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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.

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  • $\begingroup$ If $u\equiv0$ and $v\equiv1$ than the required statement is not true. $\endgroup$ – Andrew Feb 8 '16 at 14:37
  • $\begingroup$ Should ask $v$ to be mean zero as well $\endgroup$ – user70229 Feb 8 '16 at 15:15
  • $\begingroup$ @dalbrit Yes you're right, then Lax-Milgram applies for $v$. $\endgroup$ – FavorExistingPopularTags Feb 8 '16 at 15:29
  • $\begingroup$ @Andrew I changed the question $\endgroup$ – FavorExistingPopularTags Feb 8 '16 at 15:30
  • $\begingroup$ @FavorExistingPopularTags does $v$ depends on $t$? $\endgroup$ – Andrew Feb 8 '16 at 17:08
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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 $C(t)=Me^{-\lambda_2 t}$.

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  • 2
    $\begingroup$ In other words, the question reduces to proving asymptotic bounds on the heat kernel :-) $\endgroup$ – Willie Wong Feb 8 '16 at 19:39

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