Reconstruct a one-soliton solution

Can scattering data produce the wave without guessing its shape? For a reflectionless potential with one bound state, the Marchenko integral equation reduces to a single algebraic equation. You will solve it, recover the positive KdV pulse, and check its time evolution and normalization against the direct calculation. This is an exact one-bound-state inverse problem, not a derivation of inverse scattering for arbitrary initial data.

Required background. Relate a wave to auxiliary scattering defines the right-normalized bound-state coefficient $c(t)$ and eigenvalue $-\kappa^2$. Derive a KdV travelling wave supplies an independent PDE check.

Helpful background. The convention reference fixes the relation between the positive field $u$ and the attractive Schrödinger potential $-u$.

A Marchenko equation with one exponential

Keep $u_t+6u u_x+u_{xxx}=0$ on the line with real smooth decaying fields and $L=-\partial_x^2-u$. Specify reflectionless right scattering data consisting of one negative eigenvalue $-\kappa^2$, with $\kappa\gt0$, and a positive squared norming coefficient

$$c(t)=c_0e^{8\kappa^3t},\qquad c_0\gt0.$$

The input function is

$$F(s,t)=c(t)e^{-\kappa s}.$$

For each fixed $x,t$, solve for $K(x,y,t)$ on $y\geq x$ using the right Marchenko equation

$$\begin{aligned} 0={}&K(x,y,t)+F(x+y,t)\\ &+\int_x^\infty K(x,z,t)F(z+y,t)\,dz. \end{aligned}$$

Recover the field by

$$u(x,t)=2\frac{d}{dx}K(x,x,t).$$

The derivative is the total derivative along the diagonal at fixed time. Both arguments of $K$ change with $x$. The plus sign recovers our field $u$; the Schrödinger potential $q=-u$ has the usual reconstruction sign $q=-2dK(x,x,t)/dx$.

These are the right-end formulas of Aktosun 2009, §§ VII–VIII, equations (7.9) and (8.1)–(8.3), after changing his field to $q=-u$. His kernel $\Omega$ is our $F$. In more general data an additional Fourier integral of the reflection coefficient appears in $F$; it is zero in the present calculation. Left-end reconstruction uses a different kernel and integration interval, so its signs and exponentials must be translated together.

Entry check and repair

For $G(x,y)=e^{-\kappa(x+y)}$, compare $\partial_xG(x,y)$ evaluated at $y=x$ with $dG(x,x)/dx$.

Repair. Holding the second argument fixed gives $-\kappa e^{-2\kappa x}$. Moving along the diagonal gives $-2\kappa e^{-2\kappa x}$. In general

$$\frac{d}{dx}G(x,x) =\left.(\partial_xG+\partial_yG)\right|_{y=x}.$$

For reconstruction, substitute $y=x$ first and differentiate that one-variable expression. This avoids silently dropping half of a derivative—or more, when the two arguments enter differently.

Reduce the integral equation to algebra

Suppress $t$ temporarily and try a separable kernel

$$K(x,y)=-h(x)e^{-\kappa y}.$$

This form is forced for any solution for which the integral exists: the last two terms in the Marchenko equation are proportional to $e^{-\kappa y}$, so its first term must have the same dependence on $y$.

The integral becomes

$$\int_x^\infty K(x,z)F(z+y)dz =-h(x)c\,e^{-\kappa y}\frac{e^{-2\kappa x}}{2\kappa}.$$

Cancel $e^{-\kappa y}$ to obtain

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

Thus

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

The denominator is strictly positive for the declared data. The integral converges at its upper endpoint because $\kappa\gt0$. These two facts establish an actual regular solution for every finite $x$, not just a formal rearrangement. The forced exponential form and nonzero denominator also prove uniqueness within the class for which this integral equation is defined.

Recover the wave and its center

Define the positive dimensionless quantity

$$\eta(x,t)=\frac{c(t)}{2\kappa}e^{-2\kappa x}.$$

Then $K(x,x,t)=-2\kappa\eta/(1+\eta)$ and $\eta_x=-2\kappa\eta$. The diagonal derivative gives

$$u(x,t)=\frac{8\kappa^2\eta}{(1+\eta)^2}.$$

Write

$$x_0=\frac{1}{2\kappa}\log\frac{c_0}{2\kappa}.$$

The time dependence of $c$ implies

$$\eta=e^{-2\kappa(x-4\kappa^2t-x_0)}.$$

Using $4e^{-2z}/(1+e^{-2z})^2=\operatorname{sech}^2z$ therefore recovers

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

The eigenvalue determines the height and speed. The positive coefficient $c_0$ determines the position. The exponential growth of $c(t)$ means the pulse moves; it does not mean the field amplitude grows.

An equivalent expression uses $\tau(x,t)=1+\eta(x,t)$:

$$K(x,x,t)=\partial_x\log\tau, \qquad u=2\partial_x^2\log\tau.$$

Here $\tau$ is simply a convenient scalar function derived from the kernel. A general hierarchy of tau functions requires additional structure.

Close the direct and inverse checks

Choose $\kappa=1$, $c_0=2e^{-4}$. The reconstructed center is $x_0=-2$, and

$$u(x,t)=2\operatorname{sech}^2(x-4t+2).$$

At $t=1/2$ the center is zero and $c(t)=2$. The field’s height is still $2$. The direct spectral calculation gives a bound state at $-1$ with right-normalized norm squared $1/2$, so its inverse norm squared is $2$, agreeing with the input at that time.

There are three distinct checks. Substituting $K$ into the integral equation checks inversion. Computing the right-normalized bound-state norm checks the data normalization. Substituting the reconstructed field into KdV checks the time law. In the last step $U''=4\kappa^2U-3U^2$ and $u_t=-4\kappa^2U'$ make the PDE residual exactly zero. None of these steps is replaced by plotting the pulse.

Exercises

Guided practice: verify one row of the kernel

Take $\kappa=1$, $c=2$ at a fixed time. Find $K(0,y)$, $y\geq0$, and directly verify the Marchenko equation at $x=0$. What field value does the full diagonal formula give at $x=0$?

Hint

Here $F(s)=2e^{-s}$ and $\int_0^\infty e^{-2z}dz=1/2$. Use $\eta(0)=1$ only after obtaining the derivative formula.

Solution

The kernel is $K(0,y)=-e^{-y}$. The three terms are

$$-e^{-y}+2e^{-y} \underbrace{-2e^{-y}\int_0^\infty e^{-2z}dz}_{-e^{-y}}=0.$$

The field formula gives $u(0)=8/(1+1)^2=2$. The single value $K(0,0)=-1$ alone does not determine $u(0)$; reconstruction needs how the diagonal varies near $x=0$.

Independent practice: catch the wrong derivative

At the center $x=X$, where $\eta=1$, compute $\partial_xK(x,y)$ and $\partial_yK(x,y)$ before setting $y=x$. Show why replacing the total diagonal derivative by $\partial_xK$ would give a false field value at the peak.

Hint

At fixed time, $K=-ce^{-\kappa(x+y)}/(1+\eta)$ and $\eta_x=-2\kappa\eta$. Differentiate the numerator and denominator separately.

Solution

For independent arguments,

$$\partial_xK=\kappa\frac{\eta-1}{1+\eta}K, \qquad \partial_yK=-\kappa K.$$

At the diagonal center $K=-\kappa$ and $\eta=1$, so $\partial_xK=0$, $\partial_yK=\kappa^2$. The correct field is $2(\partial_xK+\partial_yK)=2\kappa^2$. Using only the first partial derivative would incorrectly report zero at the pulse maximum.

Transfer: test inadmissible input data

Keep $\kappa\gt0$ but replace the positive coefficient by a negative real number $c$. Locate a zero of the kernel’s denominator. Explain why this is not a smooth negative soliton of the same line problem. What happens if $c=0$ instead?

Hint

Solve $1+(c/2\kappa)e^{-2\kappa x}=0$. Recall the definition of $c$ as an inverse squared norm.

Solution

For $c\lt0$, the denominator vanishes at

$$x_* =\frac{1}{2\kappa}\log\frac{|c|}{2\kappa}.$$

The numerator is nonzero there, so the kernel is singular and the reconstructed field is not a smooth decaying pulse on the whole line. A negative number also cannot equal the inverse squared norm of a nonzero bound state. Algebraically inserting it violates the admissibility of the specified scattering data.

For $c=0$, $F=0$, the equation gives $K=0$ and hence $u=0$. That case has no retained bound state: it is not a one-bound-state data set with a vanishing norm. As $c\to0^+$ with $\kappa$ fixed, the pulse center tends to $-\infty$; the field tends to zero at each fixed $x$ while its full-line integral remains $4\kappa$. Pointwise and integrated limits need not agree.

Test the construction numerically

You have recovered a pulse from admissible scattering data and checked the diagonal derivative, time sign and normalization independently. Continue to the KdV convergence laboratory to distinguish field error, boundary truncation and invariant drift, then compare the two constructions in the course project. The Library article collects the proof and its scope.

References

• Aktosun, Tuncay. “Inverse Scattering Transform and the Theory of Solitons.” In Encyclopedia of Complexity and Systems Science, 4960–4971 (2009). DOI. Author version arXiv:0905.4746v1; HTML, Open PDF. Section and equation locators above refer to the author version.