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第14讲哈密顿算子(2)张元中中国石油大学(北京)地球物理与信息工程学院z《矢量分析与场论》主要内容z4.算子运算教材:第3章z《矢量分析与场论》z4.算子运算∇z在应用这些公式的时候,设法将其中的常矢都移到的前面,而将变矢都保留在的后面。∇)()()(BACACBCBAvvvvvvrvvו=ו=וz算子的运算中,经常用到三个矢量的混合积公式,及二重矢量积公式,∇)()()(BACCABCBAvvvvvvrvv•−•=××例4:证明(14))()()()()(BAABBAABBAvvvvvvvvvv•∇+•∇−∇•−∇•=××∇)()()(ccBABABAvvvvvv××∇+××∇=××∇证:根据算子的微分性质,应用乘积的微分法则,则有,∇)()()(BACCABCBAvvvvvvrvv•−•=××)()()()()(BAABBAABBAvvvvvvvvvv•∇+•∇−∇•−∇•=××∇)()()(∇•−•∇=××∇cccABBABAvvvvvvBABAvvvv)()(∇•−•∇=)()()(ABABBAcccvvvvvv•∇−∇•=××∇)()(ABABvvvv•∇−∇•=z4.算子运算解:3=•∇rvyzxkzjyixusin3)(vvv∂∂+∂∂+∂∂=∇例5:已知,,求。yzxusin3=ruv•∇kzjyixrvvvv++=rururuvvv•∇+•∇=•∇)coscos(sin3kyzxyjyzxziyzvvv++=rururuvvv•∇+•∇=•∇rkyzxyjyzxziyzyzxvvvv•+++=)coscos(sin3sin9yzxyzyzxcos6sin12+=z4.算子运算Av×∇例6:设,求点处。)1,2,1(MkyzjyzxixzAvvvv42322+−=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−−−=34222382022430yzzyxzxxyzxzzADvkxyzjxziyxzAvvvv)04()03()]2(2[224−−+−+−−=×∇解:因为,故由的雅可比矩阵ArotAvv=×∇Avkxyzjxziyxzvvv43)22(224−++=kjiAMvvvv836−+=×∇z4.算子运算rv例7:验证,其中为常矢,为位置矢量。av∫∫∫•=•×SlSdaldravvvvv2)(证:斯托克公式为,,raAvvv×=∫∫∫•×∇=•SlSdAldAvvvv)(取∫∫∫∫•××∇=•=•×SllSdraldAldravvvvvvvv))(()()()()()()(BAABBAABBAvvvvvvvvvv•∇+•∇−∇•−∇•=××∇)()()()()(raarraarravvvvvvvvvv•∇+•∇−∇•−∇•=××∇arayaxazzyxvv30)(0+−∂∂+∂∂+∂∂−=akajaiazyxvvvv3)(+++−=aavw3+−=av2=∫∫∫•=•×SlSdaldravvvvv2)(z4.算子运算例8:验证格林第一公式与格林第二公式,证:奥氏公式为,,vuA∇=v∫∫∫∫∫ΩΔ+∇•∇=•∇dVvuuvSdvuS)()(v取∫∫∫∫∫ΩΔ−Δ=•∇−∇dVuvvuSduvvuS)()(vdVASdAS∫∫∫∫∫Ω•∇=•vvvAuAuAuvvv•∇+•∇=•∇dVvuSdvuS∫∫∫∫∫Ω∇•∇=•∇)()(vdVvuvu∫∫∫Ω∇•∇+∇•∇=)(dVvuvu∫∫∫ΩΔ+∇•∇=)(z4.算子运算证:同理可得,∫∫∫∫∫ΩΔ+∇•∇=•∇dVvuuvSdvuS)()(v两式相减,可得,∫∫∫∫∫ΩΔ−Δ=•∇−∇dVuvvuSduvvuS)()(vdVuvuvSduvS∫∫∫∫∫ΩΔ+∇•∇=•∇)(vdVvuvuSdvuS∫∫∫∫∫ΩΔ+∇•∇=•∇)(v∫∫∫∫∫ΩΔ−Δ=•∇−∇dVuvvuSduvvuS)()(v例8:验证格林第一公式与格林第二公式,z4.算子运算由于∫∫∫∫∫ΩΔ+∇•∇=•∇dVvuuvSdvuS)()(v∫∫∫∫∫ΩΔ−Δ=•∇−∇dVuvvuSduvvuS)()(vdVvuvuSdvuS∫∫∫∫∫ΩΔ+∇•∇=•∇)(v当时,可得,vu=dVuuuSduuS∫∫∫∫∫ΩΔ+∇=•∇))((2v通用的形式,∫∫∫∫∫Ω∇+∇•∇=•∇dVvuuvdSnvuS)()(2v,∫∫∫∫∫Ω∇−∇=•∇−∇dVuvvudSnuvvuS)()(22vdVuuudSnuuS∫∫∫∫∫Ω∇+∇=•∇))((22v例8:验证格林第一公式与格林第二公式,z4.算子运算;)(aarvvv=•∇(1)例:设为常矢,,,求证(习题7第6题)bavv,rrv=kzjyixrvvvv++=(2));(1)(arrarvvv•=•∇(3));(1)(arrarvvv×=×∇(4);])[(babarvvvvv×=•×∇(5)].)()[(2)(2araraaravvvvvvvv•−•=×∇z4.算子运算证明:(1)例:设为常矢,,,求证(习题7第6题)bavv,rrv=kzjyixrvvvv++=)(arvv•∇(2)))((arkzjyixvvvvv•∂∂+∂∂+∂∂=))((zyxzayaxakzjyix++∂∂+∂∂+∂∂=vvv.av=)(arv•∇).(1arrvv•=ararvv•∇+•∇=)(arrvv•=(3))(arv×∇).(1arrvv×=ararvv×∇+×∇=)(arrvv×=(4)])[(barvvv•×∇barbarvvvvvvו∇+×∇•=)]([)(bavv×=z4.算子运算证明:(5)例:设为常矢,,,求证(习题7第6题)bavv,rrv=kzjyixrvvvv++=)(2ravv×∇))(())(()()(cbdadbcadcbavvvvvvvvvvvv••−••=ו×)]()[(raravvvvו×∇=)(2ravv×∇)])(())([(rararraavvvvvvvv••−••∇=22)()()(raraavvvv•∇−∇•=)()(2)(2rararaarvvvvvv•∇•−∇•=ararraarvvvvvv)(2)(2•−•=])()[(2araraavvvvvv•−•=z4.算子运算求证满足以下方程组(习题7第7题)AuBvv×∇+∇=例:已知函数和无源场分别满足,证明:因代入,Avu),,,(zyxFu=Δ),,(zyxGAvv−=Δ),,,(zyxFB=•∇v),,(zyxGBvv=×∇AuBvv×∇+∇=有)(AuBvv×∇+∇•∇=•∇)(Auv×∇•∇+∇•∇=)(Auv×∇•∇+Δ=,0)(=×∇•∇Av),,,(zyxFu=Δ)(AuBvv×∇•∇+Δ=•∇0),,(+=zyxF),,(zyxF=z4.算子运算求证满足以下方程组(习题7第7题)AuBvv×∇+∇=例:已知函数和无源场分别满足,证明:因代入,Avu),,,(zyxFu=Δ),,(zyxGAvv−=Δ),,,(zyxFB=•∇v),,(zyxGBvv=×∇AuBvv×∇+∇=有)(AuBvv×∇+∇×∇=×∇)(Auv×∇×∇+∇×∇=,0)(=∇×∇u),,,(zyxGAvv−=Δ),,(zyxGBvv=×∇AAuvvΔ−•∇∇+∇×∇=)(0=•∇Avz4.算子运算证明:(1)格林第一公式为,(1)∫∫=∂∂SdSnf0例:设为区域的边界曲面,为的向外单位法矢,与均为中的调和函数,证明SΩnvSfgΩ(2)∫∫∫∫∫Ω∇=∂∂SdVfdSnff2(习题7第8题)(3)∫∫∫∫∂∂=∂∂SSdSnfgdSngf∫∫∫∫∫ΩΔ+∇•∇=•∇dVvuuvSdvuS)()(vz4.算子运算证明:(1)∫∫∫∫∫ΩΔ+∇•∇=•∇dVvuuvSdvuS)()(v∫∫∫∫∫ΩΔ+∇•∇=•∇dVffSdfS)11()1(v令,可得,fvu==,1为调和函数,故,故有,f0=Δf0)(=•∇∫∫SSdfv∫∫•∇=SdSnfv)(∫∫∂∂=SdSnf0=∂∂∫∫SdSnf为中调和函数的充要条件是,fΩz4.算子运算证明:(2)∫∫∫∫∫ΩΔ+∇•∇=•∇dVvuuvSdvuS)()(v∫∫∫∫∫ΩΔ+∇•∇=•∇dVffffSdffS)()(v令,可得,fvfu==,为调和函数,故,故有,f0=Δf∫∫∫∫∫Ω∇=•∇dVfSdffS2)(v∫∫∫∫•∇=•∇SSdSnffSdffvv)()(∫∫∂∂=SdSnff∫∫∫∫∫Ω∇=∂∂SdVfdSnff2z4.算子运算证明:(3)令,可得,gvfu==,为调和函数,故,故有,gf,00,=Δ=Δgf∫∫∫∫∫ΩΔ−Δ=•∇−∇dVuvvuSduvvuS)()(v∫∫∫∫∫ΩΔ−Δ=•∇−∇dVfggfSdfggfS)()(v0)(=•∇−∇∫∫SSdfggfv∫∫∫∫•∇=•∇SSSdfgSdgfvv∫∫∫∫•∇=•∇SSdSnfgdSngfvv∫∫∫∫∂∂=∂∂SSdSnfgdSngfz4.算子运算例:证明(14))()()()()(BAABBAABBAvvvvvvvvvv•∇+•∇−∇•−∇•=××∇)()()(BABABABAvvvvvvvv××∇+××∇=××∇)()()(BACCABCBAvvvvvvrvv•−•=××)()()()()(BAABBAABBAvvvvvvvvvv•∇+•∇−∇•−∇•=××∇BAABBAAAAvvvvvvvvv)()()(•∇−∇•=××∇证:算子首先是算子,对分别起作用,BAvv,∇算子又是普通矢量,与构成双重叉积,BAvv,∇上式右端第一项算子为,不能写成,BAvv•∇Av∇ABBAABBABBBBvvvvvvvvvvvv)()()()(•∇+∇•−=××−∇=××∇z4.算子运算
本文标题:第14讲哈密顿算子2
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