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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:

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.