advanced35 min studyLesson 413 in Mathematics

d'Alembert, Kirchhoff, and fundamental solutions

Compare explicit free-wave formulas in one and three dimensions and explain where dimension-dependent tails enter.

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What you’ll understand

  • Evaluate the one-dimensional wave solution from displacement and velocity data.
  • Interpret the three-dimensional spherical-mean formula with correct normalization.
  • Separate finite propagation from the stronger Huygens property.

Explicit wave formulas reveal more than the existence of a propagation cone. They show which parts of the initial data contribute to an observation inside that cone. In one dimension, initial velocity contributes through an interval integral. In three dimensions, the free homogeneous wave equation uses spherical data at one retarded radius. The contrast explains why a compact pulse may leave a tail in one setting and clear out after passage in another, despite finite propagation in both.

Two traveling parts and a velocity integral

For a whole-line wave equation with positive constant speed, prescribe initial displacement f and initial velocity g. The solution is the mean of two shifted displacement values plus a scaled integral of g over the dependence interval. Differentiating the formula checks both initial conditions and the PDE for smooth data. Less regular finite-energy data can be handled by approximation, with derivatives interpreted weakly. A formula for a classical point value should not be applied indiscriminately to arbitrary representatives of rough data.

The two shifted f values travel in opposite directions, but the velocity contribution involves every point between their feet. Thus knowing only the interval endpoints is insufficient for generic one-dimensional wave data. If g vanishes and f is compactly supported, the displacement consists of two departing profiles; if g has nonzero integral, a persistent interior displacement can remain. That interior constant carries no gradient or kinetic energy after the fronts have passed locally, so it is consistent with energy being transported away.

u(x,t)=f(x−ct)+f(x+ct)2+12c∫x−ctx+ctg(s) dsu(x,t)=\frac{f(x-ct)+f(x+ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}g(s)\,ds
d'Alembert's formula solves u_tt=c²u_xx on the whole line with u(x,0)=f(x), u_t(x,0)=g(x). The factor 1/(2c) ensures the correct initial velocity.↔ Scroll wide equations horizontally.

Spherical means in three dimensions

In three spatial dimensions, define the spherical mean by averaging over unit directions after scaling the sphere radius. The Kirchhoff formula differentiates time times the spherical mean of f and adds time times the spherical mean of g, both at radius ct. For smooth compactly supported data these operations are classical and can be justified directly. At small time, the first term approaches f while the derivative of the second supplies g, making the normalization check transparent.

Only values on the expanding sphere occur in this three-dimensional homogeneous constant-coefficient formula. Once that sphere has moved beyond a compact data support, the observation becomes zero. This is the sharp Huygens behavior for the free wave equation in three dimensions, with analogous behavior in odd spatial dimensions at least three. Even dimensions have interior-cone tails, and dimension one has the velocity-integral behavior just described. Boundaries, potentials, curvature, or altered equations can change this conclusion even in a dimension where the free-space formula has it.

Mrh(x)=14π∫S2h(x+rω) dω,u(x,t)=∂t ⁣(tMctf(x))+tMctg(x)M_rh(x)=\frac1{4\pi}\int_{S^2}h(x+r\omega)\,d\omega,\qquad u(x,t)=\partial_t\!\left(tM_{ct}f(x)\right)+tM_{ct}g(x)
This is the three-dimensional Kirchhoff formula for smooth whole-space data. The mean uses unit-sphere surface measure, whose total is 4π.↔ Scroll wide equations horizontally.

Fundamental solutions encode support as well as amplitude

A retarded fundamental solution is a distribution supported at nonnegative times whose wave operator gives a space-time point source. Convolution with initial data or a forcing term provides a systematic representation of propagation. Its singular support and its full support are different: a wavefront can be concentrated on a cone boundary while the fundamental solution also has a tail inside. The explicit formulas provide concrete instances of this distinction without requiring a pointwise interpretation of the source distribution.

Finite propagation is the common statement that the retarded influence stays inside or on the forward cone. Huygens behavior says more: in the relevant free-space dimensions the propagation kernel for positive times is supported on the cone surface. Numerical plots of a single sinusoid cannot distinguish these support properties, because a sinusoid is nonzero across the domain from the start. Compactly supported test data and a clearly specified equation are necessary when interpreting a simulation as evidence about tails or propagation fronts.

(∂t2−c2Δ)E=δ(0,0),supp⁡E⊆{(t,x):t≥0, ∣x∣≤ct}(\partial_t^2-c^2\Delta)E=\delta_{(0,0)},\qquad\operatorname{supp}E\subseteq\{(t,x):t\ge0,\ |x|\le ct\}
The retarded fundamental solution has finite propagation. Whether positive-time support is confined to the cone boundary depends on spatial dimension and the operator.↔ Scroll wide equations horizontally.

Checking both initial data

Use d'Alembert's formula for c=2, f(x)=x², and g(x)=2x. Find u and evaluate it at x=1,t=0.5. These smooth polynomial data need not have finite total energy on the whole line.

  1. The shifted-displacement mean is ((x−2t)²+(x+2t)²)/2=x²+4t².

  2. The velocity contribution is one quarter of the integral of 2s from x−2t to x+2t. It equals ((x+2t)²−(x−2t)²)/4=2xt.

  3. Therefore u=x²+4t²+2xt and u(1,0.5)=3. Direct differentiation gives u_tt=8 and 4u_xx=8; at t=0 the displacement is x² and the velocity is 2x.

u(x,t)=x2+4t2+2xt,u(1,1/2)=3u(x,t)=x^2+4t^2+2xt,\qquad u(1,1/2)=3
The explicit formula is valid for these smooth polynomial data even though the global energy integral diverges.↔ Scroll wide equations horizontally.
The result

The solution is three at the requested point, and its PDE and both initial traces agree with the specified data.

Explore the model

Interactive lab

Interactive lab · waves

A traveling-mode formula with fixed speed

Pause and Reset, then set phase difference 0° and amplitude ratio 0. Predict the value at x=π/4,t=0. After one Step to t=0.5, predict the new crest location by solving 2x−t=π/2. Compare the moving profile with this characteristic relation.

A traveling-mode formula with fixed speedTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe crest starts at π/4 and moves to π/4+0.25. This is one right-traveling sinusoidal component at speed one half; it does not display compact data, spherical means, or Huygens tails.

Check your understanding

Which assertion correctly distinguishes the generic free-wave formulas on the whole line and in three-dimensional space?

Try it yourself

For c=1, f=0, and g equal to one on [−1,1] and zero elsewhere, use the weak d'Alembert formula to find u(0,0.5) and u(0,2). What does the later value show?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The displacement is half the length of the overlap between [−t,t] and [−1,1].

Compare with the solution

At t=0.5 the overlap has length one, so u(0,0.5)=0.5. At t=2 it has length two, so u(0,2)=1. The later nonzero value shows an interior displacement tail from initial velocity in one dimension. It does not violate finite propagation: the observation lies inside the forward dependence cone, and locally the eventual constant displacement can have zero derivative energy.

Take it with you

  • The initial-velocity normalization in d'Alembert's formula is 1/(2c).
  • Three-dimensional Kirchhoff propagation uses expanding spherical means.
  • Finite cone support holds generally for this free wave, while sharp Huygens behavior is dimension-specific.

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