Notes On the Lax1 equation

   Cf. [Drazin 1989], p. 121.

   Let L(t) be a bounded linear operator that acts on a Hilbert space 
   and depends on a parameter t.  Define the adjoint A of any bounded
   operator A as usual by

          (u, Av)  =  (A u, v)

   for all u and v in .

   If L(t) is unitarily equivalent to L(0), i.e.,

          there exists a U(t) such that U(t) L(t) U(t)  =  L(0)

          U(t) U(t)  =  I

   then there also exists a skewsymmetric M which may depend on t

          M(t)  =  - M(t)

   such that

          dU(t)/dt  =  M(t) U(t)

   Clearly, in this case, [M(t), U(t)] = 0.

   Now, if L is selfadjoint, then

          dL(t)/dt + [L(t), M(t)]  =  0.

   If L(t) and M(t) are nxn complex matrices, and L(t) is Hermitean,

          dL(t)/dt + [L(t), M(t)]  =  0.

   then the eigenvalues of L(t) are independent of t.

   This "derives" a Lax equation; now, in reverse to solve it:

          dU(t)/dt  =  M U(t)

   and M is NOT dependent on t, this equation integrates immediately to

          U(t)  =  exp( M t )

   If instead M = M(t) is a continuous function of t, and M(t) is a
   commutative family of operators, [M(t), M(t')] = 0, then

          dU(t)/dt  =  M(t) U(t)

   integrates to

             U(t)  =  exp(  M(s) ds )

   in the sense of a Riemann integral.  The sum of skew-symmetric
   matrices is skew-symmetric, and so U is indeed unitary, and the
   Riemann integral is easily and well defined.


   Checking this solution by computing the first derivative according
   to definition:

   dU(t)/dt  =

                     t+Δ(t)               t
   lim (1/Δ(t)) [exp(   M(s) ds ) - exp(  M(s) ds )]
                      0                   0

                        t+Δ(t)                  t
   =  lim (1/Δ(t)) [exp(  M(s) ds ) - 1 ] exp(  M(s) ds )
                         t                      0

   Using the exp power series expansion about zero

   =  lim (1/Δ(t)) X

        t+Δ(t)                               t
   [ 1 +   M(s) ds + O( Δ²(t) ) - 1 ] exp(  M(s) ds )
         t                                   0

   =  lim (1/Δ(t)) X

       t+Δ(t)                          t
      [  M(s) ds + O( Δ²(t) ) ] exp(  M(s) ds )
        t                              0


       dU(t)/dt  =  M(t) U(t)

   When M(t) is skewsymmetric => U(t) is unitary, and also then

       dU(t)/dt  =  + U(t) M(t)

       dU(t)/dt  =  - U(t) M(t)

       (d/dt)² U(t)  =  d( M(t) U(t) )/dt

                   =  d( M(t) )/dt U(t) + M(t) d( U(t) )/dt

                   =  d( M(t) )/dt U(t) + M²(t) U(t)

                   =  ( dM(t)/dt + M²(t) ) U(t)

       (d/dt)3 U(t)  =

	d( dM(t)/dt + M²(t) )/dt U(t) +  ( dM(t)/dt + M²(t) ) dU(t)/dt

        =  d²M(t)/dt² U(t)  + 2 M(t) dM(t)/dt U(t)
         + dM(t)/dt dU(t)/dt +  M²(t) dU(t)/dt

       = d²M(t)/dt² U(t)  + 2 M(t) dM(t)/dt U(t)
         + dM(t)/dt M(t) U(t) +  M3(t) U(t)

       = ( d²M(t)/dt² + 2 M(t) dM(t)/dt + dM(t)/dt M(t) +  M3(t) ) U(t)


             U(t) L(t) U(t)  =  L(0)

   L(0) being independent of t, then taking a derivative

   dU(t)/dt L(t) U(t) + U(t) dL(t)/dt U(t) + U(t) L(t) dU(t)/dt  =  0

   - U(t) M(t) L(t) U(t) + U(t) dL(t)/dt U(t) + U(t) L(t) M(t) U(t)  =  0

   multiplying on the left by U(t)

        - M(t) L(t) U(t) + dL(t)/dt U(t) + L(t) M(t) U(t)  =  0

   multiplying on the right by U(t)

        - M(t) L(t) + dL(t)/dt + L(t) M(t)  =  0

        L(t) M(t) - M(t) L(t) + dL(t)/dt  =  0

             [L(t), M(t)] + dL(t)/dt  =  0

             dL(t)/dt  =  [M(t), L(t)]



   But, M(t) being a commutative family implies that M(t) commutes with
   with all its derivatives w.r.t. t, if they exist.  If M(t) is polynomial,

             M(t)  =  Σ mk tk

   with matrix coefficients mk, then [mk, mj] = 0, for all k, j.
   The requirement then of M(t) being a commutative family is a fairly
   strong requirement.  Suppose M(t) is not a commutative family, is there
   anything that can be done to express a solution to the Lax equation?

   If one approaches anew the idea of the Riemann integral, there is a
   method possibly suitable for approximation in the case of continuous t,
   and also possibly useful as a definition in an analogous case of a
   discrete parameter t.  The method of "Product Integrals" was apparently
   first introduced in [Volterra 1938].  Cf. [Dollard 1977].

   Where the Riemann integral is defined as a limit of successively
   approximating Riemann sums, the product integral is defined as a
   limit of approximating products.  Start as in the Riemannian case
   by partitioning the interval [ta, tb] with a
   sequence of tk ∈ [ta, tb], with

     ta = t0 < t1 < ... < tk-1 < tk < tk+1 < ... < tn-1 = < tb

   making n cuts and n-1 subintervals contained in [ta, tb],

             Δtk  :=  tk - tk-1

   If the covering of [ta, tb] can be taken with n large enough, and
   the Δtk small enough so that M(tk) can reasonably be
   taken as constant within Δtk, then we can write an approximating

	U(tk+1)  =  exp( M(tk+1) Δtk+1 ) U(tk) 

   Collecting this recursiveness of expression,

	U(t)  =  Π exp( M(tk) Δtk ) U(0) 

   where, of course, the ordering of the exp factors is important, because
   it is generally true that [M(tk), M(tj)] ≠ 0.  This is the product
   analog of the Riemannian sum.

   Convergence of the Riemann Product

   Let m{t}  :=  sup Δtk

   With M(t) a continuous function of t, the limit will actually exist
   in the limit as the implied mesh goes to zero.  [Volterra 1938]
   Since the limit is actually independent of the initial U(0), we can
   write then, for an arbitrary such interval [y, x] of the domain of M(t),

	Π exp( M(t) dt )  :=

	     lim     Π exp( M(tk) Δtk )
            m{t}→0  k=0

   Properties of the Product Integral

   In [Dollard 1977], eleven properties of the product intergal
   are listed and proved.  Restating these properties:

    1. For M(t) continuous and and [y, x] in the domain of M(t)

             x                          x
	d/dx Π exp( M(t) dt )  =  M(x) Π exp( M(t) dt )
             y                          y


        x                     z                x
	Π exp( M(t) dt )  =  Π exp( M(t) dt ) Π exp( M(t) dt )
        y                     y                z

      for z ∈ [y, x]


	Π exp( M(t) dt )  is non singular.

        Use det exp(A) = exp( Tr A ) to prove.

    4.  Relying on 3 and defining,

	Π exp( M(t) dt )  :=  I,  we have

        x                       y
	Π exp( M(t) dt )  =  [ Π exp( M(t) dt ) ]-1
        y                       x

    5. If the operator family {M(t)} is commutative, then

        x                         x
	Π exp( M(t) dt )  =  exp(  M(t) dt )
        y                         y


          x                            x
	‖ Π exp( M(t) dt ) ‖  ≤  ‖ exp(  M(t) dt ) ‖
          y                            y

       for norms on the Hilbert space.

          x                              x
	‖ Π exp( M(t) dt ) - I ‖  =  exp(  ‖ M(t) ‖ dt ) - 1
          y                              y

    8.  A "sum rule" for A(t) and B(t) both continuous on their common
        domain, and [y, x] as before within the domain, define

	P(x)  :=  Π exp( A(t) dt )


        x                                   x
	Π exp( (A(t) + B(t)) dt )  =  P(x) Π exp( P-1(t) B(t) P(t) dt )
        y                                   y

   9a.  For a constant (independent of t) operator T with inverse,

                    x                      x   
	       T-1  Π exp( A(t) dt ) T  =  Π exp( T-1 A(t) T ) dt
                    y                      y

   9b.  The "similarity rule".  For M as usual, and T with continuous
        first derivative, T'(t)

	       T-1(x)  Π exp( A(t) dt ) T(y)

	  =  Π exp( (T-1(t) A(t) T(t) - T-1(t) T'(t)) dt )

   10, 11  deal with the definitions of improper product integrals when
   x → ∞; all is well there as well.  Define,

	∞                          y
	Π exp( A(t) dt )  :=  lim Π exp( A(t) dt )
	y                     y→∞  x

   provided the limit exists.  The integral can be singular.  E.g., take
   A = -I for all t, then from 5., the limit of the product integral
   vanishes.  Interest is then in the conditions for which the limit exists
   and is nonsingular.

   10. For nxn A(t) defined on [y, ∞), continuous A(t) ∈ L1(a, ∞), i.e.,

	 ‖ A(t) ‖ dt  <  ∞,

   The limit of the product integral exists and is nonsingular.
   Also, if the A and B are both continuous and in L1(a, ∞), then
   the product integral exists for A + B exists and is nonsingular.

   11.  If the improper integral

	          ∞               x
	h(y)  :=   A(t)  =  lim   A(t) dt
	          y          x→∞  y

   exists (without A having to be in L1), then if (h A) is in L1),
   the product integral of A exists and is nonsigular.


   Let T be a normal operator acting on a finite dimensional complex
   Hilbert space.  T is then diagonalizable and possesses an orthonormal
   eigenbasis |tμ> that spans the Hilbert space, and there exists a cyclic
   operator CT, easily constructible:

		CT |tμ>  =  |tμ-1>

   with μ understood as discrete and (mod n), n being the dimension of the
   Hilbert space.  Both Hermitean and unitary operators are normal, and
   the spectrum of T is in general complex.  T can always be diagonalized
   by a unitary operator, and it also has a generator that is not
   necessarily Hermitean, but is necessarily normal, i.e., exp maps normal
   operators to normal operators.  Also, then,

		CTk  |tμ>  =  |tμ-k>

		CT†k  |tμ>  =  |tμ+k>

   Now define as an analytic continuation from integral k,

	M(s)  := CTs M CT†s

   for arbitrary complex s.  CT has a generator K so that we can

	CT  =  exp( +K )

	CT  =  exp( -K )

   so K is skewhermitean; then also express M(s) by

	M(s)  :=  exp( +K s ) M exp( -K s )

   For such M(s),  [M(s), M(s')]  =  0.

   If s is real, M(s) is skewhermitean if M is; for general s, it is not
   skewhermitean even if M is.  Taking a derivative w.r.t. s,

	dM(s)/ds  =  exp( +K s ) [K, M] exp( -K s )

	dM(s)/ds  =  [K, M(s)]

   Suggesting that such a construction is repeatable if K is now actually
   a function of s, and that the general Lax equation can be decomposed
   into a collection of Lax equations that are of a type simpler, so as to
   be solvable, and that this can be organized by successive polynomial


   Obviously for arbitrary A

             A    S-1 A S

   leaves the spectrum of A invariant for any nonsingular S, therefore
   trace and determinant are invariant.
   If S = U is unitary, then U-1 = U
             A    U A U

   leaves the spectrum invariant, but not so for nonsingular and nonunitary S
   where the mapping is

             A    S A S

   We have

        Tr(S A S)  =  Tr(S S A)  X=  Tr( A )

   Since Det( S )  =  Det*( S )

        Det( S A S )  = Det( S ) Det( A ) Det( S )

                       =  |Det( S )|² Det( A )

   If S is special ( Det( S ) = 1 ), so then  Det( S ) = 1,

        Det( S A S )  =  Det( A )


   1. See also The Toda Equation



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Created: 1997
Last Updated: May 28, 2000
Last Updated: July 6, 2004