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Let $y$ solve the Airy ordinary differential equation $y^{\prime \prime }(s)=sy(s)$. For $t>0$, set $s=x/(3t{)}^{1/3}$ and define the displayed similarity solution. Differentiating the profile equation gives $y^{\prime \prime \prime }=y+s{y}^{\prime }$, so the chain rule yieldsTheir sum is zero, proving the Airy equation $u_t+u_{xxx}=0$. Every real linear combination of the two standard Airy functions gives such a solution. This is a solution of the linear dispersive equation; the nonlinear Korteweg-De Vries equation includes an extra term and does not follow from the same arbitrary profile construction.

The scalar linear partial differential equation $u_t+u_x+u_{xxx}=0$ is an Airy equation with a first-order transport term. It is a linearization of the Korteweg-De Vries equation with an advective term retained. It is distinct from the viscous incompressible-flow Stokes equation. On the whole line its spatial Fourier transform evolves by $e^{i(k^3-k)t}$.

Direct differentiation givesConsequently this transformation takes solutions of the modified Korteweg-De Vries equation to solutions of the Korteweg-De Vries equation in these sign conventions. Both summands in $u$ have the same scaling weight, so the modified-equation scaling induces $(x,t,u)\mapsto (e^{s}x,e^{3s}t,e^{-2s}u)$.

A field $u(x,t)$ has infinitely many degrees of freedom: its initial state is a function rather than a finite list of coordinates. For a Hamiltonian evolution, the analogue of the finite-dimensional construction seeks an infinite sequence of independent conserved functionals $H_j[u]$ whose Poisson brackets vanish. Their Hamiltonian vector fields then define commuting time evolutions. For example, a Poisson operator $J$ produces the field equation $u_t=J\delta H/\delta u$, where $\delta H/\delta u$ is the variational derivative.

A Lax equation packages many conservation laws into one operator identity. If $L_t=[A,L]$, cyclicity of an appropriate trace givesFor a formal pseudodifferential operator the relevant trace is the Adler trace; suitable fractional powers of a monic differential operator produce further conserved functionals. A Lenard-Magri recursion can prove that such functionals are Poisson-commuting functions, rather than merely conserved quantities. A Lax equation by itself does not establish all the independence and completeness properties needed for integrability.

The Korteweg-De Vries equation illustrates the mechanism. In one time normalization take $L={\partial }^2+u$ and $A={\partial }^3+\frac{3}{2}u\partial +\frac{3}{4}{u}_x$. Direct composition givesIts spectral problem, together with the inverse scattering transform for decaying data or suitable spectral coordinates for periodic data, turns nonlinear evolution into simple evolution of spectral data. Solitary waves arise from discrete spectral data, while continuous data describe dispersive radiation.

Thus infinite-dimensional Hamiltonian integrability is more than an infinite list of formulas: one wants sufficiently complete commuting invariants and a reconstruction mechanism that solves the evolution. Boundary conditions and the function space are essential. A formal hierarchy need not converge, and an infinite family of conserved quantities does not automatically give a compact infinite-dimensional torus or an unrestricted analogue of the Arnold-Liouville theorem.

Insert the finite exponential expansion into the Gelfand-Levitan-Marchenko equation. The integral contribution isComparing coefficients of $e^{-c_{n}y}$ givesFor this coefficient comparison take the $c_n$ distinct; repeated exponents can first be combined. The linear system supplies the coefficients whenever its matrix is invertible.

In the inverse scattering transform convention $u_t-6u{u}_x+u_{xxx}=0$, the Korteweg-De Vries equation solution is $u=-2dK(x,x)/dx$. The reflectionless scattering data evolve as ${\beta }_n(t)={\beta }_n(0)e^{8c_n^{3}t}$, with the $c_n$ fixed. For one exponential,Differentiating its diagonal givesFor $\beta >0$, this is the smooth solitonIt travels at speed $4c^2$. If $\beta =0$ the solution is zero; if $\beta <0$ the denominator vanishes somewhere, so the formula is not a globally smooth one-soliton. Positivity is the necessary qualification for a regular soliton.

Integrating the vector-field equations gives the one-parameter transformationsUnder this scaling, $v_t,v^2{v}_x,v_{xxx}$ have weights $\gamma -\beta ,3\gamma -\alpha ,\gamma -3\alpha$. Equality of all three weights givesThus the nontrivial Lie point symmetry generator can be normalized to $x{\partial }_x+3t{\partial }_t-v{\partial }_v$.

For the Miura transformation $u=v^2+v_x$, direct differentiation gives the operator identityFor example $u_x=2v{v}_x+v_{xx}$ and $u_{xxx}=6v_x{v}_{xx}+2v{v}_{xxx}+v_{xxxx}$; inserting them makes the $v_x{v}_{xx}$ terms cancel and leaves the right side. Hence $u$ solves $u_t-6u{u}_x+u_{xxx}=0$ whenever $v$ solves the mKdV equation. Both $v^2$ and $v_x$ have weight $-2\alpha$, so the corresponding Korteweg-De Vries equation symmetry is

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

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.

Putand seek the rank-one Gelfand-Levitan-Marchenko kernel in the separable formSubstitution into the Gelfand-Levitan-Marchenko equation givessoand hence

WriteThe reconstruction formula for this inverse scattering transform convention yieldsIf $q=e^{-2s}$, then $4q/(1+q{)}^2={\mathrm{sech}}^{2}s$. Takingtherefore produces the required one-soliton of the Korteweg-De Vries equation, withHere the usual bound-state norming constant has ${\beta }_0>0$ and $\chi >0$, so $\varphi$ is real.

Put $\xi =x-vt$ and $U(\xi )=A{\mathrm{sech}}^2\xi$. For $A\ne 0$, differentiation gives $U^{\prime \prime }=4U-6U^2/A$ and therefore $U^{(3)}=4U^{\prime }-(12/A)U{U}^{\prime }$. The Korteweg-De Vries equation reduces toSince $U$ is nonconstant, both coefficients must vanish. Thus the nonzero solitary wave isThere is also the trivial solution $A=0$, for which $v$ has no effect. The negative amplitude follows from the minus sign of the nonlinear term; it agrees with the sech-squared solitary wave of the KdV equation.

For the sign convention $u_t-6u{u}_x+u_{xxx}=0$ and any $\kappa >0$, the displayed negative solitary wave travels at speed $4{\kappa }^2$. To verify it, put $\xi =x-ct$ and write $u=U(\xi )$. The equation becomes $(-cU-3U^2+U^{\prime \prime }{)}^{\prime }=0$. A decaying profile has zero integration constant, so $U^{\prime \prime }=cU+3U^2$. Direct differentiation of the displayed profile gives $U^{\prime \prime }=4{\kappa }^{2}U+3U^2$, proving the result. Its depth is $2{\kappa }^2$, so narrower and deeper pulses travel faster. Changing the sign of the nonlinear term reverses the amplitude sign. This wave belongs to the soliton family of the Korteweg-De Vries equation.