112年高考三級-應用數學詳解

112年公務人員高等考試三級考試試題

類 科：天文、氣象
科 目：應用數學（包括微積分、微分方程與向量分析）

解答： $$\mathbf{(一)}\;f(x,y)=x^3-3y^2+6xy-x-9y+10 \Rightarrow \nabla f=(f_x,f_y)=(3x^2+6y-1, -6y+6x-9) \\\quad \Rightarrow \nabla f(1,-1)=(3-6-1,6+6-9)=(-4,3)\\ \quad \Rightarrow 最大方向導數=||\nabla f(1,-1)|| = \sqrt{(-4)^2+3^2} =\bbox[red, 2pt] 5 \\\mathbf{(二)}\;f(x,y)=x^2+y^2+2x-2y+12 = (x+1)^2+(y-1)^2+10 \ge 10\\ 當x=-1及 y=1時(符合x^2+y^2=2 \le 4)，f(x,y)有最小值10\\當x^2+y^2=4時\Rightarrow f(x,y)=2x-2y+16，利用\text{ Lagrange 算子求極值}\\ \cases{f=2x-2y+16\\ g=x^2+y^2-4} \Rightarrow \cases{f_x=\lambda g_x \\ f_y=\lambda g_y\\ g=0} \Rightarrow \cases{2=2x \lambda \\ -2=2y \lambda} \Rightarrow -1={x\over y} \Rightarrow x=-y\\ \Rightarrow g(x,-x)=0 \Rightarrow x^2=2 \Rightarrow x=\pm \sqrt 2 \Rightarrow f(\sqrt 2,-\sqrt 2)=16+4\sqrt 2為最大值\\ 因此\bbox[red, 2pt]{最大值=16+4\sqrt 2, 最小值=10}$$

解答： $$\mathbf{(一)}\; A=\begin{bmatrix}1 & 2 \\2 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow (\lambda+1)( \lambda-3)=0 \Rightarrow \lambda_1=-1, \lambda_2=3\\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I) x=0 \Rightarrow x_1+x_1=0，取v_1=\begin{bmatrix}-1 \\1 \end{bmatrix}\\ \lambda_2=3 \Rightarrow (A-\lambda_2 I)x=0 \Rightarrow x_1=x_2,最v_2= \begin{bmatrix}1 \\1 \end{bmatrix}\\ \Rightarrow P=[v_1\; v_2] =\bbox[red, 2pt]{\begin{bmatrix}-1 & 1 \\1 & 1 \end{bmatrix}}\\ \mathbf{(二)}\; \vec x'(t)=A \vec x(t) \Rightarrow \vec x(t)=c_1e^{\lambda_1 t}v_1+ c_2e^{\lambda_2 t}v_2 =c_1e^{-t}\begin{bmatrix}-1 \\1 \end{bmatrix} +c_2 e^{3t} \begin{bmatrix}1 \\1 \end{bmatrix}\\ 初始值\vec x(0)=\begin{bmatrix}1 \\1 \end{bmatrix} =c_1 \begin{bmatrix}-1 \\1 \end{bmatrix} +c_2   \begin{bmatrix}1 \\1 \end{bmatrix} \Rightarrow \cases{-c_1+c_2=1\\ c_1+c_2=1} \Rightarrow \cases{c_1=0\\ c_2=1}\\ \qquad \Rightarrow \bbox[red, 2pt]{\vec x(t)=e^{3t} \begin{bmatrix}1 \\1 \end{bmatrix}}$$

解答： $$\mathbf{(一)}\; f(x)=x^2 \Rightarrow f(x)為偶函數 \Rightarrow b_n=0,\forall n\\ a_0={1\over 2\pi}\int_{-\pi}^\pi f(x)\,dx ={1\over 2\pi}\left. \left[ {1\over 3}x^3 \right] \right|_{-\pi}^\pi ={\pi^2\over 3} \\ a_n= {1\over \pi}\int_{-\pi}^\pi f(x) \cos(nx)\,dx = {1\over \pi} \left. \left[ {1\over n^3}((n^2x^2-2)\sin(nx)+2nx\cos(nx)) \right] \right|_{-\pi}^\pi  ={4\over n^2}(-1)^n \\ \Rightarrow \bbox[red, 2pt]{f(x)={\pi^2\over 3}+ \sum_{n=1}^\infty {4\over n^2}(-1)^n \cos(nx)}\\ \mathbf{(二)}\; 假設u(x,t)=X(x+\pi)T(t),則u_x(-\pi,t) =u_x(\pi,t)=0 \Rightarrow X'(0)T(t)=X'(2\pi)T(t)=0\\ \Rightarrow X'(0)=X'(2\pi)=0\\ 又u_t(x,t)=u_{xx}(x,t) \Rightarrow X(x+ \pi)T'(t)= X''(x+\pi) T(t) \Rightarrow {T'(t)\over T(t)} ={X''(x+\pi) \over X(x+\pi)} =k為常數\\若k\gt 0 \Rightarrow x''(x+\pi)-kX(x+\pi)=0 \Rightarrow X(x+ \pi)=c_1e^{\sqrt k(x+\pi)} +c_2e^{-\sqrt k(x+\pi)}\\ \qquad \Rightarrow X'(x+\pi) =c_1\sqrt ke^{\sqrt k(x+\pi)} -c_2\sqrt ke^{-\sqrt k(x+\pi)} \\\qquad \Rightarrow 初始值\cases{X'(0)=c_1\sqrt k=0 \cdots(1)\\ X'(2\pi) =c_1\sqrt ke^{2\sqrt k\pi} =c_2\sqrt k e^{-2\sqrt k\pi} \cdots(2)},由(1)得c_1=0代入(2) \\ \qquad \Rightarrow c_2=0 \Rightarrow X(x+\pi)=0 \Rightarrow u(x,t)=0為明顯解,不討論\\ 若k=0 \Rightarrow X''(x+\pi)=0 \Rightarrow X(x+\pi) =c_1x+c_2 \Rightarrow X'(x+\pi)=c_1 \\\qquad \Rightarrow 初始值X'(0)=X'(2\pi)=0 \Rightarrow c_1=0 \Rightarrow X(x+\pi)=c_2\\ \qquad 同時T'=0 \Rightarrow T=c_3 \Rightarrow u(x,t)=c_2c_3 \Rightarrow u(x,t)=c 為一常數\\　若k=-\rho^2 \lt 0 \Rightarrow X''(x+\pi)+\rho^2X(x+\pi)=0 \Rightarrow X(x+\pi)= A\cos(\rho(x+\pi)) +B\sin(\rho(x+\pi)) \\ \qquad \Rightarrow X'(x+\pi)= -A\rho \sin(\rho(x+\pi)) +B\rho \cos(\rho(x+\pi))\\\qquad \Rightarrow 初始值\cases{X'(0)=B \rho=0 \cdots(3) \\ X'(2\pi) =-A\rho \sin 2\rho \pi+ B\rho\cos(2\rho \pi)=0 \cdots(4)},由(3)得B=0代入(4)\\ \qquad \Rightarrow \cases{A=B=0 \Rightarrow u(x,t)=0明顯解\\ \sin 2\rho \pi=0 \Rightarrow \rho=n/2,n\in \mathbb N} \Rightarrow X_n(x+\pi)= \cos(n(x+\pi)/2), n\in \mathbb N\\ \qquad 同時T'+\rho^2 T=0 \Rightarrow T=c_4e^{-\rho^2 t} \Rightarrow u_n(x,t)=c_4 e^{-n^2t/4} \cos({n\over 2}(x+\pi)),n\in \mathbb N\\ 綜合以上討論,u(x,t)= a_0+ \sum_{n=1}^\infty a_ne^{-n^2t/4} \cos(n(x+\pi)/2)\\ \Rightarrow u(x,0)=x^2= a_0+ \sum_{n=1}^\infty a_n\cos(n(x+\pi)/2) ,其中 a_0= {1\over 2\pi}\int_{-\pi}^\pi x^2\,dx = \pi^2/3,\\ a_n={1\over \pi} \int_{-\pi}^\pi x^2 \cos(n(x+\pi)/2)dx ={8\pi \over n}(1+(-1)^n) ={16\pi \over n},n為偶數 \\ \Rightarrow \bbox[red, 2pt]{u(x,t)={\pi^2 \over 3}+ \sum_{k=1}^\infty {8\pi \over k}e^{-k^2t} \cos(k(x+\pi))}$$

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