二维三温热传导方程组的分数步隐式差分格式

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1、S27LS1?N?Vol. 27 No. 1 2004?1HACTA MATHEMATICAE APPLICATAE SINICAJan., 2004?wy?v?z?r?u?(Z?O?F266071)?o?cg?l?Hk?iNoM?A?o?o?hkf?o?Nl?H1iM?rY?x?l?Hk?i?oM?o?Dc?rY?1?K?GKJ?(ICF)MbfLUw?Y?p?UVd?G?Zd?G?d?Gj?Mq?n?a?j?h1,?YA?gT?M?k?Gj?hM?M?MjV?r?wP2,3X?j?B?bfV?X?a?MHT?n?n?nL?H46,?bf?k?Gj?hMnL?n?r?n?j?n?MCb?qX

2、?LJk?MZ?H1hLz?qX?nL?je?Q?bftY?e?ftY?B?R?Xx?R?q?tYML?w?A?Ip?m?bnj?hM?UtYCveTe t=1 div(K(,Te) gradTe) + ei(Ti Te) + er(Tr Te),(1.1)CviTi t=1 div(K(,Ti) gradTi) ei(Ti Te),(1.2)CvrTr t=1 div(K(,Tr) gradTr) er(Tr Te),(1.3)?e T(x,y,0) = T0(x,y), = e,i,r,(1.4)?YTe,Ti,TrAiS?L?n?Vdk?Zdk?dk?ei?ern?VdCZd?VdC?d

3、M?g?L?e?XMt?p2001?6I4?EK?2003?5I14?EK?uz? ?e?P?A(19871043?)?e?P?A(Q98A07115)c?y?28?M?27Lej?De = xy0,T, xy=?(x,y) | 0 x 1; 0 y 1?,TeP?L?nBOpR?Y?M?T?T= 0, = e,i,r,(x,y,t) xy 0,T.(1.5)n?0 C Cv C,0 K K(,T) K,(1.6) ?K T(,T)? D, = e,i,r.(1.7)?C,C,K,K,DMeP?L?tYM?j?M?LK?BEM?2?qt?mh =1 N, xi= ih, yj= jh; t =

4、T L, tn= nt, Wnij= W(xi,yj,tn).?e,i,rYM?Th?TM?nDQ?kn,i+12,j=?K(,Tnh,i+1,j) + K(,Tn h,ij)?2,Kn,i+12,j=?K(,Tn,i+1,j) + K(,Tn ,ij)?2,?kn,i,j+12,Kn,i,j+12YQX?x(knxTn+1 h)ij= h2?kn,i+12,j(Tn+1 h,i+1,j Tn+1h,ij) kn ,i12,j(Tn+1 h,ij Tn+1h,i1,j)?,(2.1)y(knyTn+1 h)ij= h2?kn,i,j+12(Tn+1h,i,j+1 Tn+1 h,ij) kn ,

5、i,j12(Tn+1h,ij Tn+1h,i,j1)?,(2.2)h (knhTn+1 h)ij= x(knxTn+1 h)ij+ y(knyTn+1 h)ij.(2.3)Vd?Gj?(1.1)MnL?n?eCn+1ve,ijTn+12 eh,ij Tneh,ij t=1 x(kn exTn+12 eh)ij+1 y(kn eyTn eh)ij+ n+1ei,ij(Tn ih Tn eh)ij+ n+1 er,ij(Tn rh Tn eh)ij,1 i N 1,(2.4a)Cn+1ve,ijTn+1eh,ij Tn+12 eh,ij t=1 y(kney(Tn+1 eh Tneh)ij,1 j

6、 N 1. (2.4b)Zd?Gj?(1.2)MnL?n?eCn+1vi,ijTn+12 ih,ij Tnih,ij t=1 x(kn ixTn+12 ih)ij+1 y(kn iyTn ih)ij n+1 ei,ij(Tn ih Tn eh)ij,1 i N 1,(2.5a)Cn+1vi,ijTn+1ih,ij Tn+12 ih,ij t=1 y(kn iy(Tn+1 ih Tnih)ij,1 j N 1. (2.5b)1?J?Q?cg?l?Hk?iNoM?o?29?d?Gj?(1.3)MnL?n?eCn+1vr,ijTn+12 rh,ij Tnrh,ij t=1 x(kn rxTn+12

7、rh)ij+1 y(kn ryTn rh)ij n+1 er,ij(Tn rh Tn eh)ij,1 i N 1,(2.6a)Cn+1vr,ijTn+1rh,ij Tn+12 rh,ij t=1 y(kn ry(Tn+1 rh Tnrh)ij,1 j N 1. (2.6b)?De T0eh= T0 e(x,y),T0ih= T0 i(x,y),T0rh= T0 r(x,y).(2.7)nL?n?M?R?A?ETneh,ij,Tn ih,ij,Tn rh,ij?S?F?(2.4a), (2.5a)?(2.6a)?xj?bve?R?U?Tn+1 2 eh,ij,Tn+12 ih,ij,Tn+12

8、rh,ij?,J?(2.4b), (2.5b)?(2.6b)?yj?bve?R?Tn+1eh,ij,Tn+1 ih,ij,Tn+1 rh,ij.?PX?(1.6)?QS?n?(2.4)2.6M?d?3?a?Lv = vij, w = wij,X?Z?L2?hLn?e?v,w? =N1?i,j=1vijwijh2,?v?h= ?v,v?1 2.(3.1)?a xv,xw? =N1?i=0N1?j=1ai+1 2,jxvijxwijh2,?a yv,yw? =N1?i=1N1?j=0ai,j+1 2yvijywijh2,?ahv, hw? = ?a xv,xw? + ?a yv,yw?,(3.2)

9、?hv, hw? =N1?i=0N1?j=1xvijxwijh2+N1?i=1N1?j=0yvijywijh2.(3.3)X?Z?H1?hL?hLn?e|v|21= ?hv, hv?,?v?1=?v?2h+ |v|2 1?12.(3.4)X?k?UhL?hLn?e?v?0,=max 0i,jN|vij|,|v|1,=max 0i,jN|hvij|.?v?1,= max?v? 0, |v|1,?.?Lu(x,y,t)X?hL?u?m,c= max 0msup (x,y,t)?u x1y2?,(3.5)30?M?27L? = 1+ 2, 1 0, 2 0.Em = 0?u?0,ce?u?c.?R?

10、n?1?a?Lvij, wi,jK?T?el?L?x(ax)v, w? = ?axv, xw?,?y(ay)v, w? = ?ayv, yw?,?h (ahv), w? = ?ahv, hw?.K?MVrY?M(,)?CU?YMg?MP?L?Mj(j =1,2,), n?Ct, hv?MP?L?MP?L?F?GVd?Gj?(2.4a)?(2.4b)?Tn+12 ehQLP?M?nj?Cn+1ve,ijdtTn eh,ij1 h (knehTn+1 eh)ij+(t)2 2x?knex?Bn+1e(y(kneydtTn e)?ij=n+1ei,ij(Tn ih Tn eh)ij+ n+1 er,

11、ij(Tn rh Tn eh)ij,(3.6)?YdtTneh,ij=1 t(Tn+1 eh,ij Tneh,ij), Bn+1e= (Cn+1ve)1.?= T Th, = e,i,r.K(1.1)?Y?t = tn+1,C(3.6)?QLz?j?Cn+1ve,ijdtn e,ij1 h (knehn+1 e)ij+(t)2 2x?knex?Bn+1e(y(kneydtn e)?) ij=Cn+1ve,ij? dtTne,ijTe,ij t? t=tn+1? +1 h?(Kne kne)hTn+1 e?ij+1 ? (KneTn+1 e)ij h (KnehTn+1 e)ij?+1 (Kn+

12、1 e Kne)Tn+1 e)ij+?n+1ei,ij(dtTn i,ij dtTn e,ij)t + n+1 er,ij(dtTn r,ij dtTn r,ij)t?+?n+1ei,ij(n i ne) + n+1 er,ij(n r ne)?+(t)2 2x?knex?Bn+1e(y(kneydtTn e)?ijGn1,ij+ Gn 2,ij+ Gn 3,ij+ Gn 4,ij+ Gn 5,ij+ Gn 6,ij+ Gn 7,ij.(3.7)?R|Gn1,ij| M? ?Cve?c,?2Tet2? c) t,(3.8)|Gn3,ij| M?K? 3,c,?Te?4,c? h2,(3.9)|

13、Gn4,ij| M? ?K?1,c,?Te t? 1,c+ ?Te?2,c? t,(3.10)|Gn5,ij| M? ?wei?c,?wer?c,?Tet? c,?Tit? c,?Trt? c? t,(3.11)|Gn6,ij| M?w ei?c,?wer?c?|ne,ij| + |n i,ij| + |n r,ij|?.(3.12)(3.7)?e?Cn+1e ne= dtnetnZ?L2?m?Rb?n?1?J?Q?cg?l?Hk?iNoM?o?31?(3.8)(3.12)B?QL?Cn+1vedtne,dtn e?t +1 ?kn ehn+1 e, h(n+1e ne)?+(t)3 2?x?knex?Bn+1e(y(kneydtn e)?,d tne?Gn2,dtn e?t + ?Gn 7,dtn e?t + M 1?|ne| + |n i| + |n r| + t + h2,dtn e?t.(3.13)?(1.6)?QS ?Cn+1vedtne,dtn e?t C?dtn e?2 ht.(3.14)?Cauchy?P?QL1 ?kn ehn+1 e,h(n+1e ne)? 1 2?kn ehn+

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