PhD thesis in English

where n = 1, 2, 3, . . . is an integer, and5. BEC excitation by modulation of scattering lengthm 11 = 4, m 12 = P/u 3 ρ0 u2 z0 , m 21 = 2 P/u 3 ρ0 u2 z0 , m 22 = λ 2 z + 3/u4 z0 + 2 P/u2 ρ0 u3 z0 .The n-th order functions f ρn (t) and f zn (t) depend only on the solutions u ρi (t) andu zi (t) of lower orders, i < n. For n = 1 we haveand for n = 2 we get correspondinglycos Ωt cos Ωtf ρ1 (t) = −u 3 ρ0 u , f z1 (t) = − ,z0u 2 ρ0 u2 z0f ρ2 (t) =f z2 (t) =3cos Ωt uu 4 ρ1 (t) − 6 uρ0u z0 u 5 ρ1 (t) 2 −6P uρ0 u 5 ρ1 (t) 2ρ0u z0+ 1 cos Ωt uu 3 ρ0u 2 z1 (t) −3P uz0u 4 ρ0u 2 ρ1 (t)u z1 (t) −P uz0 u 3 ρ0u 3 z1 (t) 2 ,z02cos Ωt u ρ1 (t) −3P u ρ1 (t) 2 −3P u z1 (t) 2u 3 ρ0 u2 z0u 4 ρ0 u2 z0+ 2 cos Ωt uu 2 z1 (t) − 6 uρ0 u3 z0u 5 z1 (t) 2 −z0u 2 ρ0 u4 z04Pu 3 ρ0 u3 z0u ρ1 (t)u z1 (t).The linear transformationu ρn (t) = x n (t) + y n (t), (5.27)u zn (t) = c 1 x n (t) + c 2 y n (t), (5.28)with the coefficientsc 1,2 = m 22 − m 11 ± √ (m 22 − m 11 ) 2 + 4m 12 m 212m 12,decouples the system at the n-th level and leads to the equations of the form:ẍ n (t) + ω 2 Q0 x n(t) + c 2 f ρn (t) − f zn (t)c 2 − c 1= 0, (5.29)ÿ n (t) + ω 2 B0 y n(t) + c 1 f ρn (t) − f zn (t)c 1 − c 2= 0. (5.30)Now it is clear how to proceed: we first solve Eqs. (5.29) and (5.30) for x 1 (t) andy 1 (t), and then using Eqs. (5.27) and (5.28) we obtain u ρ1 (t) and u z1 (t). In the128