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Define a function $\varphi$ on $\mathbb{R}_{+}$ by $$\varphi(t):=\begin{cases}e^{-1/t^{2}}, & {t>0}\\ 0, & {t\leq 0}\end{cases}\tag{1}$$

It is well-known that $\varphi$ is $C^{\infty}(\overline{\mathbb{R}}_{+})$ and $\varphi^{(k)}(0)=0$ for all integers $k\geq 0$. Tychonoff famously showed that the function $u:\mathbb{R}_{+}\times\mathbb{R}\rightarrow\mathbb{R}$ defined by

$$u(t,x):=\sum_{k=0}^{\infty}\dfrac{\varphi^{(k)}(t)}{(2k)!}x^{2k}, \qquad (t,x)\in\mathbb{R}_{+}\times\mathbb{R}\tag{2}$$

belongs to $C^{\infty}(\mathbb{R}_{+}\times\mathbb{R})$ and satisfies the Cauchy problem

$$\begin{cases}\partial_{t}u-\Delta u=0 & {(t,x)\in\mathbb{R}_{+}\times\mathbb{R}},\\ u(0,\cdot)=0 & {}\end{cases}\tag{3}$$

Question. Motivated by this question, I am trying to show that for $t>0$ fixed, $u(t,\cdot)$ does not define a tempered distribution. I am looking for a "lower bound" on the growth of $u$ for $t>0$ fixed which I can use to construct a sequence of test functions $\varphi_{m}\in\mathcal{S}(\mathbb{R}^{n})$ which tend to zero in the Schwartz topology but

$$\left|\langle{u(t,\cdot),\varphi_{m}}\rangle\right|=\left|\int_{\mathbb{R}^{n}}u(t,x)\varphi_{m}(x)dx\right|\geq c,\quad\forall m\in\mathbb{N}\tag{4}$$

for some $c>0$.

I know that the function $u$ does not satisfy the growth condition

$$\sup_{0\leq t\leq T}\left|u(x,t)\right|\leq Ae^{c\left|x\right|^{2}},\quad\forall x\in\mathbb{R}^{n}\tag{5}$$

where $T>0$ is fixed and $A,c>0$ are constants depending on $T$; however, I fail to see how this helps in the task described above.

Edit: Einar Rødland has presented some graphical evidence to suggest that for fixed $t>0$, $u(t,\cdot)$ is "well-behaved" and defines a tempered distribution. I am seeking a proof of disproof of this conjecture. Note that a "wild solution" can be "well-behaved" for $t>0$ fixed. For example, in this paper, the authors present an example of nonuniqueness for the Cauchy problem which is continuous on $\mathbb{R}\times [0,\infty)$ and satisfies

$$\left|u(x,t)\right|\leq C e^{\epsilon/t},\qquad (x,t)\in\mathbb{R}\times\mathbb{R}_{+}$$

where $C=C(\epsilon)$, for any $\epsilon>0$. Also, at the end of the paper the authors remark that all other (i.e. besides theirs) nonuniqueness solutions are unbounded in $x$, which would seem to contradict Einar's suggestion.

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Matt Rosenzweig
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  • I'm not very familiar with the field, and with tempered distributions in particular, so I may be missing something. However, for $t>0$, it seems to me $u(t,x)\rightarrow 1$ as $|x|\rightarrow\infty$, otherwise $u(t,x)$ is perfectly smooth in $x$, So it should form a tempered distribution, shouldn't it? – Einar Rødland May 18 '15 at 15:45
  • @EinarRødland: I don't follow your argument. For $t>0$, $u(t,x)$ is smooth in $x$. This follows from the uniform convergence on compact sets of the partial sums defining $u(t,\cdot)$. – Matt Rosenzweig May 18 '15 at 18:18
  • As I warned, I'm not well versed in this topic, but as I understand if $u(t,x)$ is smooth in $x$ and doesn't grow too fast as $x$ increases, it gives a tempered distribution. In this case, $u(t,x)$ doesn't grow at all: i.e., there is a bound $|u(t,x)|<C_t$ for each $t>0$. By the way, I'm not claiming to have a proof of this...it's just an experimental observation. – Einar Rødland May 18 '15 at 18:53
  • @EinarRødland: You are correct in that if $u(t,x)$ and all its partial space derivatives grow at most polynomially, $u(t,x)$ defines a tempered distribution. But I don't see how you are getting the bound $\left|u(t,x)\right|<C_{t}$; would you please give a proof of this? It was my understanding that $u$ grows quite large as $\left|x\right|\rightarrow\infty$. – Matt Rosenzweig May 18 '15 at 18:59
  • Correct me if I'm wrong, but as I understand bounded partial derivatives defines Schwartz functions $\mathcal{S}$, while the tempered distributions are their dual space $\mathcal{S}'$, where a continuous function that grows at most polynomially will yield a tempered distribution. The observation that $u(t,x)$ is bounded for each $t$ is simply based on plotting the function for some $t$ and limited ranges of $x$. – Einar Rødland May 18 '15 at 19:55
  • @EinarRødland: A Schwartz function and all its partial derivatives decay faster than any polynomial. You are correct that any continuous function with polynomial growth is a tempered distribution. I confused that with a sufficient condition for a product with a Schwartz function to be Schwartz. Your observation doesn't square with comments I have read about $u(t,x)$. Maybe you are right that for some $t>0$, $u(t,x)$ is a tempered distribution. In any case, $C_{t}$ blows up as $t\rightarrow 0$. – Matt Rosenzweig May 18 '15 at 20:14
  • The bound on $u(t,x)$ referred to there should be global, i.e. for all $t$. As you note, as $t\rightarrow 0^+$, the bounds blow up, but for each $t$ they can be met. I've tried to expand a little on this in the answer. Not sure how to prove my claims formally, however. – Einar Rødland May 20 '15 at 07:34

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Here's a summary of, and extending remarks and explanations to, the comments. Note that my description of $u(t,x)$ is based on experimental observations, not formal proofs.

I plotted the sum for $u(t,x)$ for a few values of $t$: within a moderate range for $x$, it would converge, although I did need extended accuracy (used Maple for this). My understanding is largely based on what these plots looked like.

Basically, for $t\le0$ we have $u(t,x)=0$. But then, for $t>0$, the $u(t,x)$ looks like a wave packet moving in from infinity towards the origin. For a given $t>0$, beyond the region containing the wave packet, the function levels off quickly to $1$ as $x$ increases. As $t$ increases, this wave moves towards the origin and dampens out, with $u$ eventually converging to constant $1$.

However, if we move back in time towards $t=0^+$, the wave pattern is more violent, spread out, and further away from the origin: i.e., higher amplitude and wave spreads further out to higher values of $x$. As we lower $t$ towards 0, and the wave packet is further away from the origin, $u(t,x)$ converges towards $0$ for any fixed $x$.

Here's a plot showing $y=u/[(1+u^2)/2]^{1/4}$ (had to compress the scale to enable large and small amplitudes to show), for $t=1$, $1/2$, $1/3$, $1/4$, $1/5$, $1/6$, and $1/7$, with $t=1$ having the smoothest curve and $t=1/7$ the most variable. Plot of u (transformed to reduce extrema) for t=1/1,...,1/7.

Thus, for any given $t>0$, we have a nice solution with no exotic behaviour. Moreover, for $t\le0$, we have $u(t,x)=0$ which means the solution for $t>0$ does not arise from prior conditions. The extreme behaviour is only apparent as $t\rightarrow 0^+$. Thus, for any given $t>0$ the solution is a tempered distribution (and should also be a Schwartz function), but $u(t,x)$ is wildly unbounded as $t\rightarrow 0^+$.

NB: The solution of $u$ expressed from $\phi(t)$ works for any $C^\infty$ function, as far as I can tell. What makes $\phi(t)=e^{-1/t^2}$ for $t>0$, $\phi(t)=0$ for $t\le0$, special is that all derivatives are zero for $t\le0$ allowing $u(t,x)$ to be defined as continuous in both $t$ and $x$, while identically zero for $t\le0$.

Einar Rødland
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