111年成大土木碩士班-工程數學詳解

國立成功大學111學年度碩士班招生考試

系所: 土木工程系
科目: 工程數學

解答: $$\textbf{(1)}\; \text{div}(f\textbf{u})= (\nabla f)\cdot \textbf{u}+f(\nabla \textbf{u}) =(4x\textbf{i}+2y\textbf{j} +2z\textbf{k}) \cdot (xz\textbf{i}+yz\textbf{k}) +(2x^2+y^2+z^2)(z+y) \\\qquad =(4x^2z+2yz^2) +(2x^2z+2x^2y+ y^2z+y^3+ z^3+yz^2) \\\qquad= \bbox[red, 2pt]{6x^2z+ 3 yz^2+2x^2y+ y^2z+y^3+z^3} \\\textbf{(2)}\; \nabla^2 f =4+2+2=\bbox[red, 2pt]8 \\ \textbf{(2)}\; \text{curl(grad }f) =\text{curl}(4x\textbf{i} +2y \textbf{j}+ 2z\textbf{k}) =\begin{vmatrix}\textbf{i} & \textbf{j} &\textbf{k} \\\frac{\partial }{\partial x} & \frac{\partial  }{\partial y} & \frac{\partial  }{\partial z} \\ 4x& 2y& 2z\end{vmatrix} =\bbox[red,2pt]0 \\ \textbf{(4)}\; \text{div(curl }\textbf u) = \text{div}\left( \begin{vmatrix} \textbf{i} & \textbf{j} &\textbf{k} \\\frac{\partial }{\partial x} & \frac{\partial  }{\partial y} & \frac{\partial  }{\partial z} \\ xz& 0& yz\end{vmatrix}\right) =\text{div}(z\textbf i+x\textbf j) =\bbox[red, 2pt]0 \\\textbf{(5)}\; \text{grad}(\textbf u\cdot \textbf u) =\text{grad}(x^2z^2+y^2z^2) = \bbox[red, 2pt]{2xz^2\textbf i+2yz^2 \textbf j+2z(x^2+y^2)\textbf k}$$

解答: $$\textbf{(1)}\; x'+{x\over t}=2\cos(t) \Rightarrow tx'+x=2t\cos(t) \Rightarrow (tx)'=2t\cos(t) \Rightarrow tx= \int 2t\cos(t)\,dt\\\quad \Rightarrow tx=2t\sin(t)+2\cos(t)+c_1 \Rightarrow \bbox[red, 2pt]{x(t)=2\sin(t)+{2\over t}\cos(t)+{c_1\over t}} \\\textbf{(2)}\; {dy\over dx} =9y^2x^4 \Rightarrow {1\over y^2}dy=9x^4dx \Rightarrow -{1\over y}={9\over 5}x^5+c_1 ={9x^5+c_2\over 5} \Rightarrow y(x)=-{5\over 9x^5+c_2} \\\quad \Rightarrow y(2)=-{5\over 288+c_2}=6 \Rightarrow c_2=-{1733\over 6} \Rightarrow \bbox[red, 2pt]{y(x)={30\over 1733-54x^5}}$$

解答: $$\textbf{(1)}a_0={1\over 2\pi} \int_0^\pi \sin(2t)\,dt = {1\over 2\pi}\cdot \left. \left[ -{1\over 2} \cos(2t)\right] \right|_0^\pi =0\\ a_n ={1\over \pi} \int_0^\pi \sin(2t)\cos(nt)\,dt ={2 \over (n^2-4) \pi} \left( (-1)^{n }-1 \right) \\ b_n={1\over \pi} \int_0^\pi \sin(2t)\sin(nt) \,dt =0 \\ \Rightarrow \bbox[red, 2pt]{f(x)={2\over   \pi} \sum_{n=1}^\infty  {1\over n^2-4}\left( (-1)^n- 1\right)\cos(nx))} \\\textbf{(2)} A(\omega) =\int_{-\infty}^\infty f(x)\cos(\omega x)\,dx = \int_0^\pi \sin(2x) \cos(\omega x)\,dx = {2\over \omega^2-4}(\cos(\omega \pi)-1) \\ B( \omega)= \int_{-\infty}^\infty f(x)\sin(\omega x)\,dx =\int_0^\pi \sin(2x) \sin(\omega x)\,dx ={2\sin(\omega \pi)\over \omega^2-4} \\ \Rightarrow \bbox[red, 2pt]{f(x)={1\over \pi} \int_0^\infty \left({2\over \omega^2-4}(\cos(\omega \pi)-1) \cos(\omega x) + {2\sin(\omega \pi) \over \omega^2-4} \sin(\omega x)\right)d\omega}$$

解答: $$\det(A) =-(\lambda-1)(\lambda-2)(\lambda-3)=0 \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: }1,2,3} \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \left(\begin{matrix} 0 & 0 & 0 \\1 & 1 & 0 \\1 & 2 & 2 \end{matrix} \right) \begin{pmatrix} x_1 \\ x_2 \\x_3 \end{pmatrix} =0 \Rightarrow \cases{x_1=2x_3\\ x_2=-2x_3} \Rightarrow v=x_3\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} ,\\ \qquad 取v_1=\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \\\lambda_2=1 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \left( \begin{matrix} -1 & 0 & 0 \\1 & 0 & 0 \\1 & 2 & 1\end{matrix} \right)  \begin{pmatrix} x_1 \\ x_2 \\x_3 \end{pmatrix} =0 \Rightarrow \cases{x_1=0\\ 2x_2=-x_3} \Rightarrow v= x_3\begin{pmatrix} 0 \\ -{1\over 2} \\ 1 \end{pmatrix} ,\\ \qquad 取v_2 = \begin{pmatrix} 0 \\ -{1\over 2} \\ 1 \end{pmatrix} \\ \lambda_3=3  \Rightarrow (A-\lambda_3 I)v =0 \Rightarrow \left( \begin{matrix}-2 & 0 & 0 \\1 & -1 & 0 \\1 & 2 & 0\end{matrix} \right) \begin{pmatrix} x_1 \\ x_2 \\x_3 \end{pmatrix} =0 \Rightarrow \cases{x_1=0\\ x_2=0} \Rightarrow v=x_3 \begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}\\ \qquad 取v_3=\begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{ \text{eigenvectors: }\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ -{1\over 2} \\ 1 \end{pmatrix} ,\begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}}$$

解答: $$\textbf{(1)}\; u_x-72u_t-10u=0 \Rightarrow \cases{a=1\\ b=-72} \\ 令\cases{i=x\\ s=-bx+at=72x+t} \Rightarrow \cases{u_x=u_ii_x +u_ss_x =u_i+72u_s\\ u_t=u_ii_t+ u_ss_t =0+u_s} 代回原式\\\Rightarrow u_i+72u_s=72u_s+10u \Rightarrow u_i-10u=0 \Rightarrow \text{integration factor: }e^{-10i} \\ \Rightarrow e^{-10i}u_i-10e^{-10i}u=0 \Rightarrow \left( e^{-10i}u\right)'=0 \Rightarrow e^{-10i}u=\rho(s) \Rightarrow u(i,s)=e^{10i}\rho(s) \\ \Rightarrow u(x, t)=e^{10x} \rho(72x+t) \Rightarrow u_x=10e^{10x} \rho(72x+t) +72e^{10x} \rho'(72x+t) \\ \Rightarrow u_x(t=0)=10e^{10x} \rho(72x) +72e^{10x} \rho'(72x) =6e^{-2x} \Rightarrow \left( e^{10x} \rho(72x) \right)'=6e^{-2x} \\ \Rightarrow e^{10x} \rho(72x) =-3e^{-2x} +c_1 \Rightarrow \rho(\omega)=-3e^{-\omega/ 6}+ c_1e^{-5\omega/36} \\ \Rightarrow u(x, t)=e^{10x} \rho(72x+t)= u(x, t)=e^{10x} \left( -3e^{-12x-t/6 +c_1 e^{-10x-5t./36}}\right) \\ \Rightarrow \bbox[red, 2pt]{u(x,t)=-3e^{-2x-t/6}+ c_1e^{-5t/36}}\\ \textbf{(2)}\; \bbox[cyan,2pt]{題目有誤}, u(0,y)=5e^{2/x}不應該有變數x!!$$

 

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