## Related questions with answers

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

VerifiedThe solution $u(x,y,t)$ is represented via double Fourier series

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

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

$\boxed{B_{mn}=\frac{4}{ab}\int_0^b\int_0^a f(x,y)\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\dd{x}\dd{y}.}$

We now proceed to compute the coefficients $B_{mn}$.

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