112年成功大學土木工程碩士班-工程數學詳解

國立成功大學112學年度碩士班招生考試試題

系所:土木工程學系
考試科目:工程數學

解答: $$\mathbf{(a)}\;Ａ=\begin{bmatrix} 1& -1& 1\\ -1& -1& 1\\ 1& 1& 1\end{bmatrix} \Rightarrow \det(A-\lambda I)= -\lambda^3+ \lambda^2+4 \lambda-4 =-(\lambda-1)(\lambda-2)(\lambda+2)=0 \\ \Rightarrow \lambda= 1 ,2-2\\ \lambda_1=1  \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow  \begin{bmatrix}0& -1 & 1\\ -1 & -2 & 1\\ 1 & 1& 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow v= \begin{bmatrix} -k\\ k\\ k\end{bmatrix},取v_1=\begin{bmatrix} -1\\ 1\\ 1 \end{bmatrix} \\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow  \begin{bmatrix} -1 & -1 & 1\\ -1 & -3 & 1\\ 1 & 1& -1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow v= \begin{bmatrix} k\\ 0\\ k \end{bmatrix},取v_2=\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} \\ \lambda_3=-2\Rightarrow (A-\lambda_3 I)v =0 \Rightarrow  \begin{bmatrix}3 & -1 & 1\\ -1 & 1 & 1\\  1 & 1& 3 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow v= \begin{bmatrix} -k\\ -2k\\ k \end{bmatrix},取v_3= \begin{bmatrix} -1\\ -2\\ 1 \end{bmatrix}\\ 因此特徵值為\bbox[red, 2pt]{1,2,-2}相對應的特徵向量為\bbox[red, 2pt]{\begin{bmatrix} -1\\ 1\\ 1 \end{bmatrix},\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} ,\begin{bmatrix} -1\\ -2\\ 1 \end{bmatrix}} \\ \mathbf{(b)}\; \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& -2& 2 & 1 & 1& 0\\ 0& 2& 0 & -1 & 0 &1\end{array} \right] \\ \xrightarrow{R_2/2,R_3/2} \left[ \begin{array}{rrr|rrr} 1& -1& 1 & 1& 0 & 0\\ 0& -1& 1 & 1/2 & 1/2& 0\\ 0& 1& 0 & -1/2 & 0 &1/2\end{array} \right] \xrightarrow{R_3+ R_1\to R_1,R_3+R_2 \to R_2} \left[ \begin{array}{rrr|rrr} 1& 0& 1 & 1/2& 0 & 1/2\\ 0& 0& 1 & 0 & 1/2& 1/2\\ 0& 1& 0 & -1/2 & 0 &1/2\end{array} \right] \\ \xrightarrow{-R_2+R_1 \to R_1}  \left[ \begin{array}{rrr|rrr} 1& 0& 0 & 1/2& -1/2 & 0\\ 0& 0& 1 & 0 & 1/2& 1/2\\ 0& 1& 0 & -1/2 & 0 &1/2\end{array} \right] \xrightarrow{R_2 \leftrightarrow R_3}  \left[ \begin{array}{rrr|rrr} 1& 0& 0 & 1/2& -1/2 & 0\\ 0& 1& 0 & -1/2 & 0& 1/2\\ 0& 0& 1 & 0 & 1/2 &1/2\end{array} \right] \\ \Rightarrow \bbox[red, 2pt]{A^{-1}= \begin{bmatrix} 1/2& -1/2 & 0\\ -1/2& 0 & 1/2\\ 0 & 1/2& 1/2\end{bmatrix}}$$

解答: $$\cases{x=r\cos \theta\\ y= r\sin \theta} \Rightarrow Ｔ_{xx}+T_{yy} =T_{rr}+{1\over r}T_r =-2 \Rightarrow rT_{rr}+T_{r} =-2r \\ \Rightarrow (rT_r)'=-2r \Rightarrow rT_r=-r^2+c_1 \Rightarrow T_r=-r+{c_1\over r} \Rightarrow T=-{1\over 2}r^2+c_1\ln r+c_2\\ \Rightarrow T(x,y) =-{1\over 2}(x^2+y^2)+{c_1\over 2} \ln(x^2+y^2) +c_2 \\\Rightarrow \cases{T(0,y)=-{1\over 2}y^2+ c_1\ln y+c_2= 0\\ T(\pi, y) =-{1\over 2}(\pi^2+y^2) +{c_1\over 2}\ln(\pi^2+y^2) +c_2=1\\ T(x,0)= -{1\over 2}x^2+ c_1\ln x+c_2=0}....\text{to be continued}$$

解答: $$假設u(x,t)= X(x)T(t), 則u_{tt}=u_{xx} \Rightarrow XT''=X''T \Rightarrow {T''\over T}={X''\over X}=k為一常數\\ 又\cases{u(0,t)=X(0)T(t)=0 \\ u(1,t)=X(1)T(t) =0} \Rightarrow X(0)=X(1)=0\\ \text{Case I: }k=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow \cases{X(0)=c_2=0\\ X(1)=c_1+ c_2 =0} \Rightarrow X=0 \Rightarrow u=0\\ \text{Case II:}k\gt 0,假設k＝\lambda^2, X''-\lambda^2X=0 \Rightarrow X=c_1e^{\lambda x}+ c_2 e^{-\lambda x} \Rightarrow \\\qquad  \cases{X(0)=c_1+ c_2=0\\ X(1)=c_1e^\lambda + c_2e^{-\lambda} =0} \Rightarrow c_1e^\lambda-c_1e^{-\lambda} =c_1e^{\lambda}(1-e^{-2\lambda})=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \\\qquad \Rightarrow X=0 \Rightarrow u=0\\ \text{Case III:}k \lt 0, 假設k=-\lambda^2 \Rightarrow X''+\lambda^2 X=0 \Rightarrow X=c_1\cos(\lambda x)+ c_2\sin(\lambda x)\\ \qquad \Rightarrow \cases{X(0)=c_1=0\\ X(1)=c_2 \sin(\lambda)=0} \Rightarrow \sin \lambda =0 \Rightarrow \lambda =n\pi \Rightarrow X=c_2\sin(n\pi x)\\ T''+\lambda^2 T=0 \Rightarrow T=c_3\cos(\lambda t)+ c_4 \sin(\lambda t)\\ \Rightarrow u_n(x,t)=X(x)T(t) = c_2\sin(n\pi x)(c_3\cos(n\pi t)+c_4 \sin(n\pi t)) \\\qquad =\sin(n\pi x)(A_n\cos(n\pi t) +B_n\sin(n \pi t))\\\qquad  \Rightarrow u(x,t)= \sum_{n=1}^\infty \sin(n\pi x)(A_n\cos(n\pi t) +B_n\sin(n \pi t)) \\ 再由u(x,0)=0 \Rightarrow \sum_{n=1}^\infty A_n \sin(n\pi x)=0 \Rightarrow A_n=0  \Rightarrow u(x,t)= \sum_{n=1}^\infty B_n\sin(n\pi x)  \sin(n \pi t)\\ \Rightarrow \left. {\partial u\over \partial t} \right|_{t=0} =\left. \sum_{n=1}^\infty n\pi B_n\sin(n\pi x)  \cos(n \pi t)\right|_{t=0} =\sum_{n=1}^\infty n\pi B_n\sin(n\pi x) =\sin(\pi x)+2\sin (3\pi x) \\ \Rightarrow \cases{B_1=1/\pi \\ B_3=2/3\pi \\ B_n=0,n=2,4,5,\dots} \Rightarrow \bbox[red, 2pt]{u(x,t)={1\over \pi}\sin(\pi x)\sin(\pi t)+ {2\over 3\pi} \sin(3\pi x) \sin(3\pi t)}$$

解答: $$f(x)=\pi^4-x^4 \Rightarrow f(x)=f(-x) \Rightarrow f\text{ is even} \Rightarrow b_n=0, n\in \mathbb N \\ a_0={1\over 2\pi} \int_{-\pi}^\pi (\pi^4-x^4)\,dx ={1\over 2\pi}  \left. \left[ \pi^4x -{1\over 5}x^5\right] \right|_{-\pi}^\pi ={4\over 5}\pi^4\\ a_n = {1\over \pi}\int_{-\pi}^\pi (\pi^4-x^4) \cos(nx)\,dx= {1\over \pi} \left(-{8 \pi\over n^4}(n^2\pi^2-6)(-1)^n \right) ={8\over n^4}(6-n^2\pi^2)(-1)^n \\ \Rightarrow f(x)={4\over 5}\pi^4+ \sum_{n=1}^\infty {8\over n^4}(6-n^2\pi^2)(-1)^n \cos(n x)={4\over 5}\pi^4+ \sum_{n=1}^\infty {8\over n^4}(6-n^2\pi^2) \\ \Rightarrow \bbox[red, 2pt]{f(x)= {4\over 5}\pi^4+ \sum_{n=1}^\infty {8\over n^4}(6-n^2\pi^2) }$$

解答: $$\vec F=(x,y,z) \Rightarrow curl (\vec F) =\begin{vmatrix}\vec i &\vec j& \vec k\\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ x & y & z \end{vmatrix} =0 \Rightarrow \iint_S curl(\vec F) \cdot  \,ds =0\\ 2x+y+2z=6與座標軸交點\cases{A(3,0,0) \\ B(0,6,0)\\ C(0,0,3)} \\ \Rightarrow 逆時針三曲線\cases{C_1(A\to C):\{(3-3t,0,3t) \mid 0\le t\le 1\} \\C_2(C\to B): \{(0,6t,3-3t) \mid 0\le t\le 1\} \\C_3(B\to A):\{(3t, 6-6t, 0) \mid 0\le t\le 1\} \\}\\ \int_C\vec F\cdot dr =\int_{C_1} \vec F\cdot dr + \int_{C_2} \vec F\cdot dr +\int_{C_3} \vec F\cdot dr \\=\int_0^1 (3-3t,0,3t) \cdot (-3,0,3)\,dt +\int+0^1 (0,6t,3-3t) \cdot(0,6,-3) \,dt +\int_0^1 (3t,6-6t,0) \cdot (3,-6,0)\,dt \\= \int_0^1 108t-54\,dt = 0\\ 因此 \text{Stoke's} 定理:　\int_C \vec F\cdot dr = \iint_S curl(\vec F)\cdot ds 成立, \bbox[red, 2pt]{Q.E.D.}$$

解答: $$\mathbf{(a)}\; 先求齊次解,y''+4y=0 \Rightarrow \lambda^2+4=0 \Rightarrow \lambda =\pm 2i \Rightarrow y_h=c_1\cos(2x)+ c_2\sin(2x)\\ 利用\text{ variation of parameters }求y_p, 令\cases{y_1=\cos (2x)\\y_2=\sin (2x)\\ r(x)=16\cos(2x)} \Rightarrow W=\begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} =\begin{vmatrix} \cos(2x) & \sin(2x)\\ -2\sin(2x)' & 2\cos(2x)' \end{vmatrix} =2\\ y_p = -y_1\int{y_2 r\over W}dx +y_2 \int{y_1r\over W}dx =-\cos(2x) \int{ 8\sin(2x)\cos(2x)}\,dx +\sin(2x) \int{8\cos^2(2x)}\,dx \\=-\cos(2x) \int 4\sin(4x)\,dx +\sin(2x) \int 4\cos(4x)+4\,dx= \cos(2x)\cos(4x)+ \sin(2x) \sin(4x)+ 4x\sin(2x)\\= \cos(4x-2x)+ 4x\sin(2x)= \cos(2x)+ 4x\sin(2x) \\ \Rightarrow y=y_h+y_p = c_1\cos(2x)+ c_2\sin(2x)+\cos(2x)+ 4x\sin(2x)=c_3\cos(2x)+ c_2\sin(2x)+ 4x\sin(2x) \\ \Rightarrow y'=-2c_3\sin(2x)+ 2c_2\cos(2x) +4\sin(2x)+ 8x\cos(2x) \\ \cases{y(0)=0\\ y'(0)=0} \Rightarrow \cases{c_3=0\\ 2c_2=0} \Rightarrow \bbox[red, 2pt]{y= 4x\sin(2x)}$$ $$\mathbf{(b)}\; 先求齊次解,y''+2y'+10y=0 \Rightarrow \lambda^2+2\lambda+10=0 \Rightarrow \lambda =-1\pm 3i \\ \Rightarrow y_h=c_1e^{-x}\cos(3x) +c_2e^{-x}\sin(3x)\\接著令y_p =A\cos x+B \sin x+C\cos(3x)+ D\sin(3x) \\\Rightarrow \cases{y_p'=-A\sin x+B \cos x-3C\sin(3x)+ 3D\cos(3x) \\ y_p''= -A\cos x-B\sin x-9C\cos(3x)-9D\sin(3x)} \\ \Rightarrow y_p''+2y_p'+10y_p\\=(9A+ 2B )\cos x+( -2A+ 9B)\sin  x+(C+6D )\cos(3x)+ ( -6C+ D)\sin (3x) \\=17\sin x-37\sin(3x) \Rightarrow \cases{9A+2B=0 \\ -2A+9B=17\\ C+6D=0\\ -6C+D=-37} \Rightarrow \cases{A=-0.4\\ B=1.8\\ C=6\\ D=-1} \\ \Rightarrow y_p= -0.4\cos x+1.8\sin x+6\cos(3x)-\sin(3x) \\ \Rightarrow y=y_h+y_p \\ \Rightarrow y=c_1e^{-x}\cos(3x) +c_2e^{-x}\sin(3x) -0.4\cos x+1.8\sin x+6\cos(3x)-\sin(3x) \\ \Rightarrow y=c_1e^{-x}\cos(3x) +c_2e^{-x}\sin(3x) -0.4\cos x+1.8\sin x+6\cos(3x)-\sin(3x) \\ \Rightarrow y'=-c_1e^{-x}\cos(3x) -3c_1e^{-x} \sin(3x)-c_2e^{-x}\sin(3x) +3c_2e^{-x}\cos(3x)\\ \quad +0.4\sin x+1.8\cos x-18\sin(3x)-3\cos(3x)\\ \Rightarrow \cases{y(0)=c_1-0.4+6=6.6 \Rightarrow c_1=1\\ y'(0)= -c_1+3c_2+1.8-3=-2.2 \Rightarrow c_2=0} \\ \Rightarrow \bbox[red, 2pt]{y=e^{-x}\cos(3x) -0.4\cos x+1.8\sin x+6\cos(3x)-\sin(3x)}$$