115年台大碩士班-工程數學A詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 工程數學(A)

$$\boxed{\textbf{第一題:向量微積分(20%)}}$$

解答:$$利用散度定(\text{Divergence Theorem}):\iint_S \mathbf F\cdot \mathbf n\;dS = \iiint_V (\nabla \cdot \mathbf F)dV = \iiint_V (2x+2y+2z)\,dV \\=\int_0^{2\pi} \int_0^2 \int_1^4 (2r\cos \theta+ 2r\sin \theta+ 2z)r\,dz\,dr\,d\theta =\int_0^{2\pi} \int_0^2 \left( 6r^3(\cos\theta+\sin \theta)+15r \right)\,dr \,d\theta \\=\int_0^{2\pi} (24(\cos\theta+ \sin \theta)+30)\,d\theta= \bbox[red, 2pt]{60\pi}$$

$$\boxed{\textbf{第二題:矩陣特徵值與線性常微分方程組(20%)}}$$

解答:$$ \det(A-\lambda I) =(\lambda+1)^2(5-\lambda)=0 \Rightarrow \lambda=5, -1 \\ \qquad \lambda_1=5 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-4&2& 2\\ 2&-4& 2\\2& 2& -4 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1=x_2=x_3 \\ \qquad \Rightarrow v= x_1 \begin{bmatrix}1\\1\\1 \end{bmatrix}, 取v_1= \begin{pmatrix}1\\1\\1 \end{pmatrix} \\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I )v=0 \Rightarrow \begin{bmatrix}2& 2& 2\\2& 2& 2\\2& 2& 2 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=0 \Rightarrow x_1+x_2+x_3=0 \\\qquad \Rightarrow v= \begin{bmatrix}-x_2-x_3\\ x_2\\ x_3 \end{bmatrix} =x_2 \begin{bmatrix}-1\\1\\0 \end{bmatrix} +x_3 \begin{bmatrix}-1\\0\\1 \end{bmatrix}, 取v_2= \begin{pmatrix} -1\\1\\0\end{pmatrix}, v_3= \begin{pmatrix} -1\\ 0\\1\end{pmatrix} \\ \Rightarrow 特徵值為\bbox[red, 2pt]{5,-1}, 相對應的特徵向量為 \bbox[red, 2pt]{\begin{pmatrix}1\\1\\1 \end{pmatrix} ,\begin{pmatrix} -1\\1\\0\end{pmatrix},   \begin{pmatrix} -1\\ 0\\1\end{pmatrix}}$$

解答:$$ \mathbf x(t)=c_1 e^{\lambda_1 t}v_1+ c_2 e^{\lambda_2 t}v_2 + c_3e^{\lambda_2 t}v_2  \Rightarrow \bbox[red, 2pt]{\mathbf x(t) =c_1e^{5t}\begin{pmatrix}1\\1\\1 \end{pmatrix}+c_2 e^{-t}\begin{pmatrix} -1\\1\\0\end{pmatrix} +c_3 e^{-t}\begin{pmatrix} -1\\ 0\\1\end{pmatrix}}$$

$$\boxed{\textbf{第三題:常微分方程式(15%)}}$$

解答:$$y''-4y'+4y=0 \Rightarrow r^2-4r+4=0 \Rightarrow (r-2)^2=0 \Rightarrow r=2 \Rightarrow y_h(t)=c_1e^{2t}+ c_2te^{2t} \\ 取\cases{y_1=e^{2t} \\y_2=te^{2t} \\g(t)=e^{2t}/t} \Rightarrow W = \begin{vmatrix} y_1& y_2\\ y_1'& y_2'\end{vmatrix} = \begin{vmatrix} e^{2t}& te^{2t}\\ 2e^{2t}&  e^{2t} +2te^{2t}\end{vmatrix} =e^{4t} \\ 使用 \text{Variation of Parameters: }y_p =-e^{2t} \int 1\,dt +te^{2t} \int {1\over t}\,dt =-te^{2t}+te^{2t}\ln t \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y(t)= c_1e^{2t}+ c_2te^{2t}-te^{2t}+te^{2t}\ln t}$$
$$\boxed{\textbf{第四題:初始值問題(15%)}}$$

解答:$$L\{y''\}+2L\{y'\}+2L\{y\}= L\{\delta(t-\pi)\} \Rightarrow s^2Y(s)+2sY(s)+2Y(s)=e^{-\pi s} \\ \Rightarrow Y(s)={e^{-\pi s}\over s^2+2s+2} ={e^{-\pi s} \over (s+1)^2+1} \Rightarrow y(t) =L^{-1} \{Y(s)\} =L^{-1} \left\{ {e^{-\pi s} \over (s+1)^2+1} \right\} \\=u(t-\pi)L^{-1} \left\{ {1 \over (s+1)^2+1} \right\}(t-\pi) \Rightarrow \bbox[red, 2pt]{y(t)= -u(t-\pi) e^{-(t-\pi)} \sin t}$$

$$\boxed{\textbf{第五題:波動方程式(Wave equation) (15%)}}$$

解答:$$u(x,t) =X(x)T(t) \Rightarrow u_{tt} =c^2 u_{xx} \Rightarrow XT''=c^2X''T \Rightarrow {T''\over c^2T} ={X''\over X} = \lambda \Rightarrow \cases{X''=\lambda X\\ T''=\lambda c^2T} \\ \text{boundary conditions: }\cases{u(0,t)=0\\ u(L,t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\ \text{Solving for }X(x): \cases{\lambda=0 \Rightarrow \text{trivial solution }X=0\\ \lambda \gt 0 \Rightarrow \text{trivial solution }X=0\\ \lambda=-\beta^2 \lt 0 \Rightarrow X''+\beta^2X=0 \Rightarrow X=c_1\cos \beta x+ c_2\sin \beta x} \\\qquad X(0)=c_1=0 \Rightarrow X(L)=c_2\sin \beta L=0 \Rightarrow \beta L=n\pi \Rightarrow \beta={n\pi\over L} \Rightarrow X_n=\sin {n\pi x\over L},n=1,2,\dots \\ \text{Solving for }T(t): T''=-\beta^2c^2T  \Rightarrow T_n= \cos {n\pi ct\over L}+ \sin {n\pi c t\over L}, n=1,2,\dots \\ \Rightarrow u(x,t) =\sum_{n=1}^\infty \sin{n\pi x\over L}(A_n\cos {n\pi ct\over L}+ B_n\sin {n\pi c t\over L})   \Rightarrow u(x,0) = \sin {3\pi x\over L}   \\ \Rightarrow B_n=0 \text{ and }A_n=\cases{1,n=3\\ 0, \text{otherwise}} \Rightarrow \bbox[red, 2pt]{u(x,t)=\sin{3\pi x\over L} \cos{3\pi c t\over L}}$$

$$\boxed{\textbf{第六題:擴散方程式(Diffusion equation) (15%)}}$$

解答:$$\text{Solving the problem by Fourier transform: } \mathcal F\{u_t\} =\mathcal F\{ku_{xx}\} \Rightarrow {\partial \over \partial t}\hat u(\omega ,t) =-k\omega^2 \hat u(\omega ,t)\\ \text{initial condition: }u(x,0)= \delta(x) \Rightarrow \mathcal F\{u(x,0)\} = \mathcal F\{\delta(x)\} \Rightarrow \hat u(\omega,0) =1 \\ \Rightarrow \cases{{d\over dt}\hat u(\omega, t)=-k\omega^2 \hat u(\omega ,t) \\ \hat u(\omega ,0)=1} \Rightarrow \hat u(\omega ,t)=A(\omega )e^{-k\omega^2 t} \Rightarrow \hat u(\omega,0)=1 =A(\omega) \Rightarrow \hat u(\omega ,t)=e^{-k\omega^2 t} \\ \Rightarrow u(x,t)= \mathcal F^{-1}\{e^{-k\omega^2 t} \} = {1\over 2\pi} \int_{-\infty}^\infty e^{-k\omega^2 t} \cdot e^{i\omega x} \,d\omega ={1\over 2\pi}e^{-x^2/4kt}  \int_{-\infty}^\infty  e^{-kt(\omega-ix/2kt)^2}\,d\omega \\= {1\over 2\pi}e^{-x^2/4kt}  \sqrt{\pi\over kt} \Rightarrow \bbox[red, 2pt]{u(x,t) ={1\over \sqrt{4\pi kt}}e^{-x^2/4kt}}$$

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解題僅供參考，碩士班歷年試題及詳解