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： $$\mathbf{(a)}\;A=\begin{bmatrix}1 & -1&-1 \\-1 & 1& -1\\ -1& -1&1 \end{bmatrix} \Rightarrow \det(A)=1-1-1-1-1-1=-4 \\ \Rightarrow \det(A^5)= (\det(A))^5=(-4)^5=\bbox[red,2pt]{-1024}\\ \mathbf{(b)}\;\det(A-\lambda I)=0 \Rightarrow -(\lambda+1)(\lambda-2)^2=0 \Rightarrow A的特徵值-1，2，2\\ \Rightarrow A^5的特徵值：(-1)^5,2^5,2^5 \Rightarrow 特徵值的和＝-1+32+32=\bbox[red,2pt]{63} \\ \mathbf{(c)}\;\left[\begin{array}{rrr|rrr}1 & -1 & -1& 1& 0 & 0 \\-1 & 1 & -1 & 0 & 1& 0 \\ -1 & -1 & 1 & 0 & 0 & 1\end{array}\right] \xrightarrow{r_1+r_2\to r_2,r_1+r_3\to r_3} \left[\begin{array}{rrr|rrr}1 & -1 & -1& 1& 0 & 0 \\0 & 0 & -2 & 1 & 1& 0 \\ 0 & -2 & 0 & 1 & 0 & 1\end{array}\right]\\ \xrightarrow{r_2 \leftrightarrow{r_3}}\left[\begin{array}{rrr|rrr}1 & -1 & -1& 1& 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 1\\0 & 0 & -2 & 1 & 1& 0 \end{array}\right] \xrightarrow{-r_2/2\to r_2,-r_3/2 \to r_3}\left[\begin{array}{rrr|rrr}1 & -1 & -1& 1& 0 & 0 \\ 0 & 1 & 0 & -1/2 & 0 & -1/2\\0 & 0 & 1 & -1/2 & -1/2& 0 \end{array}\right]\\ \xrightarrow{r_1+r_2+r_3 \to r_1} \left[\begin{array}{rrr|rrr}1 & 0 & 0& 0& -1/2 & -1/2 \\ 0 & 1 & 0 & -1/2 & 0 & -1/2\\0 & 0 & 1 & -1/2 & -1/2& 0 \end{array}\right] \\ \Rightarrow A^{-1}=\bbox[red, 2pt]{\begin{bmatrix}0 & -1/2 & -1/2 \\-1/2 & 0& -1/2\\ -1/2 & -1/2 & 0 \end{bmatrix}}\\ B=\begin{bmatrix}0 & -1/2 & -1/2 \\-1/2 & 0& -1/2\\ -1/2 & -1/2 & 0 \end{bmatrix} \Rightarrow B^2=\begin{bmatrix}1/2 & 1/4 & 1/4 \\1/4 & 1/2 & 1/4\\ 1/4 & 1/4 & 1/2 \end{bmatrix} \Rightarrow B^4=\begin{bmatrix}3/8 & 5/16 & 5/16 \\5/16 & 3/8 & 5/16\\ 5/16 & 5/16 & 3/8 \end{bmatrix} \\ \Rightarrow B^5=\begin{bmatrix}-5/16 & -11/32 & -11/32 \\-11/32 & -5/16 & -11/32 \\ -11/32 & -11/32 & -5/16 \end{bmatrix};\\由(b)知：A的特徵值為-1,2 \Rightarrow B=A^{-1}的特徵值為-1,1/2 \Rightarrow B^5的特徵值為-1,1/32\\ \lambda_1=-1 \Rightarrow (B^5-\lambda_1 I)\mathbf x=0 \Rightarrow \begin{bmatrix}11/16 & -11/32 & -11/32 \\-11/32 & 11/16 & -11/32 \\ -11/32 & -11/32 & 11/16 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\x_3 \end{bmatrix}=0\\ \qquad \Rightarrow \cases{x_1=x_3\\ x_2=x_3},取v_1=\begin{bmatrix}1 \\ 1 \\1 \end{bmatrix}\\ \lambda_2=1/32 \Rightarrow (B^5-\lambda_2 I)\mathbf x=0 \Rightarrow \begin{bmatrix}-11/16 & -11/32 & -11/32 \\-11/32 & -11/16 & -11/32 \\ -11/32 & -11/32 & -11/16 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\x_3 \end{bmatrix}=0 \\ \qquad \Rightarrow x_1+x_2+x_3=0,取v_2=\begin{bmatrix}-1 \\ 1 \\0 \end{bmatrix},v_3=\begin{bmatrix}-1 \\ 0 \\1 \end{bmatrix}\\ 因此B^5的特徵向量為\bbox[red,2pt]{\begin{bmatrix}1 \\ 1 \\1 \end{bmatrix}, \begin{bmatrix}-1 \\ 1 \\0 \end{bmatrix},\begin{bmatrix}-1 \\ 0 \\1 \end{bmatrix}}$$

解答：(a) $$F(x,y,z)=x^2+y^2-z^2 \Rightarrow \vec n=(F_x,F_y,F_z)=(2x,2y,-2z) =(2,0,-1)\\ \Rightarrow 單位法向量{\vec n\over |\vec n|}=\bbox[red, 2pt]{({2\over \sqrt 5},0,-{1\over \sqrt 5})}$$ (b) $$C:x^2+y^2=1,z=1 \Rightarrow r(t)=(x(t),y(t),z(t)),其中\cases{x(t)=\cos t\\ y(t)=\sin t\\ z(t)=1} \\\Rightarrow r'(t)=( -\sin t,\cos t,0),0\le t\le 2\pi  \Rightarrow \vec F=(-\sin t,\cos t,\cos^3t \sin^3t)\\ \Rightarrow 線積分\int_C \vec F\cdot dr = \int_0^{2\pi} (-\sin t,\cos t,\cos^3t \sin^3t)\cdot ( -\sin t,\cos t,0)\,dt \\=\int_0^{2\pi} \sin^2t +\cos^2 t\,dt =\int_0^{2\pi} 1\,dt =\bbox[red, 2pt]{2\pi}$$

解答： $$\mathbf{(a)}\;先求齊次解y'''-3y''+3y'-y=0 \Rightarrow \lambda^3-3\lambda^2 +3\lambda -1=0 \Rightarrow (\lambda-1)^3=0\\ \Rightarrow \lambda=1 \Rightarrow y_h=(C_1+C_2x+ C_3x^2)e^x\\ 接著令y_p=Ax^3e^x \Rightarrow y_p'=3Ax^2e^x+ Ax^3e^x \Rightarrow y_p''= 6Axe^x +6Ax^2e^x+ Ax^3e^x\\ \Rightarrow y_p'''= 6Ae^x+18 Axe^x +9Ax^2e^x + Ax^3e^x \\ \Rightarrow y_p'''-3y_p''+3y_p'-y_p=6Ae^x=e^x \Rightarrow A={1\over 6} \\ \Rightarrow y=y_h+yp \Rightarrow \bbox[red, 2pt]{y=(C_1+C_2x+ C_3x^2)e^x+ {1\over 6}x^3e^x} \\\mathbf{(b)}\;L\{ xy'' \}- L\{xy'\} +L\{y\} =L\{1\} \Rightarrow -{d\over ds}(s^2Y(s)-s-1)+{d\over ds} (sY(s)-1)+ Y(s)={1\over s}\\ \Rightarrow (2-2s)Y(s)+(s-s^2)Y'(s)+1={1\over s} \Rightarrow Y'(s)+{2\over s}Y(s)={1\over s^2}\\ 積分因子\mu(s)=e^{\int 2/s\,ds} =s^2 \Rightarrow s^2Y'(s)+2sY(s)=1 \Rightarrow (s^2Y(s))'=1 \\ \Rightarrow s^2Y(s)=s+C \Rightarrow Y(s)={1\over s}+{C\over s^2} \Rightarrow y(x) =L^{-1}\{Y(s)\} =L^{-1}\{{1\over s}\}+L^{-1}\{ {C\over s^2}\} \\ \Rightarrow y(x)=1+Cx \Rightarrow y'(0)=C=1 \Rightarrow C=1 \Rightarrow \bbox[red,2pt]{y(x)=x+1}$$

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) $$u(x,0)=f(x) \Rightarrow f(x)= \sum_{n=1}^\infty A_n  \sin(n\pi x), n\in \mathbb N \Rightarrow A_n= 2\int_0^1 f(x)\sin(n\pi x)\,dx\\ u_t(x,0)=g(x)= \left. \sum_{n= 1}^\infty (n\pi B_n \cos(n\pi t) -n\pi A_n \sin(n\pi t)) \sin(n\pi x)\right|_{t=0} \\\Rightarrow g(x)=\sum_{n= 1}^\infty (n\pi B_n ) \sin(n\pi x) \Rightarrow B_n={2\over n\pi}\int_0^1 g(x)\sin(n\pi x)\,dx\\ 因此\bbox[red,2pt]{u(x,t)= \sum_{n=1}^\infty (A_n\cos(n\pi t) +B_n\sin(n\pi t))\sin (n\pi x), 其中\cases{A_n=2\int_0^1 f(x)\sin(n\pi x)\,dx\\ B_n={2\over n\pi}\int_0^1 g(x)\sin(n\pi x)\,dx},n \in \mathbb{N}}$$