Loading [MathJax]/jax/element/mml/optable/GeneralPunctuation.js

網頁

2023年5月18日 星期四

112年台大碩士班-工程數學(A)詳解

 國立臺灣大學112學年度碩士班招生考試

題號:194
科目:工程數學(A)




解答(a)A=[111111111]det(A)=111111=4det(A5)=(det(A))5=(4)5=1024(b)det(AλI)=0(λ+1)(λ2)2=0A122A5(1)5,25,251+32+32=63(c)[111100111010111001]r1+r2r2,r1+r3r3[111100002110020101]r2r3[111100020101002110]r2/2r2,r3/2r3[1111000101/201/20011/21/20]r1+r2+r3r1[10001/21/20101/201/20011/21/20]A1=[01/21/21/201/21/21/20]B=[01/21/21/201/21/21/20]B2=[1/21/41/41/41/21/41/41/41/2]B4=[3/85/165/165/163/85/165/165/163/8]B5=[5/1611/3211/3211/325/1611/3211/3211/325/16];(b)A1,2B=A11,1/2B51,1/32λ1=1(B5λ1I)x=0[11/1611/3211/3211/3211/1611/3211/3211/3211/16][x1x2x3]=0{x1=x3x2=x3,v1=[111]λ2=1/32(B5λ2I)x=0[11/1611/3211/3211/3211/1611/3211/3211/3211/16][x1x2x3]=0x1+x2+x3=0,v2=[110],v3=[101]B5[111],[110],[101]
解答(a)F(x,y,z)=x2+y2z2n=(Fx,Fy,Fz)=(2x,2y,2z)=(2,0,1)n|n|=(25,0,15)(b)C:x2+y2=1,z=1r(t)=(x(t),y(t),z(t)),{x(t)=costy(t)=sintz(t)=1r(t)=(sint,cost,0),0t2πF=(sint,cost,cos3tsin3t)CFdr=2π0(sint,cost,cos3tsin3t)(sint,cost,0)dt=2π0sin2t+cos2tdt=2π01dt=2π

解答(a)y

解答(a)令u(x,t)=F(x)G(t) \Rightarrow \cases{u_{tt}=FG''\\ u_{xx} =F''G} ,依題意FG''=F''G \Rightarrow {F'' \over F}= {G'' \over G}=k為常數\\ 邊界條件\cases{u(0,t) = F(0)G(t)=0\\ u(1,t) =F(1)G(t)=0},若G(t)=0,則u=0為明顯解,不討論\\ 因此邊界條件變為F(0)= F(1)=0\\ 現在對常數k進行討論:\\ 若k=0,則F''=0 \Rightarrow F(x)=ax+b,再將邊界條件代入\Rightarrow \cases{F(0)=b=0\\ F(1)=a+b=0} \\ \qquad\Rightarrow F=0,則u=0為明顯解,不討論\\ 若k\gt 0,假設k=c^2 \Rightarrow F''-c^2F=0 \Rightarrow F=C_1e^{cx}+C_2 e^{-cx},代入邊界條件 \\ \qquad \Rightarrow \cases{F(0)= C_1+C_2=0\\ F(1)=C_1e^c+C_2e^{-c}=0} \Rightarrow C_1=C_1=0,則u=0為明顯解,不討論\\因此只能假設k\lt 0, 假設k=-c^2 \Rightarrow F''+c^2F=0 \Rightarrow F=A\cos (cx)+B \sin(cx)\\ 再代入邊界條件:\cases{F(0)=A=0\\ F(1)=A\cos(c)+ B\sin(c)=0} \Rightarrow  \sin(c)=0 \Rightarrow c=n\pi \\ \Rightarrow F_n(x)= \sin(n\pi x),n\in \mathbb N\\ 接著求G(t),{G''\over G}=k=-c^2=-n^2\pi^2 \Rightarrow G''+n^2\pi^2 G=0\\ \Rightarrow G_n(t)=A_n \cos n\pi t+B_n\sin n\pi t \Rightarrow u_n(x,t)=(A_n \cos n\pi t+B_n\sin n\pi t)\sin(n\pi x)\\ \Rightarrow \bbox[red, 2pt]{u(x,t)=\sum_{n=1}^\infty (A_n \cos n\pi t+B_n\sin n\pi t)\sin(n\pi x), n\in \mathbb N} (b)u(x,0)=f(x) \Rightarrow f(x)= \sum_{n=1}^\infty A_n  \sin(n\pi x), n\in \mathbb N \Rightarrow A_n= 2\int_0^1 f(x)\sin(n\pi x)\,dx\\ u_t(x,0)=g(x)= \left. \sum_{n= 1}^\infty (n\pi B_n \cos(n\pi t) -n\pi A_n \sin(n\pi t)) \sin(n\pi x)\right|_{t=0} \\\Rightarrow g(x)=\sum_{n= 1}^\infty (n\pi B_n ) \sin(n\pi x) \Rightarrow B_n={2\over n\pi}\int_0^1 g(x)\sin(n\pi x)\,dx\\ 因此\bbox[red,2pt]{u(x,t)= \sum_{n=1}^\infty (A_n\cos(n\pi t) +B_n\sin(n\pi t))\sin (n\pi x), 其中\cases{A_n=2\int_0^1 f(x)\sin(n\pi x)\,dx\\ B_n={2\over n\pi}\int_0^1 g(x)\sin(n\pi x)\,dx},n \in \mathbb{N}}

沒有留言:

張貼留言