Relate a wave to auxiliary scattering

How does a nonlinear wave become a linear scattering problem? Freeze a KdV profile and use its negative as a Schrödinger potential. For one soliton, you can calculate the bound state and the scattering wave explicitly: the discrete eigenvalue determines the pulse’s width and speed, while a normalization coefficient records its position. The scattering wave is an auxiliary mathematical object, not the original water-wave displacement or its small perturbation.

Required background. Use the pulse formula from Derive a KdV travelling wave and distinguish a function from the differential operator acting on it. You need complex exponentials and square-integrability; the entry check below distinguishes bound and scattering states.

Helpful background. Verify KdV conservation laws explains why conserved numbers alone do not specify the evolving field. The Library derivation expands the operator calculation.

The auxiliary Schrödinger operator

For the real smooth rapidly decaying line field in

$$u_t+6u u_x+u_{xxx}=0,$$

define

$$L(t)=-\partial_x^2-u(x,t),\qquad L\psi=\lambda\psi.$$

At fixed time this is a Schrödinger operator on $L^2(\mathbb R,dx)$ with domain $H^2(\mathbb R)$ for the bounded smooth potential considered here. The Sobolev domain means that the function and its first two weak derivatives are square-integrable. A positive pulse $u$ makes an attractive potential $-u$. A negative discrete eigenvalue can support a square-integrable bound state; positive $\lambda=k^2$ describes oscillatory scattering behavior at spatial infinity. These conventions match Grunert and Teschl 2009, § 2, author PDF pp. 4–5, equations (2.3)–(2.10) after setting their $q=-u$.

The distinction between $u$ and $\psi$ matters. The field $u$ solves a nonlinear time-dependent PDE. At each fixed time, $\psi$ solves a linear equation whose coefficient is that field. A physical perturbation $\delta u$ instead obeys the linearized KdV equation

$$(\delta u)_t+6u(\delta u)_x+6u_x\delta u+(\delta u)_{xxx}=0.$$

This is a different equation.

Entry check and repair

Which of $e^{ikx}$ for real $k$ and $\operatorname{sech}(\kappa x)$ for $\kappa\gt0$ has finite $\int_{\mathbb R}|\psi|^2dx$?

Repair. The plane wave has constant modulus, so its integral diverges. In contrast,

$$\int_{\mathbb R}\operatorname{sech}^2(\kappa x)dx=\frac2\kappa.$$

A scattering wave can still be useful without being a normalizable bound state. Its normalization is specified by the amplitudes of incoming and outgoing waves, not by setting its full-line norm to one.

Why the spectrum is relevant to evolution

Introduce the differential expression

$$A=-4\partial_x^3-6u\partial_x-3u_x.$$

The last term multiplies the function by $-3u_x$; it is not another derivative acting to the right. The commutator $[A,L]=AL-LA$ compares the two orders of composition: on a function $f$, it means $A(Lf)-L(Af)$. On a smooth compactly supported test function, the product rule gives

$$[A,L]=\bigl(u_{xxx}+6u u_x\bigr),$$

where the right-hand side is multiplication by the displayed scalar. All derivative terms acting on the test function cancel. Because $L_t=-u_t$, the compatibility equation $L_t=[A,L]$ is exactly KdV.

Formally, if $L\psi=\lambda\psi$ and $\psi_t=A\psi$, differentiating the first equation yields $\lambda_t\psi=0$. This motivates conserved spectral values. The differential-expression identity alone does not establish existence of the propagator, preservation of operator domains, or a general spectral theorem. Lax’s operator construction is given in Lax 1968, report pp. 6–9, equations (1.11)–(1.16), PDF; his field is $6u$ and his spatial operator is $-L$. Here the explicit pulse calculations below let us verify the relevant eigenvalue directly at every time.

Find the pulse’s bound state

Write the pulse center as $X(t)=4\kappa^2t+x_0$ and put $z=\kappa(x-X)$. Since

$$\frac{d^2}{dx^2}\operatorname{sech}z =\kappa^2\operatorname{sech}z -2\kappa^2\operatorname{sech}^3z,$$

the function

$$\phi(x,t)=\sqrt{\frac\kappa2}\operatorname{sech}z$$

satisfies $L\phi=-\kappa^2\phi$ and $\int|\phi|^2dx=1$. The eigenvalue stays fixed while the eigenfunction translates. It encodes amplitude $2\kappa^2$ and speed $4\kappa^2$, but not the center $X$.

One can also exclude additional negative eigenvalues for this particular potential. Define $Q=\partial_x+\kappa\tanh z$. Its formal adjoint for decaying functions is $Q^\dagger=-\partial_x+\kappa\tanh z$, and

$$Q^\dagger Q=L+\kappa^2, \qquad QQ^\dagger=-\partial_x^2+\kappa^2.$$

The kernel of $Q$ is the one-dimensional span of $\operatorname{sech}z$. Any different negative-energy bound state of $L$ would map under $Q$ to a nonzero square-integrable eigenfunction of $-\partial_x^2$ at the same negative energy. This is impossible because $\langle g,-g''\rangle=\int|g'|^2dx\geq0$. The smooth exponentially decaying eigenfunctions here justify these integrations. Thus this pulse has exactly one negative bound-state eigenvalue.

Calculate a reflectionless scattering wave

For real $k\gt0$, specify the right-normalized Jost solution by $f_+(k,x)\sim e^{ikx}$ as $x\to+\infty$. For this pulse an explicit solution is

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

To check it without a long second derivative, apply $Q^\dagger$ to the free wave $e^{ikx}$. The factorization gives $LQ^\dagger=Q^\dagger(-\partial_x^2)$, and normalizing the result at $+\infty$ produces the displayed formula.

At the opposite end,

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

There is no $e^{-ikx}$ term. A wave with unit incoming amplitude from the left is $f_+/a$, so its left-incident reflection and transmission amplitudes are

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

“Reflectionless” does not mean the potential is absent: $T$ has a nontrivial phase. For example, at $k=\kappa$, $T=i$, and $|T|^2=1$ as expected for this real potential with zero reflection. The formula is stated for positive real $k$; threshold $k=0$ and analytic continuation have separate meanings.

The data that record position

Continue the explicit Jost formula to $k=i\kappa$. It gives

$$f_+(i\kappa,x,t) =\frac12e^{-\kappa X(t)}\operatorname{sech}z.$$

Its coefficient is fixed by the right-tail condition $f_+\sim e^{-\kappa x}$, so it is not the same normalization as $\phi$. Define the squared norming coefficient

$$c(t)\equiv \left(\int_{\mathbb R}|f_+(i\kappa,x,t)|^2dx\right)^{-1}.$$

Then

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

This is the square of the norming constant $\gamma_+$ used in Grunert and Teschl 2009, author PDF p. 5, equations (2.10) and (2.13). The pulse’s right scattering data are therefore zero reflection, one eigenvalue $-\kappa^2$, and this positive coefficient $c(t)$. Although the reflection calculation above used left incidence, both reflections vanish for this explicit potential; the norming coefficient remains explicitly right-normalized.

Exercises

Guided practice: normalize the bound state

For $\kappa=2$ and $X=0$, write $u$, $L$, and the unit-normalized bound state. Find its eigenvalue and verify the normalization integral.

Hint

The coefficient $\sqrt{\kappa/2}$ is one in this example. Use $y=2x$ in the integral.

Solution

Here $u=8\operatorname{sech}^2(2x)$, $L=-\partial_x^2-8\operatorname{sech}^2(2x)$, and $\phi=\operatorname{sech}(2x)$. The derivative identity gives $L\phi=-4\phi$. Also $\int\operatorname{sech}^2(2x)dx=(1/2)\int\operatorname{sech}^2y\,dy=1$. Replacing the potential by $+8\operatorname{sech}^2(2x)$ would invalidate the eigenvalue calculation.

Independent practice: recover a center

Two reflectionless pulses have the same eigenvalue $-1$ and transmission amplitude, but their right squared norming coefficients at $t=0$ are $2$ and $2e^6$. Find their centers and explain why the eigenvalue and transmission amplitude alone cannot distinguish them.

Hint

Solve $c_0=2\kappa e^{2\kappa x_0}$ for $x_0$.

Solution

Both have $\kappa=1$. Their centers are

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

namely $0$ and $3$. Their shape, amplitude, speed and transmission coefficient agree because those depend only on $\kappa$. Translation changes the coefficient required to keep the bound-state tail normalized against the same coordinate function $e^{-\kappa x}$. Retaining $c_0$ restores the missing positional information.

Transfer: reverse the sign of the potential well

At a fixed time replace the positive pulse by $\widetilde u=-2\kappa^2\operatorname{sech}^2(\kappa x)$ and keep $\widetilde L=-\partial_x^2-\widetilde u$. Can $\widetilde L$ have a negative eigenvalue? Does negating the original travelling pulse automatically give a solution of the same KdV equation?

Hint

Evaluate $\langle f,\widetilde Lf\rangle$. Separately track how the linear and quadratic terms of KdV change under $u\mapsto-u$.

Solution

For an admissible normalized test function,

$$\langle f,\widetilde Lf\rangle =\int_{\mathbb R}\left(|f'|^2+ 2\kappa^2\operatorname{sech}^2(\kappa x)|f|^2\right)dx\geq0.$$

Thus there is no negative-energy bound state. Moreover, if $u$ solves KdV, substitution of $-u$ gives residual $12u u_x$, which is generally nonzero. Negating the field changes the nonlinear-sign convention unless other changes are made. A snapshot may define a valid scattering problem without being the claimed travelling KdV solution.

Reverse the construction

You have calculated the pulse’s bound state, reflection coefficient and positional data, with their normalizations explicit. Next, reconstruct the one-soliton solution from $\kappa$ and $c(t)$ without assuming its profile in advance.

References

• Grunert, Katrin, and Gerald Teschl. “Long-Time Asymptotics for the Korteweg–de Vries Equation via Nonlinear Steepest Descent.” Mathematical Physics, Analysis and Geometry 12, 287–324 (2009). DOI. Open author PDF. Locators above use the author PDF’s printed pages.
• 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. Lax page and equation locators above refer to the report.