北大数学物理方法期终考试试题2及参考答案

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¡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(b)2b-34(4-)@b-34(4-)op?;Bj-D(2-)s~Ssj-(2-)15,(20-)G(x,t;x0,t0);G(x,p;x0,t0),yB@=0pG(x,p;x0,t0)−κd2dxG(x,p;x0,t0)=e−pt0δ(x−x0),G(x,p;x0,t0) x=0=0,G(x,p;x0,t0) x→∞→0.@G(x,p;x0,t0)=Asinhrpκx,xx0,Bexp−rpκx,xx0.i78G(x,p;x0,t0) x=x0+0x=x0−0=0,−κddxG(x,p;x0,t0) x=x0+0x=x0−0=e−pt0Bexp−rpκx0−Asinhrpκx0=0,Bexp−rpκx0+Acoshrpκx0=1√κpe−pt0,TB~DA=1√κpe−pt0exp−rpκx0,B=1√κpe−pt0sinhrpκx0.RG(x,p;x0,t0)=1√κpe−pt0exp−rpκx0sinhrpκx,xx0,1√κpe−pt0exp−rpκxsinhrpκx0.xx0,VRG(x,t;x0,t0)=12pκπ(t−t0)exp−(x−x0)24κ(t−t0)−exp−(x+x0)24κ(t−t0)η(t−t0).Laplace0Rb-34B@=(4-)@34xx0(2-)xx0(2-)i78E(4-)(2-)η(t−t0)(2-)VR(4-)16

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