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Question

Find the deflection u(x, y, t) of the square membrane of side π and c²=1 for initial velocity 0 and initial deflection 0.1 xy(π-x)(π-y)

Solution

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Answered 2 years ago
Answered 2 years ago
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The solution u(x,y,t)u(x,y,t) is represented via double Fourier series

u(x,y,t)=m=1+n=1+(Bmncosλmnt+Bmnsinλmnt)sinmπxasinnπyb,\boxed{u(x,y,t)=\sum_{m=1}^{+\infty}\sum_{n=1}^{+\infty}\left(B_{mn}\cos\lambda_{mn}t+B_{mn}^*\sin\lambda_{mn}t\right)\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b},}

where a=b=πa=b=\pi, \ c2=1c^2=1 and λmn=πm2a2+n2b2\lambda_{mn}=\pi\sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}. The initial deflection f(x,y)f(x,y) is given as f(x,y)=0.1xy(πx)(πy)f(x,y)=0.1xy(\pi-x)(\pi-y), while the initial velocity is g(x,y)=0=tu(x,y,t)t=0g(x,y)=0=\left.\partial_t u(x,y,t)\right|_{t=0}. This means that Bmn=0B^*_{mn}=0, while

Bmn=4ab0b0af(x,y)sinmπxasinnπyb ⁣dx ⁣dy.\boxed{B_{mn}=\frac{4}{ab}\int_0^b\int_0^a f(x,y)\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\dd{x}\dd{y}.}

We now proceed to compute the coefficients BmnB_{mn}.

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