数理方程课件.ppt

1、第二章第二章 分离变量法分离变量法一、有界弦的自由振动二、有限长杆上的热传导三、拉普拉斯方程的定解问题四、非齐次方程的解法五、非齐次边界条件的处理六、关于二阶常微分方程特征值问题的一些结论2022-12-31基本思想：首先求出具有变量分离形式且满足边界条件的特解，然后由叠加原理作出这些解的线性组合，最后由其余的定解条件确定叠加系数。适用范围：波动问题、热传导问题、稳定场问题等特点：a.物理上由叠加原理作保证，数学上由解的唯一性作保证；b.把偏微分方程化为常微分方程来处理，使问题简单化。22222,0,0(0,)0,(,)0,0(,0)(,0)(),(),0uuaxl ttxutu l ttu

2、xu xxxxlt 2022-12-32令(,)()()u x tX x T t带入方程：2()()()()X x Tta Xx T t2()()()()XxTtX xa T t 令2()()0()()0XxX xTta T t带入边界条件(0)()0,()()0XT tX l T t(0)0,()0XX l22222,0,0(0,)0,(,)0,0(,0)(,0)(),(),0uuaxl ttxutu l ttu xu xxxxlt 1 求两端固定的弦自由振动的规律一 有界弦的自由振动2022-12-33()()0(0)0,()0XxX xXX l特征（固有）值问题：含有待定常数常微分方程在

3、一定条 件下的求解问题特征（固有）值：使方程有非零解的常数值特征（固有）函数：和特征值相对应的非零解分情况讨论：01）()xxX xAeBe 00llABAeBe 00ABX02）()X xAxB00ABX()cossinX xAxBx0sin0ABl03）令 ，为非零实数 2(1,2,3,)nnl222(1,2,3,)nnnl222nl()sin(1,2,3,)nnnXxBxnl2022-12-342222()()0nna nTtT tl()cos sin(1,2,3,)nnnn atn atT tCDnll(,)(cossin)sin(1,2,3,)nnnn an anux tCtDtxn

4、lll11(,)(,)(cossin)sin(1,2,3,)nnnnnu x tux tn an anCtDtxnlll2()()0()()0XxX xTta T t22222,0,0(0,)0,(,)0,0(,0)(,0)(),(),0uuaxl ttxutu l ttu xu xxxxlt 222(1,2,3,)nnnl()sin(1,2,3,)nnnXxBxnl2022-12-3501(,)(,0)sin()ntnnu x tu xCxxl10(,)sin()nntu x tn anDxxtll1sin)sincos(nnnxlntlanDtlanCu2001 cos 2/sindd22

5、llnlnlx xxl001sinsindcoscosd02llnmnmnmxx xxxxllll xxlmxlnCxxlmxlnnldsinsindsin)(010 mCl2lmxxlmxlC0dsin)(2lnxxlnxanD0dsin)(2lnxxlnxlC0dsin)(22022-12-36)()(),(tTxXtxu2/lnnxlnBxXnnsin)(tlanDtlanCTnnnsincos1sin)sincos(nnnxlntlanDtlanC11nnnnnTXuulnxxlnxanD0dsin)(2lnxxlnxlC0dsin)(20 XX02 TaT分离变量求特征值和特征函数求

6、另一个函数求通解确定常数分离变量法可以求解具有齐次边界条件的齐次偏微分方程。lxxtxuxxuttlututlxxuatu0),()0,(),()0,(0,0),(,0),0(0,0,222222022-12-372 解的性质 x=x0时：(,)(cossin)sinnnnn an anux tCtDtxlll其中：22arctannnnnnnnDn aACDlC00(,)sincos()nnnnnux tAxtlcos()sinnnnnAtxlxlnsin驻波法 2nlnlt=t0时：22nnnaflnnvfnllna 22Ta 00(,)cos()sinnnnnnux tAtxl(1,2,

7、3,)n 2022-12-38例1：设有一根长为10个单位的弦，两端固定，初速为零，初位移为 ，求弦作微小横向振动时的位移。()(10)1000 xxx)()(),(tTxXtxuTXTX 410TTXX 41010 XX0104 TT0)()0(),0(tTXtu 0)10(,0)0(100,0XXxXX0)0(X0)()10(),10(tTXtu0)10(X100,0)0,(,1000)10()0,(0,0),10(),0(0,100,1022422xtxuxxxuttututxxutu解：2022-12-39 0)10(,0)0(100,0XXxXX20 02 XX1010(0)0()0

8、XABX lAeBe0 BA0)(xXxxBeAexX)(0BAxxX)(0 BA0)(xX0 X20(0)0(10)sin100XAXB,3,2,1,10/nnn100/22nnxnBxXnn10sin)(xBxAxXsincos)(02 XX2022-12-310,3,2,1,100/22nnnxnBxXnn10sin)(0104 TT010022 nnTnTtnDtnCTnnn10sin10cos1110sin)10sin10cos(nnnnnxntnDtnCuunnnTXu)10sin10cos(10sintnDtnCxnBnnnxntnDtnCnn10sin)10sin10cos(1

9、00,0)0,(,1000)10()0,(0,0),10(),0(0,100,1022422xtxuxxxuttututxxutu0 XX0104 TT2022-12-311110sin)10sin10cos(nnnxntnDtnCu1000)10(10sin)0,(1xxxnCxunn0sin)0,(1nnxlnlanDtxu0nD100d10sin1000)10(102xxnxxCn13310)12(sin)12(10cos)12(54nxntnnu100d10sin)10(50001xxnxx)cos1(5233nn为奇数，为偶数，nnn33540100,0)0,(,1000)10()0

10、,(0,0),10(),0(0,100,1022422xtxuxxxuttututxxutu2022-12-312弦的振动振幅放大100倍，红色、蓝色、绿色分别为n=1,2,3时的驻波。2022-12-313)()(),(tTxXtxu2XTa X T21XTXaT0 XX20Ta T0)()0(),0(tTXtu0,010(0)0,()0XXxXX l0)0(X(,)()()0u l tX l T tx()0X l222222,0,0(,)(0,)0,0,0(,0)(,0)2,0,0uuaxl ttxu l tuttxu xu xxlxxlt解：例2求下列定解问题2022-12-3140,0

11、(0)0,()0XXxlXX l20 02 XX(0)0()0llXABX lA eB e0 BA0)(xXxxBeAexX)(0BAxxX)(0 BA0)(xX0 X20(0)0()cos0XAX lBl(21)/2,1,2,3,nnln222(21)/4nnl(21)()sin2nnnXxBxlxBxAxXsincos)(02 XX2022-12-315222(21)/4nnl(21)()sin2nnnXxBxl20Ta T2222(21)04nnnaTTl(21)(21)cossin1,2,3,22nnnnanaTCtDtnll11(21)(21)(21)(cossin)sin222nn

12、nnnnananuuCtDtxlllnnnTXu(21)(21)(21)(cossin)sin222nnnananCtDtxlll222222,0,0(,)(0,)0,0,0(,0)(,0)2,0,0uuaxl ttxu l tuttxu xu xxlxxlt0 XX20Ta T2022-12-3161(21)(21)(21)(cossin)sin222nnnnananuCtDtxlll21(21)(,0)sin22nnnu xCxxlxl1(,0)(21)(21)sin022nnu xnanDxtll0nD202(21)(2)sind2lnnCxlxx xll2331321(21)(21)c

13、ossin(21)22nlnanutxnll 23332(21)ln 2(,0)(,0)2,0u xu xxlxt初始条件2022-12-317222222,0,0(,)(0,)0,0,0(,0)(,0)2,0,0uuaxl ttxu l tuttxu xu xxlxxlt2331321(21)(21)cossin(21)22nlnanutxnll 若l=1,a=10时的震动。2022-12-318)()(),(tTxXtxuTXTX TTXX 0 XX0 TT0)()1(),1(0)()0(),0(tTXtutTXtu0)1(,0)0(XX 0)1(,0)0(10,0XXxXX10,0)0,

14、(,sin)0,(0,0),1(),0(0,10,2222xtxuxxuttututxxutu例3 求下列定解问题解：2022-12-319 0)1(,0)0(10,0XXxXX0202 XX(0)0(1)0XABXAeBe0 BA0)(xXxxBeAexX)(0BAxxX)(0 BA0)(xX0 X02xBxAxXsincos)(0sin)1(,0)0(BXAX,3,2,1,nnn22nnxnBxXnnsin)(02 XX10,0)0,(,sin)0,(0,0),1(),0(0,10,2222xtxuxxuttututxxutu2022-12-320,3,2,1,22nnnxnBxXnnsi

15、n)(0 TT022 nnTnTtnDtnCTnnnsincos11sin)sincos(nnnnnxntnDtnCuunnnTXu)sincos(sintnDtnCxnBnnnxntnDtnCnnsin)sincos(xxnCxunnsinsin)0,(10sin)0,(1nnxnnDtxu0nD1011nnCn，xtusincos10,0)0,(,sin)0,(0,0),1(),0(0,10,2222xtxuxxuttututxxutu2022-12-32110,0)0,(,sin)0,(0,0),1(),0(0,10,2222xtxuxxuttututxxutuxtusincos2022

16、-12-322lxtxuxxuttlhuxtlututlxxuatu0,0)0,(),()0,(0,0),(),(,0),0(,0,0,22222)()(),(tTxXtxuTXaTX 2TTaXX 210 XX02 TaT0)()0(),0(tTXtu0)()(,0)0(lhXlXX 0)()(,0)0(0,0lhXlXXlxXX0)()()()()()()(),(),(tTlhXlXtTlhXtTlXtlhuxtlu例4 求下列定解问题令带入方程：解：2022-12-323 0)()(,0)0(0,0lhXlXXlxXX02xxBeAexX)(0)()(0)0(llllBhehAeeBeA

17、lhXlXBAX0 BA0)(xX02 XX0BAxxX)(0)()(hAlAlhXlX0A0)(xX0 X0)0(BX02xBxAxXsincos)(0sincos)()(,0)0(lBhlBlhXlXAXhl/tan,3,2,1,nn2nnxBxXnnnsin)(02 XX2022-12-324lxtxuxxuttlhuxtlututlxxuatu0,0)0,(),()0,(0,0),(),(,0),0(,0,0,22222,3,2,1,n2nnxBxXnnnsin)(02 TaT022 nnnTaTatDatCTnnnnnsincosnnnTXu 11sinsincosnnnnnnnnx

18、atDatCuuatDatCxBnnnnnnsincossinxatDatCnnnnnsinsincos0 XX02 TaT2022-12-325lxtxuxxuttlhuxtlututlxxuatu0,0)0,(),()0,(0,0),(),(,0),0(,0,0,222221sinsincosnnnnnnxatDatCu0sin)0,(1xaDtxunnnn0nD)(sin)0,(1xxCxunnnlmmlmxxxxxC020dsindsin)(1sincosnnnnxatCuxxxxxxClmlmnnndsin)(dsinsin001 lmmxxC02dsin2022-12-326nmn

19、mxxxnlm00dsinsin0nmnmnmnmll)sin()sin(21nmnmnmnmnmnmllllllsincoscossinsincoscossin21llllnmnnmmnmnmcossinsincos)(1mmnnnmnmnmnmlllltantancoscos1)(10 xxxlnmnmd)cos()cos(210hl/tan2022-12-327二 有限长杆上的热传导222,0,0,(,)(0,)0,(,)0,0(,0)()0uuaxl ttxu l tuthu l ttxu xxxl)()(),(tTxXtxu2XTa X T 21XTXaT0 XX20Ta T0)()

20、0(),0(tTXtu0)()(,0)0(lhXlXX0)()()()()()()(),(),(tTlhXlXtTlhXtTlXtlhuxtlu令带入方程：解：2022-12-328 0)()(,0)0(0,0lhXlXXlxXX02xxBeAexX)(0)()(0)0(llllBhehAeeBeAlhXlXBAX0 BA0)(xX02 XX0BAxxX)(0)()(hAlAlhXlX0A0)(xX0 X0)0(BX02xBxAxXsincos)(0)0()()cossin0XAX lhX lBlBhlhl/tan,3,2,1,nn2nnxBxXnnnsin)(02 XX2022-12-329

21、,3,2,1,n2nnxBxXnnnsin)(20Ta T220nnnTa T22na tnnTC ennnTXu 2211sinna tnnnnnuuC ex22sinna tnnnC B ex22sinna tnnC ex222,0,0,(,)(0,)0,(,)0,0(,0)()0uuaxl ttxu l tuthu l ttxu xxxl2022-12-33000sinsind0lmnmnxx xmn20sindlnnxxL令)(sin)0,(1xxCxunnnlmmlmxxxxxC020dsindsin)(1sincosnnnnxatCuxxxxxxClmlmnnndsin)(dsin

22、sin001 lmmxxC02dsin221sinna tnnnuC ex222,0,0,(,)(0,)0,(,)0,0(,0)()0uuaxl ttxu l tuthu l ttxu xxxlhl/tan2022-12-331lxxxuttlututlxxuatu0),()0,(0,0),(,0),0(0,0,222)()(),(tTxXtxuXTaXT 2002 TaTXX 0)(,0)0(00lXXlxXXXXTaT 20)()(),(0)()0(),0(tTlXtlutTXtu0)(,0)0(lXX令带入方程：令例5 求下列定解问题解：2022-12-332 0)(,0)0(00lXX

23、lxXX0202 XXxxBeAeX0X00 XBAxX0X 0202 XXxBxAXsincoslnnxlnBXnnsin0)0(BAX0)(llBeAelX0 BA0)0(AX0sin)(lBlX,3,2,1,22nlnnn2022-12-333lxxxuttlututlxxuatu0),()0,(0,0),(,0),0(0,0,22202TaT02222nnTlnaTtlnanneAT2222nnnTXu 11sin2222ntlnannnxlneCuuxlneBAtlnannsin22222222sina ntlnnnuC exlxlnBXnnsin,3,2,1,22nlnnn1sin

24、)()0,(nnxlnCxxuxxlnxlClndsin)(202022-12-334lxxxutxtluxtutlxxuatu0),()0,(0,0),(,0),0(0,0,222)()(),(tTxXtxuXTaXT 2XXTaT 2002 TaTXX 0)(,0)0(00lXXlxXX0)()(),(0)()0(),0(tTlXxtlutTXxtu0)(,0)0(lXX例6 求下列定解问题解：2022-12-335 0)(,0)0(00lXXlxXX0202 XXxxBeAeX0X0)0(BAXlleBeAlX)(0 BA00 XBAxX0BX 0202 XXxBxAXcossinlnn

25、xlnBXnncos0)0(AX0sin)(lBlX,3,2,1,22nlnnn2022-12-336么么么么方面 Sds绝对是假的lxxxutxtluxtutlxxuatu0),()0,(0,0),(,0),0(0,0,22200BX xlnBXnncos,3,2,1,2nlnn02TaT000T00TA002222nnTlnaTtlnanneAT2222nnnTXu xlneBAtlnanncos2222xlneCtlnancos2222000CAB100cos2222ntlnannnxlneCCuu000TXu 2022-12-338lxxxutxtluxtutlxxuatu0),()0

26、,(0,0),(,0),0(0,0,222100cos2222ntlnannnxlneCCuu10cos)()0,(nnxlnCCxxuxxlCld)(100 xxlnxlClndcos)(20()1x若 则u为多少？为什么会出现这样的现象？思考2022-12-339(),10,10 xx al若001()d2llCx xl022()cosd2(1)1()lnnnCxx xllln2222001cosnna ntlnnuuCnC exl有界杆上的热传导（杆的两端绝热）2022-12-340 xxtuau20|0 xx luu)(|0 xut)()(xXtTu0)()0(LXXXXTaT/)/(

27、2220TaT20XX22exp()T Aatsin,nlXx)()(xXtTukkkkkXTu),(txuu分离变量流程图2022-12-341三 拉普拉斯方程的定解问题axxbxuxxubyyauyubyaxyuxu0),(),(),()0,(0,0),(),0(0,0,02222XYu 0 YXYXYYXX 0 XX0 YY 0)()0(0,0aXXaxXX0)()(),(0)()0(),0(yYaXyauyYXyu0)(,0)0(aXX1 直角坐标系下的拉普拉斯问题解：2022-12-342axxbxuxxubyyauyubyaxyuxu0),(),(),()0,(0,0),(),0(

28、0,0,02222 0)()0(0,0aXXaxXX0202 XXxxBeAeX0X0)0(BAX0)(aaBeAeaX0 BA00 XBAxX0X0202 XXxBxAXcossinannxanAXnnsin0)0(BX0sin)(aAaX,3,2,1,22nannn2022-12-343axxbxuxxubyyauyubyaxyuxu0),(),(),()0,(0,0),(),0(0,0,02222xanAXnnsin,3,2,1,2nann0 YY0222 nnYanYyannyannneDeCYnnnYXu 1nnuu1sinnyannyannxaneDeCxaneDeCyannyan

29、nsinsinnnyyaannnnnuC eD eAxa2022-12-344axxbxuxxubyyauyubyaxyuxu0),(),(),()0,(0,0),(),0(0,0,022221sinnyannyannxaneDeCuxanDCxxunnn1sin)()0,(xaneDeCxbxunabnnabnn1sin)(),(xxanxaDCnndsin)(2a0 xxanxaeDeCabnnabnndsin)(2a0022()()sind1n baann banx exx xaaCe022()()sind1n baann banx exx xaaDe2022-12-34522220,0

30、,0(0,)(,)0,0(,0)(),(,)(),0uuxaybxyuyu a yybxxu xx u x bxxaXYu 0 YXYXYYXX 0 XX0 YY 0)()0(0,0aXXaxXX0)()(),(0)()0(),0(yYaXxyauyYXxyu0)(,0)0(aXX例7 求下列定解问题解：2022-12-346axxbxuxxubyxyauxyubyaxyuxu0),(),(),()0,(0,0),(),0(0,0,02222 0)()0(0,0aXXaxXX0202 XXxxBeAeX0X0)(aaeBeAaX0)(BAaXa0 BA00 XBAxX00BX0202 XXxB

31、xAXcossinannxanBXnncos0)0(AX0sin)(aBaX,3,2,1,22nannn2022-12-347axxbxuxxubyxyauxyubyaxyuxu0),(),(),()0,(0,0),(),0(0,0,02222xanBXnncos,3,2,1,22nannn000BX0 YY0000DyCY0 Y00222 nnYanYyannyannneDeCYnnnYXu 000YXu 00000C yDBC yDxaneDeCyannyanncosxanBeDeCnyannyanncosxaneDeCDyCuunyannyannnn1000cos2022-12-348a

32、xxbxuxxubyxyauxyubyaxyuxu0),(),(),()0,(0,0),(),0(0,0,02222xaneDeCDyCunyannyann100cosxanDCDxxunnn10cos)()0,(xaneDeCDbCxbxunabnnabnn100cos)(),(xxanxaDCnndcos)(2a0 xxanxaeDeCabnnabnndcos)(2a01dcos)()(22a0abnabnnexxanxexaC1dcos)()(22a0abnabnnexxanxexaDxxaDd)(1a00 xxaDbCd)(1a000 xxxabCd)(-)(1a002022-12-3

33、49axxuxxuyyauyuyaxyuxu0,0),(),()0,(0,0),(),0(0,0,02222XYu 0 YXYXYYXX 0 XX0 YY 0)()0(0,0aXXaxXX0)()(),(0)()0(),0(yYaXyauyYXyu0)(,0)0(aXX例8 求下列定解问题解：2022-12-350axxuxxuyyauyuyaxyuxu0,0),(),()0,(0,0),(),0(0,0,02222 0)()0(0,0aXXaxXX0202 XXxxBeAeX0X0)(aaBeAeaX0)0(BAX0 BA00 XBAxX0X0202 XXxBxAXcossinannxanB

34、Xnnsin0)0(BX0sin)(aAaX,3,2,1,22nannn2022-12-351axxuxxuyyauyuyaxyuxu0,0),(),()0,(0,0),(),0(0,0,02222xanBXnnsin,3,2,1,2nann0 YY0222 nnYanYyannyannneDeCYyanneD11sinnyannnnxaneDuunnnYXu xanBeDnyannsinxaneDyannsinxanDxxunn1sin)()0,(xxanxaDandsin)(202022-12-3522 圆域内的拉普拉斯问题22222yuxuu22,arctanyxxysin,cos221

35、cos,sin/1122222yxyxxyxyxyxuu2222222222222sincoscos2sinsinuuuuuyuxuxuxu2222222222222sinsinsin2sincosuuuuuxuuuuyuxu11222222222cossinuuyuyuyusincosuu22211uu2022-12-35320),(),(20,01100222fuuu),0(u)2,()0,(uu)()(),(u0112 0112 欧拉方程 21102 0)2()0()2()0(20,0)()2()()0(例9 求下列定解问题解：2022-12-35420),(),(20,01100222

36、fuuu)2()0(20,00202 BeAe000 AB00A02sincosBAnn,3,2,1,22nnnnnBnAnnnsincos02 欧拉方程 lntet令ddd1 ddd ddPPtPtt 222d1 d1dd()()ddddPPPttt 0 2022-12-35520),(),(20,01100222fuuu000A,3,2,1,2nnnnBnAnnnsincos02 00002 ln000DC 0C02 n022 nnnnnnnnDCnnC000unnnu100sincosnnnnnnnFnEEuu000ECAnnnnnnnnFnECnBnAsincossincos1000s

37、incos)(),(nnnnnFnEEfu222000000111()d,()cosd,()sind2nnnnEfEfnFfn 0 2022-12-35620,cos),(20,01100222uuu)2,()0,(uu),(u)()(),(u0112 0112 21102 0)2()0()2()0(20,0)()2()()0(例10 求下列定解问题解：2022-12-35720,cos),(20,01100222uuu)2()0(20,00202 BeAe000 AB00A02sincosBAnn,3,2,1,22nnnnnBnAnnnsincos02 欧拉方程 lntet令ddd1 ddd

38、 ddPPtPtt 222d1 d1dd()()ddddPPPttt 0 2022-12-35820,cos),(20,01100222uuu000A,3,2,1,2nnnnBnAnnnsincos000000lnCD tCD 0C02 nntntnnnnnnnC eD eCD nnC000unnnu100sincosnnnnnnnFnEEuu000ECAnnnnnnnnFnECnBnAsincossincos1000sincoscos),(nnnnnFnEEu01Ecos0u其它为零0 02 2022-12-35920,1),(,0),(20,011222buaubauu)2,()0,(uu

39、)()(),(u0112 0112 21102 0)2()0()2()0(20,0)()2()()0(例12 求下列定解问题解：2022-12-36020,1),(,0),(20,011222buaubauu)2()0(20,00202 BeAe000 AB00A02sincosBAnn,3,2,1,22nnnnnBnAnnnsincos2022-12-36120,1),(,0),(20,011222buaubauu000A,3,2,1,2nnnnBnAnnnsincos02 欧拉方程 00002 ln000DC 02 n022 nnnnnnnnDC000unnnulnln00000FEDCA

40、nnnnnnDCnBnAsincosnHGnFEnnnnnnnnsincos1000sincoslnnnnnnnnnnnnnHGnFEFEuu2022-12-36220,1),(,0),(20,011222buaubauu1000sincoslnnnnnnnnnnnnnHGnFEFEuu0sincosln),(100nnnnnnnnnnaHaGnaFaEaFEau1sincosln),(100nnnnnnnnnnbHbGnbFbEbFEbu0ln00aFE0nnnnaHaG0nnnnaFaE1ln00bFE0nnnnbHbG0nnnnbFbEabaElnln0abFln10其他为零ababau

41、lnlnlnlnabalnln2022-12-3631,0)3/,()0,(3/0,6sin),1(3/0,1,011222uuuuu),0(u)()(),(u0112 0112 21102 0 0)3/()0(0)3/()0(3/0,00)()3/()()0(例13 求下列定解问题解：2022-12-3641,0)3/,()0,(3/0,6sin),1(3/0,1,011222uuuuu 0)3/()0(3/0,00202 BeAe000 AB002sincosBAnn3,3,2,1,922nnnnnBnn3sin0)0(A03/sin)3/(B2022-12-3651,0)3/,()0,(

42、3/0,6sin),1(3/0,1,011222uuuuunBnn3sin,3,2,1,92nnn02 0922 nnnnnnnnDC33nnC3nnnu1313sinnnnnnnEuu13sin6sin),1(nnnEu2,0,12nEEn66sinunnnnnnECnB333sin3sin2022-12-366qypxtpxyxyxutqxutxutqytyputyutqypxyuxuatu0,00,0),()0,(,0),(),0,(0,0,0),(),0(0,0,0,22222)()()(),(tTyYxXtyxuTYXYTXaTXY 2012 TTaYYXX XX YY)(12TTa

43、0 XX0 YY0)(2TaT例13 求下列定解问题解：2022-12-367qypxtpxyxyxutqxutxutqytyputyutqypxyuxuatu0,00,0),()0,(,0),(),0,(0,0,0),(),0(0,0,0,222220 XX0 YY0)(2TaT0)()0(pXX0)()0(qYY0202 XX00 XBAxX0X0202 XXxxBeAeX0XxBxAXsincospnn,3,2,1,2222npnnnxpnBXnnsin0)0(AX0sin)(pBpX,3,2,1,222mqmmyqmCYmmsin2022-12-368qypxtpxyxyxutqxut

44、xutqytyputyutqypxyuxuatu0,00,0),()0,(,0),(),0,(0,0,0),(),0(0,0,0,222220)(2TaT,3,2,1,222npnnxpnBXnnsin,3,2,1,222mqmmyqmCYmmsin0)(2222222mnmnTaqmpnTtaqmpnmnmneDT2222222)(mnmnmnTYXu2222222()sinsinnma tpqnmmnnmBx Cy D epq2222222()sinsinnma tpqmnnmExyepq2222222()1111sinsinnma tpqmnmnmnmnnmuuExyepq11(,0)(

45、,)sinsinmnmnnmu x yx yExypq004(,)sinsind dqpmnnmEx yxy x ypqpq 2022-12-369lxxxuttlututlxuxuatu0),()0,(0,0),(),0(0,0,222veutvexvaevetvetttt222222xvatv0),0(),0(tvetut)0,()0,(xvxu0),0(tv0),(tlv)()0,(xxv0),(),(tlvetlut例14 求下列定解问题解法一：令2022-12-370lxxxuttlututlxuxuatu0),()0,(0,0),(),0(0,0,222XTu XTTXaTX 2X

46、XaTaT 2210 XX012 TaT0)()(),(0)()0(),0(tTlXtlutTXtu0)(,0)0(lXX 0)(,0)0(0,0lXXlxXX解法二：令2022-12-371lxxxuttlututlxuxuatu0),()0,(0,0),(),0(0,0,222 0)(,0)0(0,0lXXlxXX0202 XXxxBeAeX0X0)(aaBeAeaX0)0(BAX0 BA00 XBAxX0X0202 XXxBxAXcossinannxlnBXnnsin0)0(BX0sin)(lAlX,3,2,1,22nlnnn2022-12-372lxxxuttlututlxuxuatu

47、0),()0,(0,0),(),0(0,0,222xlnBXnnsin,3,2,1,2nlnn012222nnTlnaT012 TaTtlnanneAT22221nnnTXu 111sin2222ntlnannnxlneCuuxxlnxlClndsin)(20 xlneCtlnansin22221xlnBeAntlnansin22221)(sin)0,(1xxlnCxunn2022-12-37320(0)()0,/,1,2,sinkkXXXX lkl kXx2120(0)()0,()/,0,1,2,sinkkXXXX lkl kXx2120(0)()0,()/,0,1,2,coskkXXXX lkl kXx20(0)()0,/,0,1,2,coskkXXXX lkl kXx常用本征方程 齐次边界条件2022-12-374