3
$\begingroup$

Consider the heat equation $$ (\partial_t + \Delta + V)u = 0$$ on a complete (open) Riemannian manifold with bounded geometry, where $V$ is a smooth and bounded potential.

Consider the semigroup generated by $A:=\Delta + V$ on the spaces $X = C_0(M)$ or $X = L^p(M)$ for $1\leq p < \infty$. The domain of $A$ is in each case $$ \mathrm{dom}(A) =\{ u \in X \mid Au \in X \},$$ where $Au$ denotes the distributional derivative.

My question is the following: It is clear by parabolic regularity that $e^{-tA}u$ is smooth for any $u \in X$. But is it also true that solutions $e^{-tA}u$ satisfy $$ |\nabla^ku| \in X ~~~~~~~\forall ~k$$ for $u \in \mathrm{dom}(A)$? The local statement is clear (i.e. solutions are in $W^{k, p}_{\mathrm{loc}}(M)$ and in $C^k(M)$ for every $k$), but is it also true that we have this decay at infinity?

What about the special case of vanishing $V$?

Improve this question
$\endgroup$
3
  • $\begingroup$ One trick to keep in mind is if $V$ is constant you can do the change of variables $ u(x,t)=e^{\gamma t} v(x,t)$ and write the equation in terms of $v$. This will give estimates... for the case of non-constant $V$ and am not sure if this helps one do anything or not... $\endgroup$
    – Craig
    Commented Jul 22, 2014 at 0:31
  • $\begingroup$ What is a vanishing $V$? Does it mean that $v\equiv0$? $\endgroup$
    – Andrew
    Commented Jul 23, 2014 at 8:24
  • $\begingroup$ $V$ is the potential in the thread above. That it vanishes means that $V(x) = 0$. $\endgroup$
    – Matthias Ludewig
    Commented Jul 23, 2014 at 10:51

1 Answer 1

Reset to default
2
$\begingroup$

For $C$ It follows from the interior Schauder estimates. Namely, for $r>0$ denote $H_r(x_0,t_0)=\{|x-x_0|<r^2,\ 0<t_0-t<r\}$. Then for $0<\alpha<1$ $$ \|u\|_{C^{2+\alpha}(H_r(x,t))}\le C\|u\|_{C(H_{2r}(x,t))}. $$ Taking $r=t/2$ for $t\le 1$ and $r=1/2$ for $t>1$ gives $|\nabla^ku| \in CM)$, $k\le2$. Differentiating the equation w.r.t. $x$ one can get analogous estimate for $|\nabla^3 u| \in C(M)$. Bootstrapping gets the required estimate for any $k$.

For the case $1<p<\infty$ the result follows from the similar theory in the anisotropic Sobolev spaces $W^{2,p}$.

Another way is to get estimates for the fundamental solution $\Gamma$. Then denoting $u_0=u(x,0)$ we have

$$\|\nabla^ku(\cdot,t)\|_{L_p}= \|\nabla^k\Gamma(\cdot,t)*u_0\|_{L_p}\le \|\nabla^k\Gamma(\cdot,t)\|_{L_1}\|u_0\|_{L_p}. $$

Improve this answer
$\endgroup$
5
  • $\begingroup$ What about the vanishing condition at infinity? You only showed smoothess, right? $\endgroup$
    – Matthias Ludewig
    Commented Jul 23, 2014 at 10:59
  • $\begingroup$ What exactly is meant under vanishing of solutions at infinity? $\endgroup$
    – Andrew
    Commented Jul 23, 2014 at 11:50
  • $\begingroup$ A function $u$ is in the space $C_0(\mathbb{R}^n)$ if for each $\varepsilon >0$, there exists a compact set $K$ such that $|u|< \varepsilon$ outside of $K$. $\endgroup$
    – Matthias Ludewig
    Commented Jul 23, 2014 at 12:41
  • $\begingroup$ Denote $d(x,y)$ the distance function, $d(x)=d(x,0)$ and $B_R=\{x\in M|d(x)<R\}$. If an estimate for $\Gamma$ like $$ |\Gamma(x,y,t)|\le Ct^{-n/2}e^{-c_1d^2(x,y)/t+c_2t} $$ holds then it is straightforward to obtain the required property. For $\mathbb R^n$ such an estimate is known and for $M$ can be done (I think) in the same way. $\endgroup$
    – Andrew
    Commented Jul 23, 2014 at 15:34
  • $\begingroup$ Namely dividing integral $\int_M \Gamma(x,y,t)u_0(x)\,dx$ into two: over $B_R$ and $M\backslash B_R$. Taking $x$ with $d(x)$ large enough the first part will be small due to the factor $e^{-c_1d^2(x,y)/t}$ and the second part can be made small taking $R$ s.t. $|u_0|\le \varepsilon$ on $M\backslash B_R$. $\endgroup$
    – Andrew
    Commented Jul 23, 2014 at 15:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .