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2022年1月29日 星期六

PDE utt = c^2uxx 詳細求解過程

 求一維波動方程式 one-dimentional wave equation 之解,即2t2u=c22x2u滿{u(0,t)=0u(L,t)=0,t{u(x,0)=f(x)ut|t=0=g(x)

解答u(x,t)=F(x)G(t){utt=FGuxx=FGFG=c2FGGc2G=FFGc2GtFFxkGc2G=FF=k:{u(0,t)=F(0)G(t)=0u(L,t)=F(L)G(t)=0G(t)=0u(x,t)=0F(0)=F(L)=0FkF=0F=c1ekx+c2ekx{F(0)=c1+c2=0(1)F(L)=c1ekL+c2ekL=0(2)(1)c2=c1(2)c1(ekLekL)=0c1=0ekLekLc1=0c2=0F=0;ekLekLekLekL,k>0;k=0F=0k<0,F(x)=c1(ei|k|xei|k|x)=2ic1sin(|k|x)F(L)=2ic1sin(|k|L)=0|k|L=nπ,nN|k|=nπ/LF(x)=2ic1sin(nπx/L)=Asin(nπLx),Ak=n2π2L2G+n2π2L2c2G=0G(t)=c3cos(nπLct)+c4sin(nπLct)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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