作者kagato (包)
看板Grad-ProbAsk
标题Re: [理工] [工数] [拉氏] 拉式中的PDE
时间Mon Dec 28 19:03:51 2009
※ 引述《Tonyy (Cool)》之铭言：
: Find the solution p(x,t) of the following partial differential
: equation :
: d^2 p 1 d^2 p
: ------- - --- ------ = δ(x-at) ; c > a > 0 ; 0 < x < ∞ ;
: d x^2 c^2 d t^2 =
: 0 < t < ∞
: =
: p(x,0) = 0 ; dp(x,0) / dt = 0 ; p(0,t) = 0 ; p(x,t) < ∞ as x → ∞
: ps . 上面的符号 d 皆是偏微分的意思
L{p(x,t)}=P(x,s)
d^2P(x,s) 1
_________ - ___[s^2*P - sp(x,0)-p'(x,0)] = e^(-sx/a)
d x^2 c^2
P''(x,s) - (s^2/c^2)P(x,s) = e^(-sx/a)
P (x,s)= c1exp^(sx/c)+c2exp^(-sx/c)
h
1 exp^(-sx/a)
P (x,s)= _____________exp^(-sx/a) = _________________
y D^2-(s/c)^2 (s/a)^2-(s/c)^2
exp^(-sx/a)
P(x,s) = c1exp^(sx/c)+c2exp^(-sx/c) + ______________
(s/a)^2-(s/c)^2
L{p(∞,t)} = P(∞,t)有限 ,c1=0
-1
L{p(0,t)} = P(0,s) = 0 , c2 = __________________
(s/a)^2-(s/c)^2
-exp^(-sx/c) exp^(-sx/a) (ac)^2*[exp^(-sx/a)-exp^(-sx/c)]
p(x,s) =__________________ + _________________= ______________________
(s/a)^2-(s/c)^2 (s/a)^2-(s/c)^2 s^2[c^2-a^2]
(ac)^2*[(t-x/a)*H(t-x/a) - (t-x/c)*H(t-x/c)]
p(x,t) = _________________________________________________
c^2-a^2
: 解答 :
: p = { ac^2 [- H(t-x/c) (t-x/c) + H(t-x/a) (t-x/a) ] } / ( c^2 -a^2 )
: 这题不知该从何下手 希望会的人可以帮忙一下 谢谢 !!
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