** I simplified the question: **

On bounded domains, the maximum principle implies that the solution to the heat equation is (strictly) positive, if the initial and boundary data is positive.

I would like to know whether this result is also true if applied on an unbounded domain:

Let $a>0$ and $g$ a continuous function with $g(a,t)>0$ for all $t \in [0,T]$ and $u(x,0) = u_0(x) >0$ for all $x \in (a,\infty)$ with $u_0 \in C^{\infty} \cap L^{\infty}.$

Let $u(x,t)$ be the solution to the heat equation $\left(\partial_t - \partial_{x}^2 \right)u=0$ with the above boundary data.

Does there exist then a version of the maximum principle saying that $u(x,t)>0$ for $(x,t) \in [a,\infty) \times [0,T]$?

  • $\begingroup$ Just to make sure that I'm not making some stupid mistake: As the solution of the heat equation depends monotonically on the boundary conditions, it suffices to prove this for $g = 0$, right? $\endgroup$ Commented Mar 16, 2019 at 23:41
  • $\begingroup$ @JochenGlueck i'd think so too. $\endgroup$
    – Sascha
    Commented Mar 17, 2019 at 1:51
  • $\begingroup$ I believe that the answer is no. Tychonov's examples, www-bcf.usc.edu/~lototsky/MATH445/NonUniqueness-HeatEq.pdf can be modified to construct a temperature with zero boundary and initial conditions which is not zero in the exterior of a ball. $\endgroup$ Commented Mar 17, 2019 at 12:52
  • $\begingroup$ As Alexandre Eremenko's answer below shows, my comment above is not correct - which seems to be essentially due to the non-uniqueness of the solution of the heat equation (while I still think it's correct on an appropriately chosen function space). $\endgroup$ Commented Mar 17, 2019 at 15:03
  • $\begingroup$ @JochenGlueck so you think you had something in mind that would've worked if the solution was unique?-I'd be very interested to hear about it. $\endgroup$
    – user121558
    Commented Mar 17, 2019 at 15:19

2 Answers 2


The answer is no. A counterexample has the form $$u(x,t)=\sum_0^\infty \frac{g^{(n)}(t)}{((2n+1)!)^2}(r-R)^{2n+1},$$ where $r>R$ is the distance from the origin in $R^2$. First one shows that this function satisfies the heat equation formally. Second, there exists an infinitely differentiable $g\not\equiv 0$ such that $g^{(n)}(0)=0$ for all $n\geq 0$, and such that the above series converges. So our function is not zero, satisfies the heat equation in the exterior of the ball, and has zero boundary and initial conditions. For the existence of such $g$ one can refer to http://www-bcf.usc.edu/~lototsky/MATH445/NonUniqueness-HeatEq.pdf

WLOG, $u$ is negavive somewhere, otherwise replace it by $-u$. Then add a small positive constant to $u$, the boundary and initial conditions will be positive, but $u$ is still negative somewhere.

  • 2
    $\begingroup$ +1. It might be worthwhile to add, though, that the solution $u(t)$ in this example is not contained in any classical function space such as, say, $L^1$ (with respect to the Lebesgue measure). This makes the counterexample somewhat non-physical. $\endgroup$ Commented Mar 17, 2019 at 14:57
  • $\begingroup$ As Korner describes it in the reference that I gave, "the heat wave comes from a source at infinity":-) $\endgroup$ Commented Mar 18, 2019 at 12:18

As a companion to Alexandre Eremenko's answer, the answer to your question is "yes" if we assume in addition that $u$ satisfies the growth condition $|u| < Ae^{M|x|^2}$ for some $A,\,M > 0$.

To see this assume for simplicity that $a = 0$ and $T = 1$. Let $h_R = -t^{-1/2}e^{-\frac{(x-R)^2}{4t}}$ and let $b_R = Ae^{MR^2}h_R$. By the comparison principle on bounded domains we have $u \geq b_R$ on $[0,\,R] \times [0,\,1]$. A short computation shows that $b_R \geq -8A\sqrt{M}e^{-3MR^2}$ on $[0,\,R/2] \times [0,\, (64M)^{-1}]$. Taking $R \rightarrow \infty$ we conclude that $u \geq 0$ for $t \in [0,\,(64M)^{-1}]$, and repeating the argument finitely many times we get $u \geq 0$ on $[0,\,T]$. That $u > 0$ follows from the strong maximum principle.


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