Long time behavior of the heat equation on R - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:57:58Z http://mathoverflow.net/feeds/question/69884 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69884/long-time-behavior-of-the-heat-equation-on-r Long time behavior of the heat equation on R Anand 2011-07-09T16:01:55Z 2011-07-09T19:57:37Z <p>Let $\mu\in\mathcal{S}'(R)$ be a Schwartz distribution. The solution of a heat equation with $\mu$ as the initial data is</p> <p>$$u(t,x)= \int_R \frac{e^{-\frac{(x-y)^2}{2t}}}{\sqrt{2\pi t}} \mu(d y)$$</p> <p>You can assume that $\mu$ is non-negative, i.e., a measure on $R$.</p> <p>The problem is how $u(t,x)$ behaves for $x$ fixed as $t\rightarrow\infty$. I guess that it might not increase too fast for large $t$, e.g., it does not increase like $e^t$. Do anyone have any idea? </p> <p>Thank you very much in advance!</p> <p>EDIT: Here is one try: If we smooth $\mu$ by a test function to get say $\mu_n$, then $\mu_n$ is a smooth function with at most certain polynomial increase. However, the degree of the domination polynomial might depend on $n$...</p> http://mathoverflow.net/questions/69884/long-time-behavior-of-the-heat-equation-on-r/69902#69902 Answer by Andrew for Long time behavior of the heat equation on R Andrew 2011-07-09T19:34:14Z 2011-07-09T19:57:37Z <p>Denote $\Gamma(x,t)$ the fundamental solution of the heat equation form the integral. By the theorem of L. Schwarz for any $\mu\in S'(S)$ there is a number $m\in \mathbb N$ and $C>0$ such that $$|u(x,t)|=|(\mu,\Gamma(t,\cdot))|\le C\|\Gamma(t,\cdot)\|_m,$$ where $\|\varphi\|_m=$</p> <p>$\sup_{\alpha \le m,\ x \in \mathbb R}(1+|x|)^m |\partial^\alpha \varphi(x)|.$ </p> <p>It is straightforward to obtain, as it is said in the comments, that the $u$ increases at most polynomially. Namely, it is known that $$|\partial_x^k \Gamma(t,x)|\le C_k t^{-(k+1)/2}e^{-c_k x^2/t}.\$$ From here it is easy to get $$|x^m \partial_x^k \Gamma(t,x)|\le C_{k,m} t^{(m-k-1)/2}e^{-c_{k,m} x^2/t},$$ since $|y|^m e^{-c y} \le C_m$. Putting $y=x^2/t\$ we have $|x|^m e^{-c x^2/t}\le t^{m/2}e^{-c x^2/2t}$.</p>