[試題] 105-2 林太家 偏微分方程導論 期末考

作者: Mathmaster (^_^)   2017-06-20 19:19:40
課程名稱︰偏微分方程導論
課程性質︰數學系必修
課程教師︰林太家
開課學院:理學院
開課系所︰數學系
考試日期(年月日)︰2017/6/20
考試時限(分鐘):110分鐘+延長n分鐘
試題 :
Final June 20th
Total: 120 points.
1.(10%)
Solve the diffusion problem u = ku in 0<x<L, with thw mixed boundary
t xx
conditions u(0,t) = u (L,t) = 0.
x
2.(20%)
Suppose u is a smooth solution of
╭ u - u +c(x)u = 0 for x ∈ (0,1), t>0,
│ t xx

│ u(0,t) = u(1,t) = 0 for t>0,

╰ u(x,0) = g(x) for x ∈ (0,1),
and the function c = c(x) sarisfies c(x)≧γ>0 for x ∈ (0,1), where γ>0
-σt
is a constant. Prove if |g(x)|≦1 for x ∈ (0,1), then ∥u∥ ≦ Ce for
2k
k ∈ N, t>0, where C, σ are positive constants independent of k and ∥‧∥
2k
2k 1 2k
is defined by ∥u∥ = ∫ u dx.
2k 0
3.(10%)
2
Solve u = c u for 0<x<π, with the boundary conditions u (0,t) =
tt xx x
2
u (π,t) = 0 and the initial conditions u(x,0) = 0, u (x,0) = cos (x).
x t
4.(20%)
Solve u + u = 1 in the annulus a<r<b with u(x,y) vanishing on both
xx yy
parts of the boundary r = a and r = b.
5.(20%)
Find the general solution of 3u + 10u + 3u = sin(x+t).
tt xt xx
6.(20%)
-1 3 ∞ 3
Let u(x) = ∫ |x-y| f(y) dy for x ∈ R , where f ∈ C ( R ). Prove
3 0
R
3 3
that Δu = cf in R , where Δ is the Laplace operator on R and c is a
constant. Calculate the constant c.
7.(20%)
2 4 4
Assume u ∈ C ( R \ {0} ), and u≧0 is a harmonic in R \{0}. Prove that u
-2 4
has the form that u(x) = C |x| + C for x ∈ R \{0}, where C , C ≧0
1 2 1 2
are constants.
作者: jack7775kimo (阿龐)   2017-06-21 14:00:00
6就Evans第二章翻一下。7比較困難,可以翻Axler他們合寫的Harmonic function theory第三章最後一段。

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