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Lax's equation in the convention used here isLet $\mathcal{L}\psi =\lambda \psi$ with $(\psi ,\psi )=1$. Since $\mathcal{L}$ is self-adjoint and $\mathcal{A}$ is anti-self-adjoint, differentiating $\lambda =(\psi ,\mathcal{L}\psi )$ givesThe first and third terms equal $\lambda ,d(\psi ,\psi )/dt=0$. For the middle term,Thus $\dot{\lambda }=0$, proving the Isospectral Lax equation property: every eigenvalue is independent of $t$.

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 $(†)$.

When $u$ is independent of $y$ and $z=x+iy$,PutThe zero-curvature equation becomesSincewe obtain the Isospectral Lax equationTherefore, for every positive integer $n$, the cyclic property of the matrix trace givesThese are the trace invariants of a Lax equation.

The PDE from part (b) reduces toWith $\varphi =2u$, this isTo extract a nontrivial invariant, use $n=2$. Writing $p=u_x$, direct multiplication givesThe terms involving $\lambda$ are constant, and $p={\varphi }^{\prime }/2$. Henceis a first integral. Direct differentiation verifies it:

The Lax equation $L_t=[A,L]$ withreduces, after expanding the commutator, to the KdV equationThus the evolution is isospectral: the eigenvalues of $L$ are constant. The fourth derivative shown in the converted question is a transcription error.

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 $H={\lambda }_{+}^2/2$.

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.

If $Lf=\lambda f$ and $f_t+Af=0$, differentiation and the Lax equation give $(L-\lambda )f_t=-A(L-\lambda )f$, so ${\lambda }_t=0$ after projection onto the normalized eigenfunction for a nondegenerate eigenvalue.

For $\Psi =(f,f_x{)}^T$,The compatibility ${\Psi }_{xt}={\Psi }_{tx}$ is precisely ${\partial }_t{M}_L-{\partial }_x{M}_A-[M_A,M_L]=0$. If all coefficients are $x$-independent, this becomes ${\dot{M}}_L=[M_A,M_L]$, and cyclicity of trace gives $d\mathrm{Tr}(M_L^k)/dt=0$.