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Let $M$ be a $C^3-$compact manifold and $v \in W^{2,1}_p(M\times [0,T])$ ($1\le p<\infty$) be the solution of:

$\begin{cases} \partial_t v-\Delta_{M} v=f(v), \quad M\times [0,T]\\ v(x,0)=v_0, \quad M \end{cases}$

for some $f(v)\in L^{\infty}(M\times [0,T])$ and initial data $v_0$ such that ${\vert \vert v_0 \vert \vert}_{L^\infty(M)} \le C$

I would like to prove that ${\vert \vert v(\cdot,t)-v_0(\cdot) \vert \vert}_{L^{\infty}(M)} \le Ct (*)$

However applying the Hölder inequality, and using the regularity of the time derivative of $v$, I obtain: ${\vert \vert v(\cdot,t)-v_0(\cdot) \vert \vert}_{L^p(M)} \le \hat C t^{1/q}$ without using at all the $L^{\infty}$ assumption on the initial data.

EDIT: More precisely I find applying the Hölder inequality, ${\vert \vert v(\cdot,t)-v_0(\cdot) \vert \vert}_{L^p(M)} \le \int_0^t{\vert \vert \partial_sv(\cdot,s)\vert\vert}_{L^p}\;ds \le {\vert \vert \partial_sv(\cdot,s)\vert\vert}_{L^p(L^p)} t^{1-1/p}$ I noticed that letting $p\to \infty$ then it seems reasonable to expect $(*)$. However this could be a proof if I could show that ${\vert \vert \partial_sv(\cdot,s)\vert\vert}_{L^p(L^p)}$ is bounded also for $p=\infty$ Yet I don't know how to prove this...

I think I will have to use a maximum principle technique but at this point I don't see from where to start.

I would appreciate any help or hints so I can fill in the details on my own.

DISCLAIMER: I already asked this in MSE but since I got no answer, I thought it 'd be better to post it also here.

Thanks a lot in advance!

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Maybe I'm missing something, but it doesn't seem that your desired inequality can be true (even with the right-hand side of $(*)$ replaced by $C t^{\epsilon}$). It would imply uniform convergence of $v(\cdot,t)$ to $v_0(\cdot)$ as $t \rightarrow 0$, which wouldn't be possible if $v$ is continuous (say $f=0$) and $v_0$ is taken to be discontinuous.

Another way of looking at this is to test on eigenfunctions: If $v_0$ is an eigenfunction of the Laplacian with eigenvalue $-\lambda^2$ and $f = 0$, then $v(\cdot,t) - v_0(\cdot) = (e^{-\lambda^2 t} - 1) v_0(\cdot)$, but the coefficient $e^{-\lambda^2 t} - 1$ can be made bigger than $1/2$ for any positive time by simply taking $\lambda$ large enough.

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I agree with the previous answer. Maybe the following observation for the global problem on $\mathbb R^n$ can help. Using the fundamental solution $H(t)(4π t)^{-n/2} e^{-{\vert x\vert^2/4t}}$ of the heat equation, one gets for $f=0$, \begin{multline} v(t,x)-v_0(x)=H(t) (4π t)^{-n/2}\int e^{-\vert y\vert^2/4t} \bigl(v_0(x+y)-v_0(x)\bigr) dy \\=H(t)\int e^{-π \vert z\vert^2}\bigl(v_0(x+2z\sqrt{π t})-v_0(x)\bigr) dz, \end{multline} and thus $$ v(t,x)-v_0(x)= H(t)(4π t)\int e^{-π \vert z\vert^2}\int_0^1(1-\theta)(\nabla^2 v_0)(x+\theta 2z\sqrt{π t}) z^2 dz d\theta, $$ so that, thanks to Jensen's inequality and to the translation invariance of the $L^p$-norm in $\mathbb R^n$ $$ \Vert v(t)-v_0\Vert_{L^p}\le H(t) 4π t\int e^{-π \vert z\vert^2}\vert z\vert^2\Vert\nabla^2 v_0\Vert_{L^p} dz/2=c_n t\Vert \nabla^2v_0\Vert_{L^p}, $$ where the estimates holds true for $1\le p\le +\infty$.

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  • $\begingroup$ Maybe I should had written this in advance but $f$ depends on $v$ somehow, yet is still uniformly bounded in $L^{\infty}(M\times [0,T])$. This is why I didn't think of the fundamental solution. However, what is wrong in my approach? $\endgroup$ – kaithkolesidou Jul 11 at 15:36
  • $\begingroup$ @kaithkolesidou If you want in fact boundlessness of $u_t$ the initial function $v_0$ should be from $C^{1,1}$ (first order derivatives are Lipschitz) as illustrated by this post. $\endgroup$ – Andrew Jul 11 at 16:36
  • $\begingroup$ @Andrew Will I need this kind of regularity for my initial data since I want $\partial_t u$ to be Lipschitz...? Is there any theorem I miss in this case? $\endgroup$ – kaithkolesidou Jul 12 at 11:56
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    $\begingroup$ @kaithkolesidou Consider the Cauchy problem for the heat equation $u_t-\Delta u=0$ in $\mathbb R^n$. The bounded solution will be smooth for $t>0$. So uniforme Lipschitz $u_t$ wrt $t$ means boundedness of $u_{tt}$. From the equation $u_{tt}=\Delta^2u$. At the initial moment $t=0$ it turns into $u_{tt}=\Delta^2v_0$. So for $u_{tt}$ to be bounded in all half space $t>0$ it is necessary that $\Delta^2v_0\in L_\infty(\mathbb R^n)$. $\endgroup$ – Andrew Jul 12 at 14:14

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