Re: [工數] 一題偏微 有提示
※ 引述《poo123456 (poo)》之銘言：
: 想了很久想不出來
: 想來請問大家一下 可以的話可以寫仔細一點嗎
: 十分感謝。
: http://ppt.cc/Rd2k
First, use the method of separation of variables.
That is, u(x,t)=f(x)T(t).
Then,
@^2 u d^2 T(t)
－－－＝f(x)－－－－
@t^2 dt^2
@^2 u d^2 f(x)
－－－＝T(t)－－－－
@x^2 dx^2
insert this two relations and divid by 9f(x)T(t) on both sides
=>
1 d^2 T(t) 1 d^2 f(x)
－－－*－－－－＝－－－*－－－－＝-k (k>0)
9T(t) dt^2 f(x) dx^2
where k must be greater than zero, or it will be a trivial solution.
Second, with two boundary conditions, solve the Sturm-Liouville problem.
f''(x) + kf(x)=0
u(0,t)=f(0)T(t)=0 => f(0) = 0
u(L,t)=f(L)T(t)=0 => f(L) = 0
=> f(x) = A cos(√k x) + B sin(√k x)
f(0) = A = 0
f(L) = B sin(√k L) = 0 , B does not be equal to zero.
∴ sin(√k L) = 0 , √k L = nπ n=1,2,3,....
k=(nπ/L)^2 ... eigenvalue
fn(x)= B sin(nπx/L) ... eigenfunction of x part.
T''(t) + 9kT(t)=0
IC:@u/@t (t=0) = 0.
=> T(t) = C cos(3√k t) + D sin(3√k t)
=> T'(t) = -3√k C sin(3√k t) + 3√k D cos(3√k t)
T'(0) = 3√k D = 0
∴ Tn(t)= C cos(3nπt/L) ... eigenfunction of t part.
Third, use the Fourier series with another initial condition
to obtain the solution.
u (x,t)= fn(x)*Tn(t)
n
u(x,t)=ΣC* cos(3nπt/L)sin(nπx/L) (n from 1 to infinity)
u(x,0)=ΣC* sin(nπx/L) = e^(-x) ...
C* is the coefficient of Fourier Sine series
2 L
C* =－∫ e^(-x) sin(nπx/L) dx
L 0
∴u(x,t)=ΣC* cos(3nπt/L)sin(nπx/L) (n from 1 to infinity)
where
2 L
C* =－∫ e^(-x) sin(nπx/L) dx
L 0 ＃＃
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◆ From: 140.118.234.20
※ 編輯: theye 來自: 140.118.234.20 (06/21 16:26)
※ 編輯: theye 來自: 140.118.234.20 (06/21 16:38)
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