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)F(x,y,z)=x2+y2−z2⇒→n=(Fx,Fy,Fz)=(2x,2y,−2z)=(2,0,−1)⇒單位法向量→n|→n|=(2√5,0,−1√5)(b)C:x2+y2=1,z=1⇒r(t)=(x(t),y(t),z(t)),其中{x(t)=costy(t)=sintz(t)=1⇒r′(t)=(−sint,cost,0),0≤t≤2π⇒→F=(−sint,cost,cos3tsin3t)⇒線積分∫C→F⋅dr=∫2π0(−sint,cost,cos3tsin3t)⋅(−sint,cost,0)dt=∫2π0sin2t+cos2tdt=∫2π01dt=2π
解答:(a)先求齊次解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}}
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