推 jvrmusic:很詳細 謝謝 02/17 19:45
※ 引述《jvrmusic (圈x2)》之銘言：
: 題目↓
: http://www.library.ntut.edu.tw/ezfiles/7/1007/img/805/98MME001.pdf
: 第四題&第五題
: 不知道怎麼下手
: 麻煩解惑一下
: 謝謝
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4.
原 P.D.E. 的 characteristic eq. 為:
du
dx = dy = ────
2(x+y)u
由 dx = dy 解得 x - y = c1
du [(x+y)^2]/2
由 d(x+y) = ─── 解得 u = c2*e
(x+y)u
[(x+y)^2]/2
所以通解為 u(x,y) = f(x-y)*e , where f(.) is any
function of notation "."
根據 b.c. ┌ u(0,0) = 1 = f(0)
└ u(1,0) = exp(2) = f(1)*exp(1/2)
解得 ┌ f(0) = 1
└ f(1) = exp(3/2)
[(x+y)^2]/2
因此 u(x,y) = f(x-y)*e with ┌ f(0) = 1
└ f(1) = exp(3/2)
5.
令 C1: straight line from z=0 to z=R
C2: counterclockwise 1/8 circle from z=R to z=R*exp(iπ/4)
C3: straight line from z=R*exp(iπ/4) to z=0
C = C1 + C2 + C3 being a closed contour
exp(iz^2) - 1
考慮 ∮ f(z) dz = 0 with f(z) = ─────── and defining proper
C z^2
branch cut such that 0≦ Arg(z)＜2π
( since f(z) is analytic in the region C )
π/4 │exp[i*R^2 *exp(2θ)] - 1│
│∫ f(z) dz│ ≦ ∫ ────────────── dθ by M-L ineq.
C2 0 R
π/4 1 + exp[-R^2 *sin(2θ)]
≦ ∫ ──────────── dθ
0 R
π/4 1 + exp(-R^2 *mθ)
≦ ∫ ───────── dθ for some m 屬於 R+
0 R
π 1 - exp(-R^2 *mπ/4)
= ── + ─────────── →0 as R→∞
4R m*R^3
所以 lim ∮ f(z) dz = 0
ε→0 C
R→∞
∞ exp(ix^2) - 1 0 exp(-x^2) - 1 iπ/4
→ ∫ ─────── dx + ∫ ──────── e dx = 0
0 x^2 ∞ exp(iπ/2) * x^2
∞ exp(ix^2) - 1 -iπ/4 ∞ exp(-x^2) - 1
→ ∫ ─────── dx = e *∫ ─────── dx
0 x^2 0 x^2
-iπ/4
= -√π * e ____ Note (1)
∞ sin(x^2) √π
即 ∫ ───── dx = ──
0 x^2 √2
Note (1):
∞ exp(-x^2) - 1 exp(-x^2) - 1 x→∞ ∞ -x^2
∫ ─────── dx = ─────── │ - 2∫ e dx
0 x^2 -x x=0 0
= √π
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