KdV two-soliton scattering

How can two nonlinear waves interact and recover their original amplitudes and speeds? For the decaying KdV equation, a two-by-two reconstruction matrix produces an exact solution whose faster pulse advances and whose slower pulse lags. This article derives that matrix, fixes its norming constants, verifies the nonlinear equation, and extracts both position shifts. The result concerns two distinct positive spectral parameters on the real line; it is not a completeness theorem for arbitrary initial data.

Required background. Differentiate a determinant and a logarithm, solve a two-by-two linear system, and use the one-pole Marchenko construction. Helpful background. Measure a KdV soliton collision develops the moving-frame limits, while the convention reference fixes the field and spectral signs.

Two scattering levels on the line

Use real normalized coordinates $x,t$, zero background, and

$$u_t+6uu_x+u_{xxx}=0, \qquad L=-\partial_x^2-u.$$

We construct smooth real fields that decay exponentially at both spatial ends at each fixed time. The Schrödinger operator acts in $L^2(\mathbb R,dx)$ on $H^2(\mathbb R)$, as in the one-pole article. Choose

$$\kappa_1\gt\kappa_2\gt0, \qquad c_j(0)\gt0, \qquad c_j(t)=c_j(0)e^{8\kappa_j^3t}.$$

The proposed bound-state energies are $-\kappa_j^2$. The coefficient $c_j$ is defined using a Jost solution normalized at positive infinity:

$$\begin{aligned} Lf_+(x,k,t)&=k^2f_+(x,k,t),\\ f_+(x,k,t)e^{-ikx}&\longrightarrow1\quad(x\to+\infty),\\ c_j(t)&=\left[\int_{\mathbb R}f_+(x,i\kappa_j,t)^2\,dx\right]^{-1}. \end{aligned}$$

We verify this normalization below. Aktosun calls this Jost solution $f_{\mathrm l}$ and writes his field as the Schrödinger potential $q=-u$. His norming coefficient is the square of the corresponding inverse $L^2$ norm; see Aktosun 2009, § VI, equation (6.1) and the definition preceding (6.3); § VII, equation (7.9). Specifying the normalization endpoint avoids ambiguity about the words “left” and “right.”

Reduce the Marchenko equation to a matrix

For reflectionless data with these two levels, set

$$F(s,t)=\sum_{j=1}^2c_j(t)e^{-\kappa_js}.$$

The reconstruction equation and our positive-field recovery rule are

$$\begin{aligned} K(x,y)+F(x+y) +\int_x^\infty K(x,z)F(z+y)\,dz&=0,\qquad y\geq x,\\ u(x,t)&=2\frac{d}{dx}K(x,x,t). \end{aligned}$$

Time is suppressed in the first line. The second line uses the total derivative along the diagonal. The source has the opposite recovery sign because its field is $q=-u$; compare Aktosun, § VIII, equations (8.1)–(8.3), and § IX, equations (9.1)–(9.3).

Define a real column and its Gram matrix by

$$w_j(x,t)=\sqrt{c_j(t)}e^{-\kappa_jx}, \qquad G(x,t)=I+\int_x^\infty w(z,t)w(z,t)^{\mathsf T}\,dz.$$

Then $F(x+y)=w(x)^{\mathsf T}w(y)$, and direct integration gives

$$G_{jl}=\delta_{jl} +\frac{\sqrt{c_j(t)c_l(t)}e^{-(\kappa_j+\kappa_l)x}} {\kappa_j+\kappa_l}.$$

For every nonzero real column $v$,

$$v^{\mathsf T}Gv =\|v\|_2^2+\int_x^\infty[v^{\mathsf T}w(z)]^2\,dz\gt0.$$

Thus $G$ is positive definite and invertible for all finite $x,t$. This is a proof of solvability, not just a numerical determinant check.

Every term in the integral equation other than $K(x,y)$ is a linear combination of the two exponentials in $y$. Hence any solution for which the integrals exist must have the form $K(x,y)=h(x)^{\mathsf T}w(y)$. Substitution gives

$$h^{\mathsf T}G=-w(x)^{\mathsf T}, \qquad \boxed{K(x,y)=-w(x)^{\mathsf T}G^{-1}w(y).}$$

The two exponentials are independent because the $\kappa_j$ are distinct, and invertibility makes this solution unique in the stated integral class. The source’s integral matrix $\Gamma=I+\int YX$ is similar to $G$ through $S=\operatorname{diag}(\sqrt{c_1},\sqrt{c_2})$: $G=S^{-1}\Gamma S$. Deriving the entries from the integral fixes where the exponential factors belong.

Since $G_x=-ww^{\mathsf T}$, Jacobi’s determinant identity gives

$$\begin{aligned} \partial_x\log\det G &=\operatorname{tr}(G^{-1}G_x)\\ &=-w^{\mathsf T}G^{-1}w=K(x,x). \end{aligned}$$

Therefore the positive KdV field is

$$\boxed{u=2\partial_x^2\log D,\qquad D=\det G.}$$

Check what the norming coefficients normalize

The matrix construction also verifies the bound states, without assuming a general inverse theorem. At fixed $t$, write

$$M=G^{-1},\qquad a=Mw, \qquad \Lambda=\operatorname{diag}(\kappa_1,\kappa_2).$$

The identities

$$\begin{aligned} M_x&=aa^{\mathsf T},\\ \Lambda G+G\Lambda&=2\Lambda+ww^{\mathsf T},\\ a_x&=(I-2M)\Lambda a \end{aligned}$$

follow by differentiating $G$ and multiplying the second identity by $M$. In particular,

$$\begin{aligned} (I-2M)\Lambda(I-2M)&=\Lambda-2aa^{\mathsf T},\\ u&=4a^{\mathsf T}\Lambda a,\\ a_{xx}&=\Lambda^2a-u a. \end{aligned}$$

Thus each scalar component $a_j$ solves $La_j=-\kappa_j^2a_j$. To determine its norm, note that $M(+\infty)=I$ and $M(-\infty)=0$. For the latter limit, write $G=I+WCW$, where $W=\operatorname{diag}(w_1,w_2)$ and $C_{jl}=1/(\kappa_j+\kappa_l)$. The matrix $C$ is positive definite because it is the Gram matrix of two independent decaying exponentials on $[0,\infty)$. Both diagonal entries of $W$ diverge at the left end, so the smallest eigenvalue of $G$ diverges. Integrating $M_x$ therefore proves

$$\int_{\mathbb R}a_j(x)a_l(x)\,dx=\delta_{jl}.$$

At the right end $a_j\sim\sqrt{c_j}e^{-\kappa_jx}$, so the Jost-normalized bound solution is $f_+(x,i\kappa_j)=a_j/\sqrt{c_j}$. Its squared norm is $1/c_j$, exactly the convention stated above. The formula $u=4\sum_j\kappa_ja_j^2$ also proves $u\geq0$ and yields the mass

$$\int_{\mathbb R}u(x,t)\,dx=4(\kappa_1+\kappa_2).$$

Check the reflectionless scattering statement

For real $k\ne0$, the same matrix gives an explicit solution:

$$f_+(x,k)=e^{ikx}\left[1-\sum_{j=1}^2 \frac{a_j(x)w_j(x)}{\kappa_j-ik}\right].$$

To verify its equation, set $b_j=a_jw_j$ and $h=\sum_jb_j/(\kappa_j-ik)$. The identities above give

$$b_j''=-2\kappa_jb_j'-ub_j, \qquad h''+2ikh'=u(1-h),$$

where $-2\sum_jb_j'=u$ was used in the second equality. Substitution now yields $f_+''+(k^2+u)f_+=0$. At $+\infty$ its bracket tends to one. At $-\infty$, using $G=I+WCW$ and inverting the two-by-two matrix $C$, the bracket tends to

$$1-\sum_{j=1}^2\frac{(\mathbf1^{\mathsf T}C^{-1})_j}{\kappa_j-ik} =\prod_{j=1}^2\frac{k-i\kappa_j}{k+i\kappa_j}, \qquad \mathbf1=(1,1)^{\mathsf T}.$$

There is no reflected $e^{-ikx}$ term. For $k\gt0$, with unit incidence from the left, the transmission amplitude is

$$T(k)=\prod_{j=1}^2\frac{k+i\kappa_j}{k-i\kappa_j}, \qquad R_{\mathrm{left}}(k)=0.$$

Its two upper-half-plane poles occur at the input spectral levels, and $|T(k)|=1$ for real $k$. This calculation checks reflectionlessness for the constructed potential directly. The Jost and transmission conventions are those of Aktosun, § VI, equations (6.1)–(6.2), with the field sign already translated.

Expand the determinant and convert the phases

Set

$$r_j=\frac{c_j(t)}{2\kappa_j}e^{-2\kappa_jx}, \qquad A=\left(\frac{\kappa_1-\kappa_2}{\kappa_1+\kappa_2}\right)^2.$$

Expanding the two-by-two determinant gives

$$\begin{aligned} D&=(1+r_1)(1+r_2) -\frac{4\kappa_1\kappa_2}{(\kappa_1+\kappa_2)^2}r_1r_2\\ &=1+r_1+r_2+A r_1r_2. \end{aligned}$$

The interaction coefficient comes from the off-diagonal matrix entries. Setting it to one would discard their contribution.

For studying moving pulses, growing exponentials are convenient. Define real phase parameters $x_j$ by

$$\boxed{c_j(0)=\frac{2\kappa_j}{A}e^{2\kappa_jx_j},} \qquad \eta_j=2\kappa_j(x-4\kappa_j^2t-x_j).$$

Then $e^{\eta_j}=1/(Ar_j)$, so

$$\boxed{ \tau=\frac{D}{Ar_1r_2} =1+e^{\eta_1}+e^{\eta_2}+Ae^{\eta_1+\eta_2}, \qquad u=2\partial_x^2\log\tau. }$$

The multiplier $1/(Ar_1r_2)$ has a logarithm affine in $x$; its second $x$ derivative vanishes. Such a multiplier changes the tau function without changing $u$. Both coefficients $c_j(0)$ contain $1/A$: the one-pole conversion $c_0=2\kappa e^{2\kappa x_0}$ cannot be reused unchanged.

All four terms of $\tau$ are positive because $0\lt A\lt1$. Hence its logarithm is smooth for all real $x,t$. At either spatial end, one exponential term dominates, and the remaining ratios and their derivatives decay exponentially. This establishes the stated spatial decay. If physical dimensions are retained, $[\kappa]=[x]^{-1}$, $[t]=[x]^3$, $[u]=[x]^{-2}$ and $[c_j]=[x]^{-1}$; every $r_j$ and $\eta_j$ is dimensionless.

Verify the KdV equation directly

Reconstruction motivates the formula; a separate calculation checks the PDE. For any positive smooth $\tau$, define

$$\mathcal B[\tau] =\tau\tau_{xt}-\tau_x\tau_t +\tau\tau_{xxxx}-4\tau_x\tau_{xxx}+3\tau_{xx}^2.$$

Writing $g=\log\tau$ and applying the product rule gives

$$\frac{\mathcal B[\tau]}{\tau^2} =g_{xt}+g_{xxxx}+6g_{xx}^2.$$

With $u=2g_{xx}$, a further $x$ derivative yields the exact identity

$$u_t+6uu_x+u_{xxx} =2\partial_x\left(\frac{\mathcal B[\tau]}{\tau^2}\right).$$

Thus $\mathcal B[\tau]=0$ is sufficient. In Hirota derivative notation it is half of $(D_xD_t+D_x^4)\tau\mathbin{\cdot}\tau$, but no additional formalism is needed to check the displayed polynomial.

To expand it economically, put $z_1=e^{\eta_1}$ and $z_2=e^{\eta_2}$. Derivatives act on any polynomial in $z_1,z_2$ as

$$\begin{aligned} \partial_x&=2\kappa_1z_1\partial_{z_1}+2\kappa_2z_2\partial_{z_2},\\ \partial_t&=-8\kappa_1^3z_1\partial_{z_1}-8\kappa_2^3z_2\partial_{z_2}. \end{aligned}$$

For $\tau=1+z_1+z_2+Az_1z_2$, substitution and collection of like powers gives

$$\mathcal B[\tau] =48z_1z_2\kappa_1\kappa_2 \left[A(\kappa_1+\kappa_2)^2-(\kappa_1-\kappa_2)^2\right].$$

The bracket vanishes for the determinant’s value of $A$. This proves the KdV equation for every real $x,t$, independently of an inverse-scattering existence theorem or a discretized evolution. It also explains why the coefficient governing the collision is constrained by the nonlinear PDE.

Derive the incoming and outgoing trajectories

Let $v_j=4\kappa_j^2$ and use the moving coordinate $\xi_j=x-v_jt$. The ordering gives $v_1\gt v_2$. A solitary profile of parameter $\kappa_j$ and intercept $b_j$ is

$$U_j(\xi_j;b_j) =2\kappa_j^2\operatorname{sech}^2[\kappa_j(\xi_j-b_j)].$$

In the fast frame, $\eta_1$ stays bounded while

$$\eta_2=2\kappa_2[(v_1-v_2)t+\xi_1-x_2]$$

goes to $-\infty$ in the past and $+\infty$ in the future. Consequently $\tau$ reduces to $1+e^{\eta_1}$ in the past, and to $e^{\eta_2}(1+Ae^{\eta_1})$ in the future. The latter prefactor disappears under $\partial_x^2\log$. The new center solves $\eta_1+\log A=0$.

In the slow frame, $\eta_2$ stays bounded while $\eta_1$ instead goes to $+\infty$ in the past and $-\infty$ in the future. Its reduced tau functions are $e^{\eta_1}(1+Ae^{\eta_2})$ and $1+e^{\eta_2}$, respectively. We obtain

Soliton	Incoming intercept $b_j^-$	Outgoing intercept $b_j^+$
Fast, $\kappa_1$	$x_1$	$x_1-\log A/(2\kappa_1)$
Slow, $\kappa_2$	$x_2-\log A/(2\kappa_2)$	$x_2$

Precisely, for fixed distinct $\kappa_1,\kappa_2$ and fixed $x_1,x_2$,

$$u(v_jt+\xi,t)-U_j(\xi;b_j^\pm)\longrightarrow0 \qquad(t\to\pm\infty),$$

uniformly for $\xi$ in each bounded interval. The same statement holds for each fixed number of spatial derivatives: the discarded exponential ratios and their derivatives tend to zero, and the limiting denominators stay positive. These are asymptotic profile limits in separate moving frames, not an assumption that two identifiable maxima persist throughout the interaction.

Define the signed position change as outgoing minus incoming intercept. With

$$\ell=\log\frac{\kappa_1+\kappa_2}{\kappa_1-\kappa_2}\gt0,$$

the shifts are

$$\boxed{\Delta x_1=\frac\ell{\kappa_1},\qquad \Delta x_2=-\frac\ell{\kappa_2}.}$$

The fast pulse advances; the slow pulse lags. Each recovers amplitude $2\kappa_j^2$, width scale $1/\kappa_j$ and speed $4\kappa_j^2$. These asymptotic properties are the elastic-scattering statement established here. Their mass-weighted shifts cancel:

$$\kappa_1\Delta x_1+\kappa_2\Delta x_2=0.$$

For example, $\kappa_1=1$, $\kappa_2=1/2$ gives $A=1/9$, speeds $4,1$, amplitudes $2,1/2$, and shifts $\log3,-2\log3$. If both phase parameters are zero, the incoming slow trajectory has intercept $2\log3$, not zero. This is why a phase parameter cannot automatically be labeled an incoming position.

Two limits that require care

A sum of pulses is different initial data

Let $u_1,u_2$ be two independently traveling one-soliton solutions. Their sum obeys

$$(u_1+u_2)_t+6(u_1+u_2)(u_1+u_2)_x+(u_1+u_2)_{xxx} =6\partial_x(u_1u_2).$$

For finite separation the tails overlap, so this residual is not identically zero. At large separation the sum can be a useful approximation. It is not an exact collision solution.

There is also an initial-profile distinction. At a fixed time, the sum of two isolated pulses with parameters $\kappa_1,\kappa_2$ has a tau function $(1+\alpha e^{2\kappa_1x})(1+\beta e^{2\kappa_2x})$, with $\alpha,\beta\gt0$. Its mixed coefficient is the product of its two single coefficients. The exact two-soliton tau requires $A$ times that product. Equality of the fields would make the logarithms differ by an affine function of $x$; normalizing both tau functions to approach one at $-\infty$ removes that freedom. Independence of the four exponentials then forces $A=1$, impossible here. Thus even adjusting both centers cannot make that sum the exact two-soliton initial profile with these same spectral parameters. Initialize an exact-solution benchmark from the full tau function.

Equal spectral parameters do not describe this collision

The derivation assumes strict inequality. As $\kappa_1-\kappa_2\to0$, the speed difference vanishes and $\ell$ diverges. The time needed to separate the profiles cannot be bounded uniformly in that limit.

For a concrete limit, hold $x_1,x_2,x,t$ fixed and let both parameters tend to $\kappa\gt0$. Then $A\to0$ and

$$\begin{aligned} \tau&\longrightarrow1+Ce^{2\kappa(x-4\kappa^2t)},\\ C&=e^{-2\kappa x_1}+e^{-2\kappa x_2}. \end{aligned}$$

The resulting local limit is a single pulse with intercept $-(2\kappa)^{-1}\log C$. Its mass is $4\kappa$, whereas the masses before the limit approach $8\kappa$; this convergence is not convergence in $L^1(\mathbb R)$. The fixed-space-time limit and the separated-pulse long-time limits cannot be interchanged. Holding the norming coefficients fixed is a different limit because their conversion to the phase parameters contains $1/A$.

For a real decaying scalar Schrödinger potential, a bound eigenvalue is simple; see Aktosun, § VI, the bound-state discussion preceding (6.3). At the ODE level, two bound eigenfunctions at the same energy have a constant Wronskian; decay makes it zero, so they are dependent. A repeated label therefore does not supply two independent bound states. Other spectral problems can have different multiplicities, but this two-pole construction does not establish their limits.

Exercises

Recover the mass from the tau function

Compute $\int_{\mathbb R}u\,dx$ from the behavior of $\tau$ at both spatial ends, without using the normalized bound functions. Explain why this agrees with two asymptotic solitary pulses.

Solution

At $-\infty$, $\partial_x\log\tau\to0$. At $+\infty$, the mixed term dominates and $\partial_x\log\tau\to2(\kappa_1+\kappa_2)$. Therefore

$$\int_{\mathbb R}u\,dx =2[\partial_x\log\tau]_{-\infty}^{+\infty} =4(\kappa_1+\kappa_2).$$

Each solitary pulse has mass $4\kappa_j$. The sum agrees with the independently obtained determinant result. This integral is conserved even when the overlapping field cannot be separated into two individual pulse shapes.

Collapse equal exponentials before assigning spectral labels

Set $\kappa_1=\kappa_2=\kappa$ directly in the Marchenko kernel while keeping finite positive $c_1,c_2$. What kernel and field result? Which step of the two-bound-state normalization argument no longer applies?

Solution

The kernel becomes $F(s)=(c_1+c_2)e^{-\kappa s}$, a rank-one kernel. Reconstruction gives one pulse with

$$X(t)=\frac1{2\kappa}\log\frac{c_1(t)+c_2(t)}{2\kappa}.$$

Both coefficients evolve with $e^{8\kappa^3t}$, so this center has speed $4\kappa^2$. The two exponentials used to form $C$ are now identical; $C$ has rank one, and $G^{-1}$ does not tend to the zero matrix at the left end. The earlier proof of two orthonormal bound functions therefore fails at exactly the assumption that has changed. There is one bound state, not two equal-energy states.

Scope of the reconstruction

The finite-rank equation, bound-state normalization, determinant, PDE identity and asymptotic shifts agree for this explicit two-level family. These calculations establish more than a plotted collision, but they concern a specified reflectionless sector. They do not prove stability against perturbations, soliton resolution for arbitrary decaying data, or corresponding results on a circle, half-line or nonzero background. The terminology reference separates these statements from broader claims of integrability.

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], 28 May 2009. Open HTML. Cited locators use this version: §§ VI–VII for the normalization and its evolution; § VIII, equations (8.1)–(8.3), and § IX, equations (9.1)–(9.3), for finite-rank reconstruction. The source field is $q=-u$.