Archivo: Membrane exampleA

Descripción: Graphical representation of the solution to the wave equation for 2D membrane bounded by a rectangular region given by: ∇2u=1c2utt (0≤x≤π), (0≤y≤π), (t≥0){\displaystyle \nabla ^{2}u={\frac {1}{c^{2}}}u_{tt}\ (0\leq x\leq \pi ),\ (0\leq y\leq \pi ),\ (t\geq 0)} where c = 1 subject to the boundary conditions: u(0,y,t)=0, u(π,y,t)=0u(x,0,t)=0, u(x,π,t)=0{\displaystyle {\begin{aligned}u(0,y,t)=0,&\ u(\pi ,y,t)=0\\u(x,0,t)=0,&\ u(x,\pi ,t)=0\end{aligned}}} with the initial displacement of the membrane given by: u(x,y,0)=x(π−x)y(π−y)ut(x,y,0)=0{\displaystyle {\begin{aligned}u(x,y,0)&=x(\pi -x)y(\pi -y)\\u_{t}(x,y,0)&=0\end{aligned}}} The solution is: u(x,y,t)=∑n=1∞∑m=1∞Cmnsin⁡[(2n−1)x]sin⁡[(2m−1)y]cos⁡(ωmnt){\displaystyle u(x,y,t)=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{mn}\sin[(2n-1)x]\sin[(2m-1)y]\cos(\omega _{mn}t)} where: ωmn=m2+n2{\displaystyle \omega _{mn}={\sqrt {m^{2}+n^{2}}}} Cmn=4π2∫0πsin⁡(mx)∫0πx(x−π)y(y−π)sin⁡(ny) dy dx=4×4(−1+(−1)m)(−1+(−1)n)π2m3n3; n,m=1,2,…=64π2(2m−1)3(2n−1)3; n,m=1,2,…{\displaystyle {\begin{aligned}C_{mn}&={\frac {4}{\pi ^{2}}}\int _{0}^{\pi }\sin(mx)\int _{0}^{\pi }x(x-\pi )y(y-\pi )\sin(ny)\ dy\ dx\\&={\frac {4\times 4\left(-1+(-1)^{m}\right)\left(-1+(-1)^{n}\right)}{\pi ^{2}m^{3}n^{3}}};\ n,m=1,2,\ldots \\&={\frac {64}{\pi ^{2}(2m-1)^{3}(2n-1)^{3}}};\ n,m=1,2,\ldots \end{aligned}}}
Título: Membrane exampleA
Créditos: Trabajo propio
Autor(a): User:Wtt~commonswiki
Términos del Uso: Creative Commons Attribution-Share Alike 3.0
Licencia: CC-BY-SA-3.0
Enlace de Licencia: http://creativecommons.org/licenses/by-sa/3.0/
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