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Let $u(x,y,t)$ be the solution of wave equation $u_{tt}=u_{xx}+u_{yy}, 0\lt x\lt 1, 0\lt y\lt 1, t\ge 0,$ $u(x,y,0)=(x-x^2)(y-y^2), u_t(x,y,0)=0$ and $ u(x,y,t)=0$ on the boundary of the square. Then $$u(1/2,1/2,t)=f(t)=\frac{64}{\pi^6}\sum_{n=0}^\infty\sum_{m=0}^\infty \frac{(-1)^{m+n}}{(2m+1)^3(2n+1)^3}\cos\left(\pi t \sqrt{(2m+1)^2+(2n+1)^2}\right)$$ It is unlikely that the function has a closed form expression. My question is: Is there an analytical way to find $\sup_{t\ge 0} f(t)$ and $\limsup_{t\to\infty} f(t)$?

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What is $f$?... –  Steve Huntsman Apr 17 '13 at 14:13
    
$f(t)=u(1/2,1/2,t)$ –  TCL Apr 17 '13 at 14:15
    
Of course, it is easy to find bounds with arbitrary precision, but that is not what you are asking. Why do you think if should be a tabulated constant? –  Jung Wen Chen Oct 28 '13 at 12:39

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An easy to find upper bound for the expression without the constant term is $$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{1}{(2m+1)^3(2n+1)^3}\approx 1.1063.$$ For each pair of a pair of odd numbers $(n_1,m_1)$ and $(n_2,m_2)$ with $n_1^2+m_1^2=n_2^2+m_2^2$ and $n_1+m_1 \not \equiv n_2+m_2\ (\mod 4)$ (e.g. (7,1) and (5,5)) we can improve this bound a little bit by $$2 \min\left(\frac{1}{n_1^3m_1^3},\frac{1}{n_2^3m_2^3}\right)$$ as $$(-1)^{\frac{n_1+m_1-2}{2}}\cos(\pi t \sqrt{n_1^2+m_1^2})=-(-1)^{\frac{n_2+m_2-2}{2}} \cos(\pi t \sqrt{n_2^2+m_2^2}).$$ Of course this is far from being a solution.

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Good observation. –  TCL Apr 28 '13 at 1:31

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