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1、湘潭大学 硕士学位论文 依赖于时间的薛定锷方程的半离散两层网格方法 姓名：王建云 申请学位级别：硕士 专业：计算数学 指导教师：金继承 20080428 ? ? ? ? ? ? ?k k1? ? ? I Abstract We fi rstly introduce the relevant background of the two-grid method and the Schrodinger equation in this dissertation. Then we give a review of the results made by many researchers. In the
2、 second chapter, to a kind of the time-dependent Schrodinger equation, we use the fi nite element method to construct a two-grid algorithm in the semi-discrete form. In the third chapter, we have a strictly theo- retical analysis to the algorithm and prove that the semi-discrete solution has the sam
3、e error order with the solution solved by the standard fi nite element method under k k1-norm. Finally, we summarize the dissertation and make a prospect to the future work. Keywords: Schr odinger equations, fi nite element method,two-grid method, error estimate. II ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
4、 ? ? ? ? ?1992? ?32? ?(?)? ?(? ?)? ?2,26,31,34? ?36,37,38?20? ?6,7,15,22,24,28? 1992? ?32? ? ? ? ?35? ?GMRES(?)? ? ?33? ? 1994? ?31? H1?O(H5)? ?AAO. Ammi?M. Marion? ?Galerkin?Navier Stokes?2? ? ?(XH?Xh)?Mh ? ?Galerkin?Galerkin? ?O(H2) (?H1?L2?)? 1995?M. Marion? ?Galerkin?26? ?Galerkin?H1? ?O(H3)?DM.
5、 Bedivan?Galerkin? ?8? ? ? ? 1996? 1 ?34? ? ? ? ?(?VH? ?Vh?H = O(h),0 0, wh Sh,(2.5) uh(x,y,0) = Rhu0(x,y),(2.6) ? uh(x,y,0) = uI 0(x,y), (2.7) ?uI 0(x,y) S h ?u0(x,y)? ?(2.5)?(2.6) ?(2.5)?(2.7) ? ?SH?Sh? ?H?h? ?SH Sh H1 0() ? ?2.1 ?uH SH? i(uH)t,wH) = 1 2(uH,wH) + (V uH,wH) + (f,wH), wH SH,t 0, uH|
6、t=0= RHu0(x,y) SH. (2.8) ?uh Sh? 1 2(u h,wh) = i(uH)t,wh) (V uH,wh) (f,wh), wh Sh,t 0, uh|t=0= Rhu0(x,y) Sh. (2.9) 7 ? ?2.2 ?uH SH? i(uH)t,wH) = 1 2(uH,wH) + (V uH,wH) + (f,wH), wH SH,t 0, uH|t=0= uI 0(x,y) S H. (2.10) ?uh Sh? 1 2(u h,wh) = i(uH)t,wh) (V uH,wh) (f,wh), wh Sh,t 0, uh|t=0= uI 0(x,y) S
7、 h. (2.11) ?2.1?2.2? ?3.2?3.3?2.1? ?2.2?uh Sh?uh Sh?H1? ?2.1?2.2 ? 8 ? ?2.1?2.2? ?.?x . y?C? x Cy. ?3.1?t 0,T, w(x,y,t),wt(x,y,t) H2()? Rhw(x,y,t) Sh? k w Rhw ks. h2sk w k2,s = 0,1 ,(3.1) k (w Rhw)tks. h2s(k w k2+ k wtk2),s = 0,1 .(3.2) ?(3.1) ?19?(5.40)?(5.41)? ?(3.2)? ? at(u,v) = Z Vt(x,y,t)u vdxd
8、y,(3.3) E(x,y,t) = w(x,y,t) Rhw(x,y,t). Et(x,y,t) = E1(x,y,t) + E2(x,y,t), ? E1(x,y,t) = wt(x,y,t) Rhwt(x,y,t), E2(x,y,t) = Rhwt(x,y,t) tRhw(x,y,t). ?t 0,T? Sh?(2.4)? a(E2,)=a(Rhwt tRhw,) =a(wt tRhw,) = d dta(E,) a(E, t ) at(E,) =at(E,), ? = E2? 1 2 | E2|2 1 +V0k E2k2a(E2,E2) =at(E,E2) .k E kk E2k .
9、 9 ? k E2k2.k E kk E2k,(3.4) | E2|2 1.k E kk E2 k .(3.5) ?(3.4)? k E2k.k E k,(3.6) ?(3.5)? | E2|1.k E k .(3.7) ?(3.6)?(3.7)?(3.1)? k E2ks. h2k w k2,s = 0,1.(3.8) ?(3.1)?wt? k wt Rhwtks. h2sk wtk2,s = 0,1,(3.9) ?(3.8)?(3.9)? k Etksk E1ks+ k E2ks. h2s(k w k2+ k wtk2),s = 0,1, (3.2)? ?3.2?t 0,T?w(x,y,t),wt(x,y,t),wtt(x,y,t) H2()? ?Rhw(x,y,t) Sh? k (w Rhw)ttks. h2s(k w k2+ k wtk2+ k wttk2), s = 0,1.(3.10) ?
