Deflection $u(x,y,t)$ of the square membrane of side $\pi$ and $c^2=1$ for initial velocity $0$ and initial deflection $0.01 \sin x \sin y$ I'v run through this problem several times, and I come up with $u(x,y,t)=0.01 \cos(t\sqrt2)\sin x \sin y$, but Kreyszig's Advanced Mathematics, 9th edition says it is $0.01 \sin(t\sqrt2) \sin x \sin y$. 
I know $B_{mn}^*=0$ and $B_{mn}=0.01$ when $m=1,n=1$, and $\lambda_{11}=\sqrt{2}$  
When I plug this into the double Fourier Series, I get the answer with the $\cos(...)$ not the one with $\sin(...)$. Is there something I am missing?
 A: It really helps if you post a precise formulation of the equation and conditions. I'm assuming you want to solve
$$
            \frac{\partial^2u}{\partial t^2}=c^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right) \\
                 0 \le x \le \pi,\;\;\; 0 \le y \le \pi \\
                u(t,0,y)=0,\;\; u(t,1,y)=0,\;\; 0 \le y \le \pi \\
                u(t,x,0)=0,\;\; u(t,x,1)=0,\;\; 0 \le x \le \pi, \\
 u(0,x,y)=0.01\sin(x)\sin(y),\;\; 0\le x\le 1,\;0\le y\le 1, \\
              u_{t}(0,x,y) = 0, \;\; 0\le x \le 1,\; 0\le y \le 1.
$$
Assuming $u(t,x,y)=T(t)X(x)Y(y)$ leads to integer separation parameters $n,m$ because of the $0$ endpoint conditons:
$$
        \frac{1}{c^2}\frac{T''}{T}=\frac{X''}{X}+\frac{Y''}{Y} \\
        \frac{X''}{X}=n^2,\;\; \frac{Y''}{Y}=m^2,\;\; \frac{1}{c^2}\frac{T''}{T}=n^2+m^2 \\
       X_n(x)=\sin(nx),\;\; Y_m(y)=\sin(my).
$$
The time solutions have the form
$$
       T_{0,0}(t)=1\\
       T_{n,m}(t)=\cos(c\sqrt{n^2+m^2}t)\\
      n,m=1,2,3,\cdots.
$$
The general solution is
$$
      u(t,x,y) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
   A_{n,m}T_{n,m}(t)X_n(x)Y_m(y)
$$
To match the initial condition,
$$
         u(0,x,y)=0.01\sin(x)\sin(y)=\sum_{n,m}A_{n,m}\sin(nx)\sin(my),
$$
which implies that only $A_{1,1}$ is non-zero, and must be $0.01$. So,
$$
         u(t,x,y) = 0.01\sin(x)\sin(y)\cos(c\sqrt{2}t).
$$
Based on what you've written, I think you are correct. The initial velocity is non-zero for the solution with $\sin(\sqrt{2}ct)$, which is easily checked. And the initial displacement is wrong, too.
