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Assume we have $u \in C^3(\mathbb{R}^n \times (0,\infty))$ satisfying the heat equation $$ \Delta u(x,t) = \partial_t u(x,t)$$ and a function $u_0:\mathbb{R}^n \to \mathbb{R}$ with unknown regularity (at least two times differentiable if necessary) such that $$ \sup_{x \in B(0,R)} \left\lvert u(x,t) - u_0(x) \right\rvert \xrightarrow{t \to \infty} 0 \quad \forall R>0.$$

I need to show that $\Delta u_0 = 0$ on the whole $\mathbb{R}^n$.

My first approach was the mean value theorem for functions satisfying the heat equation. Then, it is enough to show that $$ \frac{1}{4r^n} \int_{E(x,t;r)} u(y,s) \frac{\left| x-y \right|^2}{(s-t)^2} \, dy \, dt \xrightarrow{t \to \infty} \frac{1}{\left| B(x,R) \right|} \int_{B(x,R)} u_0(y) \, dy$$ for every $R$ and some $r$ (may be the same), where $$E(x,t;r) = \{ (y, s) \in \mathbb{R^{n+1}}\ :\ s \le t,\ \Phi(x-y,t-s)\ge\frac{1}{r^n}\}$$ are the famous heat balls and $\Phi$ the fundamental solution of the heat equation.

I'm not sure whether this approach leads to the desired result. Even though, I have no idea how to proceed with the heat balls here.

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3 Answers 3

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Let us fix $R>0$ first. Let $\phi \in C_c^\infty(B(0,R))$ be an arbitrary function. Then multiplying both sides of the equation, integrating in $B(0,R)$ and using one of Green's formulas, one has $$ \int_{B(0,R)}(u(x,t)-u_0(x))\Delta\phi(x)dx+\int_{B(0,R)}\Delta u_0(x)\phi(x)dx=\partial_t\int_{B(0,R)}(u(x,t)-u_0(x))\phi(x)dx. $$ Integrating from $s$ to $2s$, one has $$ \int_s^{2s}\int_{B(0,R)}(u(x,t)-u_0(x))\Delta\phi(x)dxdt+s\int_{B(0,R)}\Delta u_0(x)\phi(x)dx=\int_{B(0,R)}(u(x,2s)-u_0(x))\phi(x)dx-\int_{B(0,R)}(u(x,s)-u_0(x))\phi(x)dx. $$ Letting $s\to\infty$, one can easily derive $$ \int_{B(0,R)}\Delta u_0(x)\phi(x)dx=0. $$ By the arbitraryness of $\phi$, one has $$ \Delta u_0(x)=0 \text{ in }B(0,R) $$ for any $R>0$. So $\Delta u_0(x)=0$ for all $x$ in $\mathbb{R}^n$

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    $\begingroup$ Wouldn't it be sufficient to integrate from $s$ to $s+1$? All the other summands will still converge to zero. Yet, a very nice answer! Thanks! :) $\endgroup$
    – YoungMath
    Dec 17, 2018 at 9:33
  • $\begingroup$ Yes, it would be better. $\endgroup$
    – xpaul
    Dec 17, 2018 at 13:41
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I answer in the affirmative to this question by using the stated hypotheses on $u(x,t)$ and the properties of the Laplace transform $\mathscr{L}$, instead of using general properties of caloric functions: this is perhaps a "soft analysis" approach, which nevertheless has its advantages, briefly described in the notes.

Lemma 1. Under the above hypotheses on $u$ and $u_0$, $$ \lim_{t\to \infty} u(x,t)=u_0(x)\: \text{ and this implies that}\: \lim_{t\to \infty} u(x,t)\text{ is finite} \quad \forall x \in \mathbb{R}^n\label{1}\tag{1} $$ Indeed, since $\sup_{x \in B(0,R)} \left\lvert u(x,t) - u_0(x) \right\rvert \xrightarrow{t \to \infty} 0$ for all $R>0$, we have $$ 0\le |u(x,t)-u_0(x)|\le\sup_{x \in B(0,R)} \left\lvert u(x,t) - u_0(x) \right\rvert \xrightarrow{t \to \infty} 0 $$ by the sandwich theorem: then the first part of \eqref{1} a consequence of a standard real analysis argument (see for example [1], §3.7, prob. 7.7, pp. 127). For the second part of \eqref{1}, $u_0\in C^2(\mathbb{R}^n)$ implies that $|u_0(x)|<\infty$ for each $x \in \mathbb{R}^n$ therefore the limit exists and is finite for each point. $\blacksquare$

Since $u\in C^3\big(\mathbb{R}^n\times (0,+\infty)\big)$, \eqref{1} implies that $u(x,t)$ is bounded on every half-closed half line $[\varepsilon,+\infty)$ for any $\varepsilon>0$.

Definition 1. For each $\varepsilon >0$, the $\varepsilon$-shift of $u(x,t)$ is defined as $$ u^\varepsilon(x,t)=u(x,t+\varepsilon) $$ Lemma 2. For all $\varepsilon>0$, the $\varepsilon$-shift $u^\varepsilon(x,t)$ of a function $u(x,t)$ satisfying all the hypotheses of the problem under analysis has the following properties:

  1. $\lim_{t\to \infty} u(x,t)=\lim_{t\to \infty} u^\varepsilon(x,t)=u_0(x)$ for all $x \in \mathbb{R}^n$,
  2. $\Delta u^\varepsilon(x,t) = \partial_t u^\varepsilon(x,t)$,
  3. $u^\varepsilon(x,t)$ is $\mathscr{L}$(aplace)-transformable respect to $t\in[0,+\infty)$.

The first two properties follow trivially from the definition of $\varepsilon$-shift above while the third one follows from the continuity of $u$ on every half line $[0,+\infty)$. $\blacksquare$

Putting $U^\varepsilon(x,s)\triangleq \mathscr{L}\{u^\varepsilon\}(x,s)$, by properties 1 and 3 of lemma 2 and by the final value theorem (remember that $u^\varepsilon$ is bounded and has a finite limit for $t\to +\infty$ for for all $x \in \mathbb{R}^n$) for the Laplace transform we have $$ \lim_{t\to \infty} u(x,t)=\lim_{t\to \infty} u^\varepsilon(x,t) =\lim_{s\to 0} U^\varepsilon(x,s) =u_0(x)\quad\forall\, \varepsilon >0,\,x \in \mathbb{R}^n\label{2}\tag{2} $$ Now, taking the Laplace transform of equation 2 of lemma 2 and multiplying each side by $s\in\mathbb{C}$, $s\neq 0$ we have $$ \Delta\big[s U^\varepsilon(x,s)\big] = s^2U^\varepsilon(x,s)\label{3}\tag{3} $$ and using \eqref{2} for calculating the limit for $s$ going $0$ of each side of \eqref{3} $$ \begin{split} \lim_{s\to 0^+}\Delta\big[s U^\varepsilon(x,s)\big]&= \Delta u_0(x)\\ \lim_{s\to 0^+}s^2 U^\varepsilon(x,s)=&\lim_{s\to 0}s\cdot u_0(x)=0 \end{split}\implies \Delta u_0(x)=0\quad \forall x \in \mathbb{R}^n $$ i.e. $u_0$ is harmonic on the whole $ \mathbb{R}^n$.

Notes

  • The above argument could be straightforward if it could be assumed that $u\in C^3\big(\mathbb{R}^n\times [0,+\infty)\big)$: the continuity of $u$ on the whole $[0,+\infty)$ half line implies directly its $\mathscr{L}$-transformability and thus make superfluous the introduction of $\varepsilon$-shifts and related analysis. However this assumption has its advantages: the initial value $u|_{t=0^+}$ can be assumed to be very general, for example in the class of analytic functionals and hyperfunctions.
  • The real difficult step in this kind of results is the proof of the existence of a limit for $t\to\infty$ of the function $u(x,t)$ which, however, in this case is assumed as hypothesis.
  • The assumptions can probably be further weakened: $u(x,t)$ can be chosen to be a weak or very weak solution of the heat equation, and be only (locally) integrable. This approach requires to extend the proof of the final value theorem for the Laplace transform as given in the Wikipedia to functions which are only locally integrable (apart from obviously having a finite limit for $t \to +\infty$).

[1] Emanuel Fischer (1983), "Intermediate Real Analysis", Undergraduate Texts in Mathematics, Berlin-Heidelberg-New York: Springer-Verlag, ISBN 0-387-90721-1, xiv+770, MR681692 (84e:26004), 0506.26002.

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    $\begingroup$ I appreciate the generality of your answer. But I needed this for my undergraduated students in my exercise class. That would had been too much. Thank you anyway. $\endgroup$
    – YoungMath
    Dec 22, 2018 at 10:11
  • $\begingroup$ @YoungMath: you're welcome. I appreciate very much your effort in teaching in an optimal way to your undergraduate students, and also I am here only to help and learn, so I am nevertheless glad to be of some help with my answer. $\endgroup$ Dec 22, 2018 at 11:35
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Set $v(x,t) = \int_t^{t+1} u(x,s) ds$. Then $$ \Delta v(x,t) = \frac{\partial}{\partial t} v(x,t) = u(x,t+1) - u(x,t) $$ and also $v(x,t) \to u_0(x)$ as $t \to \infty$, uniformly on ball $B_R$. Then also $\Delta v(x,t) \to 0$, uniformly on balls $B_R$.

Let $ y \in \mathbb{R}^n$. The above implies that the mean value of $v(\cdot,t) - v(y,t)$ over any ball about $y$ converges to $0$. Therefore the mean value of $u_0 - u_0(y)$ is zero for any such ball and any $y$ and $u_0$ must be harmonic.

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  • $\begingroup$ Is it obvious that from $\Delta v (x,t) \to 0$ follows that the mean value of $v(\cdot,t)-v(y,t)$ converges to zero? The rest seems clear. It's pretty elegant! $\endgroup$
    – YoungMath
    Dec 16, 2018 at 23:11
  • $\begingroup$ If $|\Delta w| \le \varepsilon$ on $B_r(y)$ for some function $w$, then $w(x) \pm \frac{\varepsilon}{2n} (r^2 - |x-y|^2)$ is sub / superharmonic on this ball. Hence $w(y)$ is within $O(\varepsilon)$ of its mean over this ball. Now you can make $\varepsilon$ arbitrarily small as $t$ becomes large. $\endgroup$ Dec 17, 2018 at 2:17

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