# “Decay” of $L^2$-norm of the solution of heat equation with mixed boundary conditions

I'm considering the heat equation $$u_t = u_{xx}$$ on the interval $[0, \pi]$ with mixed boundary conditions $$u(0,t)=0 \quad \text{and} \quad u_x (\pi,t)=0,$$and smooth initial data $u(x,t)=u_0 (x)$ which satisfy the same boundary conditions. I would like to prove that $\| u( \cdot, t) \|_{L^2} \to 0$ when $t \to +\infty$.

Some hints are given - Poincaré's and Grönwall's inequalities - but the idea I had fails because it seems that $u$ doesn't belong to $H^1 _0 ([0,\pi])$. Indeed I firstly considered the function defined by $$e(t) = \int_0 ^ \pi u^2 (x,t) \, dx$$ and I noticed that $$\begin{split} e'(t) & = 2 \int_0^\pi u(x,t) u_t (x,t) \, dx = 2 \int_0^\pi u(x,t) u_{xx} (x,t) \, dx \\ & =2 \underbrace{[u(x,t) u_x (x,t) ]^\pi _0}_{=0} - 2 \int_0^\pi u^2 _x \, dx < 0\end{split}$$and then $e(t)$ is strictly decreasing. Moreover even if I assume $u$ regular enough to have $u$ and $u_x \in L^2 ([0,\pi])$, I can't use Poincaré's inequality because I don't know the value of $u(\pi,t)$ (?). Indeed if it was $u \in H^1_0 ([0,\pi])$ I would have $$e'(t) = - \int_0^\pi u_x ^2 \, dx \le - \int_0^\pi u^2 \, dx = - e(t)$$ and I might apply Grönwall's inequality in order to get $$e(t) \le e(0) e^{-t}.$$

Any idea/hint?

The solutions of the heat equation are smooth (even up to the boundary, because one can reflect across the boundary). So, $u_x\in L^2([0,\pi])$ is assured.
Reflection also helps with the Poincaré inequality. Extend $u_0$ to $[0,2\pi]$ so that $u_0(2\pi -x)=u_0(x)$. Consider the function $v$ that solves the heat equation on $[0,2\pi]\times [0,\infty)$ with this initial condition, and $u(0,t)=u(2\pi,t)=0$. By symmetry, $v_x(\pi,t)=0$. By uniqueness, $v=u$ on $[0,\pi]\times [0,\infty)$.
Then work with $v$ instead of $u$; the Poincaré inequality applies and the argument concludes as you expected.