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Use the rapidly decaying real-potential normalization $u_t-6u{u}_x+u_{xxx}=0$. Let $D={\partial }_x$, $L=-D^2+u$ and $A=4D^3-3(uD+Du)=4D^3-6uD-3u_x$. Expanding the commutator cancels its derivative terms and givesThus the KdV equation is the compatibility condition of the Lax pair $L\psi =\lambda \psi$, ${\psi }_t=-A\psi$. The operator $A$ is skew-adjoint on decaying functions. Its evolution transports normalized eigenfunctions unitarily, so the spectrum of $L$ is fixed; alternatively differentiating $L\psi =\lambda \psi$ and using the two equations gives ${\lambda }_t\psi =0$.

At each time solve the KdV Schrodinger spectral problem. For $\lambda =k^2>0$, use the left Jost solution $f_{-}$, normalized as $e^{-ikx}$ at $x\to -\infty$, and write $f_{-}\sim a(k,t)e^{-ikx}+b(k,t)e^{ikx}$ at $+\infty$. The right reflection coefficient is $R=b/a$, and transmission is $1/a$. Because $-A$ tends to $-4D^3$ at infinity, maintaining this Jost normalization gives $f_{-,t}=-A{f}_{-}+4i{k}^3{f}_{-}$. Substitution of its two asymptotic waves yields $a_t=0$ and $b_t=8i{k}^{3}b$. The negative discrete eigenvalues are $-{\kappa }_j^2$, fixed in time. A unit-normalized bound state with right tail $c_j(t)e^{-{\kappa }_{j}x}$ obeys $c_j^{\prime }=4{\kappa }_j^3{c}_j$. Consequently the Time evolution of KdV scattering data isThese signs follow from the specified right-reflection convention; reversing which infinity defines reflection reverses the continuous-data phase convention.

The inverse scattering transform now solves the nonlinear evolution in three steps. Compute $R(k,0)$, the discrete eigenvalues and the norming constants from $u(x,0)$; evolve this KdV scattering data by the simple formulas above; reconstruct $u$ by the Gelfand-Levitan-Marchenko equation. Explicitly setand solve, for $y\ge x$,The recovered potential is $\boxed{u(x,t)=-2dK(x,x,t)/dx}$. Each fixed discrete eigenvalue produces a soliton component, whereas the continuous scattering data encode the dispersive part. The inverse transform, rather than a linear superposition of the potentials, accounts for their nonlinear interaction.

The Korteweg-De Vries equation is the compatibility condition $L_t=[P,L]$ for the Lax pairFor rapidly decaying initial data, first solve the direct KdV Schrodinger spectral problemComparing its Jost solutions at the two spatial infinities produces the KdV scattering data: the reflection coefficient $R(k)$ on the continuous spectrum, discrete eigenvalues $-{\kappa }_n^2$, and norming amplitudes $c_n$ for the bound states.

The Lax evolution is isospectral. More precisely, the Time evolution of KdV scattering data isThis is the simple linear evolution that makes the nonlinear initial-value problem tractable.

To invert the data, formSolve the Gelfand-Levitan-Marchenko equationand reconstructDirect scattering at $t=0$, linear evolution of the data, and this inverse step constitute the inverse scattering transform.

The spectral equation $(S)$ isThis is the KdV Schrodinger spectral problem. Differentiating it with respect to $t$ givesFora direct differentiation, followed by substitution of the two preceding identities, yieldsThereforeComparison with $(\ast )$ givesThis is the reusable KdV Schrodinger spectral problem Wronskian identity.

Now take $u$ to satisfy the Korteweg-De Vries equation and $\varphi ={\phi }_n$, where ${\lambda }_n=-{\kappa }_n^2$. Both ${\phi }_n$ and its derivative decay at infinity, as does $u$, so integration of $(\ast )$ over the real line givesThe normalization was used in the last equality. Hencewhich is the Isospectrality of the KdV discrete spectrum.

With the KdV equation and ${\dot{\lambda }}_n=0$, the Wronskian identity saysDecay at infinity makes the constant zero. Thus $(Q/{\phi }_n{)}_x=0$ between zeros, and continuation across the isolated zeros givesMultiplying by ${\phi }_n$ and integrating,The first integral is zero by differentiating the normalization. Integration by parts turns the last integral into $\int u_x{\phi }_n^{2}dx$, soTo evaluate this, differentiate $(S)$ in $x$ and use it to writeIts integral is $-2\int {\phi }_{n,x}{\phi }_{n,xx}dx=0$ by decay. Hence

As $x\to +\infty$, rapid decay of $u$ and $u_x$, together withreduces $Q=0$ toSince ${\kappa }_n$ is constant,as stated by the Evolution of a KdV discrete norming constant.