Derive a KdV travelling wave

How can a nonlinear wave move without changing shape? For the Korteweg–de Vries equation, decay at both ends of the line turns this question into an ordinary differential equation with a distinguished positive pulse. You will derive its profile and show that its height, width and speed cannot be chosen independently. This constructs one family of solutions; it does not solve arbitrary initial data or prove a general integrability theorem.

Required background. Differentiate a function of $x-vt$, integrate a second-order ODE once, and use decay to determine an integration constant. The entry check below repairs the chain-rule step. The KdV course introduction explains how this calculation fits the sequence.

Helpful background. The model and convention reference describe the line problem and its alternative signs.

A decaying wave of permanent shape

We use the normalized equation

$$u_t+6u u_x+u_{xxx}=0, \qquad x\in\mathbb R,$$

for real smooth fields. In this lesson $u$ and its required derivatives tend to zero as $x\to\pm\infty$. Write a travelling wave as

$$u(x,t)=U(\xi),\qquad \xi=x-vt-x_0,$$

where $v$ is its speed and $x_0$ its center at $t=0$. Then

$$-vU'+6UU'+U'''=0.$$

Primes mean derivatives with respect to $\xi$. Integration gives

$$U''-vU+3U^2=C_1.$$

Since $U,U''\to0$ in either tail, $C_1=0$. Multiply by $U'$ and integrate again:

$$\frac12(U')^2-\frac v2U^2+U^3=C_2.$$

The decay of $U$ and $U'$ sets $C_2=0$, so

$$(U')^2=U^2(v-2U).$$

The boundary conditions have done real work: discarding the constants without them would silently exclude other waves. This is the travelling-wave reduction of Lax 1968, report pp. 2–3, equations (1.5)–(1.8), PDF, after converting his field $w$ to $w=6u$.

Entry check and repair

For $f(x,t)=\operatorname{sech}^2(x-3t)$, is $f_t$ equal to $3f_x$ or $-3f_x$? Where is the maximum at $t=2$?

Repair. The chain rule gives $f_t=-3f_x$. The maximum occurs where $x-3t=0$, so at $t=2$ it is at $x=6$. A minus sign in $x-vt$ describes motion to the right for $v\gt0$. Check a moving center before differentiating a longer expression.

Determine the pulse and its speed

A nonzero smooth decaying pulse has an extremum with $U'=0$ and $U\ne0$. The first integral forces its value there to be $U=v/2$. A localized smooth pulse requires $v\gt0$: near $U=0$ the first integral has $(U')^2\simeq vU^2$, allowing exponential tails only for positive $v$. At $v=0$, a negative branch can decay algebraically on one side but develops a singularity instead of returning smoothly to zero. For $v\lt0$ even sufficiently small nonzero tails are excluded by the sign of the right-hand side.

Set $v=4\kappa^2$ with $\kappa\gt0$. To find the positive branch, try $U=a\operatorname{sech}^2(\kappa\xi)$. Since

$$\frac{d^2}{d\xi^2}\operatorname{sech}^2(\kappa\xi) =4\kappa^2\operatorname{sech}^2(\kappa\xi) -6\kappa^2\operatorname{sech}^4(\kappa\xi),$$

the ODE $U''=vU-3U^2$ requires $a=2\kappa^2$. Thus

$$u(x,t)=2\kappa^2\operatorname{sech}^2 \!\left[\kappa(x-4\kappa^2t-x_0)\right].$$

Its amplitude is $2\kappa^2$, speed is $4\kappa^2$, and characteristic width is $1/\kappa$. Larger pulses are faster and narrower in this normalization. The pulse is conventionally called a one-soliton solution. Establishing collision properties requires more than this single-wave calculation.

The reduction also explains uniqueness within the smooth positive pulse family: at its maximum $U=v/2$ and $U'=0$, the second-order ODE has a unique solution for these initial values. Translating the maximum supplies $x_0$. The zero solution is separate.

Verify the result without reintegrating

For a worked example choose $\kappa=1/2$ and $x_0=-1$. Then

$$u(x,t)=\frac12\operatorname{sech}^2 \!\left[\frac{x-t+1}{2}\right].$$

Its center is $x=t-1$, its speed is $1$, and its height is $1/2$. The profile satisfies $U''=U-3U^2$, hence $U'''=U'-6UU'$. Because $u_t=-U'$, substitution into KdV gives

$$u_t+6u u_x+u_{xxx} =-U'+6UU'+U'-6UU'=0.$$

This direct residual verifies the time sign as well as the profile shape. Checking only a snapshot would not detect a wave translated at the wrong speed.

The normalization also has a scaling check. If $[x]=\ell$, consistency assigns $[t]=\ell^3$, $[u]=\ell^{-2}$, $[\kappa]=\ell^{-1}$ and $[v]=\ell^{-2}$. These are scaling dimensions for the normalized model, not a claim that every physical realization uses these laboratory units.

Exercises

Guided practice: read the parameters

A decaying KdV pulse has maximum height $8$ at $x=3$ when $t=0$. Find $\kappa$, its speed, and its center at $t=1/4$. Write the solution.

Hint

Use height $2\kappa^2$ first; the speed is twice the height.

Solution

The height condition gives $\kappa=2$, and therefore $v=16$. With $x_0=3$,

$$u(x,t)=8\operatorname{sech}^2[2(x-16t-3)].$$

The center at $t=1/4$ is $x=3+16/4=7$. Its characteristic width is $1/2$, which is consistent with the factor $2$ in the argument.

Independent practice: detect a wrong speed

Let $U(\xi)=2\kappa^2\operatorname{sech}^2(\kappa\xi)$, but translate it at an arbitrary speed $w$: $u_w(x,t)=U(x-wt)$. Find its KdV residual. Explain why inspecting only the residual at the pulse center is a bad check.

Hint

Use $U'''=4\kappa^2U'-6UU'$. What is $U'$ at the maximum?

Solution

The residual is

$$\mathcal R[u_w] \equiv (u_w)_t+6u_w(u_w)_x+(u_w)_{xxx} =(4\kappa^2-w)U'(x-wt).$$

It vanishes everywhere only when $w=4\kappa^2$, apart from the trivial zero profile. At the center $U'=0$, so every choice of $w$ produces zero residual there. Test the flanks, or a norm over an interval, to detect a wrong speed. A visually correct shape is insufficient.

Transfer: add a constant background

Replace the zero background by a fixed real number $b$. Seek $u(x,t)=b+V(x-wt-x_0)$ with $V\to0$ in both tails. Determine the speed of a pulse whose height above the background is $2\kappa^2$. Which step of the zero-background derivation must change?

Hint

Substitute $b+V$ into the PDE before integrating. The effective speed in the equation for $V$ is $w-6b$.

Solution

The reduced equation is

$$-(w-6b)V'+6VV'+V'''=0.$$

Thus $V=2\kappa^2\operatorname{sech}^2(\kappa\xi)$ requires $w=6b+4\kappa^2$, giving

$$u(x,t)=b+2\kappa^2\operatorname{sech}^2 \!\left[\kappa\bigl(x-(6b+4\kappa^2)t-x_0\bigr)\right].$$

For $U=b+V$, the first integration constant is $3b^2-wb$, not zero. Subtracting the background before applying decay recovers the correct constant. This solution belongs to a different boundary regime: the integrals of $u$ and $u^2$ over the whole line generally diverge. The zero-background conservation and scattering formulas cannot be transferred unchanged.

From one pulse to conserved quantities

You can now derive a smooth decaying wave, recover its parameter relations and reject a false travelling speed. Next, verify three conserved integrals for an entire class of decaying solutions, then evaluate them on this pulse.

References

• Lax, Peter D. Integrals of Nonlinear Equations of Evolution and Solitary Waves. Courant Institute report NYO-1480-87, January 1968. Open report PDF. Published version: Communications on Pure and Applied Mathematics 21, 467–490 (1968), DOI. Page and equation locators above refer to the report.