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※ 編輯: iyenn 來自: 123.193.214.165 (11/30 23:46)
※ 引述《winer8 (快來明星3 缺1 )》之銘言：
: 1.solve the partial differential equation 2Ux-3Uy+2U=2x
: inttial condition U(x,y)=x^2 for the line 2y+x=0
: y+0.5x 1 2 1
: ANS:U=e [---(2y+3x) - ----(2y+3x)+1 ]+x-1
: 4 2
: 2.Utt=4Uxx -∞<x<∞ t>=0
: U(x,0)=0 Ut(x,0)=δ(x) solve bu fourier transform
: ANS:U=0.25[H(x+2t)-H(x-2t)]
: 3.solve
: dx dy dz
: -----------=----------- =---------
: x^2+y^2-yz -x^2-y^2+xz (x-y)z
: ANS: x^2+y^2=c1z^2 與x+y-z=c2 2曲面交線
1.solve the partial differential equation 2Ux-3Uy+2U=2x
inttial condition U(x,y)=x^2 for the line 2y+x=0
let Uh=e^ζx+ηy
(2ζ-3η+2)e^ζx+ηy=0
=>Uh=e^-xΦ(1.5x+y)
let Up=Ax+B
=>A=1 B=-1
U(x,y)=e^-xΦ(1.5x+y) +x-1
U(-2y,y)=x^2=e^2yΦ(-2y)-2y-1
=>Φ(-2y)=Φ(x)=e^x(x^2-x+1)
Φ(1.5x+y)=e^(1.5x+y)((1.5x+y)^2-(1.5x+y)+1)
=>U(x,y)=e^(0.5x+y)((1.5x+y)^2-(1.5x+y)+1) + x-1
2.Utt=4Uxx -∞<x<∞ t>=0
U(x,0)=0 Ut(x,0)=δ(x) solve bu fourier transform
let F(U(x,t))=U(w,t)
U''+4w^2U=0
L(U(w,t))=u(w,s)
s^2u-1+4w^2u=0
u=1/(s^2+(2w)^2)
=>
1
U=---sin2wt
2w
U(x,t)=F^-1(U(w,t))
1 ∞ 1
=---∫ ---e^i(x+2t)w-e^i(x-2t)wdw
2π-∞ 4iw
1
=----[H(x+2t)-H(x-2t)]
4
1
recall: F(u(t))=--- + πδ(w)
iw
3. dx dy dz
-----------=----------- =---------
x^2+y^2-yz -x^2-y^2+xz (x-y)z
dx+dy dz
=>---------=------
(x-y)z (x-y)z
dx+dy=dz =>x+y-z=c2 (1)
xdx+ydy dz
-------------------=-----
(x-y)(x^2+y^2) (x-y)z
0.5d(x^2+y^2) dz
-------------------=-----
x^2+y^2 z
lnx^2+y^2-2lnz=c1
x^2+y^2=c1z^2 ...(2)
Ans:x+y-z=c2 與x^2+y^2=c1z^2 的交曲面
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