tailieunhanh - Handbook of mathematics for engineers and scienteists part 105

Handbook of mathematics for engineers and scienteists part 105. Tài liệu toán học quốc tế để phục vụ cho các bạn tham khảo, tài liệu bằng tiếng anh rất hữu ích cho mọi người. | 696 Nonlinear Partial Differential Equations Here for simplicity the formulas are written out for the case of a second-order differential operator. For higher-order operators the right-hand sides of relations will contain higher-order derivatives of . The functionals and functions 1 X . X QT x . pn x together are assumed to be linearly independent and the Aj C are linearly independent functions of C1 . Cn. The basis functions are determined by solving the usually overdetermined system of ordinary differential equations j z 1 1 Q . n n v n Pj 1F1 Pj 2P2 Pj nPn j 1 . k where pj i are some constants independent of the parameters C1 . Cn. If for some collection of the constants pij system is solvable in practice it suffices to find a particular solution then the functions pi pi x define a linear subspace invariant under the nonlinear differential operator . In this case the functions appearing on the right-hand side of are given by fi C1 . Cn P1 iA1 C1 . Cn P2 iA2 C1 . Cn PkA C1 . Cn Bi C1 . Cn . Remark. The analysis of nonlinear differential operators is useful to begin with looking for twodimensional invariant subspaces of the form Z2 1 q x . Proposition 1. Let a nonlinear differential operator F w admit a two-dimensional invariant subspace of the form C 2 1 p x where p x Pp1 x qp2 x P and q are arbitrary constants and the functions 1 p1 x P2 x are linearly independent. Then the operator F w also admits a three-dimensional invariant subspace 2 2 1 p1 x p2 x . Proposition 2. Let two nonlinear differential operators F1 w and F2W admit one and the same invariant subspace 2 n 1 x . pn x . Then the nonlinear operator PF1 w qF2 w where p and q are arbitrary constants also admits the same invariant subspace. Example 3. Consider the nonlinear differential operator . We look for its invariant subspaces of the form iZ2 1 q x . We have F C1 C2Q x F. . kQ2 C2 a A kC2 bC c bC2 2kC C2 q. Here l i X qX 2 kQ2 and E X a Xx. .