advanced32 min studyLesson 74 in Mathematics

Wave equation and energy

Follow finite-speed propagation and use energy flux to understand reflection, uniqueness, and stable initial data.

Before you begin

On this page

The question for today

What you’ll understand

Use the one-dimensional d'Alembert formula with two initial conditions.
Derive energy conservation and identify its boundary-flux term.
Distinguish traveling waves, standing modes, and dissipative boundaries.

Displace a stretched string and release it: the disturbance spreads while energy moves between kinetic motion and elastic deformation. Unlike the heat equation, the wave equation requires both initial displacement and initial velocity. Its information travels along characteristic cones, and its natural energy measures those two kinds of data differently. Keeping track of the boundary flux explains why an isolated ideal string conserves energy while a driven or absorbing string need not.

Two traveling families encode the initial data

For u_tt=c²u_xx on the line with c>0, a right-moving profile F(x−ct) and a left-moving profile G(x+ct) each solve the equation. Their sum provides the general sufficiently regular solution. Matching u(x,0)=f(x) and u_t(x,0)=g(x) determines the two components and gives d'Alembert's formula. The velocity contributes an integral over the interval [x−ct,x+ct], while the displacement contributes endpoint values. A sign error in the velocity term can be found by differentiating at t=0.

The formula also reveals the domain of dependence: changing initial data outside that interval cannot alter the solution at (x,t). For compactly supported initial data, the support spreads no faster than c. This finite propagation property differs sharply from the Gaussian tails of diffusion. In higher dimensions the geometry and exact representation change, although finite-speed energy estimates remain central. Some dimensions have special sharp-front properties, but one should not generalize a one-dimensional drawing to every dimensional wave kernel.

u(x,t)=\frac{f(x-ct)+f(x+ct)}{2}+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds

This formula solves the unforced one-dimensional wave equation on the whole line with initial displacement f and velocity g, under sufficient regularity or the corresponding weaker interpretation.↔ Scroll wide equations horizontally.

Energy leaves through a boundary term

Define energy as one half the integral of u_t²+c²u_x². Differentiate in time, substitute the equation, and integrate by parts once. The interior terms cancel, leaving c²u_tu_x evaluated at the endpoints. With fixed homogeneous Dirichlet endpoints, u_t is zero there for a smooth solution. With homogeneous Neumann endpoints, u_x is zero. Either condition makes the flux vanish. Periodic endpoints cancel as well. Without a boundary condition, saying that the wave equation always conserves energy would omit a possible energy transfer.

This calculation is also a uniqueness argument. Subtract two solutions with the same data and compatible homogeneous boundary conditions. Their difference has zero initial energy, hence zero velocity and zero spatial derivative at all times. The initial displacement or boundary condition then fixes the remaining spatial constant. In the energy topology, displacement belongs naturally to H1 and velocity to L2. Additional regularity may be needed to interpret pointwise endpoint derivatives, but the estimate extends by approximation to energy solutions without pretending those solutions are classically twice differentiable.

E(t)=\frac12\int_0^L\left(u_t^2+c^2u_x^2\right)dx,\qquad E'(t)=c^2[u_tu_x]_{0}^{L}

For a sufficiently smooth unforced solution, the energy changes only by this boundary contribution. Zero Dirichlet, zero Neumann, or periodic conditions make the net flux vanish; driven or absorbing boundaries can change the energy.↔ Scroll wide equations horizontally.

Modes and interference require different interpretations

On an interval with fixed ends, sin(nπx/L) supplies the spatial modes. Their amplitudes oscillate at frequency cnπ/L rather than decay. A standing wave is a superposition of oppositely traveling pieces and repeatedly trades kinetic and gradient energy. A plot at a moment of zero displacement may still carry nonzero energy through velocity. Conversely, a momentarily stationary displaced string stores elastic energy. Tracking only the height of the graph loses half of the state.

The waves lab sums sin(2x−t) and a phase-shifted, amplitude-scaled copy traveling in the same direction. Set a=1 and compare phase differences zero and 180 degrees to observe constructive and destructive interference. These are exact traveling solutions with speed one half, but the lab has no counterpropagating component and therefore does not depict a fixed-end standing wave. The amplitude slider changes the chosen solution; it is not a physical energy source acting during an otherwise closed evolution. Use the worked example to track an actual conserved standing-mode energy.

Work it through

A string passes through zero without losing energy

For c=2 and 0<x<π with fixed ends, take $u(x,0)=\sin x$ and u_t(x,0)=0. Compute the motion and its energy.

The first Dirichlet spatial mode has second derivative −sin x. Writing u=a(t)sin x gives a''+4a=0, a(0)=1, and a'(0)=0.
Thus $u(x,t)=\cos(2t)\sin x$ , with $u_t=-2\sin(2t)\sin x$ and $u_x=\cos(2t)\cos x$ . The boundary displacement is zero at every time.
Because each squared sine or cosine integrates to π/2, $E=(1/2)[4\sin^2(2t)(\pi/2)+4\cos^2(2t)(\pi/2)]=\pi$ .
At t=π/4 the displacement vanishes, but the velocity is −2sin x and all the energy is kinetic. At t=0 the velocity is zero and all the energy is in the displacement gradient.

The result

The standing mode has constant total energy π even when its visible displacement becomes identically zero.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Interfere two traveling solutions

Set amplitude ratio a to one and vary phase difference from zero toward 180 degrees in sin(2x−t)+a sin(2x−t+phase). Both components move right at speed 1/2. Their sum illustrates linear interference; at equal amplitude and opposite phase both summed displacement and velocity vanish. This lab does not demonstrate a standing mode or compute boundary energy flux.

SumFirst componentSecond component

Sum at x = 0: 0.866
Resultant amplitude: 1.732
Phase difference: 60°

Phase difference60°Amplitude ratio1

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 forExact cancellation here removes both displacement and velocity of the summed field. It does not demonstrate a standing mode exchanging kinetic and elastic energy.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The energy identity has a boundary flux. Homogeneous fixed endpoints make u_t vanish there for smooth solutions; positive speed or zero initial velocity alone does not remove later energy exchange.

Your turn

Try it yourself

On the line with c=1, f(x)=0, and g(x)=2 for all x, find u(x,t). Why does this example not have finite total energy on the line?

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

Give me a hint

Integrate the constant velocity datum over an interval of length 2t.

Compare with the solution

D'Alembert gives $u(x,t)=\frac12\int_{x-t}^{x+t}2\,ds=2t$ . Thus u_t=2 and u_x=0. It is a classical solution but its energy density is two at every point, whose integral over the entire line is infinite. Existence of a formula does not automatically place it in the finite-energy function class.

Keep the idea

Take it with you

A wave state needs displacement and velocity.
Energy conservation follows only after accounting for boundary flux.
A traveling-wave interference display and a fixed-end standing mode represent different problems.

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