research35 min studyLesson 427 in Mathematics

KdV evolution and dispersive balancing

Derive the Korteweg–de Vries traveling profile, fix its dispersion sign, and distinguish nonlinear balance from diffusion or linear superposition.

On this page

What you’ll understand

  • Derive and verify the decaying KdV traveling-wave profile equation.
  • Compute its amplitude, width, speed, and conserved integral.
  • Compare cubic linear dispersion with nonlinear wave balance and formal critical scaling.

Dispersion can spread a packet, while nonlinear transport can steepen it. KdV balances these mechanisms without a viscous damping term, allowing a localized traveling pulse. Its famous soliton formula is more informative when derived from the equation: the profile fixes a relationship between height and speed, explains its width, and makes the sign convention testable. The same calculation also warns against treating a sum of two individually valid nonlinear waves as another exact solution.

Fix the cubic dispersion relation

Use u_t+6u u_x+u_xxx=0 for a real-valued field on the whole line. Smooth rapidly decaying solutions permit the integrations by parts used below. For the linearized equation, insert exp(i(kx−omega t)); the dispersion relation is omega=−k cubed. Thus phase velocity is minus k squared and group velocity is minus three k squared. These signs belong to this particular convention. Reversing the spatial coordinate or the sign of the third derivative changes the interpretation.

The linear Fourier multiplier is exp(i k cubed t), which has modulus one. It changes phase rather than damping each Fourier mode. The nonlinear term is the derivative of three u squared and moves different amplitudes at different local speeds. Dimensional balance compares amplitude squared over width with amplitude over width cubed, suggesting amplitude proportional to inverse width squared. That scaling predicts the shape relation, while the traveling-wave equation determines the exact coefficient and profile.

ut+6uux+uxxx=0,ω(k)=−k3,vg(k)=ω′(k)=−3k2u_t+6uu_x+u_{xxx}=0,\qquad \omega(k)=-k^3,\qquad v_{\mathrm g}(k)=\omega'(k)=-3k^2
The plane-wave convention is exp(i(kx−omega t)). A linear dispersive mode keeps its Fourier amplitude, unlike a heat mode.↔ Scroll wide equations horizontally.

Integrate the traveling-wave equation once

Set u(t,x)=Q(x−ct−x₀), with positive speed c and a profile decaying together with its derivatives. Substitution gives minus c Q prime plus six Q Q prime plus Q triple prime equal to zero. Integrating once and using decay sets the integration constant to zero, leaving Q double prime minus cQ plus three Q squared equal to zero. The localized solution is c over two times sech squared of square root of c times the traveling coordinate divided by two.

Consequently, larger positive pulses are taller, narrower, and faster. Their positive rightward travel does not contradict the leftward group velocity of the linearized zero-background equation; nonlinear transport changes the balance. Integrating the profile gives two times square root of c, and its squared L² integral is two thirds c to the three halves. These are whole-line quantities. Periodic KdV has different traveling families and integration constants, so imposing zero at infinity would not be justified for every periodic profile.

Qc(z)=c2sech⁡2 ⁣(c2z),Qc′′−cQc+3Qc2=0,∫RQc dx=2cQ_c(z)=\frac c2\operatorname{sech}^2\!\left(\frac{\sqrt c}{2}z\right),\qquad Q_c''-cQ_c+3Q_c^2=0,\qquad \int_{\mathbb R}Q_c\,dx=2\sqrt c
The formula assumes c>0 and a decaying zero-background profile. Its amplitude is c/2 and its width scale is 2/sqrt(c).↔ Scroll wide equations horizontally.

Conservation and scaling do not imply superposition

For sufficiently regular decaying KdV solutions, integrating the equation preserves the spatial integral of u. Multiplying by u and integrating preserves the squared L² norm because both the transport and dispersive contributions become boundary terms. A Hamiltonian is the integral of one half u_x squared minus u cubed under this normalization. Deriving these invariants requires the same control of boundary terms as the traveling-wave integration. The identities do not mean the profile remains a soliton for arbitrary initial data.

KdV scales as lambda squared times u at time lambda cubed t and position lambda x. In one dimension the homogeneous H to the s norm gains lambda to the s plus three halves, giving formal critical order minus three halves. This dimensional value is not a sharp low-regularity well-posedness theorem; derivative nonlinearities and resonant interactions create additional difficulties. Likewise a linear superposition argument fails because squaring a sum creates cross-interactions. Exact multisoliton constructions use more structure than adding pulses, and a plot of separated pulses is not proof that those interaction terms vanish.

Compare two traveling pulses

Take two separate zero-background KdV solitons, each centered at zero at t=0, with speeds c=1 and c=4. Compare their heights, width scales, positions at t=1/2, and spatial integrals.

  1. For c=1, the height is one half and the width scale is two. For c=4, the height is two and the width scale is one, so the faster pulse is four times taller and half as wide.

  2. Their centers would be ct for the two separate exact solutions: one half and two at t=1/2. Their formulas are (1/2)sech²((x−t)/2) and 2sech²(x−4t), respectively.

  3. The spatial integrals are 2sqrt(1)=2 and 2sqrt(4)=4. These describe two individually valid solutions; adding them is not an exact two-soliton solution because the nonlinear product generates cross terms.

The result

Height, width, speed, and conserved integral are linked by the nonlinear profile. The comparison does not invoke a false superposition principle.

Explore the model

Interactive lab

Interactive lab · heat

Use damping as a contrast to dispersion

Pause and reset, then set both controls to spatial mode n = 1 and diffusivity κ = 1. Predict heat amplitude exp(−t). For a fresh comparison, pause and reset, then set n = 3 and κ = 1; predict exp(−9t), with much faster damping. On paper compare the linear KdV multiplier exp(i n³t), whose magnitude remains one for either frequency. The instrument solves heat diffusion, not KdV; it makes the damping contrast visible and must not be read as a soliton or an Airy solver.

Use damping as a contrast to dispersionThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. 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 forDiffusion changes Fourier magnitudes; linear dispersion changes phases, and nonlinear KdV balance adds another mechanism.

Check your understanding

For the convention u_t+6u u_x+u_xxx=0, which statement is correct?

Try it yourself

For c=9, compute the KdV soliton's height, width scale, speed, and spatial integral. Check that it is obtained from the c=1 profile by the KdV scaling with lambda=3.

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

Give me a hint

Apply amplitude factor lambda², space factor lambda, and time factor lambda³ to the entire traveling solution.

Compare with the solution

The height is 9/2, width scale two thirds, speed nine, and integral six. Rescaling the c=1 solution gives 9 times (1/2)sech²((3x−27t)/2), equal to (9/2)sech²((3/2)(x−9t)). The integral scales by amplitude factor nine divided by spatial factor three, multiplying the original integral two by three. Its speed scales by lambda², consistent with the profile parameter. This exact scaling verification concerns a single traveling wave, not arbitrary interactions between several waves.

Take it with you

  • The sign of KdV dispersion depends on the stated plane-wave convention.
  • A decaying traveling profile links speed, height, width, and conserved integrals.
  • Formal critical scaling and exact single-wave formulas do not replace low-regularity or multisoliton analysis.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.