KdV Lax pair and one-soliton reconstruction

A KdV soliton can be constructed from a nonlinear travelling-wave equation or from linear scattering data. This article connects the two constructions: it derives the Lax commutator, proves the one-soliton potential has exactly one negative eigenvalue and no reflected wave, and reconstructs that potential by solving a rank-one integral equation. The statements concern smooth real fields on the line with zero background. They do not assert a general inverse-scattering existence theorem or the asymptotic completeness of arbitrary KdV evolution.

Required background. Use the product rule for differential operators, integration by parts, and second-order linear ODEs. The wave-to-scattering lesson develops the bound-state and scattering meanings. Helpful background. Derive the travelling wave to compare the two constructions; the nonlinear-wave reference collects the sign choices.

Differential operators for KdV

We use

$$u_t+6uu_x+u_{xxx}=0, \qquad L=-D^2-u, \qquad A=-4D^3-6uD-3u_x,$$

with $D=\partial_x$ and real $u(\cdot,t)$ in the Schwartz class on $\mathbb R$. Multiplication by $u$ or $u_x$ is an operator; for example, $uD$ means “differentiate, then multiply,” whereas $Du=uD+u_x$. Our Schrödinger potential is $q=-u$. Lax’s original construction uses the field $w=6u$ and the operator $D^2+w/6=-L$; see Lax 1968, report PDF, pp. 6–9, equations (1.14)–(1.16) and the third-order construction.

Calculate the commutator

For a smooth coefficient $f$, the product rule gives

$$\begin{aligned} D^2f&=fD^2+2f_xD+f_{xx},\\ D^3f&=fD^3+3f_xD^2+3f_{xx}D+f_{xxx}. \end{aligned}$$

Apply these identities to an arbitrary smooth test function. The four nonzero contributions to $[A,L]=AL-LA$ are

$$\begin{aligned} [-4D^3,-u]&=12u_xD^2+12u_{xx}D+4u_{xxx},\\ [-6uD,-D^2]&=-12u_xD^2-6u_{xx}D,\\ [-6uD,-u]&=6uu_x,\\ [-3u_x,-D^2]&=-6u_{xx}D-3u_{xxx}. \end{aligned}$$

The coefficients of $D^2$ and $D$ cancel. Hence

$$[A,L]=u_{xxx}+6uu_x, \qquad L_t=-u_t.$$

The remaining terms are multiplication operators. The equality $L_t=[A,L]$ is therefore equivalent to the stated KdV equation as an identity of differential expressions. In particular, replacing $uD$ by $Du$ without also changing the zeroth-order term would spoil the cancellation.

What the operator statement requires

At fixed time, bounded real $u$ is a bounded symmetric perturbation of $-D^2$. Thus $L$ is self-adjoint on $H^2(\mathbb R)$ in $L^2(\mathbb R,dx)$. This is the domain used for the decaying Schrödinger problem in Grunert and Teschl 2009, author PDF, § 2, p. 4, equation (2.3). Here $H^2$ consists of square-integrable functions whose first two weak derivatives are square-integrable.

On Schwartz test functions, integration by parts gives $A^*=-A$ as a formal adjoint identity. The two products $AL$ and $LA$ involve higher derivatives; writing their difference is not an assertion that each product is defined on every vector in $H^2$. A proof of unitary evolution for these unbounded operators needs suitable time regularity, domains, and domain preservation. Those analytic steps are separate from the cancellation above.

For a smooth normalized eigenfunction branch $L\psi=\lambda\psi$, assume the differentiations and integrations by parts are valid and $A\psi$ lies in the required domain. The Lax identity then implies

$$\dot\lambda=\langle\psi,L_t\psi\rangle =\langle\psi,[A,L]\psi\rangle=0.$$

The last equality moves $L$ onto $\psi$ in both terms. This explains spectral conservation under the stated hypotheses. It neither counts independent Hamiltonian integrals nor proves the completeness of a spectral transform.

Factorizing the one-soliton operator

Fix $\kappa\gt0$ and a center $X$. At this stage $X$ is a fixed spatial parameter. Let

$$\xi=x-X, \qquad u(x)=2\kappa^2\operatorname{sech}^2(\kappa\xi), \qquad W(x)=\kappa\tanh(\kappa\xi).$$

Define $Q=D+W$ and its formal adjoint $Q^*=-D+W$. Since $W'=\kappa^2\operatorname{sech}^2(\kappa\xi)$,

$$\begin{aligned} Q^*Q&=-D^2+W^2-W'=L+\kappa^2,\\ QQ^*&=-D^2+W^2+W'=-D^2+\kappa^2. \end{aligned}$$

These identities also hold as quadratic-form identities on the appropriate Sobolev domains. The equation $Q\psi_0=0$ has the normalized solution

$$\psi_0(x)=\sqrt{\frac\kappa2}\operatorname{sech}(\kappa\xi), \qquad L\psi_0=-\kappa^2\psi_0.$$

The normalization uses $\int\operatorname{sech}^2(\kappa\xi)\,dx=2/\kappa$.

There are no other negative eigenvalues. Positivity of $Q^*Q$ first excludes $E\lt-\kappa^2$. If $L\psi=E\psi$ with $-\kappa^2\lt E\lt0$, then $Q\psi\ne0$ and the factorization gives

$$(-D^2)(Q\psi)=E(Q\psi).$$

But the free operator has no nonzero $L^2(\mathbb R)$ eigenfunction at negative energy: its exponentially growing and decaying solutions cannot decay at both ends while solving the same ODE smoothly. For this smooth bounded potential, the eigenfunction regularity needed to apply $Q$ follows from its differential equation. Finally, the first-order equation $Q\psi=0$ has a one-dimensional solution space, so $-\kappa^2$ is simple.

This is an explicit bound-state proof for this potential. A plot of the well, or the mere presence of a pole in a guessed amplitude, would not establish the eigenvalue count.

Jost solution and reflectionless scattering

For real $k\ne0$, the right-end normalized Jost solution obeys

$$Lf_+(x,k)=k^2f_+(x,k), \qquad f_+(x,k)e^{-ikx}\longrightarrow1 \quad(x\longrightarrow+\infty).$$

The intertwining identity $LQ^*=Q^*(-D^2)$ follows from the same factorization. Applying it to $e^{ikx}$, and normalizing at $+\infty$, gives

$$f_+(x,k)=e^{ikx} \frac{k+i\kappa\tanh(\kappa\xi)}{k+i\kappa}.$$

At the left end,

$$f_+(x,k)\sim a(k)e^{ikx}, \qquad a(k)=\frac{k-i\kappa}{k+i\kappa}.$$

There is no $e^{-ikx}$ term. For $k\gt0$, a unit-amplitude wave incident from the left is $f_+/a$; the reflection and transmission amplitudes are

$$R_{\mathrm{left}}(k)=0, \qquad T(k)=\frac1{a(k)}=\frac{k+i\kappa}{k-i\kappa}.$$

For real $k$, $|T|=1$: the well produces a phase shift without reflected flux. The same calculation from the other end gives zero reflection for incidence from the right. $T$ has a pole at $k=i\kappa$, corresponding to energy $k^2=-\kappa^2$. The bound-state proof above independently identifies that state.

For $k\ne0$, $f_+$ and its complex conjugate are independent, as their nonzero Wronskian at $+\infty$ shows. Every positive-energy solution is a combination of their oscillating asymptotes, so none is in $L^2$. At zero energy the factorization again maps a putative eigenfunction to an $L^2$ solution of the free zero-energy equation, which does not exist. Thus the negative state already found is the only bound state; the bounded but nonnormalizable threshold solution is examined in the exercises.

The center $X$ does not appear in $T$. Consequently, an eigenvalue and the transmission amplitude do not distinguish all translates of this potential. Position is encoded in the bound-state normalization.

Define the norming coefficient

Analytically continue the displayed Jost solution to $k=i\kappa$. Then

$$f_+(x,i\kappa)=\frac{e^{-\kappa X}}2 \operatorname{sech}(\kappa\xi).$$

Its right-end amplitude is still fixed by $f_+(x,i\kappa)\sim e^{-\kappa x}$. Define

$$c\equiv\left[\int_{\mathbb R}f_+(x,i\kappa)^2\,dx\right]^{-1} =2\kappa e^{2\kappa X}.$$

Thus $c\gt0$ and $X=(2\kappa)^{-1}\log[c/(2\kappa)]$. This $c$ is the square of the norming constant $\gamma_+$ in Grunert and Teschl 2009, author PDF, p. 5, equation (2.10). Using the same symbol for a normalized eigenfunction’s tail amplitude, its square, and its reciprocal would produce different evolution laws; the definition must travel with the formula.

Solve a rank-one Marchenko equation

Take reflectionless data consisting of one $\kappa\gt0$ and one coefficient $c\gt0$. For a kernel normalized at the $+\infty$ end, define

$$F(s)=ce^{-\kappa s}.$$

Consider the integral equation

$$K(x,y)+F(x+y) +\int_x^\infty K(x,z)F(z+y)\,dz=0, \qquad y\geq x.$$

Here $x$ is a parameter when solving for the function of $y$. This is the reflectionless, single-eigenvalue specialization of Aktosun 2009, § VIII, equations (8.1)–(8.3), whose Schrödinger potential is $q=-u$ in our notation. The integration interval and the exponential’s sign belong together. The following calculation establishes the reconstruction for this one-pole data directly; it does not invoke an inverse-scattering theorem for arbitrary $F$.

Because $F(z+y)$ is separable, the equation forces $K(x,y)=B(x)e^{-\kappa y}$. Its scalar coefficient satisfies

$$B(x)+ce^{-\kappa x} +\frac{c}{2\kappa}B(x)e^{-2\kappa x}=0.$$

The denominator $1+[c/(2\kappa)]e^{-2\kappa x}$ is strictly positive, so the unique solution within the class for which the integral exists is

$$K(x,y)= -\frac{ce^{-\kappa(x+y)}}{1+\dfrac{c}{2\kappa}e^{-2\kappa x}}.$$

Set $r=[c/(2\kappa)]e^{-2\kappa x}$. The reconstructed KdV field is

$$\begin{aligned} u(x)&=2\frac{d}{dx}K(x,x),\\ K(x,x)&=-\frac{2\kappa r}{1+r},\\ u(x)&=\frac{8\kappa^2r}{(1+r)^2} =2\kappa^2\operatorname{sech}^2[\kappa(x-X)]. \end{aligned}$$

The derivative of $K(x,x)$ is a total derivative along the diagonal, $K_x(x,x)+K_y(x,x)$. Differentiating only its first argument is a different operation. Also, the Schrödinger potential is $q=-u=-2\,dK(x,x)/dx$, which explains the opposite sign often used in scattering texts.

One can check that the integral equation and the spectral solution describe the same object. Substitute this $K$ into the transformation formula

$$f_+(x,k)=e^{ikx}+\int_x^\infty K(x,y)e^{iky}\,dy.$$

The integral is elementary for real $k$ and equals

$$-e^{ikx}\frac{2\kappa r}{(1+r)(\kappa-ik)}.$$

Using $\tanh[\kappa(x-X)]=(1-r)/(1+r)$ reproduces the Jost solution above. Its eigenvalue equation, reflection coefficient, and bound-state normalization have therefore all been checked without assuming the answer from a general inverse theorem.

Time evolution becomes translation

Keep $\kappa$ fixed and evolve the norming coefficient by

$$c(t)=c_0e^{8\kappa^3t}, \qquad c_0=2\kappa e^{2\kappa x_0}.$$

This is the squared-norming-constant version of Grunert and Teschl 2009, author PDF, p. 5, Lemma 2.2, equation (2.13). The center reconstructed from it is

$$X(t)=\frac1{2\kappa}\log\frac{c(t)}{2\kappa} =x_0+4\kappa^2t.$$

The resulting field is exactly

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

It satisfies $u_{xx}=4\kappa^2u-3u^2$ and $u_t=-4\kappa^2u_x$. Differentiating the first identity and substituting the second verifies $u_t+6uu_x+u_{xxx}=0$ for every real $x,t$.

There is a second way to check the factor eight. Since $A\to-4D^3$ at $+\infty$, the time equation that preserves the Jost normalization is

$$\partial_tf_+=(A-4ik^3)f_+.$$

At $k=i\kappa$ this is $\partial_tf_+=(A-4\kappa^3)f_+$. Formal skew-adjointness of $A$, with decaying bound-state functions and justified integrations, gives

$$\frac{d}{dt}\|f_+(\cdot,i\kappa)\|_2^2 =-8\kappa^3\|f_+(\cdot,i\kappa)\|_2^2.$$

Taking the reciprocal recovers $\dot c=8\kappa^3c$. Normalizing the bound state in $L^2$ instead would remove this amplitude change and hide the position information carried by the right-end normalization.

What has and has not been reconstructed

For this reflectionless one-pole problem, the spectral parameter, transmission amplitude, norming coefficient, integral kernel, and KdV field agree by explicit calculation. The eigenvalue controls scale; the norming coefficient controls location. The two are distinct pieces of data.

A general decaying potential can have reflection as well as several negative eigenvalues. Its reconstruction includes a continuous spectral contribution, and the analytic work includes admissibility, invertibility, regularity, and compatibility with time evolution. A periodic, nonzero-background, or half-line problem changes those questions. None follows solely from the rank-one algebra on this page. Likewise, showing that one exact pulse translates does not prove soliton stability, collision phase shifts, or the long-time decomposition of an arbitrary initial profile.

The two-construction project compares travelling-wave and scattering reconstructions with independent residuals. The KdV model record fixes the conservation laws, Hamiltonian, scaling, and alternative boundary regimes.

Exercises

Recover the commutator coefficients

Start with $A=aD^3+buD+du_x$, where $a,b,d$ are real constants. Require that $[A,-D^2-u]$ have no $D^2$ or $D$ terms. Then choose the scale so that $L_t=[A,L]$ gives the KdV equation above.

Solution

The $D^2$ coefficient is $(-3a+2b)u_x$, and the $D$ coefficient is $(-3a+b+2d)u_{xx}$. Thus $b=3a/2$ and $d=3a/4$. The multiplication term is

$$(-a+d)u_{xxx}-buu_x =-\frac a4u_{xxx}-\frac{3a}{2}uu_x.$$

Taking $a=-4$ gives $b=-6$, $d=-3$, and $[A,L]=u_{xxx}+6uu_x$. Since $L_t=-u_t$, the signs match the desired PDE.

Translate the scattering data

Replace a one-soliton field $u(x)$ by $u(x-a)$ for a real shift $a$. Determine the changes to its eigenvalue, transmission amplitude, and coefficient $c$. Why cannot its transmission amplitude alone locate the pulse?

Solution

The center changes from $X$ to $X+a$, so $\kappa$ and $T(k)$ are unchanged, while $c$ changes to $ce^{2\kappa a}$. Translation is unitary on $L^2(\mathbb R)$, so it also preserves the spectrum. The transmission amplitude determines the phase difference across the well, which is unchanged by shifting both ends relative to the potential. The norming coefficient supplies the missing location.

Separate a threshold solution from a bound state

Set $k=0$ in the explicit Jost formula. Verify the resulting zero-energy solution of $Lf=0$ and decide whether it is another bound state.

Solution

The limit is $f(x)=\tanh[\kappa(x-X)]$. Direct differentiation gives $-f''-uf=0$. It is bounded but tends to $\pm1$ at the two ends, so it is not square-integrable. It is a zero-energy resonance (a bounded threshold solution), not an additional $L^2$ eigenfunction. The unique negative bound state remains $-\kappa^2$.

References

• Aktosun, Tuncay. “Inverse Scattering Transform and the Theory of Solitons.” In Robert A. Meyers (ed.), Encyclopedia of Complexity and Systems Science. Springer, 2009, 4960–4971. DOI. Author version arXiv:0905.4746v1 [nlin.SI], 2009. Open HTML.
• Grunert, Katrin, and Gerald Teschl. “Long-time asymptotics for the Korteweg–de Vries equation via nonlinear steepest descent.” Mathematical Physics, Analysis and Geometry 12 (2009), 287–324. DOI. Open PDF. Locators above use the author PDF’s printed pages.
• Lax, Peter D. “Integrals of nonlinear equations of evolution and solitary waves.” Communications on Pure and Applied Mathematics 21 (1968), 467–490. DOI. Open report PDF, NYO-1480-87, January 1968. Locators above use the report’s printed pages, not the journal pagination.