北大数学物理方法期终考试试题3及参考答案

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¡¢¢£,-⁄301¥”ƒ01.§¤43xx2+y2−iyx2+y2F'“!⁄ø«8!3¥-F'“MN2.Z1I|z|=2coszz3dz3.«43f(z)=coszz−1F8¶«sŁW⁄‹›∞W¥“Ph34.«43f(z)=lnzsinπzF8¶«sŁW⁄‹›∞W¥“Ph35.«'fizd2wdz2+dwdz−w=0F8¶YŁW“Pfl~-⁄251¥MM0∂2u∂t2−a2∂2u∂x2=0,0xl,t0∂u∂x x=0=∂u∂x x=l=0,t≥0u t=0=x,∂u∂t t=0=0,0≤x≤l8a5P–xXt†‡P‚3·-⁄151¥M¶LPM0∇2u=0,arb,0θπ,0ϕ2πu r=a=0,u|r=b=cos2θ,0≤θ≤π,0≤ϕ≤2π-⁄151¥MLM0∂u∂t−κ∇2u=0,0ra,0ϕ2π,t0u r=0,u r=a=sinϕ,0≤ϕ≤2π,t≥0u t=0=0,0≤r≤a,0≤ϕ≤2π8κ5P–rϕXt†‡P‚3·-⁄151¥,•—Helmholtz'fi‚„PGreen43G(x,x0).”»…d2G(x,x0)dx2+k2G(x,x0)=−δ(x−x0)0x,x0∞G x=0=0,GFx→∞«‰⁄˛–¿:Oe−iωt¥8k5P–xXx0†‡P‚3·18`´ˆ˜¯˘˙,-⁄301¥1.xx2+y2−iyx2+y2=1zF¨z=0P˚¸P‹⁄‹›∞W¥!-MN1z0=−1z2!JK(21)MNJK(21)!3(21)2.I|z|=2coszz3dz=−πi(61)3.z=15f(z)=coszz−1PŁWresf(1)=−sin1?(31)z=∞5f(z)=coszz−1PMNWresf(∞)=sin1(31)4.z=0Xz=∞5lnzP˙Wß˝˛z=0¿z=∞˜¯Ulnz z=1=0Y˜¯U-˘71˙(11)z=15ˇŁWresf(1)=0(21)z=k,k=−1,±2,±3,···5,œW(11)resf(k)=(−1)kπlnk,k=2,3,4,···ik=−1(−1)kπln|k|−(−1)ki,k=−2,−3,−4,···(11)(11)lnz z=1=2nπi,n=±1,±2,±3,···Yz=k,k=±1,±2,±3,···†5,œWresf(k)=−2nik=1(−1)kπlnk+(−1)k2ni,k=2,3,4,···−(2n−1)ik=−1(−1)kπln|k|+(−1)k(2n−1)i,k=−2,−3,−4,···5.z=05'fizd2wdz2+dwdz−w=0P—,YŁW(31)'fizd2wdz2+dwdz−w=0Fz=0WPflOρ1=ρ2=0(31)~-⁄251¥,MOu(x,t)=C0t+D0+∞Xn=1Cnsinnπlat+Dncosnπlatcosnπlx,u t=0=D0+∞Xn=1Dncosnπlatcosnπlx=x,19∂u∂t t=0=C0+∞Xn=1Cnnπlacosnπlx=0,=⇒Cn=0n=0,1,2,···D0=1lZl0xdx=l2Dn=2lZl0xcosnπlxdx=2ln2π2[(−1)n−1]n=1,2,3,···90PMOu(x,t)=l2−4lπ2∞Xn=01(2n+1)2cos2n+1lπatcos2n+1lπxλ=02„PM(41)43(41)Tn(41)(21)(21)Cn=0(11)D0(31)Dn(31)M(21)-D†'fi•Øu–ϕ‡'fiO1r2∂∂rr2∂u∂r+1r2sinθ∂∂θsinθ∂u∂θ=0,MOu(r,θ)=∞Xl=0Alrl+Blr−l−1Pl(cosθ)u r=a=∞Xl=0Alal+Bla−l−1Pl(cosθ)=0u r=b=∞Xl=0Albl+Blb−l−1Pl(cosθ)=cos2θ·Alal+Bla−l−1=0Albl+Blb−l−1=2l+12Zπ0cos2θPl(cosθ)sinθdθ=2l+1213δl,0+215δl,220=⇒A0=b3(b−a)B0=−ab3(b−a)A2=2b33(b5−a5)B2=−2a5b33(b5−a5)90PMOu(r,θ)=b3(b−a)1−ar+2a2b33(b5−a5)r2a2−a3r3P2(cosθ)u–ϕ‡(11)Laplace'fi(21),M;43(21)Rl(r)(11)'(21)Al–BlP'fi(11)Al–BlP'fiÆ(11)A0(11)B0(11)A2(11)B2(11)M(11)-Eu(r,ϕ,t)=v(r,t)sinϕYv(r,t)»…M0∂v∂t−κ1r∂∂rr∂v∂r−vr2=0,0ra,t0v r=0,v r=a=1,t0v t=0=0,0raEv(r,t)=ra+w(r,t)Yw(r,t)»…M0∂w∂t−κ1r∂∂rr∂w∂r−wr2=0,0ra,t0w r=0,w r=a=0,t0w t=0=−ra,0ra701rddrrdR(r)dr+λ−1r2R(r)=021R(0),R(a)=0PMOλi=μ1ia2,Ri(r)=J1μ1iar,i=1,2,3,···8μ1i5J1(x)PiłW·:‡w(r,t)P,MOw(r,t)=∞Xi=1CiJ1μ1iare−κ(μ1i/a)2tw t=0=∞Xi=1CiJ1μ1iar=−ra†‡Ci=Za0−raJ1μ1iarrdrZa0J21μ1iarrdr=−1aaμ1i3x2J2(x) x=μ1ix=0a22J012(μ1i)=2μ1iJ0(μ1i)¶•90PMOu(r,ϕ,t)=rasinϕ+2∞Xi=11μ1iJ0(μ1i)J1μ1iare−κ(μ1i/a)2tsinϕ.∇2Pª(21)„43(21)w(r,t)PM0(21)7–43(21)Ti(t)(11)CiPª(21)'(11)1(21)M(11)-;Łx6=x0–Ø'fiŒd2G(x,x0)dx2+k2G(x,x0)=022G(x,x0)=Asinkx+Bcoskxxx0Ceikx+De−ikxxx0(2º)(2º)G(x,x0) x=0=0=⇒B=0(2º)Gx→∞«‰⁄˛–¿Œe−iωt¥=⇒D=0(2º)G(x,x0) x=x0−=G(x,x0) x=x0+=⇒Asinkx0=Ceikx0(2º)dG(x,x0)dx x=x0+−dG(x,x0)dx x=x0−=−1=⇒ikCeikx0−Akcoskx0=−1(2º)=⇒A=1keikx0xx0C=1ksinkx0xx0G(x,x0)=1keikx0sinkxxx01ksinkx0eikxxx0(1º)(3º)(3º)x˜LaplaceæØG(x,x0);g(p,x0)dG(x,x0)dx;pg(p,x0)(2º)d2G(x,x0)dx2;p2g(p,x0)−dG(x,x0)dx x=0(2º)ıfiŒp2g(p,x0)−dG(x,x0)dx x=0+k2g(p,x0)=−e−px0(2º)g(p,x0)=1p2+k2dG(x,x0)dx x=0−e−px0(2º)łøœG(x,x0)=dG(x,x0)dx x=01ksinkxη(x)−1ksink(x−x0)η(x−x0)(2º)Gx→∞߉⁄¿Œe−iωt¥=⇒dG(x,x0)dx x=01ksinkx−1ksink(x−x0)∝eikx(2º)=⇒dG(x,x0)dx x=0−coskx0=isinkx0(1º)=⇒dG(x,x0)dx x=0=coskx0+isinkx0=eikx0(1º)ıG(x,x0)=1keikx0sinkxxx01ksinkx0eikxxx0(1º)23

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