannot be excited by a centred pluck. (L does not enter.)
  3. Using Parseval's theorem ∫0L y2 dx = (L/2) Σn bn2, find the fraction of the mean-square displacement carried by the fundamental of the centre-plucked string (odd n only, bn = 8h/n2π2 up to sign).
    Solution Modal weights bn2 ∝ 1/n4 over odd n. Fraction in the fundamental = 1 / Σodd n (1/n4). The sum Σodd 1/n4 = π4/96 = 1.01468. Fraction = 1/1.01468 = 0.9855, i.e. about 98.6% of the mean-square displacement is in the fundamental.
  4. A uniform initial displacement is approximated by y(x) = d = 2.0 mm (constant) on 0 < x < L. Find the modal amplitudes and comment on convergence.
    Solution bn = (2/L)∫0L d sin(nπx/L) dx = (2d/L)·(L/nπ)[1 − cos nπ] = (2d/nπ)[1 − (−1)n]. Odd n: bn = 4d/nπ = 4(2.0×10−3)/(nπ); so b1 = 2.55 mm, b3 = 0.849 mm, b5 = 0.509 mm. Even n: zero. The 1/n decay is slow because the constant shape violates the boundary conditions at the ends (jump), producing Gibbs overshoot near x = 0, L.
  5. The fundamental frequency of a string is f1 = 220 Hz. After a centre pluck, write the amplitude of the nth harmonic in the radiated sound relative to the fundamental, and state the frequency of the loudest overtone.
    Solution Modal amplitudes scale as |bn| = |b1|/n2 for odd n (even modes absent). Relative amplitude of harmonic n is 1/n2: the third harmonic (frequency 3f1 = 660 Hz) is the loudest overtone at 1/9 ≈ 0.111 of the fundamental; the fifth (1100 Hz) is 1/25 = 0.040. Even harmonics (440 Hz, 880 Hz) are absent for a perfectly centred pluck.