Heat equation on a rectangle with two insulated sides ($T={\sum }_{n\ge 1,m\ge 0}{A}_{nm}\sin (n\pi x/a)\cos (m\pi y/b)e^{-\lambda {\pi }^2(n^2/a^2+m^2/b^2)t}$)

On $0<x<a$, $0<y<b$, impose zero Dirichlet boundary conditions at $x=0,a$ and zero Neumann boundary conditions at $y=0,b$. Separation of variables for $T_t=\lambda \Delta T$ gives the displayed sine-cosine expansion. The zero cosine mode must be included: its norm is $b$, whereas the positive cosine modes have norm $b/2$. Thus $A_{n0}=2(ab{)}^{-1}\int \int T_0\sin (n\pi x/a)$ and $A_{nm}=4(ab{)}^{-1}\int \int T_0\sin (n\pi x/a)\cos (m\pi y/b)$ for $m\ge 1$. Expanding $x(a-x)$ gives sine coefficient $4a^2[1-(-1{)}^n]/(n^3{\pi }^3)$. Initial data $x(a-x){\sin }^2(2\pi y/b)$ therefore involve only $m=0,4$ and odd $n$. Their leading long-time mode is $4a^2\sin (\pi x/a)e^{-\lambda {\pi }^{2}t/a^2}/{\pi }^3$.