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**Additional resources for 1+1 Dimensional Integrable Systems**

**Sample text**

206) x Write an+m as a function of u, then the above equation becomes ((an+m,uu + an+m )ux )x = 0. 207) The boundary condition as x → −∞ gives an+m,uu + an+m = 0. 208) 43 1+1 dimensional integrable systems where α and β are constants. Now take a special an+m : a0n+m = 14 cos 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.

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. 18 n = 1, m = 2, α0 = −4, α1 = 0, β0 = 0, β1 = 1, then the equation becomes the equation describing one-dimensional nonlinear lattice of atoms [70] 3 uxt + u2x uxx + uxxxx − sin u = 0. 217) ⎛ ⎞ α Now we consider the Darboux transformation.

199) gives an+m,x ux x + an+m ux = 0. 206) x Write an+m as a function of u, then the above equation becomes ((an+m,uu + an+m )ux )x = 0. 207) The boundary condition as x → −∞ gives an+m,uu + an+m = 0. 208) 43 1+1 dimensional integrable systems where α and β are constants. Now take a special an+m : a0n+m = 14 cos 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.

### 1+1 Dimensional Integrable Systems

by William

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