Suppose $$u_{t}=u_{xx}+u_{yy}$$ on the region $0\leq x,y \leq 1$ and that $u=0$ on the boundary.
Show that $$\int^1_0\int^1_0u^2dxdy\xrightarrow{}0, \text{ as } t\xrightarrow{}\infty $$
I'm really lost as to begin here. Any help would be appreciated.
By Poincare's Inequality, $$ \|\nabla u\|_{L^2}^2\ge\lambda_1\|u\|_{L^2}^2. $$ Here $\lambda_1>0$ is the first eigenvalue of $-\Delta$. Multiplying both sides of the equation by $u$ and integrating in $\Omega$, one has $$ \frac12 (\|u(t)\|_{L^2}^2)'=-\|\nabla u(t)\|^2_{L^2}\le -\lambda_1\|u(t)\|_{L^2}^2$$ and hence $$ (\|u(t)\|_{L^2}^2)'+2\lambda_1\|u(t)\|_{L^2}^2\le0. $$ So $$ (e^{2\lambda_1t}\|u(t)\|_{L^2}^2)'\le0 $$ Integrating this from 0 to $t$, one has $$ \|u(t)\|_{L^2}^2\le e^{-2\lambda_1t}\|u(0)\|_{L^2}^2. $$ Now it is easy to see $$ \lim_{t\to\infty}\|u(t)\|_{L^2}^2=0. $$