作者pattrick (派特瑞克)
看板NTU-Exam
标题[试题] 96下 谢传璋 工程数学二 2nd期中考
时间Tue May 20 21:31:27 2008
课程名称︰工程数学二
课程性质︰必修
课程教师︰谢传璋/王昭男
开课学院:工学院
开课系所︰工程科学与海洋工程学系
考试日期(年月日)︰2008/5/20
考试时限(分钟):120
是否需发放奖励金:是
(如未明确表示,则不予发放)
试题 :
1. Find the general solution of the following P.D.E for u(x,y):
(i) u = u = 0
yy xx
xy
(ii) u + u = e
y
2. The 2-D Laplacian operator in Cartesian coordinate is :
2 2
2 δ δ
▽≡( ─ + ─ )
2 2
δx δy
Transform this operator in Polar coordinate:
x = rcosθ ; y = rsinθ
3. Solve the following wave equation of u(x,t)
2 2
δ u 2 δ u
----- = c ------ , 0 < x < L 0≦t -----(1)
2 2
δ t δ x
u(0,t) = 0
B.C: 0≦t<∞ -----(2)
u(L,t) = 0
3 πx
u(x,0) = sin (-----)
I.C: L ------(3)
u (x,0) = 0
t
Plot wave form at t = 1 and t = 2, assume c = 1; L = 1,
4. Use the method of Fourier transform, find the solution (you may express
the solution in double integral form) of the following heat equation:
2
δu 2δu
---- = c ---- -∞ < x < ∞ ; t>0 -----(1)
2
δt δx
-Uo when -1 < x < 0
I.C. u(x,0) = Uo when 0 < x < 1 -----(2)
0 otherwise
what is temperature at x = 0 i.e. u(0,t)?
5. Consider one dimensional wave equation
2 2
δu 2δu
---- = c ---- , -∞ < x < ∞ ; t > 0 -----(1)
2 2
δt δx
u(x,0) = f(x)
I.C. -∞ < x < ∞ -----(2a,b)
u (x,0) = g(x)
t
the D'Alembert's solution is:
1 x+ct
u(x,t) = 0.5 [ f(x+ct) + f(x-ct) ] + ---∫ g(s)ds
2c x-ct
If the initial conditions f(x) and g(x) are given as:
1 when -0.5≦x≦0.5
f(x) =
0 otherwise
1 when -1≦x≦1
g(x) =
0 otherwise
Plot the wave form u(x,t) at t = 1, and t = 2 , assume c = 1
每题20分,共100分
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