By V. J. D Baston
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Example text
114) such that D(h1 , · · · , hN r , λi )hi = 0 (i = 1, 2, · · · , N r). 79). (2) If det Fr−1 = 0, then the above Darboux matrix of degree r can be decomposed as D(h1 , · · · , hN r , λ) = D D(h1 , · · · , hN (r−1) , λN (r−1)+1 )hN (r−1)+1 , · · · , D(h1 , · · · , hN (r−1) , λN r )hN r , λ · ·D(h1 , · · · , hN (r−1) , λ). 119) 27 1+1 dimensional integrable systems On the right hand side of this equality, the first term is a Darboux matrix of degree one and the second term is a Darboux matrix of degree (r − 1).
Therefore, we can also obtain the theorem of permutability by this symmetry. 9 can be applied not only to the AKNS system, but also to many other evolution equations, especially to the Lax pairs whose U and V are polynomials of λ. On the other hand, we also show that those Darboux matrices include all the diagonalizable Darboux matrices of the form λI − S, and any non-diagonalizable Darboux matrix can be obtained from the limit of diagonalizable Darboux matrices. 138) n−j Vj (x, t)λ , j=0 Uj ’s and Vj ’s are N × N matrices.
Since r− (ζ) and r+ (ζ) are holomorphic in C+ and C− respectively, their zeros are discrete. These zeros are the eigenvalues of L. The set of all eigenvalues of L is denoted by IP σ(L). 270) has a nontrivial bounded solution. σ(L) = R ∪ IP σ(L) is called the spectrum of the operator L. Its compliment C − σ(L) is called the regular set of L. Property 4. If r− (ζ) = 0 and r+ (ζ) = 0 hold for ζ ∈ R, then IP σ(L) is a finite set. Hereafter, we always suppose r− (ζ) = 0 and r+ (ζ) = 0 when ζ ∈ R. First we consider the eigenvalues.
