Let $u$ and $v$ be the weak solutions of
$$u\_t - \Delta u = f$$
$$u(0)=u\_0$$
and
$$-\Delta v = f$$
$$|\Omega|^{-1}\int\_\Omega v =0$$
on a bounded domain $\Omega$, where $u$ and $v$ satisfy homogeneous Neumann BCs. Here we may take $f$ and $u\_0$ to have spacial mean values zero.


I'm trying to prove an estimate of the form
$$\lVert{u(t)-v}\rVert\_{L^1(\Omega)} \leq C(t)\lVert{u\_0-v}\rVert\_{L^1(\Omega)}$$
where $C(t) \to 0$ as $t \to \infty$.


I can prove this if instead of $L^1$ norms we had $L^2$, but I need $L^1$ on both sides. Does anyone know how to achieve this? The $L^2$ case can be done with testing the weak form satisfied by $u-v$ with $u-v$, Poincare's inequality and actually we will find $C(t) = e^{-Kt}$ after using it as an integrating factor.



$L^1$ convergence to equilibrium of solutions of heat equation
Here is an idea. Put $w=u-v$, $w\_0=u\_0-v$. The required inequality takes form $\lVert{w}\rVert\_{L^1(\Omega)} \leq C(t)\lVert{w\_0}\rVert\_{L^1(\Omega)}$. Denote $G(x,y,t)$ the Green function of the Neumann problem for the heat equation in $\Omega$. There is an integral representation of the solution:
$$
w(x,t)=\int\_\Omega G(x,y,t)w\_0(y)\,dy.
$$
Since $w\_0$ has mean value zero, denoting $G'(x,y,t)=G(x,y,t)-|\Omega|^{-1}$, it can be rewritten as
$$
w(x,t)=\int\_\Omega G'(x,y,t)w\_0(y)\,dy.
$$
So it is enough to prove the inequality with 
$C(t)=\max\_{x,y\in \Omega}|G'(x,y,t)|$. May be it is proven somewhere that the last quantity tends to zero as $t\to+\infty$. 


If not, let $\lambda\_1=0<\lambda\_2<\ldots$ and $\varphi\_1$, $\varphi\_2,\ldots$ be eigenvalues and eigenfunctions of the Neumann problem in $\Omega$, $\|\varphi\_n\|\_{L\_2(\Omega)}=1$. Then 
$$
G'(x,y,t)=\sum\_{n=2}^\infty \varphi\_n(x)\varphi\_n(y)e^{-\lambda\_n t}.
$$
If values $\max\_{x\in \Omega}|\varphi\_n(x)|$ grow not too quickly the result follows. Say, for one-dimensional case $\Omega=[0,\pi]$ eigenfunctions $\varphi\_n(x)=\cos nx$ are uniformly bounded and one can take the constant in the form $C(t)=Me^{-\lambda\_2 t}$.


