Shock waves, solitons, and nonlinear waves

Start with a jump and a pulse

At time zero, a supplied field is 3 m/s to the left of x = 0 and 1 m/s to the right. Its jump moves to x = 2 m after one second. In a separate model, a smooth pulse of height 1 moves from x = 0 to x = 3 m in the same time. Both profiles move, but they obey different equations and have different vertical quantities.

The first field is a velocity-valued scalar u. The second field is a dimensionless amplitude w. Neither is supplied as a measured gas pressure or water-surface elevation. We will calculate the two speeds from their respective rules, rather than infer a common physical mechanism from their shapes.

Units, standards, and physical constants prepares the quantity notation. Algebra, functions, graphs, and logarithms prepares substitution and exponentials. The exponential definition needed for the pulse is given below.

Start with the profiles and finite calculations. Derivatives, Taylor series, and local approximation and Partial differential equations and Green functions are published optional return paths for the smooth derivative check. The classical wave equation offers a separate linear-model comparison; it is not assumed to predict these nonlinear cases.

Conserve across a discontinuity

Let x be position in metres, positive to the right, and t be time in seconds. For all real x and t ≥ 0, prescribe the inviscid Burgers conservation law ∂u/∂t + ∂f(u)/∂x = 0, with f(u) = u²/2. Here u has units m/s, so the flux f has units m²/s². “Inviscid” means this supplied equation has no viscosity term.

A conservation law can be read as a boundary balance. For fixed a < b, the time rate of the integral of u over [a,b] equals f(u(a,t))−f(u(b,t)), using appropriate boundary traces. The integral has units m²/s; it has not been identified as a mass. For our moving jump this balance holds almost everywhere in time, including intervals on which the front lies inside [a,b].

Where u is smooth, the chain rule gives ∂u/∂t+u∂u/∂x = 0. At a jump, the ordinary spatial derivative does not exist. A weak solution instead satisfies the conservation law after integration against smooth test functions; the derivative acts on the test function. A discontinuity can then satisfy the balance without pretending that it is differentiable.

Suppose a nonzero jump separates constant states u_L on the left and u_R on the right and moves at speed s. Boundary balance gives the Rankine–Hugoniot condition s(u_R−u_L) = f(u_R)−f(u_L). The jump in flux and the moving jump in u must agree, with their signs retained.

Check speed and admissibility

For u_L = 3 m/s and u_R = 1 m/s, the fluxes are 9/2 and 1/2 m²/s². Thus s = (−4 m²/s²)/(−2 m/s) = 2 m/s. The front position is x_s(t) = st; the profile is 3 m/s for x < x_s(t) and 1 m/s for x ≥ x_s(t).

Conservation alone does not select every acceptable discontinuity. For this strictly convex flux, f′(u) = u is the characteristic speed: the speed along which a smooth field value is carried. A compressive jump requires f′(u_L) > s > f′(u_R). Our values satisfy 3 m/s > 2 m/s > 1 m/s: the characteristic directions approach the front from both sides.

This is the scalar entropy admissibility condition. It is a mathematical selection rule here, not a calculation of a gas temperature or thermodynamic entropy production. Reversing the states would still give s = 2 m/s from the flux balance, but it would fail this compression test. The appropriate expanding fan is worked out below.

Assigning u_R at the front is a right-continuous plotting convention. Altering that one point does not change the weak solution. It does not describe the thickness or interior structure of a physical shock; the initial data in this example already contain a discontinuity.

Define a smooth solitary profile

For a separate dimensionless field w on all real x and t ≥ 0, prescribe ∂w/∂t + c₀∂w/∂x + αw∂w/∂x + β∂³w/∂x³ = 0. The coefficients are c₀ = 1 m/s, α = 6 m/s and β = 1/2 m³/s. Each term has units 1/s. This constant-coefficient Korteweg–de Vries, or KdV-type, model has no supplied forcing or damping.

Choose the initial profile w(x,0) = A sech²(κx), with A = 1 and κ = 1 m⁻¹. Define sech(z) = 2/(exp(z)+exp(−z)) for a dimensionless real argument z. The profile peaks at 1 and decays towards zero far from its centre; it is positive at every finite position.

Try the translating shape w(x,t) = A sech²(κ(x−Vt)), where V is its unknown speed. Substitution, checked below, requires both V = c₀+4βκ² and αA = 12βκ². Our data satisfy the second condition and give V = 3 m/s. The crest position is Vt, not the field value w.

This exact supplied equation and profile are not a calibrated water-wave or acoustic model. No water depth, pressure scale or smallness parameter is given. The same named equation can arise in physical approximations with additional restrictions; those restrictions must be checked for an actual medium.

Read the three snapshots

The two upper panels show the jump at t = 0 and t = 1 s. Solid blue plateaus terminate at separate markers: a hollow circle gives the left-hand limit and a filled square gives the assigned right-hand value. The dashed vertical reference locates the jump; it does not assign intermediate field values. No vertical curve connects the plateaus. At t = 0, the guide coincides with the position-zero axis; at t = 1 s, it is separate and visibly dashed.

The lower panel shows the smooth pulse at the same two times, using violet hollow circles and orange hollow diamonds to distinguish the profiles. Their centres are 0 and 3 m. All panels share the horizontal position scale, but the upper ordinate is u in m/s and the lower ordinate is dimensionless w. Equal heights between those panels do not mean equal physical quantities.

The figure displays −3 ≤ x ≤ 7 m. The equations and profiles are defined on the whole line; the display edges are not walls or imposed zero boundaries. The two time snapshots are field profiles, not paths of tracked particles.

Check the traveling shape

The following derivative check is optional on a first reading. Let ξ = x−Vt, in metres, and F(ξ) = A sech²(κξ). Primes below mean derivatives with respect to ξ. Differentiating gives F″ = 4κ²F−6κ²F²/A. The traveling equation is (c₀−V)F′+αFF′+βF‴ = 0.

Integrating with respect to ξ gives (c₀−V)F+(α/2)F²+βF″ = C. Decay of F and the required derivatives as ξ tends to either infinity sets C = 0. Substituting F″ leaves the coefficients c₀−V+4βκ² and α/2−6βκ²/A. Since F varies continuously, both coefficients must vanish. This recovers the two speed and width conditions.

For the supplied profile, offsets ξ = −1, 0 and 1 m give the exact amplitudes sech²(1), 1 and sech²(1). Here sech²(1) is about 0.419974. The table uses the analytic profile, not values estimated from pixels.

Distinguish a solitary wave from a soliton

A solitary traveling wave has a localized profile that keeps its shape in a frame moving with it. Our verified smooth profile supplies that one-wave property. A soliton carries a stronger interaction statement: in the ideal model, well-separated waves retain their asymptotic profiles and speeds after interaction, although their positions can acquire phase shifts.

The ideal KdV equation has soliton solutions with that interaction property. The curves here are two snapshots of one pulse, so they neither demonstrate nor prove an interaction. During an interaction the combined profile need not look like two unchanged isolated pulses.

Do not use “soliton” as a label for every persistent-looking bump. A material model may have dissipation, additional dispersion, boundaries or coefficient restrictions that alter the result. Model-level shape preservation and an observed persistent wave are different kinds of evidence.

Keep the model and question together

A sharp front and a smooth pulse ask different mathematical questions. At the front, check the weak balance and admissibility. For the smooth pulse, check the equation, decay and compatible amplitude, width and speed. Retain the field’s units before comparing numbers.

An initial step in a scalar equation does not establish how a real sound shock forms, and a one-pulse solution does not establish real-world soliton collisions. To make a physical prediction, specify the medium, the meaning of the field, initial and boundary data, coefficient validity and the effects omitted from the model.

Worked examples

Try it yourself

1. For the main 3-to-1 m/s jump, what are the front position and the assigned field value exactly at the front at t = 1 s?

   Show the answer and reasoning

   x_s = 2 m; the assigned value is u_R = 1 m/s.

   The position follows s = 2 m/s. The right-continuous convention fixes the single front value; the left-hand limit remains 3 m/s.

2. For the main jump, compare the integral of u over [−3 m,7 m] at t = 0 and t = 1 s. Should it remain constant?

   Show the answer and reasoning

   It is 16 m²/s initially and 20 m²/s at 1 s; it increases by 4 m²/s.

   While the front lies inside the interval, the left and right portions give (3 m/s)(3 m+st)+(1 m/s)(7 m−st) = 16 m²/s+(4 m²/s²)t. Unequal boundary fluxes cause the increase; this is not a closed-system mass claim.

3. For the same flux f(u)=u²/2, take u_L = 2 m/s and u_R = −2 m/s. Find the jump speed and check admissibility.

   Show the answer and reasoning

   s = 0 m/s, and 2 m/s > 0 m/s > −2 m/s admits a stationary compressive jump.

   The two fluxes are equal at 2 m²/s², while the jump in u is −4 m/s. Zero speed satisfies conservation, and the separate entropy test also holds.

4. For the main smooth pulse at t = 1 s, locate the two selected offsets ξ = −1 m and ξ = 1 m and give their amplitudes.

   Show the answer and reasoning

   They are at x = 2 m and x = 4 m; both have w = sech²(1).

   The centre is at 3 m and x = 3 m+ξ. Evenness of sech² makes the two dimensionless amplitudes equal.

5. With A = 1/4 and the original KdV coefficients, someone retains κ = 1 m⁻¹ and changes only the speed. Can this profile satisfy the equation?

   Show the answer and reasoning

   No. αA = 3/2 m/s, whereas 12βκ² = 6 m/s.

   The amplitude–width condition fails regardless of the proposed speed. The compatible positive inverse width is 1/2 m⁻¹.

6. Two snapshots show one smooth pulse translating without a change of height. Do those snapshots establish the soliton interaction property?

   Show the answer and reasoning

   No. They do not show or prove how two waves behave after interaction.

   A verified solitary profile keeps its shape under translation. The stronger soliton statement concerns asymptotically separated waves after interaction and can include phase shifts; a one-profile plot supplies no such test.

What to carry forward

• At a discontinuity, conservation fixes a jump speed; entropy admissibility is a separate check.
• The supplied Burgers jump is velocity-valued, while the separate smooth KdV pulse is dimensionless.
• A decaying sech² profile needs compatible amplitude, width and speed, not a speed formula alone.
• One solitary profile does not establish soliton interactions or a physical medium’s validity.