※ 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 -- 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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