数学分析_第三版_复旦大学欧阳光中.pdf

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1、214§2.¦Eܼêêóª{K1.¦e¼êêµ2∂u∂u(1)u=f(x,y)§Ù¥x=rcosθ,y=rsin触,¶∂r∂r2222∂u∂u∂u∂u∂u(2)u=f(x,y)§Ù¥x=aξ,y=b秦,,,,∂ξ∂ξ2∂ξ∂η∂η∂η222∂u∂u∂∂u∂u222(3)u=f(x+y+z)§¦,,,,∂x∂x2∂x∂y∂y∂z!2x∂u∂u∂u(4)u=fx,§¦,,.y∂x∂x2∂y)µ∂u(1)=fxcosθ+fysinθ∂r2∂u222=fx2cosθ+fxysin2θ+fy2sinθ∂r222∂u∂u∂u∂u∂

2、u22(2)∂ξ=afx,∂ξ2=afx2,∂ξ∂η=abfxy,∂η=bfy,∂η2=bfy222∂u∂u∂0222022220022200222(3)=2xf(x+y+z),=2f(x+y+z)+4xf(x+y+z),=4xyf(x+y+z),∂x∂x2∂x∂y∂u∂u02220222=2yf(x+y+z),=2zf(x+y+z)∂y∂z2∂u1∂u21∂ux(4)=f1+f2,=f11+f12+f22,=−f2∂xy∂x2yy2∂yy22.Φ=Φ(x,y,z),x=u+v,y=u−v,z=uv§¦Φu,Φv.)µΦu=Φx+Φy+

3、vΦz,Φv=Φx−Φy+uΦz3.¦e¼ê©(Ù)µ(1)u=f(x+y)(2)u=f(x+y,x−y)222(3)u=f(ax+by+cz))µ0(1)du=f(x+y)(dx+dy)(2)du=(f1+f2)dx+(f1−f2)dy0222(3)du=2f(ax+by+cz)(axdx+bydy+czdz)4.yeªµ∂z∂z22(1)z=ϕ(x+y)§Ky−x=0¶∂x∂y∂u∂uxu22(2)u=yϕ(x−y)§Ky+x=¶∂x∂yy222∂u∂u∂u(3)u=xϕ(x+y)+yψ(x+y)§K−2+=0.∂x

4、2∂x∂y∂y2y²µ∂z∂z022022(1)Ï=2xϕ(x+y),=2yϕ(x+y)∂x∂y∂z∂zKy−x=0.∂x∂y∂u∂u022222022(2)Ï=2xyϕ(x−y),=ϕ(x−y)−2yϕ(x−y)∂x∂y∂u∂uxu22Ky+x=xϕ(x−y)=.∂x∂yy215∂u∂u0000(3)Ï=ϕ(x+y)+xϕ(x+y)+yψ(x+y),=xϕ(x+y)+ψ(x+y)+yψ(x+y)∂x∂y22∂u∂u00000000000K=2ϕ(x+y)+xϕ(x+y)+yψ(x+y),=ϕ(x+y)+ψ(x+y)+xϕ(x+y)+yψ

5、(x+y)∂x2∂x∂y2∂u00000=2ψ(x+y)+xϕ(x+y)+yψ(x+y)∂y2222∂u∂u∂uu´−2+=0.∂x2∂x∂y∂y2222∂u∂u∂u2225.¦∆u=++,u=f(x+y+z,x+y+z).∂x2∂y2∂z22∂u∂u2)µÏ=f1+2xf2§K=f11+4xf12+4xf22+2f2∂x∂x222∂u∂u22âé¡5§=f11+4yf12+4yf22+2f2,=f11+4zf12+4zf22+2f2∂y2∂z2222∂u∂u∂u222u´∆u=++=3f11+4(x+y+z)f12+4(x+y+z)f

6、22+6f2.∂x2∂y2∂z2p6.eu=f(r),r=x2+y2§Ù¥f(r)g§Áy²222∂u∂udu1du+=+∂x2∂y2dr2rdr222∂ux∂uyx0000y²µÏ=pf(r)§K=f(r)+f(r)∂x22∂x2223x2+y2x+y(x+y)2222∂uxy000âé¡5§=f(r)+f(r)∂y2223x2+y2(x+y)2222∂u∂udu1duu´+=+∂x2∂y2dr2rdr7.eu,vx,y¼ê§x=rcosθ,y=rsinθ§Ád∂u∂v∂u∂v=,=−∂x∂y∂y∂x∂u1∂v∂v1∂uy²ª=

7、,=−.∂rr∂θ∂rr∂θy²µÏu,vx,y¼ê§x=rcosθ,y=rsinθ∂u∂u∂u∂v∂v∂vK=cosθ+sinθ,=cosθ+sinθ,∂r∂x∂y∂r∂x∂y∂u∂u∂u∂v∂v∂v=−rsinθ+rcosθ,=−rsinθ+rcosθ∂θ∂x∂y∂θ∂x∂y∂u∂v∂u∂v∂u1∂v∂v1∂uq=,=−§K=,=−.∂x∂y∂y∂x∂rr∂θ∂rr∂θn8.f(tx,ty)=tf(x,y)§Kk∂f∂fx+y=nf∂x∂yp∂z∂zäkù5¼ê§¡ngàg¼ê.

8、^ù(J§éz=x2+y2§¦x+y.∂x∂y

9、nn−1y²µÏf(tx,ty)=tf(x,y)§Küàét¦§f1(tx,ty)x+f2(tx,ty)y=ntf(x,y)∂f∂f-t=1§Kf1(x,y)x+f2(x,y)y=nf(x