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**Example text**

It has a unique solution in the class of functions which tend to zero fast enough together with its derivatives as x → −∞. 212) as 1 gj −1 = Q(ggj ) = Q2 (ggj +1 ) = · · · = Qn+m−j+1 [sin u]. 212). gj −1 is not a diﬀerential polynomial of gj . 214) where βj (t)’s are arbitrary functions of t. Generally, we have the compound MKdV-SG hierarchy n pt + (p αj (t)M Mn−j [p [ ]+ j=0 = − u2x ). 17 n = 0, m = 2, β0 = 0, β1 = 1, then, g2 = the equation becomes the sine-Gordon equation 1 2 sin u, and uxt = sin u.

An+j (j = 1, 2, · · · , m − 1) can be determined as follows. 209) 0 where a− j−1 is the limit of aj−1 as x → −∞. 199) become x→−∞ 1 (ggj −1 )x + p a− j−1 + 4 x −∞ x pggj −1 dx = −∞ gj dx. 212) gj (ζ) dζ dξ. This is an integral equation of Volterra type. It has a unique solution in the class of functions which tend to zero fast enough together with its derivatives as x → −∞. 212) as 1 gj −1 = Q(ggj ) = Q2 (ggj +1 ) = · · · = Qn+m−j+1 [sin u]. 212). gj −1 is not a diﬀerential polynomial of gj . 214) where βj (t)’s are arbitrary functions of t.

3) The following limits hold uniformly for real ζ ∈ R: ⎛ lim x→+∞ R(x, ζ) − ⎝ ⎛ lim x→+∞ R(x, ζ) − ⎝ ⎛ lim x→−∞ L(x, ζ) − ⎝ ⎞ r− (ζ) r+ (ζ)e2iζx ⎠ = 0, r− (ζ)e−2iζx r+ (ζ) −r− (ζ)e−2iζx r− (ζ) ζ ∈ R, ⎞ ⎠ = 0, ζ ∈ R, ⎞ ⎠ = 0, ζ ∈ R, 54 DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS ⎛ lim x→−∞ L(x, ζ) − ⎝ ⎞ r+ (ζ) −r+ (ζ)e2iζx ⎠ = 0, ζ ∈ R. 261) becomes a spectral problem of a linear ordinary diﬀerential operator. We consider its spectrum in L2 (R) × L2 (R). 267) implies that ψr and ψl are linearly dependent.