Lax pair

A Lax pair encodes a nonlinear equation as compatibility of two linear equations depending on an auxiliary spectral parameter.

For the modified Korteweg-De Vries equation with linear part $q_t-q_{xxx}$, the Volterra integral equation kernel is $e^{i[k(x-x^{\prime })-4k^3(t-t^{\prime })]{\hat{\sigma }}_3}$. In column one its off-diagonal modulus is $e^{2\mathrm{Im}k(x-x^{\prime })-8\mathrm{Im}{k}^3(t-t^{\prime })}$; column two has the reciprocal modulus. Define $D_1$ by signs $(+,+)$, $D_2$ by $(+,-)$, $D_3$ by $(-,+)$ and $D_4$ by $(-,-)$. The normalization points $(0,T)$, $(0,0)$ and $(\infty ,t)$ give column domains $(D_4,D_1)$, $(D_3,D_2)$ and $({\mathbb{C}}_{+},{\mathbb{C}}_{-})$, respectively. Finite-path eigenfunctions can be entire at fixed coordinates while failing to be bounded outside these spectral sectors.

Let $A$ have diagonal entries $\pm ({\lambda }^2+w+z/2)$ and off-diagonal entries $u\lambda +u^{\prime }$ and $-2w\lambda /u+2u^{\prime }w/u^2$. Let $B$ have diagonal entries $\pm \lambda /2$ and off-diagonal entries $u/2,-w/u$. Substitution into the isomonodromic deformation compatibility identity gives the displayed equations on domains with $u\ne 0$. Thus $(w/u^2{)}^{\prime }=0$, so $w=C{u}^2$ and $u^{\prime \prime }+C{u}^3+zu/2=0$, a rescaled zero-parameter Painlevé II equation when $C\ne 0$. The leading coefficient $-2w/u$ is required by cancellation of the quadratic spectral terms.

For the normalization points $(0,T)$, $(0,0)$ and spatial infinity, the Volterra integral equation kernel is $e^{-i[k(x-x^{\prime })+2k^2(t-t^{\prime })]{\hat{\sigma }}_3}$. Its lower entry in column one has factor $e^{2i[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$; column two has the opposite factor. With quadrants numbered counterclockwise, the column domains are $(D_2,D_3)$, $(D_1,D_4)$ and $({\mathbb{C}}_{-},{\mathbb{C}}_{+})$, respectively.

For $D_i={\partial }_i+A_i$ and Higgs multiplication $\Phi$, setWrite $a=F_{12}-D_3\Phi$, $b=F_{31}-D_2\Phi$, $c=F_{23}-D_1\Phi$, where $D_i\Phi ={\partial }_i\Phi +[A_i,\Phi ]$. Direct expansion gives $[L_0,L_1]=-b+ic-2i\lambda a-{\lambda }^2(b+ic)$. Its vanishing for every complex $\lambda$ is therefore equivalent to $a=b=c=0$, the Bogomolny equations. It is a parameter-dependent zero-curvature representation.

For symmetric $L$ and antisymmetric $A$ satisfying $\dot{L}=LA-AL$, solve $\dot{Q}=-AQ$, $Q(0)=I$. Then $Q$ is orthogonal and $Q^{T}LQ$ is constant. Hence every eigenvalue and every coefficient of the characteristic polynomial of $L$ is conserved.

For periodic $U(x,t,\lambda )$ with period $L$, solve $W_x=UW$, $W(0)=I$. The monodromy matrix is $w=W(L)$. If $U,V$ satisfy the zero-curvature condition and are periodic, then $W_t=VW-WV(0)$, hence $w_t=[V(0),w]$. The matrix trace of each power of $w$ is a first integral.

A spectral parameter is an auxiliary complex parameter in a linear system whose compatibility encodes a nonlinear integrable equation. Its analytic dependence organizes conserved quantities and scattering data.

For covariant derivatives on Euclidean four-space, one anti-self-dual Yang-Mills Lax pair isThe condition $[L,M]=0$ for every $\zeta \in {\mathbb{CP}}^1$ is equivalent to anti-self-duality of the curvature.

For metric $2(dwd\bar{w}+dzd\bar{z})$, the commutator is $[L,M]=F_{wz}+\lambda (F_{w\bar{w}}+F_{z\bar{z}})+{\lambda }^2{F}_{\bar{w}\bar{z}}$. Vanishing for every spectral parameter $\lambda$ is exactly the Anti-self-dual Yang-Mills equations. The derivative parts span a totally null two-plane in the complexified tangent space. Thus an invertible fundamental solution of $L\Psi =M\Psi =0$ exists locally exactly when the three gauge curvature coefficients vanish; compatibility of one possibly zero vector alone is not enough.

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The Airy equation is the linear dispersive partial differential equation $q_t+q_{xxx}=0$. Under the spatial Fourier transform, each mode evolves by multiplication by $e^{i{k}^{3}t}$.

For positive time its fundamental solution is $(3t{)}^{-1/3}\mathrm{Ai}(x/(3t{)}^{1/3})$, connecting this evolution equation to the Airy function without identifying the two concepts.

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.

For dispersion $w(k)=-i(k^3+k)$, the other roots of $w(\nu )=w(k)$ satisfy ${\nu }^2+k\nu +k^2+1=0$. In $D_{+}=\{\mathrm{Im}k>0,\mathrm{Re}w(k)<0\}$ both roots lie in the lower half-plane. Interpolating the unknown linear polynomial $G_2+i\nu G_1$ at them eliminates two boundary traces. The resulting symmetric expression has removable singularities where the roots coalesce.

For $Q_T+Q_{XXX}+a{Q}_{XX}+b{Q}_X+cQ=0$, an exponential gauge removes the second derivative. With ${\kappa }^2=a^2/3-b>0$ and $\sigma =-c+ab/3-2a^3/27$, the remaining equation is $q_t+q_{xxx}-q_x=0$. Negative $a$ improves spatial decay of the transformed initial data; positive $a$ can destroy the convergence of its Half-range Fourier transform.

For the Airy equation with opposite dispersive sign, $q_t-q_{xxx}=0$ on $x>0$, let $Q$ be the initial Half-range Fourier transform and $F_j(k,t)={\int }_0^t{e}^{i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$. Three integrations by parts give the displayed global relation. The spatial transform is holomorphic for $\mathrm{Im}k<0$, whereas the finite-time spectral boundary transforms are entire functions. The symmetry $k\mapsto \alpha k$, ${\alpha }^3=1$, preserves their time exponent.

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For $q_t-q_{xxx}=0$ on $x>0$, evaluate the global relation at $k=-ir/2$, where $\mathrm{Re}p>0$ and the principal cube root $r=p^{1/3}$ has positive real part. The time exponent becomes $-p$, giving the displayed Laplace transform identity. Thus the second-derivative trace is determined once the value and first-derivative traces are prescribed. The homogeneous transformed equation $u^{\prime \prime \prime }-pu=0$ has exactly two spatially decaying roots, so two independent admissible conditions are required. One condition leaves a free decaying mode; three independent conditions generally violate the global constraint.

For $L=-d^2/d{x}^2+x$ on $(0,1)$, with homogeneous conditions $v(0)=0$, $v^{\prime }(1)=0$, integration by parts gives $⟨v,Lv⟩={\int }_0^1(v^{\prime 2}+x{v}^2)dx>0$ for nonzero $v$. The affine energy $J[u]=\frac{1}{2}{\int }_0^1(u^{\prime 2}+x{u}^2)dx$, minimized over $H^1$ functions with $u(0)=a$, yields $Lu=0$ and the natural boundary condition $u^{\prime }(1)=0$. The inequality $\Vert v{\Vert }_2^2\le \frac{1}{2}\Vert v^{\prime }{\Vert }_2^2$ on $v(0)=0$ proves that the form is a coercive bilinear form; Lax-Milgram theorem supplies a unique critical point, and $J[u+v]-J[u]=\frac{1}{2}\int (v^{\prime 2}+x{v}^2)$ proves that it is the unique minimum.

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 $e^{i(k^3-k)t}$.

For a time Laplace transform parameter with $\mathrm{Re}p>0$, the spatial characteristic roots satisfy $r^3+r+p=0$. Exactly one root $r_{-}$ has negative real part and two have positive real part. Indeed an imaginary root would force $p=-i(k-k^3)$ to be imaginary. The count is therefore constant on the connected right half-plane; for real $p>0$, the cubic has one negative real root and a conjugate pair with positive real parts. Repeated roots would satisfy $3r^2+1=0$, which again puts $p$ on the imaginary axis. Thus $r_{-}(p)$ is a nonzero holomorphic function in the right half-plane. Only $e^{r_{-}x}$ decays at positive spatial infinity.

For $\mathrm{Re}p>0$, let $r_{-}$ and $r_1,r_2$ be the characteristic roots of $r^3+r+p=0$, split by the signs of their real parts. Put $d_j=3r_j^2+1$. The real-line Green function of $p+{\partial }_x+{\partial }_x^3$ isThe reciprocal-polynomial root identities imply continuity of $R_p,R_p^{\prime }$ and jump $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. The distributional derivative therefore gives $(p+{\partial }_s+{\partial }_s^3)R_p={\delta }_0$. The transport term changes the roots: the pure Airy resolvent kernel with $r^3+p=0$ cannot be substituted unchanged.

For the linear dispersive Stokes equation on $x>0$, with sufficiently smooth decaying initial data and $\mathrm{Re}p>0$, define $V(x,p)={\int }_0^{\infty }{R}_p(x-y)u_0(y)dy$ and let $G(p)$ be the time Laplace transform of the prescribed derivative $u_x(0,t)$. Spatial decay and the one-dimensional decaying homogeneous space giveThe Neumann boundary condition follows by differentiating at zero. The initial forcing follows from the Stokes resolvent kernel with advection. The Bromwich inversion formula gives an integral representation with no unknown boundary traces. Truncating or extending the prescribed boundary signal beyond a target time does not affect earlier values, by the Laplace transform time-shift rule and causality of this Green function.

For $\mathrm{Re}p>0$, put $\lambda =p^{1/3}$ using the principal cube root and $r_{1,2}=\lambda e^{\pm i\pi /3}$. The decaying Green function for ${\partial }_s^3+p$ on the real line isThe identities among the three characteristic roots show that $R_p$ and $R_p^{\prime }$ are continuous, while $R_p^{\prime \prime }$ has jump one. Its distributional derivative therefore satisfies $({\partial }_s^3+p)R_p={\delta }_0$. This is the kernel of the resolvent operator $(p+{\partial }_s^3{)}^{-1}$, obtained by the time Laplace transform of the Airy equation.

For $u_t+u_{xxx}=0$ on $x>0$, define $V(x,p)={\int }_0^{\infty }{R}_p(x-y)u_0(y)dy$ and let $G$ be the time Laplace transform of $u_x(0,t)$. The transformed solution with spatial decay isOnly one characteristic root has negative real part, so one scalar boundary derivative fixes the remaining homogeneous mode. The Bromwich inversion formula supplies an integral representation involving only the initial and boundary data. For general linear differential equations on a half-line, the number and form of the necessary boundary data depend on the decaying spatial roots; Fokas and Wang study the corresponding boundary maps for linear dispersive equations.

For complex fields $q,r$, the standard two-by-two AKNS pair has zero-curvature equationsThe reductions $q=\overline{r}$ and $q=-\overline{r}$ give the focusing and defocusing cubic nonlinear Schrödinger equations, respectively.

For $\dot{L}=[L,A]$, the operator ${\partial }_t+A$ maps each eigenspace of $L$ into itself, so the spectrum is time-independent.

The Lax equation $L_t=[L,A]$ for these operators is equivalent toEliminating $p$ gives the Boussinesq equation

Ifthen every positive power has constant trace:The final equality is the cyclic property of the matrix trace.

With boundary values $q_0=q_n=0$, the finite Volterra lattice isup to rescaling or reversal of time. Writing $q_j=a_j^2$ gives ${\dot{a}}_j=a_j(a_{j+1}^2-a_{j-1}^2)$.

Let $L$ be the symmetric zero-diagonal Jacobi matrix with $L_{j,j+1}=L_{j+1,j}=a_j$, and let the skew-symmetric matrix $B$ haveThen $\dot{L}=[B,L]$ is equivalent to the Finite Volterra lattice. The corresponding similarity flow makes the eigenvalues of $L$, the coefficients of its characteristic polynomial, and its trace invariants first integrals.

For real periodic KdV data, translation by one period acts on a conjugate basis of scattering solutions byLax compatibility gives $\dot{T}=[\Lambda ,T]$ for the matrix of ${\partial }_t+A$ on the eigenspace. Hence $\mathrm{tr}T=2\mathrm{Re}a$ is conserved.

For ${\Phi }_x+U\Phi =0$ and ${\Phi }_y+V\Phi =0$, compatibility is $V_x-U_y+[U,V]=0$.

For the auxiliary equations ${\psi }_x=U\psi$ and ${\psi }_t=V\psi$, equality of mixed derivatives gives $U_t-V_x+[U,V]=0$. This is the compatibility condition for a Lax pair.

For $u_t=i{u}_{xx}$ on $x>0$, the half-line Fourier transform satisfiesSubtracting the relation at $-k$ cancels the unknown Neumann datum. Fourier inversion of the resulting odd combination gives the solution in terms of the initial transform and the prescribed Dirichlet datum, with boundary kernel $G(k,t)=kh(t)$.