I am currently working on a small thesis for my university. A part of the thesis is to derive a formula for the kinetic energy of a rectangular plate $(a,b)$. For a single particle, the kinetic energy equals: $$T = \dfrac{1}{2} m v^2$$ For the plate I am dealing with, I only have the deflection $w$ which is expressed as a double sum where $A_{mn}$ is unknown $$w(x,y) = \sum_{m=1}^\infty \sum_{n=1}^\infty A_{mn} \cdot \sin\left(\dfrac{m \pi x}{a}\right) \cdot \sin\left(\dfrac{n \pi y}{b}\right)$$ The kinetic energy then becomes: $$ \begin{aligned} T &= \dfrac{1}{2} \rho h \int_0^a \int_0^b \left(\frac{d}{dt}w(x)\right)^2dy \ dx \\ &= \dfrac{1}{2} \rho h \int_0^a \int_0^b \left(\sum_{m=1}^\infty \sum_{n=1}^\infty \dot{A}_{mn} \cdot \sin\left(\dfrac{m \pi x}{a}\right) \cdot \sin\left(\dfrac{n \pi y}{b}\right)\right)^2dy \ dx \end{aligned} $$ This is where I got stuck. I have no idea on how to tackle an integral like this. I was looking into the Cauchy product but I don't know how to apply it to more than one sum.
I know that the result is going to be something like: $$T = ... \sum_{m=1}^\infty \sum_{n=1}^\infty \dot{A}_{mn}^2$$ where $...$ are only a few constants.
I am very happy if someone could help me solving this integral!
Greetings,
In order to simplify, let us assign an integer $q$ to every pair of integers $(m,n)$. As a consequence, $$\sum_m\sum_n A_{mn}=\sum_q A_q$$ (with an abuse of notations here). We can say without much difficulty that $$\left(\sum_qA_q\right)^2=\sum_qA^2_q+2\sum_q\sum_{p\neq q}A_qA_p$$ Back to the initial notations, the double sum of the squared terms reads $$\frac{1}{2}\rho h \sum_m\sum_n \dot{A}^2_{mn}\int_0^a\sin^2\left(\frac{m\pi x}{a}\right)\int_0^b\sin^2\left(\frac{n\pi y}{b}\right)=\frac{1}{8}\rho h a b\sum_m\sum_n \dot{A}^2_{mn}\tag{1}$$ The double sum in $(q,p)$ is actually a quadruple sum of the form $$\frac{1}{2}\rho h \sum_m\sum_n\sum_i\sum_j \dot{A}_{mn}\dot{A}_{ij}\int_0^a\sin\left(\frac{m\pi x}{a}\right)\sin\left(\frac{i\pi x}{a}\right)\int_0^b\sin\left(\frac{n\pi y}{b}\right)\sin\left(\frac{j\pi y}{b}\right)\tag{2}$$ with either $m\neq i$ and/or $n\neq j$ since $q\neq p$. In other words, one of the integrals in (2) always vanish and the kinetic energy is given by (1).