= Solution
First decompose the initial $y$ dependence using $\sin^2(2\pi y/b)=[1-\cos(4\pi y/b)]/2$. Only the cosine indices $m=0$ and $m=4$ occur. Expand the remaining factor in a <Fourier sine series>:
$$
x(a-x)=\sum_{n=1}^\infty b_n\sin\frac{n\pi x}{a},\qquad b_n=\frac2a\int_0^ax(a-x)\sin\frac{n\pi x}{a}\,dx.
$$
For $k_n=n\pi/a$, two <integrations by parts> give
$$
\int_0^ax(a-x)\sin(k_nx)dx=\frac{2[1-(-1)^n]}{k_n^3},\qquad b_n=\frac{4a^2[1-(-1)^n]}{n^3\pi^3}.
$$
Thus even $n$ vanish, while $b_n=8a^2/(n^3\pi^3)$ for odd $n$. The nonzero double-series coefficients are $A_{n0}=b_n/2$ and $A_{n4}=-b_n/2$. Their separate heat-mode decay factors give
$$
\boxed{T(x,y,t)=\frac{4a^2}{\pi^3}\sum_{\substack{n\geq1\\n\text{ odd}}}\frac{\sin(n\pi x/a)}{n^3}e^{-\lambda\pi^2n^2t/a^2}\left[1-e^{-16\lambda\pi^2t/b^2}\cos\frac{4\pi y}{b}\right].}
$$
At $t=0$ the bracket is $2\sin^2(2\pi y/b)$, which recovers the initial profile with the displayed sine coefficients. Each term satisfies the mixed <Dirichlet boundary conditions> and <Neumann boundary conditions>.
Back to article page
