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Solve $\dot{Q}=-AQ$, $Q(0)=I$. Since $A^T=-A$, differentiation gives $Q^{T}Q=I$. To see why the Lax equation preserves the spectrum, direct differentiation givesThus $L(t)=Q(t)L(0)Q(t{)}^T$ is an orthogonal similarity, so all eigenvalues and all coefficients of its characteristic polynomial are first integrals.

A $2n$-dimensional Hamiltonian system is Liouville integrable when it has $n$ functionally independent first integrals, including the Hamiltonian, whose pairwise Poisson brackets vanish, on an open dense regular region.

For the three-particle periodic Toda lattice, take $p_i={\dot{q}}_i$. The proposed Hamiltonian haswhich are exactly the equations of motion. Differentiating the new variables givesThe transformation is not canonical. For examplerather than canonical coordinate brackets. Moreover $a_1{a}_2{a}_3=1/8$, so the map forgets the common translation of all the $q_i$ and is not an invertible coordinate change on the six-dimensional phase space.

Multiplication of the displayed matrices gives diagonal entries $2(a_i^2-a_{i-1}^2)$ and off-diagonal entries $(b_{i+1}-b_i)a_i$ in $LA-AL$. Therefore they form the required Lax pair. The three characteristic coefficients areThey are independent first integrals, as permitted without proof. Equivalently one may takeIndeed $C_1=-P/2$, $C_2=P^2/8-H/4$ and $C_3=1/4-J/8$. Their mutual Poisson commutation also follows directly: $P$ generates the common translation, and $\{H,J\}=0$ because $J$ is conserved.

Differentiatingwith respect to time givesIf $p_t=p^3{p}_{xxx}$, thenThus $L_t=[L,A]$ is an Isospectral Lax equation, and $(L,A)$ is a Lax pair for the Harry Dym equation.

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

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.

Since $L^T=L$ and $B^T=-B$,so the commutator is symmetric. If $Lv=\lambda v$ with $\Vert v\Vert =1$, symmetry permits a differentiable orthonormal eigenbasis locally, andThus a Lax pair evolution is isospectral.

For the harmonic oscillator,With the stated $L$, takeDirect calculation givesThe eigenvalues of $L$ areTheir constancy proves conservation of the oscillator energy $H={\lambda }_{+}^2/2$.

Differentiate the spatial equation with respect to $t$ and the temporal equation with respect to $x$. Substitution ofinto ${\psi }_{xt}-ik{\psi }_t$ cancels every term involving $\psi$, $u$, and $u_x$, leavingThe left side is $u_t$ by differentiating ${\psi }_x-ik\psi =u$, so consistency is exactly $u_t=i{u}_{xx}$. This is a scalar Lax pair formulation.

Taking the half-line Fourier transform and integrating twice by parts givesMultiplication by $e^{i{k}^{2}t}$ and integration in time yields

Replace $k$ by $-k$ and subtract. The unknown Neumann boundary value cancels:For $x>0$, Fourier inversion of the zero extension givesInverting the preceding identity therefore givesThusThis elimination is the global relation for the half-line free Schrodinger equation.

The stated operators form a Lax pair when the KdV equation is equivalent toIndeed, if $L\varphi =k^2\varphi$, differentiation and this identity showThus ${\partial }_t+A$ preserves every spectral eigenspace, which is the Isospectral Lax equation formulation.

Let $D_{\mu }={\partial }_{\mu }+A_{\mu }$. A Lax pair with spectral parameter $\lambda$ isThe coefficients of $[L,M]=0$ at orders $1,\lambda ,{\lambda }^2$ are respectivelywhich are precisely the anti-self-dual Yang-Mills equations.

The periodic Hamiltonian chain $H=\frac{1}{2}\sum p_i^2+\sum e^{q_i-q_{i+1}}$, with cyclic indices, is a Liouville integrable system. Exponentials of neighbouring coordinate differences lead to a Lax pair and conserved characteristic coefficients.

A complex parameter labels auxiliary solutions of a linear differential equation, such as exponential adjoint solutions used in a global relation for a linear boundary value problem. It need not be a discrete eigenvalue or a parameter of a nonlinear Lax pair. Analytic dependence on it allows transformed boundary identities to be sampled or inverted.

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