3
$\begingroup$

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 $n$...

Improve this question
$\endgroup$
10
  • $\begingroup$ Simple case: When $\mu$ is a delta function, you have an exact formula. $\endgroup$
    – Gerald Edgar
    Commented Jul 9, 2011 at 16:13
  • 2
    $\begingroup$ I don't think it needs to decrease. For instance if $\mu$ is simply Lebesgue measure, then $u(x,t)\equiv 1$, no? $\endgroup$
    – Peter Luthy
    Commented Jul 9, 2011 at 16:17
  • 1
    $\begingroup$ 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. $\endgroup$
    – Peter Luthy
    Commented Jul 9, 2011 at 16:49
  • 1
    $\begingroup$ If $\mu$ is a probability measure, then it should still decrease like $1/\sqrt{t}$. This is related to Brownian motion in the plane. $\endgroup$
    – Gerald Edgar
    Commented Jul 9, 2011 at 18:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Jul 9, 2011 at 21:11

1 Answer 1

Reset to default
4
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks Andrew. You remind me using the definition of the tempered distribution. A minor improvement is that $u(x,t)=|(\mu,\Gamma(t,x-\cdot))|\le C||\Gamma(t,x-\cdot)||_m$. I am now checking the rests. Thanks again for your help! :-) $\endgroup$
    – Anand
    Commented Jul 9, 2011 at 20:25
  • $\begingroup$ @Anand Yes, you are right. The answer for convolution is the same since $$\sup_{y\in\mathbb R}|y^m \partial_x^k \Gamma(t,x-y)|\le C \sup_{y\in\mathbb R} (|x|^m+|x-y|^m)\partial_x^k \Gamma(t,x-y)|\ldots$$ $\endgroup$
    – Andrew
    Commented Jul 9, 2011 at 21:09

You must log in to answer this question.

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