Welcome to MathOverflow

Question

3

5

Let $\mu\in\mathcal{S}'(R)$ be a Schwartz distribution. The solution of a heat equation with $\mu$ as the initial data is

$$ u(t,x)= \int_R \frac{e^{-\frac{(x-y)^2}{2t}}}{\sqrt{2\pi t}} \mu(d y) $$

You can assume that $\mu$ is non-negative, i.e., a measure on $R$.

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?

Thank you very much in advance!

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 $n$...

edited Jul 9 2011 at 18:05

asked Jul 9 2011 at 16:01

Anand
802●3●13

Simple case: When $\mu$ is a delta function, you have an exact formula. – Gerald Edgar Jul 9 2011 at 16:13

2

I don't think it needs to decrease. For instance if $\mu$ is simply Lebesgue measure, then $u(x,t)\equiv 1$, no? – Peter Luthy Jul 9 2011 at 16:17

1

If μ is absolutely continuous with respect to Lebesgue measure, then the density should be dominated by some polynomial. Certainly in that case, the integral will grow at most like some power of t: in particular I believe it should grow like $t^{d/2}$, where d is the degree of the polynomial bound for the density function. – Peter Luthy Jul 9 2011 at 16:49

1

If $\mu$ is a probability measure, then it should still decrease like $1/\sqrt{t}$. This is related to Brownian motion in the plane. – Gerald Edgar Jul 9 2011 at 18:53

1

Anand, even if your initial data is only a distribution, it is smooth for any positive time. So any conclusion about long time behavior given smooth initial data applies just as well to initial data that is only a distribution. – Deane Yang Jul 9 2011 at 21:11

show 5 more comments

Andrew · Accepted Answer · 2011-07-09 19:57:37Z UTC

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=$

$\sup_{\alpha \le m,\ x \in \mathbb R}(1+|x|)^m |\partial^\alpha \varphi(x)|. $

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 $|x|^m e^{-c x^2/t}\le t^{m/2}e^{-c x^2/2t}$.

Long time behavior of the heat equation on R

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.