[試題] 101下 李克強 工程數學二 期末考
課程名稱︰工程數學二
課程性質︰必修
課程教師︰李克強
開課學院：工學院
開課系所︰化學工程學系
考試日期（年月日）︰102/06/21
考試時限（分鐘）：120
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
(一) (30%)
The steady state temperature distribution T(r,φ) is governed by the
ðT
Laplace equation for an axisymmetric system, i.e. ── = 0, where the
ðθ
spherical coordinates are shown as below.
(圖就是一顆圓球,ψ：天頂角,θ：方位角,就不畫了)
(ð是 partial 的意思)
Find the steady-state temperature at an arbitrary point outside the
sphere with the distribution of T on the surface of the sphere:
T = T(R,ψ) = f(ψ) = 1 + 2 cosψ + 3 cos^2 ψ.
(二) (30%)
(A) Find the eigenvalues and corresponding eigenvectors for the matrix:
┌ 1 1 0 ┐
A = │ 0 2 1 │.
└ 0 0 3 ┘
(B) Find the nine elements of A^56.
(C) Find the corresponding expression of A in the coordinate system
┌ 1 ┐ ┌ 1 ┐ ┌ 0 ┐
(e1,e2,e3) with e1 =│ 0 │, e2 =│ 2 │, e3 =│ 0 │.
└ 1 ┘ └ 0 ┘ └ 1 ┘
(三) (40%)
Use the finite difference method to solve the ODE-BVP as follows:
d^2 y dy
─── + ─ + y = x, y(0) = 1, y(1) + y'(1) = 0, x 屬於 [0,1].
d x^2 dx
Use N = 1 and N = 2, repectively, where N is the total interior nodal
points. Compare your results with the exact solution:
y(1/2) = 0.5748305, y(1/3) = 0.7062534, y(2/3) = 0.4617324 with
√3 √3
y(x) = exp(-0.5x) [ c1 cos ── x + c2 sin ── x ] + x - 1, where
2 2
c1 = 2, c2 = -1.037403.
What is the percentage of deviation there?
Appendix：
The Legendre polynomial of degree n is denoted by Pn(x), where
P0(x) = 1, P1(x) = x, P2(x) = (3 x^2 - 1)/2, P3(x) = (5 x^2 - 3x)/2,
P4(x) = (35 x^4 - 30 x^2 +3)/8, P5(x) = (63 x^4 - 70 x^2 + 15x)/8,...etc
The Laplace Equation in a Sphere
ð ðT ð^2 T ð ðT
▽^2 T = ────(r^2 ──) + ───────── + ────── (──sinψ),
r^2 ðr ðr r^2 sin^2ψ ðθ^2 r^2 sinψðψ ðψ
T = T(r,θ,ψ).
One-sided derivative that is correct to O(h^2):
-3 y(xi) + 4 y(x(i+1)) - y(x(i+2))
y'(xi) = ───────────────── + O(h^2) at left boundary point
2h
y(x(i-2)) - 4 y(x(i-1)) + 3 y(xi)
y'(xi) = ───────────────── + O(h^2) at right boundary point
2h
Mean value theorem
y(x(i+1)) - 2 y(xi) + y(x(i-1)) h^2
y"(xi) = ──────────────── - ── y""(ζi),
h^2 12
where ζi 屬於 [ xi, x(i+1) ] ...centered difference formula
y'(xi) = [ y(x(i+1)) - y(x(i-1)) ]/2h - h^2 y"'(ηi)/6
where ηi 屬於 [ x(i-1), xi ]
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◆ From: 140.112.250.1
※ 編輯: hellersjoke 來自: 140.112.250.1 (06/22 18:47)
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06/22 22:53, , 1F
06/22 22:53, 1F