Coefficients of a cosine series

Question

Let $u$ have the cosine series representation $$u = \sum_{k_1=0}^{\infty} \sum_{k_2=0}^{\infty} a_\underline{k} \cos\left(\frac{2\pi k_1 x }{L_1}\right) \cos\left(\frac{2\pi k_2 y }{L_2}\right) $$

What are the coefficients $A_{\underline{k}}$ in $$u^2 = \sum_{k_1=0}^{\infty} \sum_{k_2=0}^{\infty} A_{\underline{k}} \cos\left(\frac{2\pi k_1 x }{L_1}\right) \cos\left(\frac{2\pi k_2 y }{L_2}\right) $$ in terms of the $a_\underline{k}$?

Thanks :)

Answer 1

$\displaystyle \cos mx \cos nx = \frac{\cos (m+n)x+\cos (m-n)x}{2}$

$\displaystyle 4A_{ij}= \sum_{m-n=i}\sum_{p-q=j} a_{mn} \, a_{pq}+ \sum_{m+n=i}\sum_{p-q=j} a_{mn} \, a_{pq}+ \sum_{m-n=i}\sum_{p+q=j} a_{mn} \, a_{pq}+ \sum_{m+n=i}\sum_{p+q=j} a_{mn} \, a_{pq}$

Answer 2

In $1D$,

$$\left(\sum a_k\cos(kt)\right)^2=\sum_{k=0}^\infty\sum_{j=0}^\infty a_k a_j\cos(kt)\cos(jt)\\ =\frac12\sum_{k=0}^\infty\sum_{j=0}^\infty a_ka_j\left(\cos((k-j)t)+\cos((k+j)t)\right)\\ =\frac12\sum_{k=0}^\infty\sum_{j=k}^{-\infty} a_ka_{k-j}\cos(kt)+\frac12\sum_{k=0}^\infty\sum_{j=k}^\infty a_ka_{j-k}\cos(kt)$$

so that

$$A_k=\sum_{j=k}^{-\infty} a_ka_{k-j}+\sum_{j=k}^\infty a_ka_{j-k}=\frac12 a_{k}^2+\sum_{j=1}^\infty a_ka_{j}.$$