e?kii???(bie)2?(cie)2?????ehSiee2e2?kjj????(bj)?(cj)???3?hSie?ee2e2?(bm)?(cm)???3?kmm????eeeeee?kij?kji???bibj?cicj?? ?ke?ke??bebe?cece?immiimim???hSieeeeeee?kjm?kmj???bjbm?cjcm??6??ehSieeeT??pi?0,pj?pm?2?k?????4?(2) 先推导变分公式，再将插值函数代入
?Te?f(u)?f(u)?u令u?，则??e e?x?Ti?u?Tie1??Te???Te??Te???Te???Jb?Tee???k???dxdy??0h?T?T??eSidg ??ee?e??Ti?Tie??x?Ti??x??y?Ti??y??ee?Jb?Jb同理可求得，（后续推导略） e?Tje?Tm
2.1.2 内部单元变分计算
内部单元的泛函
k???Te???Te??eeJi[T(x,y)]??????????dxdy <3>
2???x???y??e??将三角形单元内部的插值公式以及边界上的插值公式代入泛函，泛函
eJie?T??x,y???实际上变成了普通多元函数
22eJie?Tie,Tje,Tm????f(x,y,Tie,Tje,Tme)dxdy
e?Jie?Jie?Jie求e，e，e可以采用两种方法：
?Tj?Ti?Tm?Jie?Jie?Jie(1) 先将插值公式代入<3>式，然后再求e，e，e
?Tj?Ti?Tme2e2??????k?T?Tee?? Ji[T(x,y)]?????????dxdy <3>
2?x?y?????e??? 单元内部插值函数：
1eeeeeeee? Te(x,y)?(a?b?xc)y?T(?a?bxc)?yT(?iiiijjjj2??eme? a?mebxme)cyTm?k?11eeeeeeeeee2eeeeee2?JT,T,T?bT?bT?bT?cT?cT?c?i?ijm????2?iijjmm?jjmTm??dxdy 2?ii2?4?4??e?Jiek? e?2?Ti8?eeeeeeeeeeeeee??dxdy 2bT?bT?bTb?2cT?cT?cTc????iijjmmiiijjmmi????e?k4?2?????(b)ee2ieeeeeeeeeee?(cie)2?T?bb?ccT?bb?ccT????ijijjimimmdxdy ?i??keeeeeeeeeee?(be)2?(cie)2?T?bb?ccT?bb?ccT????iijijjimimm2?i?4?????dxdy
e??dxdy??ke2e2eeeeeeeeeeee?????? ??(b)?(c)T?bb?ccT?bb?ccT??????????iiiijijjimimm??4???同理
?Jiek??Tje8?2 ? ?同理：
eeeeeeeeeeeeee??2bT?bT?bTb?2cT?cT?cTc????iijjmmjiijjmmj?dxdy ???ek4?2????bbeeeijee2e2eeeeee???cieceT?(b)?(c)T?bb?ccT???jijjmjmmdxdy ?j?j?keeee2e2eeeeee?biebe(b)?(c)T?bb?ccT ???j?cicj?Ti??jjjjmjmm??4????Jiekeeeeeeeee2e2e??biebe(b)?(c)T ?j?cicj?Ti?(bjbm?cjcm)Tj??mmme???Tm4???写成矩阵形式
??Jie??e???Ti??ke???ii??Jie?e?e???kji??Tj??ke??Je??mi?ie????Tm??其中
ekijkejjekmje??Tie?kimee??e? keT?KT????jm??j?ee???kmmTm???ee2e2?kii???(b)?(c)?ii???e2e2?ke(b)?(cjj???jj)????ee2e2?kmm???(bm)?(cm)????e?kij?ke???biebe?ciecej? jij??ke?ke??bebe?cece?imim?mi?im?ke?ke??bebe?cece?jmjm?mj?jm?k????4?从形式上说，内部单元的变分结果与边界单元基本相同；从本质上说，仅需令边界单元变分结果中的Sie?0，即可得内部单元的变分结果。
(2) 先推导变分公式，再将插值函数代入
?Te?f(u)?f(u)?u令u?，则??e e?x?Ti?u?Tie??Te???Te??Te???Te???Jb???k???dxdy ??ee?e??Tie??x?Ti??x??y?Ti??y??ee?Jb?Jb同理可求得，（后续推导略） e?Tje?Tm
2.2 二维非稳态温度场单元变分计算，无内热源
二维非稳态温度场泛函变分问题尚未很好解决，目前采用的处理方法有两种：
①先用有限元法处理空间域，后用有限差分法处理时间域 ②先用有限差分法处理时间域，后用有限元法处理空间域
我们采用前一种处理方法。
?T二维非稳态温度场泛函（先固定时间变量，即先把看作常数）：
?t22???T??k??T???T????12? DJ[T(x,y)]?????????cTdxdy?hT?hTT????ds????????t??2?D??2????x???y????2.2.1第三类边界单元变分计算
无内热源二维非稳态温度场第三类边界单元泛函
e2e2???????k?T?T?Tee????1eee2e?Jb[T(x,y)]?????????cTdxdy?h(T)?hTT?????ds ????jm2??x???y???t?2?e???????无内热源二维稳态温度场第三类边界单元泛函
e2e2??k??T???T??1ee2e?Jb[Te(x,y)]??????dxdy?h(T)?hTT???ds ????jm?22?x?y???????e????Tee两者相比较，仅有?cT项是新的，并且我们在整个变分运算（实际上是普
?t?Te通函数积分运算）过程中，把作为常数处理。
?te??Te???Te??Jb?Te???Te??Te?Te???k??k?y?Te??y???c?t?Te?dxdye??Tie???x?T?x??i?i?i?e?
?Te???hT?hT??eSidg 0?Ti1e?
eeeTe?NieTie?NeT?NjjmTm
eee?Tee?Tie?Tje?Tm ?Ni?Nj?Nm?
?t?t?t?t?Te e?Nie
?Ti因此
ee?e2?Tie??Te?Teee?Tjee?Tm?cdxdy??c???(Ni)?NiNj?NiNm?dxdy e???t?T?t?t?tiee????Tie??c???te?Tm(N)dxdy?NNdxdy??????te?tee2ieiej?Tje?eeNNdxdy? im??e?可以证明
eee(N)dxdy?(N)dxdy?(N??i??j??m)dxdy?eee222? 6? 12eeeeee??NiNjdxdy???NiNmdxdy???NjNmdxdy?eeeee??Te?Te?c???Tie?Tj?Tmdxdy???? ???c?2?dxdy e???t?T12?t?t?tiee??得到：
e1??Te???Te??Jb?Te???Te??Te?Te??Tee??k??k?y?Te??y???c?t?Te?dxdy??0?hT?hT???TeSidg e??Tie???x?T?x??i?i?i?ie??keeeeeeee?(bie)2?(cie)2?Tie??biebej?cicj?Tj??bibm?cicm?Tm ??4???e??Tie?Tje?Tm??2????dxdy ??12e??t?t?t??c?同理可得：
e?Jbkeeeee?khSie?e?keehSie?e e2e2ee??bibj?cicj?Ti??4???(bj)?(cj)???3?Tj??4??bjbm?cjcm??6?Tm?Tje4?????e??Tie??Tje?TmhSie
??2?T???dxdy?12???t?t?t2e???c?e?kee?JbkeehSie?e?khSie?eeeeeee2e2?bibm?cicm?Ti??(bjbm?cjcm)?(bm)?(cm)??Tm??Tj????e???Tm4?6?3??4??4?ee??Tie?Tj??c??ThSmi ??????2?dxdy?T?
e12e??t?t?t?2写成矩阵形式
e???Jb?e???Ti??keiie????Jb??e?e???kji??Tj??ke??Je??mi?be???Tm???ekijkejjekmjee??Tie??niikim??e??ekejm??Tj???njiee?e???kmmTnmmi????enijnejjenmj??Tie???t?enim??e??pie????T???nejm??j???pej??tee???e??nmmpm???T???m??t???
??K??T?eeee??T???N?????P?
??t?ee注意：此处的?N?并非插值函数中的形状函数
其中