Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 33C (Integrable Systems)

Using the rules for the matrix transpose together with $L^T=L$ and $B^T=-B$,Thus the commutator $[B,L]$ is a symmetric matrix.

Adopt the boundary convention $a_0=a_n=0$. Since $L$ has nonzero entries only on its first off-diagonals and $B$ only on its second off-diagonals, $BL$ and $LB$ can contribute to $[B,L]$ only on the first and third off-diagonals. On the upper first off-diagonal,soThe lower first off-diagonal is equal to this by part (a). On the third off-diagonal the two products cancel:The diagonal and every remaining entry vanish. Therefore, ifthenThis is the commutator calculation in the Zero-diagonal Jacobi Lax pair for the finite Volterra lattice.

The $(j,j+1)$ entry of $dotL$ is $dot{a}_j$. Comparing it with the result of part (i) givesFor $q_j=a_j^2$, this becomes $dot{q}_j=2q_j(q_{j+1}-q_{j-1})$, the Finite Volterra lattice.

Underthe chain rule gives $dot{a}_j=(a_j/2){\dot{u}}_j$. Part (ii) and $a_j>0$ therefore implyConsequentlyThe conventions $u_0=u_n=-\infty$ give $e^{u_0}=e^{u_n}=0$, exactly matching $a_0=a_n=0$.

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 $(†)$.