115年成功大學地球科學-應用數學詳解

國立成功大學115學年度碩士班招生考試試題

系所:地球科學系
科目:應用數學

解答:$$\textbf{(a) }{dy\over dt}=-ky \Rightarrow \int{1\over y}\,dy= \int -k\,dt \Rightarrow \ln |y|= -kt+c_1 \Rightarrow y=c_2 e^{-kt} \\\quad \Rightarrow y(0)=c_2=0.5 \Rightarrow \bbox[red, 2pt]{y=0.5e^{-kt}} \\\textbf{(b) }y'=1+y^2 \Rightarrow \int {1\over 1+y^2} \,dy =\int 1\,dx \Rightarrow \tan^{-1}y= x+c_1 \Rightarrow \bbox[red, 2pt]{y=\tan(x+c_1)} \\\textbf{(c) }y'=-2xy \Rightarrow \int{1\over y}dy = \int -2x \,dx \Rightarrow \ln|y|=-x^2+c_1 \Rightarrow y=c_2e^{-x^2} \\\quad \Rightarrow y(0)=c_2=18 \Rightarrow \bbox[red,2pt]{y=18e^{-x^2}}$$

解答:$$\textbf{(a) } my''+cy'+ky=0 \Rightarrow mr^2+cr+k=0 \Rightarrow r= {-c\pm \sqrt{c^2-4km} \over 2m}  \\\qquad \left( r_1={-c+\sqrt{c^2-4km}\over 2m}, r_2={-c-\sqrt{c^2-4km}\over 2m} \right)\\\textbf{Case I: Overdamped }(c^2\gt 4mk) \Rightarrow y(t)=c_1e^{r_1t}+c_2 e^{r_2t} \\ \textbf{Case II: Critically Dampled }(c^2=4mk) \Rightarrow r={-c\over 2m} \Rightarrow y(t)=(c_1+c_2t)e^{rt} \\\textbf{Case III: Underdamped }(c^2 \lt 4mk) \Rightarrow \cases{\alpha=-c/2m\\ \beta=\sqrt{4mk-c^2}/2m} \Rightarrow y(t)=e^{\alpha t}(c_1 \cos \beta t+c_2\sin \beta t) \\\textbf{(b) } \text{Resonance cannot occur here. For kpure resonance to occur, two conditions must be met:} \\\text{ 1. An external periodic driving force must be applied} \\\text{2. The system must be undamped and driven at its natural frequency.} \\\textbf{(c) }\text{The circustance is that the damping coefficient is zero. That is, c=0.}$$

解答:$$u(x,t)= X(x)T(t) \Rightarrow XT''=c^2X''T \Rightarrow {T''\over c^2T} ={X''\over X}=-\lambda \Rightarrow \cases{X''+\lambda X=0\\ T''+\lambda c^2T=0} \\ \text{Solving for }X(x):\\\qquad BCs:\cases{u(0,t)= X(0)T(t)=0\\ u(L,t) = X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\\qquad \lambda \le 0 \Rightarrow X=0 \\\qquad \lambda=k^2 \gt 0 \Rightarrow X=A\cos(kx)+B\sin(kx) \Rightarrow X(0)=A=0 \Rightarrow X(L)=B\sin(kL)=0\\\qquad \Rightarrow kL= n\pi \Rightarrow k={n\pi\over L} \Rightarrow X_n=\sin{n\pi x\over L}, n=1,2,\dots \\\text{Solving for }T(t): T''+c^2k^2 T=0 \Rightarrow T''+{c^2n^2\pi^2 \over L^2}T=0 \Rightarrow T_n = B_n \cos {cn\pi t\over L}+D_n \sin{cn\pi t\over L} \\ \Rightarrow u(x,t)= \sum_{n=1}^\infty \left( B_n \cos {cn\pi t\over L}+D_n \sin{cn\pi t\over L} \right) \sin{n\pi x\over L} \Rightarrow u(x,0) =f(x)= \sum_{n=1}^\infty B_n \sin{n\pi x\over L} \\ \Rightarrow \bbox[red, 2pt]{B_n= {2\over L}\int_0^L f(x)\sin{n\pi x\over L}\,dx }\\ u'(x,0)=g(x) \Rightarrow \text{ If }g(x)=0 \Rightarrow D_n=0 \Rightarrow \bbox[red, 2pt]{u(x,t) = \sum_{n=1}^\infty B_n \cos{cn\pi t\over L} \sin {n\pi x\over L}}$$

解答:$$F(-x)=F(x) \Rightarrow F(x)\text{ is even} \Rightarrow b_n=0\\ a_0={1\over 4} \int_{-1}^1 k\,dx ={k\over 2} \\ a_n={2\over 4} \int_{-1}^1 k \cos{n\pi x\over 2}\,dx ={2k\over n\pi} \sin{n\pi\over 2} \\ F(x)= a_0+ \sum_{n=1}^\infty a_n\cos{n\pi x\over 2} =\bbox[red, 2pt]{{k\over 2}+ {2k\over\pi}\sum_{n=1}^\infty {1\over n}\sin {n\pi\over 2} \cos{n\pi x\over 2}}$$

解答:$$\textbf{(a) } I=\int_{-\pi}^\pi \cos (nx)\cos (mx) = {1\over 2} \int_{-\pi}^\pi [\cos(n-m)x+ \cos(n+m)x] \,dx \\\quad \textbf{Case I: }n\ne m \Rightarrow I= {1\over 2}\left. \left[ {\sin(n-m)x\over (n-m)}+ {\sin(n+m)x\over (n+m)} \right] \right|_{-\pi}^\pi =0\\ \textbf{Case II: }n=m\ne 0 \Rightarrow I=\int_{-\pi}^\pi \cos^2(nx)\,dx ={1\over 2} \int_{-\pi}^\pi(1+\cos(2nx))\,dx = {1\over 2} \left. \left[ x+{\sin(2nx) \over 2n} \right] \right|_{-\pi}^\pi =\pi \\ \textbf{Case III: }n=m=0 \Rightarrow I=\int_{-\pi}^\pi1\,dx =2\pi \\ \Rightarrow I=\bbox[red, 2pt]{ \begin{cases}0,& n\ne m\\ \pi, &n=m \ne 0\\ 2\pi, &n=m=0 \end{cases}} \\ \textbf{(b) } \cases{\cos(nx) \text{ is even} \\ \sin(mx) \text{ is odd}} \Rightarrow \cos(nx) \sin(mx) \text{ is odd} \Rightarrow \int_{-\pi}^\pi \cos(nx) \sin(mx)\,dx = \bbox[red, 2pt]0 \\ \textbf{(c) }I= \int_{-\pi}^\pi \sin(nx) \sin(mx)\,dx = {1\over 2} \int_{-\pi}^\pi [\cos(n-m)x- \cos(n+m)x]\,dx \\ \textbf{Case I: }n\ne m \Rightarrow I= {1\over 2} \left. \left[ {\sin(n-m)x\over n-m}-{\sin(n+m)x\over n+m} \right] \right|_{-\pi}^\pi =0 \\ \textbf{Case II: }n=m\ne 0 \Rightarrow I= {1\over 2}\int_{-\pi}^\pi [1-\cos(2nx)]',dx = {1\over 2} \left. \left[ x-{\sin(2nx)\over 2n} \right] \right|_{-\pi}^\pi =\pi \\\textbf{Case III: }n=m=0 \Rightarrow I=\int_{-\pi}^\pi 0\,dx =0 \\ \Rightarrow I= \bbox[red, 2pt]{\begin{cases}0,& n\ne m\\ \pi,& n=m\ne 0\\ 0,& n=m=0 \end{cases}} \\ 題意未清,\bbox[cyan, 2pt]{假設n,m皆為整數}$$

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解題僅供參考，碩士班歷年試題及詳解