北大数学物理方法期终考试试题及参考答案

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!#$%&’()*+,-./0(201)1.2343w=ln(1+z)5674389:5;(1)argz$=?(2)arg(1+z)$=?(3)z$@=ABC?(4)1+z$@=ABCD2.E43f(z)FGHIJKGLMNCOGLP1QRSTUVWOAXBYZ1ZCf(z)dz(1)[\]^_‘%a[bcdef%?(2)[\]^_f%a[bcde‘%?(3)[\]^_)bcde‘%?(4)[\]^_)bcdef%D3.g43f(z)Fz=aWMNf(a)=f0(a)=···=f(n−1)(a)=0,f(n)(a)6=0,Y43f0(z)/f(z)Fz=aWPh3O(1)1−n?(2)n−1?(3)−n?(4)nD4.z=∞5f(z)=1sinzP(1)ijc?(2)klc?(3)mnc?(4)opqlcD5.Γ(z)Γ(1−z)=πsinπzPrsJKO;(1)tuv?(2)wxyz0Rez1?(3){|uvRez0?(4)}|uvRez1D~-(101)MN43f(z)FP37O3v(x,y)=xx2+y2f(z)D2-(201)43lnz−1z+1Fz=∞PKLO3lnz−1z+1 z=∞=0D-(401)Z1;(1)Z∞−∞dx(x2+1)(x2−2xcosθ+1),0θπ,θ6=π/2.(2)Z∞−∞sinxx(x2+4)dx.-(101)f(t)=1πZπ0cos(tcosθ)dθ8P43F(p)D¡¢£⁄¥ƒ§¤,-(201)1.(2)(41)2.(4)(41)3.(4)(41)4.(4)(41)5.(1)(41)~-(101)'“1«u(x,y)P‹›1du(x,y)=∂u∂xdx+∂u∂ydy=∂v∂ydx−∂v∂xdy=−2xy(x2+y2)2dx+x2−y2(x2+y2)2dy,Z1u(x,y)=yx2+y2+C.f(z)FP37O3fifly=0–u(x,y)=0,†‡·«C=0.¶•f(z)=y+ixx2+y2=iz∗zz∗=iz.∂u∂x(21)∂u∂y(21)«u(x,y)(21)Z1‚3C=0(21)„”«f(z)(21)'“2»…„”‰«f(z):v(x,y)=xx2+y2=12z+z∗zz∗=121z∗+1z4=12iiz∗+iz=12iiz−iz∗=f(z)−f∗(z)2i†‡·»…¿f(z)=iz.(5`´`21)-(201)lnz−1z+1=ln1−1z1+1z=∞Xn=1(−)n−1n−1zn−1zn=∞Xn=1(−)n−1n[(−)n−1]1zn=−2∞Xn=012n+11z2n+1,|z|1.ˆ˜«¯U(31)˘71˙¨(31)ln1−1z(41)ln1+1z(41)„”(41)˚¸˝(21)-(1)(201)˛˝ˇO—GZ1I1(z2+1)(z2−2zcosθ+1)dz.h3I1(z2+1)(z2−2zcosθ+1)dz=ZR−R1(x2+1)(x2−2xcosθ+1)dx+ZCRI1(z2+1)(z2−2zcosθ+1)dz=2πiXres1(z2+1)(z2−2zcosθ+1)5=2πi1(z2+1)(z2−2zcosθ+1)Fz=iPh3+1(z2+1)(z2−2zcosθ+1)Fz=eiθPh3.•Z43Fz=iPh3=12i(−2icosθ)=14cosθ,Z43Fz=eiθPh3=1(ei2θ+1)2isinθ=12cosθ(cosθ+isinθ)2isinθ=−icosθ−isinθ4cosθsinθ=−i14sinθ−14cosθ.–:Olimz→∞z·1(z2+1)(z2−2zcosθ+1)=0,limR→∞1(z2+1)(z2−2zcosθ+1)dz=0.ƪ˛R→∞¿Z∞−∞1(x2+1)(x2−2xcosθ+1)dx=π2sinθ.Z1˝ˇ(21)Z43(21)h3(31)ŁWiØh3(41)ŁWeiθØh3(41)(21)ª(31)-(2)(201)˛˝ˇŒº8.7GZ1Ieizz(z2+4)dz.h3Ieizz(z2+4)dz=Z−δ−Reixx(x2+4)dx+ZCδeizz(z2+4)dz=ZRδeixx(x2+4)dx+ZCReizz(z2+4)dz6=2πie−2−8=−πi4e−2.:Olimz→∞1z(z2+4)=0,JordanlimR→∞ZCReizz(z2+4)dz=0.:Olimz→0z·eizz(z2+4)=14,limδ→0ZCδeizz(z2+4)dz=−πi4.ƪ˛R→∞,δ→0¿Z∞−∞eixx(x2+4)dx=πi41−e−2.¿æªZ∞−∞sinxx(x2+4)dx=π41−e−2.Z1˝ˇ(21)Z43(21)h3(31)ŁWz=2iØh3(41)(41)Jordan(21)ª(31)-(101)F(p)=1πZπ0pp2+cos2θdθ=12πZ2π0pp2+cos2θdθ=12πI|z|=1pp2+14(z+z−1)2dziz=4p2πiI|z|=1zz4+2(2p2+1)z2+1dz.˜ζ=z2zP|z|=1,OζP|ζ|=1ı¶•F(p)=4p2πiI|ζ|=11ζ2+2(2p2+1)ζ+1dζ=4p×1ζ2+2(2p2+1)ζ+1F˘LPh3X.71ζ2+2(2p2+1)ζ+1PŁW5ζ=−2(2p2+1)±p4(2p2+1)2−42=−(2p2+1)±2ppp2+1,¶•1ζ2+2(2p2+1)ζ+1F˘L,łŁWζ=−(2p2+1)+2ppp2+1,ø5,œWh3O12ζ+2(2p2+1) ζ=−(2p2+1)+2p√p2+1=14ppp2+1.æ¿F(p)=1pp2+1.cosωtPLaplace(21)Oߢ|z|=1PZ1(21)˜ζ=z2(11)ŁWØh3(21)ª(21)rs;(Rep0)(11)8¡Pl(x)=lXn=01(n!)2(l+n)!(l−n)!x−12n(2l+1)xPl(x)=(l+1)Pl+1(x)+lPl−1(x)P0l+1(x)=xP0l(x)+(l+1)Pl(x)P0l+1(x)=(2l+1)Pl(x)+P0l−1(x)Pl(x)=P0l+1(x)−2xP0l(x)+P0l−1(x)Jν(x)=∞Xk=0(−)kk!Γ(k+ν+1)x22k+νLaplace943f(t)43F(p)11perfcα2√t1pe−α√p1√παsin2√αt1p√pe−α/p1√πtcos2√αt1√pe−α/p1√πte−α2/4t1√pe−α√p1√πte−2α√t1√pe−α2/perfcα√p;!!#$%&’(D,-(201)«70PM;(1)y00(x)+λy(x)=0,y(0)=0,y(l)=0;(2)y00(x)+λy(x)=0,y0(0)=0,y0(l)=0;(3)y00(x)+λy(x)=0,y(x)=y(x+2π)=0;10(4)ddx(1−x2)dy(x)dx+λy(x)=0,y(±1).~-(251)MM0;∂ua∂xt∂y−2∂u0∂xx∂y=,0xl,t0,∂u∂x x=0=0,∂u∂x x=l=0,t0,u t=0=Acos2πxl,∂u∂t t=0=0,0xl,8aXAO‚3D-(201)MLPM0;∇2u=−4πrcosθ,ra,u r=a=0.-(151)Z1;Z∞0e−αx2J0(x)xdx,8α0J0(x)OœBessel43D-(201)Laplace'“—!0PGreen43;∂G(x,t;x0,t0)∂t−κ∂G(x,t;x0,t0)δ∂xx∂y=(x−x0)δ(t−t0),0x∞,t0,G(x,t;x0,t0) x=0=0,t0,G(x,t;x0,t0) t=0=0,0x∞,80x0∞,t00D0˚#11$%&’()*+,(20-)(1)λn=nπl2(2-)yn(x)=sinnπlx(2-)n=1,2,3,···(1-)(2)λn=2n+1lπ2(2-)yn(x)=sin2n+1lπx(2-)n=0,1,2,3,···(1-)(3)λn=n2(2-)yn(x)=sinnx,cosnx(2-)n=0,1,2,3,···(1-)(4)λl=l(l+1)(2-)yl(x)=Pl(x)(2-)l=0,1,2,3,···(1-)12.,(25-)u(x,t)=C0t+D0+∞Xn=1hCnsinnπlat+Dncosnπlatisinnπlxu t=0=D0+∞Xn=1Dnsinnπlx=u0cos2πlx=u021+cos2πlx⇒D0=u02,Dn=u02δn2∂u∂t t=0=C0+∞Xn=1Cnnlπa·sinnlπx⇒C0=Cn=0u(x,t)=u02+u02cos2πlxcos2πlat-/01X(x)234(2-)5678(2-)T(t)234(2-)9:;=λn(2-)Xn(x)(2-)n2;(1-)?@Tn(t),T0(t)(2-)+A@(2-)BCDC0(2-)Cn(2-)D0,D2(2-)Dn(n6=0,2)(2-)@E(2-)13F,(20-)GHIJK=29:LDMNu(r,θ)=∞Xl=0Rl(r)Pl(cosθ)1r2ddrr2dRldr−l(l+1)r2Rl=−4πrδl1Rl(0)O6Rl(a)=0Pl=1QRR1(r)=A1r+B1r−2−25πr3RR1(0)O6⇒B1=0R1(a)=0⇒A1=25πa2R1(r)=25πa2−r2Pl6=1QRRl(r)=Alrl+Blr−l−1RRl(0)O6⇒Bl=0Rl(a)=0⇒Al=0Rl(r)=0,l6=1STu(r,θ)=25πa2−r2rcosθDUVWXY(2-)@=3ZXY(2-)[\]φ^_(2-)9:LD(2-)Rl(r)‘a2b-34(2-)Rl(r)‘a25678(2-)l=1Q2@(3-)l6=1Q2@(3-)@E(2-)14c,(15-)3Z+deBesselLD2fDVgEhij-Z∞0e−αx2J0(x)xdx=∞Xn=0(−)nn!n!122nZ∞0e−αx2x2n+1dx=∞Xn=0(−)nn!n!122n12Z∞0e−αttndt=∞Xn=0(−)nn!n!122n+1n!αn+1=12α∞Xn=0(−)nn!14αn=12αe−1/4α3Z2XYkl(3-)mnj-(2-)opΓLDqrj-(5-)st(5-)3Z.uv1j-ZwxI(b)=Z∞0e−αx2J0(bx)xdxyI0(b)=−Z∞0e−αx2J1(bx)x2dx=12αe−αx2xJ1(bx) ∞0−b2αZ∞0e−αx2J0(bx)xdx=−b2αI(b)STI(b)=Aexp−b24α,z{A=I(0)=1/2αw|Z∞0e−αx2J0(x)xdx=I(1)=12αexp−14α.3Z2XYkl(3-)}~I(

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