作者doom8199 (～口卡口卡 修～)
看板Grad-ProbAsk
标题Re: [理工] [工数]-pde d'Alembert solution
时间Mon Nov 30 22:19:39 2009
※ 引述《winer8 (快来明星3 缺1 )》之铭言：
: 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
---
令此通解为 Z(u,x,y) = 0
→ ┌ Zx + Zu*Ux = 0
└ Zy + Zu*Uy = 0
→ ┌ Ux = - Zx / Zu 带回原 P.D.E.
└ Uy = - Zy / Zu
-2*Zx + 3Zy + (2U-2x)Zu = 0
dx dy dU
→ ____ = ____ = _____
-2 3 2U-2x
→ ┌ 3dx + 2dy = 0
└ U' + U = x
→ ┌ 3x + 2y = c1
└ U = (x-1) + c2*e^(-x)
所以该 PDE 通解为 U(x,y) = (x-1) + φ(3x + 2y)*e^(-x)
其中 φ(r) is any function of r
由 U(x,y)=x^2 for the line 2y+x=0 可知:
x^2 = (x-1) + φ(3x-x)*e^(-x)
→ φ(2x) = (x^2-x+1)*e^x
→ φ(x) = [ (x/2)^2 - (x/2) + 1]*e^(x/2)
因此 U(x,y) = (x-1) + φ(3x + 2y)*e^(-x)
3x + 2y 3x + 2y (3x + 2y)/2 -x
= (x-1) + [ (_______)^2 - (_______) + 1] * e * e
2 2
3x + 2y 3x + 2y (x/2 + y)
= (x-1) + [ (_______)^2 - (_______) + 1] * e
2 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)]
翻以前的文章你应该就会做了 ＝＝a
: 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 与x+y-z=c2 2曲面交线
同 1.
---
程式好多QQ
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◆ From: 140.113.141.151
※ 编辑: doom8199 来自: 140.113.141.151 (11/30 23:00)
1F：推 iyenn:推程式好多Q_q 11/30 23:45
2F：→ doom8199:好险学长是好人，不然我又要看日出了 ╰(〒皿〒)╯ 12/01 00:01
3F：推 winer8:感谢楼上二位罗!!! 12/01 00:12
4F：推 iyenn:反正你看习惯了(茶 12/01 00:26