求一維波動方程式 one-dimentional wave equation 之解,即∂2∂t2u=c2∂2∂x2u,滿足邊界條件{u(0,t)=0u(L,t)=0,∀t及初始條件{u(x,0)=f(x)∂u∂t|t=0=g(x)
解答:令u(x,t)=F(x)G(t),則{utt=FG″uxx=F″G,代回原式可得FG″=c2F″G⇒G″c2G=F″F由於G″c2G只含變數t,而F″F只含變數x,兩者等值代表同為一常數k,即G″c2G=F″F=k則邊界條件:{u(0,t)=F(0)G(t)=0u(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⇒F=c1e√kx+c2e−√kx,代入初始值⇒{F(0)=c1+c2=0⋯(1)F(L)=c1e√kL+c2e−√kL=0⋯(2)由(1)得c2=−c1代入(2)⇒c1(e√kL−e−√kL)=0⇒c1=0或e√kL=e−√kL若c1=0⇒c2=0⇒F=0為明顯解,不列入討論;因此只需考慮e√kL=e−√kL而e√kL≠e−√kL,∀k>0;若k=0⇒F=0為明顯解,不列入討論因此只有k<0,此時F(x)=c1(ei√|k|x−e−i√|k|x)=2ic1sin(√|k|x)⇒F(L)=2ic1sin(√|k|L)=0⇒√|k|L=nπ,n∈N⇒√|k|=nπ/L⇒F(x)=2ic1sin(nπx/L)=Asin(nπLx),A為常數k=−n2π2L2⇒G″+n2π2L2c2G=0⇒G(t)=c3cos(nπLct)+c4sin(nπLct)因此u(x,t)=Asin(nπLx)(c3cos(nπLct)+c4sin(nπLct))=∞∑n=1sin(nπLx)(Ancos(nπLct)+Bnsin(nπLct))
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\\初始條件2: \left. {\partial u\over \partial t}\right|_{t=0} =g(x) \Rightarrow g(x) = \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) \Rightarrow \sin({k\pi \over L}x)g(t) =\sum_{n=1}^\infty {cn\pi \over L}B_n\sin({k\pi \over L}x)\sin({n\pi \over L}x) \\ \Rightarrow \int_0^L \sin({k\pi \over L}x)g(x)\;dx = \sum_{n=1}^\infty \int_0^L {cn\pi \over L}B_n\sin({k\pi \over L}x)\sin({n\pi \over L}x)\;dx \\={ck\pi \over L}B_k\int_0^L \sin^2({k\pi \over L}x)\; dx={ck\pi \over L}B_k \cdot {L\over 2} ={ck\pi \over 2}B_k \Rightarrow B_k= {2\over ck\pi }\int_0^L \sin({k\pi \over L}x)g(x)\;dx\\ \Rightarrow B_n= {2\over cn\pi }\int_0^L \sin({n\pi \over L}x)g(x)\;dx \\ \Rightarrow \bbox[red,2pt]{ u(x,t)=\sum_{n=1}^\infty \left(A_n \cos({cn\pi\over L }t) +B_n \sin({cn\pi\over L }t)\right) \sin({n\pi \over L}x)},其中\\A_n={2\over L}\int_0^L f(x)\sin({n\pi \over L}x)\;dx,B_n= {2\over cn\pi }\int_0^L \sin({n\pi \over L}x)g(x)\;dx
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