解答:\mathbf{(a)}\;\vec r(\theta,\phi)=(2\cos \phi \sin \theta,2\sin \phi \sin \theta, 2\cos \theta) \Rightarrow \vec n= \vec r_\theta \times \vec r_\phi \\= (4\sin^2 \theta \cos \phi,4\sin^2\theta \sin \phi, 4\sin \theta \cos \theta) \Rightarrow \iint_S (\vec F\cdot \vec n)dA \\= \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi} (14\sin \theta \cos \phi,0,-2\cos\theta) \cdot (4\sin^2 \theta \cos \phi,4\sin^2\theta \sin \phi, 4\sin \theta \cos \theta) d\theta \\=\int_{\theta=0}^\pi \int_{\phi=0}^{2\pi} 56\cos^2\phi \sin^2\theta-8\cos^2\theta \sin \theta\,d\phi d\theta =\bbox[red, 2pt]{64\pi}\\ \mathbf{(b)}\;\vec F=7x\vec i-z\vec k \Rightarrow div \vec F=F_x+F_y+ F_z=7+0-1=6 \\ \Rightarrow \iint_S (\vec F\cdot \vec n)dA = \iiint_R div \vec F\,dV=\iiint_R6\,dV = 6\cdot {4\over 3}\pi \cdot 2^3= \bbox[red,2pt]{64\pi}
解答:
令 u( x,t) =F( x) G( t)，則 \begin{cases} u_{tt} =FG''\\u_{xx} =F''G \end{cases}，代回原式可得 FG''=c^{2} F''G \Rightarrow \frac{G''}{c^{2} G} =\frac{F''}{F}\\由於 \frac{G''}{c^{2} G} 只含變數 t ，而 \frac{F''}{F} 只含變數 x ，兩者等值代表同為一常數 k ，即\frac{G''}{c^{2} G} =\frac{F''}{F} =k\\則邊界條件:\cases{u(0,t) = F(0)G(t)=0\\ u(L,t)=F(L)G(t)=0}，若G(t)=0，則u(x,t)=0為明顯解，不列入討論\\因此邊界條件轉變成F(0)=F(L)=0；\\F''-kF=0 \Rightarrow F=c_1e^{\sqrt kx}+ c_2e^{-\sqrt k x} ，代入初始值\Rightarrow \cases{F(0)=c_1+c_2=0 \cdots(1)\\ F(L)= c_1e^{\sqrt kL}+ c_2e^{-\sqrt k L} =0 \cdots(2)}\\ 由(1)得c_2=-c_1代入(2) \Rightarrow c_1(e^{\sqrt kL}-e^{-\sqrt k L})=0 \Rightarrow c_1=0或e^{\sqrt kL}＝e^{-\sqrt k L}\\ 若c_1=0 \Rightarrow c_2=0 \Rightarrow F=0為明顯解，不列入討論;因此只需考慮e^{\sqrt kL}＝e^{-\sqrt k L}\\ 而e^{\sqrt kL}\ne e^{-\sqrt k L},\forall k\gt 0;若k=0 \Rightarrow F=0為明顯解，不列入討論\\ 因此只有k\lt 0,此時F(x)= c_1(e^{i\sqrt{|k|}x}-e^{-i\sqrt{|k|}x}) =2ic_1 \sin (\sqrt{|k|}x) \\ \Rightarrow F(L)=2ic_1 \sin (\sqrt{|k|}L) =0 \Rightarrow \sqrt{|k|}L=n\pi,n\in \mathbb N \Rightarrow \sqrt{|k|}=n\pi /L \\ \Rightarrow F(x)=2ic_1 \sin (n\pi x/L) =A \sin\left({n\pi \over L}x\right),A為常數\\ k=-{n^2\pi^2 \over L^2} \Rightarrow G''+{n^2\pi^2 \over L^2}c^2G=0 \Rightarrow G(t)=c_3 \cos\left({n\pi \over L}ct\right) +c_4\sin \left({n\pi \over L}ct\right)\\因此u(x,t)=A \sin\left({n\pi \over L}x\right)\left( c_3 \cos\left({n\pi \over L}ct\right) +c_4\sin \left({n\pi \over L}ct\right)\right) \\=\sum_{n=1}^\infty \sin\left({n\pi \over L}x\right)\left( A_n \cos\left({n\pi \over L}ct\right) +B_n\sin \left({n\pi \over L}ct\right)\right)
初始條件1: u(x,0)=f(x) \Rightarrow f(x)=\sum_{n=1}^\infty A_n \sin({n\pi \over L}x) \\\Rightarrow \int_0^L f(x)\sin({k\pi \over L}x)\;dx = \sum_{n=1}^\infty \int_0^L A_n\sin({k\pi \over L}x)\sin({n\pi \over L}x)\;dx =\int_0^L A_k \sin^2 ({k\pi \over L}x)\;dx ={L\over 2}A_k\\ \Rightarrow A_k={2\over L}\int_0^L f(x)\sin({k\pi \over L}x)\;dx \Rightarrow A_n={2\over L}\int_0^L f(x)\sin({n\pi \over L}x)\;dx\\ ={2\over L}\left(\int_0^{L/2} {2k\over L} x\sin ({n\pi \over L}x)\,dx +\int_{L/2}^L {2k\over L}(L-x)\sin ({n\pi\over L}x)\,dx\right) \\\Rightarrow \bbox[cyan,2pt]{A_n ={8k \over n\pi}\left( {1\over n\pi}\sin({n\pi \over 2}) -\cos ({n\pi \over 2})+ \cos(n\pi)\right)}
初始條件2: \left. {\partial u\over \partial t}\right|_{t=0} =0 \Rightarrow \sum_{n=1}^\infty \left( -{cn\pi \over L}A_n \sin({cn\pi \over L}t) +{cn\pi \over L}B_n\cos ({cn\pi \over L}t) \right)\sin({n\pi \over L}x)|_{t=0} \\ =\sum_{n=1}^\infty {cn\pi \over L}B_n\sin({n\pi \over L}x) =0 \Rightarrow B_n=0\\ \Rightarrow u(x,t)=\sum_{n=1}^\infty \sin\left({n\pi \over L}x\right)\left( A_n \cos\left({n\pi \over L}ct\right) \right) \\ \Rightarrow \bbox[red, 2pt] {u(x,t) = \sum_{n=1}^\infty {8k \over n\pi}\left( {1\over n\pi}\sin({n\pi \over 2}) -\cos ({n\pi \over 2})+ \cos(n\pi)\right)\sin({n\pi \over L}x) \cos({n\pi\over L}ct)}
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解題僅供參考,其他歷年試題及詳解
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