109年高考三級-應用數學詳解

109年公務人員高等考試三級考試試題

類科:氣象學
科目:應用數學(包括微積分、微分方程與向量分析)

解：

$$矩陣A為正定\Leftrightarrow A的特徵值均為正數，因此我們先求矩陣的特徵值:\\ A=\begin{bmatrix}a & b & b& b\\ b & a & b& b\\ b& b& a& b\\ b& b& b& a\end{bmatrix} \Rightarrow det(A-\lambda I)=0 \Rightarrow \begin{vmatrix} a-\lambda & b & b& b\\ b & a-\lambda & b& b\\ b& b& a-\lambda& b\\ b& b& b& a-\lambda \end{vmatrix}=0 \\ \Rightarrow \lambda^4-4a\lambda^3+ (6a^2-6b^2)\lambda^2-(4a^3+8b^3-12ab^2)\lambda+ (a^4-3b^4+8ab^3-6a^2b^2)=0 \\ \Rightarrow (\lambda-(a-b))^3(\lambda-(a+3b))=0 \Rightarrow 特徵值\lambda=a-b,a+3b\\ 因此該矩陣為正定的充分必要條件為\bbox[red, 2pt]{a>b 且 a+3b > 0}$$

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解：

$$\cases{u'(t)=2u(t)-v(t)+e^t\cdots(1a)\\ v'(t)=3u(t)-2v(t)-e^t\cdots(1b)\\ u(0)=1,v(0)=2}，由(1a)可得 v(t)=2u(t)-u'(t)+e^t\cdots(2a)，代入(1b)\\ \Rightarrow 2u'(t)-u''(t)+e^t=3u(t)-4u(t)+2u'(t)-2e^t-e^t \Rightarrow u''(t)-u(t)=4e^t\cdots(3a)\\ 同理，由(1b)可得 u(t)={1\over 3}(v'(t)+2v(t)+e^t)\cdots(2b)，代入(1a)\Rightarrow v''(t)-v(t)=4e^t\cdots(3b)\\\\將初始值\cases{u(0)=1\\ v(0)=2} 代入(2b)及(2a) \Rightarrow \cases{1={1\over 3}(v'(0)+4+1) \\2=2-u'(0)+1} \Rightarrow \cases{u'(0)=1\\ v'(0)=-2}\\ \cases{u''(t)-u(t)=4e^t\cdots(3a) \\v''(t)-v(t)=4e^t\cdots(3b)} \Rightarrow \cases{u(t)=C_1e^t+C_2e^{-t}+ 2te^t\cdots(4a)\\ v(t)=C_3e^t+C_4e^{-t}+ 2te^t\cdots(4b)}\\ \Rightarrow  \cases{u'(t)=C_1e^t-C_2e^{-t}+2e^t+ 2te^t\cdots(5a)\\ v'(t)=C_3e^t-C_4e^{-t}+2e^t+ 2te^t\cdots(5b)};\\最後將\cases{u(0)=1\\ v(0)=2}代入(4a)及(4b)，及\cases{u'(0)=1\\ v'(0)=-2}代入(5a)及(5b) \Rightarrow \cases{C_1+C_2= 1\\ C_3+C_4=2 \\ C_1-C_2=-1\\ C_3-C_4=-4} \\ \Rightarrow \cases{C_1=0\\C_2=1\\ C_3=-1\\C_4=3} \Rightarrow \bbox[red, 2pt]{\cases{u(t)=e^{-t}+2te^t\\ v(t)=-e^t+3e^{-t}+2te^t}}$$

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解：

$$令\vec F=\left[ \matrix{F_1\\ F_2\\ F_3}\right],\vec G=\left[ \matrix{G_1\\ G_2\\ G_3}\right] \Rightarrow \vec F\times \vec G =\left[ \matrix{F_2G_3-F_3G_2\\ F_3G_1-F_1G_3\\ F_1G_2-F_2G1}\right] \Rightarrow \text{div}(\vec F\times \vec G)= \left[ \matrix{D_x\\ D_y\\ D_z}\right] \cdot \left[ \matrix{F_2G_3-F_3G_2\\ F_3G_1-F_1G_3\\ F_1G_2-F_2G_1}\right]\\ 又(\text{curl}\vec F)\cdot \vec G-\vec F\cdot (\text{curl}\vec G)=\left(\left[ \matrix{D_x\\ D_y\\ D_z}\right]\times\left[ \matrix{F_1\\ F_2\\ F_3}\right] \right) \cdot \left[ \matrix{G_1\\ G_2\\ G_3}\right]-\left[ \matrix{F_1\\ F_2\\ F_3}\right] \cdot \left(\left[ \matrix{D_x\\ D_y\\ D_z}\right]\times\left[ \matrix{G_1\\ G_2\\ G_3}\right] \right) \\= \left[ \matrix{D_yF_3-D_zF_2\\ D_zF_1-D_xF_3\\ D_xF_2-D_yF_1}\right] \cdot \left[ \matrix{G_1\\ G_2\\ G_3}\right]-\left[ \matrix{F_1\\ F_2\\ F_3}\right] \cdot \left[ \matrix{D_yG_3-D_zG_2\\ D_zG_1-D_xG_3\\ D_xG_2-D_yG_1}\right] \\ =D_yF_3G_1-D_zF_2G_1+ D_zF_1G_2-D_xF_3G_2+ D_xF_2G_3-D_yF_1G_3 - \\ \qquad \qquad(F_1D_yG_3-F_1D_zG_2+ F_2D_zG_1-F_2D_xG_3+ F_3D_xG_2-F_3D_yG_1)\\ = (D_xF_2G_3+F_2D_xG_3-D_xF_3G_2) +(D_yF_3G_1+F_3D_yG_1-D_yF_1G_3-F_1D_yG_3) \\ \qquad \qquad +(D_zF_1G_2+F_1D_zG_2-D_zF_2G_1-F_2D_zG_1) \\ =D_x(F_2G_3-F_3G_2)+D_y(F_3G_1-F_1G_3) +D_z(F_1G_2-F_2G_1)\\ =\left[ \matrix{D_x\\ D_y\\ D_z}\right] \cdot \left[ \matrix{F_2G_3-F_3G_2\\ F_3G_1-F_1G_3\\ F_1G_2-F_2G_1}\right] =\text{div}(\vec F\times \vec G)，故得證$$

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解：

(一) $$f(x)=a_0+ \sum_{n=1}^\infty \left(a_n \cos {n\pi x\over L} +b_n\sin {n\pi x\over L} \right)，其中 \cases{a_0={1\over 2L}\int_{-L}^L f(x)\;dx\\ a_n={1\over L}\int_{-L}^Lf(x)\cos {n\pi x\over L}\;dx,n=1,2,...\\ b_n={1\over L}\int_{-L}^Lf(x)\sin {n\pi x\over L}\;dx,n=1,2,...}$$ (二) $$基底函數\phi (x)=1,\sin {n\pi x\over L}, \cos{n\pi x\over L}，其中n=1,2,...\\ \phi (x)=1\Rightarrow \int_{-L}^L \phi (x)^2\;dx =\int_{-L}^L 1\;dx = 2L;\\ \phi (x)=\sin {n\pi x\over L} \Rightarrow \int_{-L}^L \phi (x)^2\;dx =\int_{-L}^L \sin^2 {n\pi x\over L}\;dx= \left. \left[ {x\over 2}-{L\over 4n\pi} \sin{2n\pi x\over L}\right] \right|_{-L}^L={L\over 2}-(-{L\over 2})=L;\\ \phi (x)=\cos {n\pi x\over L} \Rightarrow \int_{-L}^L \phi (x)^2\;dx =\int_{-L}^L \cos^2 {n\pi x\over L}\;dx=\left. \left[ {x\over 2}+{L\over 4n\pi} \sin{2n\pi x\over L}\right] \right|_{-L}^L={L\over 2}-(-{L\over 2})=L\\ \Rightarrow \bbox[red, 2pt]{\cases{\phi (x)=1\Rightarrow \int_{-L}^L \phi (x)^2\;dx=2L\\ \phi (x)=\sin {n\pi x\over L} \Rightarrow \int_{-L}^L \phi (x)^2\;dx =L\\ \phi (x)=\cos {n\pi x\over L} \Rightarrow \int_{-L}^L \phi (x)^2\;dx =L}}$$ (三) $$f(x)=a_0+ \sum_{n=1}^\infty \left(a_n \cos {n\pi x\over L} +b_n\sin {n\pi x\over L} \right)\\將上式同乘f(x)再積分(x=-L至x=L) \Rightarrow \\\int_{-L}^Lf^2(x)\;dx = a_0\int_{-L}^Lf(x)\;dx+ \sum_{n=1}^\infty \left(a_n \int_{-L}^L\cos {n\pi x\over L} \;dx+b_n\int_{-L}^L\sin {n\pi x\over L} \;dx\right) \\ =2La_0^2+\sum_{n=1}^\infty \left(La_n^2 +Lb_n^2\right) = \bbox[red, 2pt]{L\cdot 2a_0^2+ L\cdot\sum_{n=1}^\infty(a_n^2+b_n^2) }$$

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解：

$$首先坐標轉換，將(x,y,z)\to (r,\theta,\varphi),其中\cases{r=\sqrt{x^2+y^2+z^2}\\  \theta =\arccos {z\over r} \\ \varphi=\arctan{y\over x} }\\ 因此\left({\partial^2 \over \partial x^2}+ {\partial^2 \over \partial y^2}+ {\partial^2 \over \partial z^2}\right)w=0 \equiv w_{xx}+w_{yy}+w_{zz}=0 \\\xrightarrow{坐標轉換} {1\over r^2}\cdot{\partial \over \partial r}\left( r^2{\partial w\over \partial r}\right) +{1\over r^2\sin \theta}\cdot {\partial \over \partial \theta}\left(\sin\theta {\partial w\over \partial \theta} \right)+{1\over r^2\sin^2\theta } \cdot {\partial^2 w\over \partial \varphi^2} =0\cdots(1)\\ 由於w(\sqrt{x^2+y^2+z^2}) =w(r)代表w的變數只有r,沒有\varphi及\theta，也就是\cases{{\partial w\over \partial \varphi}=0\\ {\partial w\over \partial \theta}=0}\\ 因此(1)可減化成{1\over r^2}\cdot{\partial \over \partial r}\left( r^2{\partial w\over \partial r}\right) =0 \Rightarrow {1\over r^2}\cdot\left( 2r{\partial w\over \partial r} +r^2{\partial^2 w\over \partial r^2}\right) =0 \\ \Rightarrow 2r{\partial w\over \partial r} +r^2{\partial^2 w\over \partial r^2}=0 \Rightarrow 2{\partial w\over \partial r} +r{\partial^2 w\over \partial r^2}=0(相當於求:xy''+2y'=0的ODE)\\ \Rightarrow w=c_1r^0+c_2r^{-1} \Rightarrow \bbox[red, 2pt]{w=c_1+{c_2\over r},c_1,c_2為常數}$$

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