The classical wave equation

Follow a pulse and one string point

An ideal stretched string carries a small upward pulse to the right. Its crest starts at x = 0 m and reaches x = 2 m at t = 1 s. A string point labelled x = 1 m rises and falls as the pulse passes. Does that point travel right with the crest?

Choose x along the undisturbed string, positive to the right, and transverse displacement y positive upward. In the selected approximation each material point keeps its equilibrium longitudinal label x and moves only transversely. The moving crest identifies a feature of the shape; it does not identify one travelling material point.

Units, standards, and physical constants and Algebra, functions, graphs, and logarithms provide the published preparation for units, functions and tables. Begin with the pulse formula, snapshots and table below. The needed field and rate notation is introduced here.

Derivatives, Taylor series, and local approximation and Partial differential equations and Green functions are optional return paths for the direct equation check. You can first compare the graphs and initial-data cases, then return to that check. The string values are supplied exactly for an ideal model, not measured in a laboratory.

Describe a field in space and time

The function y(x,t) assigns a displacement to each string label x and time t. Holding t fixed gives a spatial snapshot. Holding x fixed gives the transverse time history of that point. These are two different slices of the same field.

Write y_t = ∂y/∂t for the instantaneous transverse velocity at fixed x, in m/s, and y_tt = ∂²y/∂t² for its transverse acceleration, in m/s². The partial-derivative sign ∂ reminds us to hold the other variable fixed. These quantities differ from the speed of the pulse’s crest.

At fixed t, y_x = ∂y/∂x is the string’s dimensionless slope. The second spatial derivative y_xx describes how that slope changes along x and has units m⁻¹. A positive displacement alone does not decide the signs of either slope or velocity.

Build the string equation and speed

Supply constant tension T = 0.4 N and uniform equilibrium-label linear mass density μ = 0.1 kg/m. The string is a perfectly flexible Newtonian continuum: bending stiffness, gravity, distributed forcing, damping and dispersion are neglected. Retain small slopes and transverse motion to linear order.

For a short piece of equilibrium length Δx, its mass is μΔx. To this order the difference of the transverse tension components is T times the difference of the slopes at its two ends. Dividing that force balance by Δx and taking the short-piece limit gives μy_tt = Ty_xx.

Thus the selected classical wave equation is y_tt = c²y_xx, where c = √(T/μ) is the positive speed parameter. Here T/μ = 4 m²/s², so c = 2 m/s. The units agree: c² times y_xx has units m/s², just like y_tt.

This equation relates local acceleration to local spatial curvature under the stated assumptions. It does not say that every real wave has this speed, or that a particular initial shape selects its own motion without further data.

Supply a compact travelling pulse

Let A = 0.02 m and ℓ = 1 m. Define the dimensionless profile F(z) = (1 − z²)³ when |z| ≤ 1, and F(z) = 0 when |z| > 1. The notation |z| means the nonnegative magnitude of z. This profile is highest at z = 0 and joins the zero branch at z = ±1.

Supply y(x,t) = A F((x − ct)/ℓ), for every real x and 0 ≤ t ≤ 1 s. The argument (x − ct)/ℓ is dimensionless. Keeping that argument unchanged while increasing time requires x to increase by c times the time change: the shape translates right at 2 m/s.

The crest is at x = ct, with y = A. The displacement can be nonzero only between x = ct − ℓ and x = ct + ℓ; it is exactly zero at both ends of that interval and outside it. This is a compact pulse, with no stipulated period or wavelength.

Compare snapshots and one fixed label

The upper graph compares the field at t = 0 s, 0.5 s and 1 s. Its horizontal axis is a string label in metres. The three marked crests are at x = 0 m, 1 m and 2 m, all at y = 0.02 m.

The lower graph keeps x = 1 m fixed while time changes. Its horizontal axis is time in seconds. The point starts at y = 0 m, reaches y = 0.02 m at t = 0.5 s, then returns to y = 0 m at t = 1 s. Its longitudinal label remains 1 m throughout.

At t = 0.25 s and 0.75 s this point has the same displacement, 0.0084375 m, but opposite transverse velocities. At t = 0.5 s its transverse velocity is zero even though the crest continues right at 2 m/s. A zero rate at one instant does not mean that the point was motionless over an interval.

The spatial graph deliberately enlarges transverse displacement relative to longitudinal distance. Read the axis labels rather than treating the drawn angle as the physical string angle. Both panels use the same vertical displacement scale; their horizontal axes represent different quantities.

Initial shape does not fix initial velocity

At t = 0 the supplied right-moving field has shape A F(x/ℓ) and velocity y_t(x,0) = −(Ac/ℓ)F′(x/ℓ). Here F′ is the derivative with respect to its dimensionless argument. The initial velocity is a function over the whole string, not just a value at the crest.

A different solution, y_left(x,t) = A F((x + ct)/ℓ), has that same initial shape but the opposite initial velocity. Its crest moves left. Another solution is y_rest(x,t) = (A/2)[F((x − ct)/ℓ) + F((x + ct)/ℓ)]. Its initial shape is again A F(x/ℓ), but its initial velocity is zero everywhere: the two rate contributions cancel.

For this release from rest, at t = 0.5 s the two smaller pulse crests lie at x = −1 m and +1 m, each at y = 0.01 m. At x = 1 m the original right-moving solution instead has y = 0.02 m. The same initial photograph therefore does not identify which of these three evolutions was supplied.

Check the equation, including the joins

For a direct calculus check, set z = (x − ct)/ℓ. Inside the profile, F′(z) = −6z(1 − z²)² and F″(z) = −6 + 36z² − 30z⁴. A prime denotes a derivative with respect to z; F″ is the second such derivative.

The chain rule gives y_t = −(Ac/ℓ)F′(z), y_tt = (Ac²/ℓ²)F″(z) and y_xx = (A/ℓ²)F″(z). Since c² = T/μ, substitution gives y_tt = c²y_xx. Outside the pulse all three expressions are zero.

At z = ±1, F, F′ and F″ approach zero from inside and agree with the zero exterior branch. Thus the profile has continuous first and second derivatives, written C², and the classical second-derivative equation also holds at the moving joins. The third derivative jumps there: this example is not C³ or differentiable to every order.

At the crest, F″(0) = −6. At x = 1 m and t = 0.5 s, y_tt = −0.48 m/s² and y_t = 0 m/s. The point is instantaneously at its upper turn, with downward acceleration. Checking a supplied solution and its initial data establishes mathematical consistency within this model; it is not a measurement of a real string.

Keep the equation and window in scope

The whole-line model has no end at the graph’s x = −2 m or x = 4 m edge. The pulse’s entire support stays between −1 m and 3 m during the displayed time interval. A finite string would require actual endpoint conditions; this drawing supplies none.

Small slope is a dimensionless condition. For the supplied pulse its greatest slope magnitude is 48/(625√5), approximately 0.03435. That is consistent with the selected small-slope approximation, but it does not certify an error bound for an actual string, its material properties or an experimental measurement.

Finite curve segments approximate the exact supplied field in the drawing. The marked states and table come from the exact formulas. Different tension, density, forcing, damping, bending stiffness or an unsuitable continuum approximation would require checking or changing the model; the classical wave equation is a scoped physical model rather than a rule for every wave.

Worked examples

Try it yourself

1. At t = 0.75 s, where is the right-moving crest, and what is the signed transverse velocity at the fixed label x = 1 m?

   Show the answer and reasoning

   The crest is at x = 1.5 m; the fixed-label transverse velocity is −0.0675 m/s.

   Use x_crest = ct = (2 m/s)(0.75 s). The fixed point has z = −1/2, so −(Ac/ℓ)F′(z) = −27/400 m/s. The two answers describe different quantities.

2. At x = 3.5 m and t = 1 s, what is the supplied displacement? Does its zero value mean that x = 3.5 m is a fixed string end?

   Show the answer and reasoning

   y = 0 m. No string end is supplied there.

   The argument is (3.5 m − 2 m)/(1 m) = 1.5, outside |z| ≤ 1. The profile is zero there, but x remains a label on the whole-line model. A local zero displacement is not an endpoint condition.

3. At t = 0 s, a supplied photograph shows y = 0.0084375 m at x = 0.5 m. Can that value distinguish the right-moving, left-moving and release-from-rest cases? Compare their initial velocities at that label.

   Show the answer and reasoning

   No. Their initial velocities there are +0.0675 m/s, −0.0675 m/s and 0 m/s respectively.

   All three share A F(x/ℓ) initially. At z = 1/2, F′ = −27/16. The right and left derivatives have opposite signs, while their half-sum cancels. The photograph supplies displacement, not the missing velocity field.

4. Replace x − ct by x + ct in the pulse argument. Where is its crest at t = 0.5 s, and does that imply longitudinal material motion to the left?

   Show the answer and reasoning

   Its crest is at x = −1 m. It does not imply longitudinal material motion.

   The maximum requires x + ct = 0, so x = −ct. In the retained transverse approximation the string points still keep their equilibrium longitudinal labels. The sign change selects a travelling feature’s direction.

5. At x = 1 m, t = 0.5 s, use F″(0) = −6 to find y_xx and y_tt. Is zero transverse velocity there inconsistent with the equation?

   Show the answer and reasoning

   y_xx = −0.12 m⁻¹ and y_tt = −0.48 m/s². Zero velocity is consistent with nonzero acceleration.

   Compute (A/ℓ²)(−6) = −0.12 m⁻¹. Multiplying by c² = 4 m²/s² gives −0.48 m/s². The equation concerns acceleration and spatial curvature; at a turn the instantaneous velocity can be zero.

6. The profile’s third derivative jumps at z = ±1. Does that invalidate this direct classical second-derivative wave-equation check?

   Show the answer and reasoning

   No. F, F′ and F″ join continuously to zero, so the required second derivatives exist and agree there.

   Both y_tt and c²y_xx are zero at the joins. This is a C² solution check, not a claim of C³ or differentiability to every order. A discontinuity in a higher derivative must not be confused with a missing required derivative.

What to carry forward

• A moving crest and a fixed material label answer different questions; transverse point velocity is not wave speed.
• The classical string equation y_tt = (T/μ)y_xx uses explicit continuum, tension, density and small-slope assumptions.
• The supplied right-moving pulse satisfies the equation and both initial data, including its C² joins.
• The same initial shape can accompany different initial velocity fields and different later motions.
• Read units and scales, distinguish a plotted window from physical boundaries, and keep a mathematical check separate from real-system validation.