# Time-dependent moments from partial differential equations and the time-
dependent set of atoms

Raúl E. Curto<EMAIL_ADDRESS>Philipp J. di Dio  philipp.didio@uni-
konstanz.de Milan Korda<EMAIL_ADDRESS>Victor Magron<EMAIL_ADDRESS>University of Iowa, Department of Mathematics, 52246 Iowa City, Iowa, U.S.A.
University of Konstanz, Department of Mathematics and Statistics,
Universitätsstraße 10, D-78464 Konstanz, Germany  Laboratoire d’analyse et
d’architecture des systèmes (LAAS-CNRS), 7 Avenue du Colonel Roche, 31031
Toulouse, France  Faculty of Electrical Engineering, Czech Technical
University in Prague, Technická 2, CZ-16626 Prague, Czech Republic  Institut
de Mathématiques de Toulouse, 118 Route de Narbonne, 31062 Toulouse France

###### Abstract

We study the time-dependent moments $s_{\alpha}(t)=\int x^{\alpha}\cdot
f(x,t)\leavevmode\nobreak\ \mathrm{d}x$ of the solution $f$ of the partial
differential equation $\partial_{t}f=\nu\Delta f+g\cdot\nabla f+h\cdot f$ with
initial Schwartz function data $f_{0}\in\mathcal{S}(\mathds{R}^{n})$. At first
we describe the dual action on the polynomials, i.e., the time-evolution of
$f$ is completely moved to the polynomial side $s_{\alpha}(t)=\int p(x,t)\cdot
f_{0}(x)\leavevmode\nobreak\ \mathrm{d}x$. We investigate the special case of
the heat equation. We find that several non-negative polynomials which are not
sums of squares become sums of squares under the heat equation in finite time.
Finally, we solve the problem of moving atoms under the equation
$\partial_{t}f=g\cdot\nabla f+h\cdot f$ with $f_{0}=\mu_{0}$ being a finitely
atomic measure. We find that in the time evolution
$\mu_{t}=\sum_{i=1}^{k}c_{i}(t)\cdot\delta_{x_{i}(t)}$ the atom positions
$x_{i}(t)$ are governed only by the transport term $g\cdot\nabla$ and that the
time-dependent coefficients $c_{i}(t)$ have an analytic solution depending on
$x_{i}(t)$.

###### keywords:

moment, time evolution, partial differential equations, sum of squares, heat
equation

###### MSC:

[2020] Primary: 47A57, 44A60; Secondary: 30E05, 65D32

††journal: ArXiv

###### Contents

  1. 1 Introduction
    1. 1.1 Static Moments
    2. 1.2 Time-dependent Moments
  2. 2 The dual of a partial differential equation acting on polynomials
  3. 3 Non-negative polynomials and the heat equation
  4. 4 Time-dependent set of atoms
    1. 4.1 Coefficients in $C_{b}^{\infty}$
    2. 4.2 Extension to $C^{\infty}$
  5. 5 Summary and open questions
  6. Acknowledgments

##  1 Introduction

###  1.1 Static Moments

Let $\mu$ be a Borel measure on $\mathds{R}^{n}$ for some $n\in\mathds{N}$ and
$\alpha=(\alpha_{1},\dots,\alpha_{n})\in\mathds{N}_{0}^{n}$. The
$\alpha$-moment $s_{\alpha}$ of $\mu$ is

$s_{\alpha}=\int_{\mathds{R}^{n}}x^{\alpha}\leavevmode\nobreak\
\mathrm{d}\mu(x)$ (1)

with $x^{\alpha}:=x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}$. The _classical
moment problem_ is: Given finitely or infinitely many real numbers
$s_{\alpha}$ in a sequence
$s=(s_{\alpha})_{\alpha\in\mathsf{A}\subseteq\mathds{N}_{0}^{n}}$, does there
exist a Borel measure $\mu$ such that (1) holds for all $\alpha\in\mathsf{A}$?
If the answer to this question is affirmative, then $\mu$ is called a
_representing measure_ of $s$ and $s$ is called a _moment sequence_. If
$\mathsf{A}$ is finite, then $s$ is a _truncated_ moment sequence, and if
$\mathsf{A}=\mathds{N}_{0}^{n}$, then $s$ is called _full_ moment sequence.
Additionally, if $\mathrm{supp}\,\mu\subseteq K$ for some
$K\subseteq\mathds{R}^{n}$, then $s$ is called a _$K$ -moment sequence_. A
classical tool to investigate moment sequences is the _Riesz functional_
$L=L_{s}:\mathds{R}[x_{1},\dots,x_{n}]\to\mathds{R}$ defined by
$L_{s}(x^{\alpha}):=s_{\alpha}$ and linearly extended to
$\mathds{R}[x_{1},\dots,x_{n}]$ if $s$ is full or linearly extended to
$\\{x^{\alpha}\,|\,\alpha\in\mathsf{A}\\}$ if $s$ is truncated.
$L:\mathds{R}[x_{1},\dots,x_{n}]\to\mathds{R}$ is called a _moment functional_
if it is represented by some Borel measure $\mu$. We always assume measures to
be non-negative unless specifically denoted as _signed_ measures.

Moments are a classical field of research [2, 5, 13, 51, 16, 21, 22, 26, 28,
29, 30, 32, 35, 34, 36, 37, 38, 39, 52, 53, 54, 24, 25] and in modern times
they are still of interest, e.g. because of the following application in
optimization. Let $p\in\mathds{R}[x_{1},\dots,x_{n}]$ be a polynomial. Then

$\min_{x_{0}\in
K\subseteq\mathds{R}^{n}}p(x_{0})=\min_{\begin{subarray}{c}\mu:\mathrm{supp}\,\mu\subseteq
K,\\\ \mu(K)=1\end{subarray}}\int p(x)\leavevmode\nobreak\
\mathrm{d}\mu(x)=\min_{\begin{subarray}{c}s\ K\text{-moment}\\\
\text{sequence},\ s_{0}=1\end{subarray}}L_{s}(p),$ (2)

since for the first equality we have that for any $x_{0}\in\mathds{R}^{n}$ the
Dirac measure $\delta_{x_{0}}$ acts as a point evaluation in (1) and the
second formulation holds by linearity of the integral using the $s_{\alpha}$
definition in (1). See e.g. [36] for more.

A classical result for truncated moment sequences is the _Richter_ [45] (or
_Richter–Rogosinski–Rosenbloom_ [45, 47, 48]) _Theorem_ ; see [13] for a
detailed discussion about the historical development.

######  Theorem 1.1 (Richter Theorem 1957 [45, Satz 4]).

Let $d\in\mathds{N}$, $\mathcal{V}$ be a $d$-dimensional real vector space of
measurable real functions $f:\mathcal{X}\to\mathds{R}$ on a measurable space
$\mathcal{X}$, and $L:\mathcal{V}\to\mathds{R}$ be a moment functional, i.e.,
there exists a measure $\mu$ on $\mathcal{X}$ such that

$L(f)=\int_{\mathcal{X}}f(x)\leavevmode\nobreak\ \mathrm{d}\mu(x)$

for all $f\in\mathcal{V}$. Then there are $c_{1},\dots,c_{k}>0$ and
$x_{1},\dots,x_{k}\in\mathcal{X}$ with $k\leq d$ such that

$L(f)=\sum_{i=1}^{k}c_{i}\cdot f(x_{i})$

for all $f\in\mathcal{V}$, i.e., we always find a $k$-atomic representing
measure $\nu=\sum_{i=1}^{k}c_{i}\cdot\delta_{x_{i}}$ of $L$ with $k\leq d$.

The points $x_{i}$ in the $k$-atomic representing measure $\nu$ are called
_atoms_ and the minimal number $k$ of atoms for fixed $L$ is called
_Carathéodory number_ $\mathcal{C}(L)$ resp. $\mathcal{C}(s)$ for truncated
moment sequences. A $k$-atomic representing measure is also called a (_Gauss_)
_quadrature rule_ [18]. Two questions arise naturally in theory and
applications:

  1. (a)

How many atoms $\mathcal{C}(L)$ are required to represent $L$?

  2. (b)

Where are the atoms $x_{i}$ located in a representation of $L$?

Very recent studies about the Carathéodory number $\mathcal{C}(L)$ are [49,
12, 10] and the _set of atoms_ (or _core variety_) is studied in [17, 11, 4].

All the studies and references given so far have one thing in common: They
study _static_ moments, i.e., $s$ is fixed in these studies and properties are
only derived from and for $s$. Hence, the study of the _moment cone_
$\mathcal{S}_{\mathsf{A}}$ (= set of all moment sequences $s$) is only
pointwise and collecting or even connecting moment sequences with the same
properties in the moment cone is difficult and does not arise naturally.

###  1.2 Time-dependent Moments

Let $n,m\in\mathds{N}$. We denote by
$C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{m})$ the set of _smooth bounded
functions_

$C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{m}):=\left\\{f\in
C^{\infty}(\mathds{R}^{n},\mathds{R}^{m})\,\middle|\,\|\partial^{\alpha}f\|_{\infty}<\infty\
\text{for all}\ \alpha\in\mathds{N}_{0}^{n}\right\\}$

and by $\mathcal{S}(\mathds{R}^{n},\mathds{R}^{m})$ the _Schwartz functions_

$\mathcal{S}(\mathds{R}^{n},\mathds{R}^{m}):=\left\\{f\in
C^{\infty}(\mathds{R}^{n},\mathds{R}^{m})\,\middle|\,\|x^{\alpha}\partial^{\beta}f\|_{\infty}<\infty\
\text{for all}\ \alpha,\beta\in\mathds{N}_{0}^{n}\right\\}.$

By $C^{d}([0,\infty),C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{m}))$ we denote
all functions $f:\mathds{R}^{n}\times[0,\infty)\to\mathds{R}^{m}$ such that

  1. (i)

$f(\,\cdot\,,t),\partial_{t}f(\,\cdot\,,t),\dots,\partial_{t}^{d}f(\,\cdot\,,t)\in
C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{m})$ for all $t\geq 0$ and

  2. (ii)

$\partial^{\alpha}f(x,\,\cdot\,)\in C^{d}([0,\infty),\mathds{R}^{m})$ for all
$x\in\mathds{R}^{n}$ and $\alpha\in\mathds{N}_{0}^{n}$.

Let $d\in\mathds{N}_{0}$, $\nu=(\nu_{1},\dots,\nu_{n})^{T}\in[0,\infty)^{n}$,
$\nu\cdot\Delta=\nu_{1}\cdot\partial_{1}^{2}+\dots+\nu_{n}\cdot\partial_{n}^{2}$
be the anisotropic Laplace operator, $g=(g_{1},\dots,g_{n})\in
C^{d}([0,\infty),C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{n}))$ be a smooth
bounded vector field, $h=(h_{i,j})_{i,j=1}^{m}\in
C^{d}([0,\infty),C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{m\times m}))$ be a
smooth bounded matrix function,
$k=(k_{1},\dots,k_{m})^{T}\in\mathcal{S}(\mathds{R}^{n},\mathds{R}^{m})$ be a
Schwartz function vector-valued function, and $a\in\mathds{R}^{n}$ be a
vector.

Then by [9, Thm. 2.10] the initial value problem

$\begin{split}\partial_{t}f(x,t)&=\nu\Delta f(x,t)+[ax+g(x,t)]\cdot\nabla
f(x,t)+h(x,t)\cdot f(x,t)+k(x,t)\\\
f(x,0)&=f_{0}(x)\in\mathcal{S}(\mathds{R}^{n},\mathds{R}^{m})\end{split}$ (3)

with
$(ax+g)\cdot\nabla:=(a_{1}x_{1}+g_{1})\cdot\partial_{1}+\dots+(a_{n}x_{n}+g_{n})\cdot\partial_{n}$
has a unique solution $f\in
C^{d+1}([0,\infty),\mathcal{S}(\mathds{R}^{n},\mathds{R}^{m}))$. Additionally,
for $m=1$ and $f_{0}\geq 0$ we have that $f(\,\cdot\,,t)\geq 0$ for all $t\geq
0$.

For $f_{0}\in\mathcal{S}(\mathds{R}^{n})$ with $f_{0}\geq 0$ we can calculate
the moments

$s_{\alpha}(0):=\int_{\mathds{R}^{n}}x^{\alpha}\cdot
f_{0}(x)\leavevmode\nobreak\ \mathrm{d}x$

for all $\alpha=(\alpha_{1},\dots,\alpha_{n})\in\mathds{N}_{0}^{n}$. Since the
time-dependent solution $f$ of (3) is unique, Schwartz function valued, and
non-negative for $m=1$ and $f_{0}\geq 0$, we find that the solution $f$
induces unique time-dependent moments

$s_{\alpha}(t):=\int_{\mathds{R}^{n}}x^{\alpha}\cdot
f(x,t)\leavevmode\nobreak\ \mathrm{d}x$ (4)

for all $t\geq 0$.

The time-dependent moments can be defined for nonlinear partial differential
equations as well. Let us look at a (non-linear) example. In [9, Lem. 3.6] for
Burgers’ equation

$\partial_{t}f(x,t)=-f(x,t)\cdot\partial_{x}f(x,t)$

we calculated the time-dependent moments

$s_{k,p}(t):=\int_{\mathds{R}}x^{k}\cdot f(x,t)^{p}\leavevmode\nobreak\
\mathrm{d}x$

of $f(x,t)^{p}$ for all $k,p\in\mathds{N}_{0}$ and found the explicit
expression

$s_{k,p}(t)=\sum_{i=0}^{k}\frac{s_{k-i,p+i}(0)}{i!}\cdot
t^{i}\cdot\prod_{j=0}^{i-1}\frac{(p+j)\cdot(k-j)}{1+(p+j)^{2}}\quad\in\mathds{R}[t]$
(5)

which depends only on the initial values $s_{k-i,p+i}(0)$ of $f_{0}$. Hence,
despite the fact that for any non-constant $f_{0}\in\mathcal{S}(\mathds{R})$
the classical solution of Burgers’ equation breaks down in finite time, the
moments are _a priori_ known for all times $t\in\mathds{R}$. In [9, Exm. 3.7]
we then calculate for the one-tooth-function

$f_{0}(x):=\begin{cases}1+x&\text{for}\ x\in[-1,0],\\\ 1-x&\text{for}\
x\in[0,1]\\\ 0&\text{else}\end{cases}\quad\geq 0$

the time-dependent moments $s_{0,1}(t)$, $s_{1,1}(t)$, and $s_{2,1}(t)$ to
find

$\int_{\mathds{R}}(x-t)^{2}\cdot f(x,t)\leavevmode\nobreak\
\mathrm{d}x=L_{s(t)}((x-t)^{2})=\frac{1}{6}-\frac{2}{15}t^{2}\quad\xrightarrow{t\to\pm\infty}\quad-\infty.$
(6)

Applying a molifier $S_{\varepsilon}$ to $f_{0}$ gives
$S_{\varepsilon}f_{0}\in\mathcal{S}(\mathds{R})$ for all $\varepsilon>0$ and
(6) changes continuously with $\varepsilon>0$. Then (6) holds at least for the
time of the existence of the classical solution. But the classical solution
remains non-negative which contradicts (6), i.e., Burgers’ equation breaks
down in finite time. In other words, Burgers’ equation as a (non-linear)
transport equation, starting with a non-negative $f_{0}$ means that the
classical solution is also non-negative as long as it exist. While from (5) we
see that _a priori_ the moments might exist for all times, the derivation of
(5) requires that $f(x,t)$ is a classical solution, i.e., smooth and even
Schwartz. The break down of the classical solution of Burgers’ equation in
finite time is therefore not observed through the moments becoming infinite
(i.e., they stop to exist), but through the non-negativity property of the
representing measure which is encoded in the moments. For some $t>0$ we find a
non-negative polynomial $p\in\mathds{R}[x]$ such that $L_{s(t)}(p)<0$, i.e.,
(6) shows that non-negativity is not preserved for all times and therefore the
classical solution $u$ does not exist for all times.

The Burgers’ equation example has two nice features:

  1. (a)

the time-dependent moments are polynomial in time:
$s_{k,p}(t)\in\mathds{R}[t]$, and

  2. (b)

the example to contradict non-negativity of the moments only needs moment up
to degree 2, i.e., $0$th, $1$st, and $2$nd moments.

Studying (3) with general $\nu=(\nu_{1},\dots,\nu_{n})\in[0,\infty)^{n}$,
$C_{b}^{\infty}$-functions $g_{i}$ and $h_{i,j}$, and Schwartz functions
$k_{i}$ it is evident that neither (a) nor (b) need to be satisfied anymore.
I.e. we can in general not hope for the time-dependent moments $s_{\alpha}(t)$
to be polynomial in time, depending only on the initial values
$s_{\alpha}(0)$, and to observe certain properties we no longer can rely on
finitely many moments $s_{\alpha}(t)$. We have to explore the computational
and theoretical limits of time-dependent moments $s_{\alpha}(t)$. That is,
among other things, one purpose of this study. We say that a (moment) sequence
$s$ _evolves with respect to_ or _along the partial differential equation (
3)_ if a representing measure of $s$ evolves with respect to (3). Hence, in
principle the time-evolution $s(t)$ depends on the choice of representing
measure $\mu_{0}$ of $s(0)$ if $s(0)$ is _indeterminate_ , i.e., $s(0)$ has
more than one representing measure.

The time-dependent moments $s(t)$ from the heat equation $\partial_{t}f=\Delta
f$ were studied extensively in [6]. In [42, 43, 31, 33] moments also have been
applied to (non-linear) PDEs. In the present work we proceed the study [6] and
go beyond the heat equation. The paper is structured as follows.

In the next section (Section 2) we describe for (3) with $k=0$ the dual action
on the polynomials. We describe the dual action by the dual operator of (3)
and show that polynomials in general are no longer polynomials for $t>0$ but
at least remain in an algebra $\mathrm{Pol}(\mathds{R}^{n})$ (see (7) for the
definition). For $\nu\Delta+g\nabla+h$ with $g(x,t)=g(t)$ and $h(x,t)=h(t)$ we
can solve the dual action analytically (see 2.7). We find that $g$ and $h$
have no effect on non-negativity and therefore study the (dual) action of the
Laplace operator on polynomials in Section 3 more closely. We give a simple
way to solve the polynomial heat equation in 3.10. We collect several results
of this action and also present several specific results on non-negative
polynomials. We show several examples where non-negative polynomials become
sums of squares under the heat equation and also give counter examples, when a
non-negative polynomial which is not a sum of squares does not become a sum of
squares. Our observations lead to two open problems (5.32 and 5.33). In
Section 4 we investigate the application of (3) with $\nu=0$ and $k=0$ to
atomic measures. We find that since $g(x,t)\cdot\nabla$ and $h(x,t)$ in
general do not commute and therefore a solution can not be constructed from
solutions of the individual operators, for atomic measures we can at first
solve the time-evolution with respect to the transport operator
$g(x,t)\cdot\nabla$ and then apply the scaling operator $h(x,t)$. We find that
the number of atoms is unchanged. We describe the time-evolution of
$\mu_{t}=\sum_{i=1}^{k}c_{i}(t)\cdot\delta_{x_{i}(t)}$ in 4.27 and 4.28. For
the full moment problem we show that the time-dependent moment sequence
(functional) remains a boundary of the moment cone for all times. For the
truncated moment problem we show that we can enter the interior of the moment
cone and that the time-evolution in general depends on the representing
measure. In Section 5 we summarize the results, give final discussions, and
state the open problem.

##  2 The dual of a partial differential equation acting on polynomials

By [9, Thm. 2.10] we know that (3) has a unique solution $f\in
C^{d+1}([0,\infty),$ $\mathcal{S}(\mathds{R}^{n},\mathds{R}^{m}))$ and
therefore the time-dependent moments $s_{\alpha}(t)$ in (4) exist for all
times $t\in[0,\infty)$. In this section we show that like in the heat equation
[6] the action of the operator $A=\nu\cdot\Delta+g(x,t)\nabla+h(x,t)$ has a
dual action $A^{*}$ acting on the polynomials, i.e., the time-dependency of
the solution $f(x,t)$ is moved to a time-dependency of $p(x,t)$:

$\int p_{0}(x)\cdot f(x,t)\leavevmode\nobreak\ \mathrm{d}x=\int p(x,t)\cdot
f_{0}(x)\leavevmode\nobreak\ \mathrm{d}x$

for all $t\in[0,\infty)$. We will see in 2.3 that while
$p(x,0)\in\mathds{R}[x_{1},\dots,x_{n}]$, we have in general
$p(x,t)\not\in\mathds{R}[x_{1},\dots,x_{n}]$ for any $t\neq 0$. We have to
introduce the following space of _at most polynomially increasing functions
$\mathrm{Pol}(\mathds{R}^{n})$_ on $\mathds{R}^{n}$:

$\begin{split}\mathrm{Pol}(\mathds{R}^{n}):=\\{f\in
C^{\infty}(\mathds{R}^{n})\,|\,&\text{for all}\ \alpha\in\mathds{N}_{0}^{n}\
\text{there exists}\ p_{\alpha}\in\mathds{R}[x]\ \text{such}\\\ &\text{that}\
|\partial^{\alpha}f(x)|\leq p_{\alpha}(x)\ \text{for all}\
x\in\mathds{R}^{n}\\}.\end{split}$ (7)

We collect some simple properties of $\mathrm{Pol}(\mathds{R}^{n})$.

######  Lemma 2.2.

Let $n\in\mathds{N}$. Then the following hold:

  1. i)

$\mathrm{Pol}(\mathds{R}^{n})$ is an algebra.

  2. ii)

$\mathcal{S}(\mathds{R}^{n})\subsetneq\mathrm{Pol}(\mathds{R}^{n})$.

  3. iii)

$\mathds{R}[x_{1},\dots,x_{n}]\subsetneq\mathds{R}[x_{1},\dots,x_{n}]+C_{b}^{\infty}(\mathds{R}^{n})\subsetneq\mathrm{Pol}(\mathds{R}^{n})\subsetneq\mathcal{S}(\mathds{R}^{n})^{\prime}$.

###### Proof.

(i) and (ii) follow immediately from the definition of
$\mathrm{Pol}(\mathds{R}^{n})$ in (7). For (iii) all inclusions $\subseteq$
are clear, it remains to show that the inclusions are proper $\subsetneq$. For
the first $\subsetneq$ we have $\sin
x_{1}\in[\mathds{R}[x_{1},\dots,x_{n}]+C_{b}^{\infty}(\mathds{R}^{n})]\setminus\mathds{R}[x_{1},\dots,x_{n}]$,
for the second $\subsetneq$ we have $x_{1}^{2}\cdot\sin
x_{1}\in\mathrm{Pol}(\mathds{R}^{n})\setminus[\mathds{R}[x_{1},\dots,x_{n}]+C_{b}^{\infty}(\mathds{R}^{n})]$,
and the last $\subsetneq$ is clear. ∎

Let $p\in\mathrm{Pol}(\mathds{R}^{n})$ and $f$ be the unique solution of

$\displaystyle\partial_{t}f(x,t)$ $\displaystyle=\nu\Delta
f(x,t)+g(x,t)\cdot\nabla f(x,t)+h(x,t)\cdot f(x,t)$ $\displaystyle f(x,0)$
$\displaystyle=f_{0}(x)$

with $f_{0}\in\mathcal{S}(\mathds{R}^{n})$, $g=(g_{1},\dots,g_{n})^{T}\in
C([0,\infty),C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}))$, and $h\in
C([0,\infty)$, $C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}))$. Then

$\displaystyle\partial_{t}$ $\displaystyle\int p(x)\cdot
f(x,t)\leavevmode\nobreak\ \mathrm{d}x$ $\displaystyle=\int
p(x)\cdot[\nu\Delta+g(x,t)\cdot\nabla+h(x,t)]f(x,t)\leavevmode\nobreak\
\mathrm{d}x$ is by partial integration (since
$p\in\mathrm{Pol}(\mathds{R}^{n})$ and $f\in\mathcal{S}(\mathds{R}^{n})$)
$\displaystyle=\int
f(x,t)\cdot[\nu\Delta+g(x,t)\cdot\nabla-\mathrm{div}\,g(x,t)+h(x,t)]p(x)\leavevmode\nobreak\
\mathrm{d}x$

The following result shows, that the time evolution of $f$ can be shifted to
$p$. The proof uses techniques of the semi-group approximations of the
solution $f$.

######  Theorem 2.3.

Let $d\in\mathds{N}_{0}$, $\nu\geq 0$, $g=(g_{1},\dots,g_{n})^{T}\in
C^{d}([0,\infty),C_{b}^{\infty}(\mathds{R}^{n})^{n})$, and $h\in
C^{d}([0,\infty),C_{b}^{\infty}(\mathds{R}^{n}))$. Let $f\in
C^{d+1}([0,\infty),\mathcal{S}(\mathds{R}^{n}))$ be the unique solution of

$\begin{split}\partial_{f}(x,t)&=\nu\Delta f(x,t)+g(x,t)\cdot\nabla
f(x,t)+h(x,t)\cdot f(x,t)\\\ f(x,0)&=f_{0}(x)\end{split}$ (8)

with $f_{0}\in\mathcal{S}(\mathds{R}^{n})$. Let $T>0$. Then the unique
solution $p_{T}$ of

$\begin{split}\partial_{t}p_{T}(x,t)&=\nu\Delta p_{T}(x,t)-g(x,T-t)\cdot\nabla
p_{T}(x,t)\\\ &\qquad+(h(x,T-t)-\mathrm{div}\,g(x,T-t))\cdot p_{T}(x,t)\\\
p(x,0)&=p_{0}(x)\in\mathrm{Pol}(\mathds{R}^{n})\end{split}$ (9)

fulfills $p_{T}\in C^{d+1}([0,T],\mathrm{Pol}(\mathds{R}^{n}))$, and we have

$\int p_{0}(x)\cdot f(x,T)\leavevmode\nobreak\ \mathrm{d}x=\int
p_{T}(x,T)\cdot f_{0}(x)\leavevmode\nobreak\ \mathrm{d}x.$ (10)

###### Proof.

Note, the solution $f$ of (8) is unique since it is unique on any open bounded
set $U\subset\mathds{R}^{n}$, see e.g. [15, Ch. 7], and we have $f\in
C^{d+1}([0,\infty),\mathcal{S}(\mathds{R}^{n}))$, see e.g. [9]. Let
$N\in\mathds{N}$ and $\mathcal{Z}_{N}=\\{t_{0}=0<t_{1}<\dots<t_{N}=T\\}$ be a
decomposition of $[0,T]$. Then for the operator

$A(x,t)=\nu\Delta+g(x,t)\cdot\nabla+h(x,t)$

we have the dual operator

$A^{*}(x,t)=\nu\Delta-g(x,t)\cdot\nabla-\mathrm{div}\,g(x,t)+h(x,t).$

Note, of course, we actually also have to give the domain of the operators $A$
and $A^{*}$. We are working for $A$ on $\mathcal{S}(\mathds{R}^{n})$ and for
$A^{*}$ it is therefore sufficient to work on $\mathrm{Pol}(\mathds{R}^{n})$.
We only have to ensure that $p_{T}\in\mathrm{Pol}(\mathds{R}^{n})$.

The unique solution $f$ can be approximated in $\mathcal{S}(\mathds{R}^{n})$
by the semigroup approach (Trotter [55])

$f(x,T)=\lim_{N\to\infty}\prod_{i=1}^{N}\exp\left(\int_{t_{i-1}}^{t_{i}}\\!\\!\\!\\!\\!A(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)f_{0}(x).$ (11)

For any operator $B$ we have for the dual $B^{*}$ the relation $\langle
f,Bg\rangle=\langle B^{*}f,g\rangle$ and for exponentials $\exp(B)$ we have
$\langle f,\exp(B)g\rangle=\langle\exp(B^{*})f,g\rangle$. When we apply these
to (11) we have to pay attention at the order of the operators, since in
general they do not commute and are additionally time-dependent. We use the
order $\prod_{i=1}^{N}B_{i}=B_{N}B_{N-1}\cdots B_{1}$ in this formulas. Hence,
we get

$\displaystyle\int_{\mathds{R}^{n}}p_{0}(x)\cdot f(x,T)\leavevmode\nobreak\
\mathrm{d}x$
$\displaystyle=\lim_{N\to\infty}\int_{\mathds{R}^{n}}p_{0}(x)\cdot\prod_{i=1}^{N}\exp\left(\int_{t_{i-1}}^{t_{i}}\\!\\!\\!\\!\\!A(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)f_{0}(x)\leavevmode\nobreak\ \mathrm{d}x$
$\displaystyle=\lim_{N\to\infty}\int_{\mathds{R}^{n}}\left[\prod_{i=N}^{1}\exp\left(\int_{t_{i-1}}^{t_{i}}\\!\\!\\!\\!\\!A^{*}(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)p_{0}(x)\right]\cdot f_{0}(x)\leavevmode\nobreak\
\mathrm{d}x$ $\displaystyle=\int_{\mathds{R}^{n}}p_{T}(x,T)\cdot
f_{0}(x)\leavevmode\nobreak\ \mathrm{d}x.$

The last equality holds in the same way as (11). We set

$p_{T}(x,T):=\lim_{N\to\infty}\prod_{i=N}^{1}\exp\left(\int_{t_{i-1}}^{t_{i}}\\!\\!\\!\\!\\!A^{*}(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)p_{0}(x)$

and see that $p_{T}$ solves (10) since the last equality holds for all
$f_{0}\in\mathcal{S}(\mathds{R}^{n})$, i.e., especially for all test functions
$f_{0}\in C_{0}^{\infty}(\mathds{R}^{n})\subset\mathcal{S}(\mathds{R}^{n})$.
This indeed shows that $p_{T}\in\mathrm{Pol}(\mathds{R}^{n})$ and the time-
dependency of $A(x,t)$ combined with the fact that $A(x,t)$ and
$A(x,t^{\prime})$ for $t\neq t^{\prime}$ in general do not commute shows that
we have to take the reverse order in the operator product, i.e., $p_{T}(x,t)$
solves by substituting $t\mapsto T-t$ in $A^{*}$ the equation
$\partial_{t}p_{T}(x,t)=A^{*}(x,T-t)p_{T}(x,t)$. ∎

######  Remark 2.4.

2.3 holds with the same proof also for the anisotropic Laplace operator
$\nu\cdot\Delta=\nu_{1}\partial_{1}^{2}+\dots+\nu_{n}\partial_{n}^{2}$ with
$\nu=(\nu_{1},\dots,\nu_{n})\in[0,\infty)^{n}$. Additionally, let
$M\in\mathds{R}^{n\times n}$ be a symmetric positive-definite matrix. Then by
a change of coordinates the operator
$(\partial_{1},\dots,\partial_{n})M(\partial_{1},\dots,\partial_{n})$ can be
diagonalized to the anisotropic Laplace operator $\nu\cdot\Delta$. $\circ$

######  Remark 2.5.

In 2.3 we have seen that the action of (3) on $f_{0}$ is shifted to $p_{0}$.
But since the dual action on $p_{0}$ is independent on $f_{0}$, it also
provides the action (3) on measures $\mu_{0}$ instead of functions $f_{0}$.
This provides a way to study the set of atoms in Section 4. $\circ$

######  Remark 2.6.

In the proof of 2.3 note that $A=\nu\Delta+g\nabla+h$ is in general an
unbounded operator, i.e., it is not defined on all $L^{2}(\mathds{R}^{n})$.
Since $\partial_{t}f=Af$ with $f_{0}\in\mathcal{S}(\mathds{R}^{n})$ has a
unique Schwartz function solution it is therefore sufficient to take the
domain of $A$ as $\mathcal{D}(A)=\mathcal{S}(\mathds{R}^{n})$. This very
restrictive domain enables us in 2.3 to have
$\mathrm{Pol}(\mathds{R}^{n})\subseteq\mathcal{D}(A^{*})$. Here, the
restriction that $g$ and $h$ are $C_{b}^{\infty}$-functions is essential. It
is not sufficient to have $A$ such that $Af\in\mathcal{S}(\mathds{R}^{n})$ for
any $f\in\mathcal{S}(\mathds{R}^{n})$. E.g. take $\nu=0$, $g=0$ and $h=x^{4}$
with $f_{0}(x)=e^{-x^{2}}$. Then $\partial_{t}f(x,t)=x^{4}\cdot f(x,t)$ has as
an ordinary differential equation in $t$ with fixed $x$ the unique solution
$f(x,t)=e^{-x^{2}+t\cdot x^{4}}$, i.e.,
$f(\,\cdot\,,t)\notin\mathcal{S}(\mathds{R}^{n})$ for any $t>0$. Additionally
note, that in [14, Thm. 2.2] a condition on the generator $A$ is given such
that the semi-group $e^{At}$ maps the Schwartz class $\mathcal{S}$ to the
Schwartz class $\mathcal{S}$. $\circ$

We have seen in the previous proof that the time-reversal for the dual
equation acting on $p_{0}$ appears because $\Delta$, $g(x,t)\nabla$, and
$h(x,t)$ do not commute (pairwise) in general. If $g$ and $h$ are time-
independent, then of course the time-reversal disappears naturally since no
time-dependency exists. Another way the time-reversal disappears is, when
$\Delta$, $g(x,t)\nabla$, and $h(x,t)$ commute (pairwise), e.g. when $g$ and
$h$ do not depend on $x$. We then have the following explicit solution for
$p_{t}(x,t)$. We denote by $\Theta_{t}$ the heat kernel, i.e.,

$\Theta_{t}(x)=\frac{1}{(4\pi
t)^{n/2}}\cdot\exp\left(\frac{-\|x\|^{2}}{4t}\right)$

for all $t>0$.

######  Corollary 2.7.

Let $n\in\mathds{N}_{0}$, $\nu\geq 0$, $g=(g_{1},\dots,g_{n})$, and
$g_{1},\dots,g_{n},h\in C^{d}([0,\infty),\mathds{R})$. Then for the dual
action (9) and (10) in 2.3 we have

$p_{t}(x,t)=e^{H(t)}\cdot[\Theta_{\nu t}*p_{0}](x+G(t))$

with $\Theta_{\nu t}$ the heat kernel,
$G(t):=\int_{0}^{t}g(t)\leavevmode\nobreak\ \mathrm{d}s$, and
$H(t):=\int_{0}^{t}h(s)\leavevmode\nobreak\ \mathrm{d}s$.

In [6] the first and the second author studied the time-dependent moments from
the heat equation in more detail. For the dual action it was especially found
that

$\displaystyle p_{0}\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}\qquad$
$\displaystyle\Rightarrow\qquad[\Theta_{\nu
t}*p_{0}](x)\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}$ for all $t\geq 0$ and
additionally of course because of the convolution with the non-negative heat
kernel $\displaystyle p_{0}\geq 0\qquad$
$\displaystyle\Rightarrow\qquad[\Theta_{\nu t}*p_{0}](x)\geq 0$

for all $t\geq 0$. Hence, with $p_{0}\in\mathds{R}[x_{1},\dots,x_{n}]$ in 2.7
we have $p_{t}(x,t)\in\mathds{R}[x_{1},\dots,x_{n}]$ for all $t\in[0,\infty)$
(and even for all $t\in\mathds{R}$, see [6]). In the general case of 2.3 we
only can ensure $p_{t}(x,t)\in\mathrm{Pol}(\mathds{R}^{n})$, but not
$p_{t}(x,t)\in\mathds{R}[x_{1},\dots,x_{n}]$. To see that, let e.g. $\nu=0$
and $g=0$, i.e., we have the explicit solution

$p_{t}(x,t)=\exp\left(\int_{0}^{t}h(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)\cdot p_{0}(x)$

and with $h(x)=\sin(x)$ we have that
$p_{t}(x,t)\not\in\mathds{R}[x_{1},\dots,x_{n}]$ for all $t\neq 0$.

Therefore, in case of 2.7 we have that $g$ and $h$ do not alter the properties
of $p_{0}$ (significantly), but the group action induced by the Laplace
operator might give changes. We will see that in the next section.

##  3 Non-negative polynomials and the heat equation

Time-dependent moments induced by the heat equation were already studied in
[6]. There also the dual action on the polynomials was observed. We continue
this investigation. We repeat the essential definitions and results for the
convenience of the reader.

######  Definition 3.8.

Let $d\in\mathds{N}_{0}$. We define
$\mathfrak{p}_{2d},\mathfrak{p}_{2d+1}\in\mathds{R}[x,t]$ by

$\displaystyle\mathfrak{p}_{2d}(x,t)$
$\displaystyle:=\sum_{j=0}^{d}\frac{(2d)!}{(2d-2j)!\cdot j!}\cdot t^{j}\cdot
x^{2d-2j}$ and $\displaystyle\mathfrak{p}_{2d+1}(x,t)$
$\displaystyle:=\sum_{j=0}^{d}\frac{(2d+1)!}{(2d+1-2j)!\cdot j!}\cdot
t^{j}\cdot x^{2d+1-2j}.$

######  Example 3.9.

We have

$\displaystyle\mathfrak{p}_{0}(x,t)$ $\displaystyle=1$
$\displaystyle\mathfrak{p}_{1}(x,t)$ $\displaystyle=x$
$\displaystyle\mathfrak{p}_{2}(x,t)$ $\displaystyle=2t+x^{2}$
$\displaystyle\mathfrak{p}_{3}(x,t)$ $\displaystyle=6tx+x^{3}$
$\displaystyle\mathfrak{p}_{4}(x,t)$ $\displaystyle=12t^{2}+12tx^{2}+x^{4}$
$\displaystyle\mathfrak{p}_{5}(x,t)$ $\displaystyle=60t^{2}x+20tx^{3}+x^{5}$
$\displaystyle\mathfrak{p}_{6}(x,t)$
$\displaystyle=120t^{3}+180t^{2}x^{2}+30tx^{4}+x^{6}$ $\displaystyle\
\,\vdots$ $\circ$

Straightforward calculations show that $\mathfrak{p}_{k}$,
$k\in\mathds{N}_{0}$, solve the initial value heat equation

$\begin{split}\partial_{t}\mathfrak{p}_{k}(x,t)&=\partial_{x}^{2}\mathfrak{p}_{k}(x,t)\\\
\mathfrak{p}_{k}(x,0)&=x^{k}.\end{split}$ (12)

Hence, by linearity of the heat equation we have the following extension of
3.8 and the observation (12).

######  Theorem 3.10.

Let $d\in\mathds{N}_{0}$, $n\in\mathds{N}$, and
$f_{0}(x)=\sum_{\alpha\in\mathds{N}_{0}^{n}}c_{\alpha}\cdot
x^{\alpha}\in\mathds{R}[x_{1},\dots,x_{n}]$. Then

$\mathfrak{p}_{f_{0}}(x,t):=\sum_{\alpha\in\mathds{N}_{0}^{n}}c_{\alpha}\cdot\mathfrak{p}_{\alpha_{1}}(x_{1},t)\cdots\mathfrak{p}_{\alpha_{n}}(x_{n},t)\quad\in\mathds{R}[x_{1},\dots,x_{n},t]$
(13)

with $\alpha=(\alpha_{1},\dots,\alpha_{n})\in\mathds{N}_{0}^{n}$ solves the
initial value heat equation

$\displaystyle\partial_{t}f(x,t)$ $\displaystyle=\Delta f(x,t)$ $\displaystyle
f(x,0)$ $\displaystyle=f_{0}(x).$

###### Proof.

By linearity of the Laplace operator $\Delta$ it is sufficient to look at
$f_{0}(x)=x^{\alpha}$ for $\alpha\in\mathds{N}_{0}^{n}$. By (12) we already
have
$\partial_{t}\mathfrak{p}_{\alpha_{i}}(x_{i})=\partial_{i}^{2}\mathfrak{p}_{\alpha_{i}}(x)$
and hence

$\displaystyle\partial_{t}\mathfrak{p}_{f_{0}}(x,t)$
$\displaystyle=\partial_{t}[\mathfrak{p}_{\alpha_{1}}(x_{1},t)\cdots\mathfrak{p}_{\alpha_{n}}(x_{n},t)]$
$\displaystyle=[\partial_{t}\mathfrak{p}_{\alpha_{1}}(x_{1},t)]\cdot\mathfrak{p}_{\alpha_{2}}(x_{2},t)\cdots\mathfrak{p}_{\alpha_{n}}(x_{n},t)$
$\displaystyle\qquad+\dots+\mathfrak{p}_{\alpha_{1}}(x_{1},t)\cdots\mathfrak{p}_{\alpha_{n-1}}(x_{n-1},t)\cdot[\partial_{t}\mathfrak{p}_{\alpha_{n}}(x_{n},t)]$
$\displaystyle=[\partial_{1}^{2}\mathfrak{p}_{\alpha_{1}}(x_{1},t)]\cdot\mathfrak{p}_{\alpha_{2}}(x_{2},t)\cdots\mathfrak{p}_{\alpha_{n}}(x_{n},t)$
$\displaystyle\qquad+\dots+\mathfrak{p}_{\alpha_{1}}(x_{1},t)\cdots\mathfrak{p}_{\alpha_{n-1}}(x_{n-1},t)\cdot[\partial_{n}^{2}\mathfrak{p}_{\alpha_{n}}(x_{n},t)]$
$\displaystyle=\Delta\mathfrak{p}_{f_{0}}(x,t).\qed$

Note, in 3.8 we defined $\mathfrak{p}$ for the monomials $x^{d}$,
$d\in\mathds{N}_{0}$, and in (13) we define $\mathfrak{p}$ by linearity for
any $f_{0}\in\mathds{R}[x]$. Hence, if $f_{0}(x)=x^{d}$ then both definitions
coincide: $\mathfrak{p}_{f_{0}}=\mathfrak{p}_{d}$. The same shall hold for the
multivariate case to keep the notation simple.

######  Example 3.11 (Motzkin polynomial [41]).

Let

$f_{\text{Motz}}(x,y)=1-3x^{2}y^{2}+x^{4}y^{2}+x^{2}y^{4}\in\mathds{R}[x,y]$

be the Motzkin polynomial. Then by 3.8 (resp. 3.9) we have the substitutions

$\displaystyle x^{2}$ $\displaystyle\mapsto 2t+x^{2},\qquad x^{4}\mapsto
12t^{2}+12tx^{2}+x^{4},$ $\displaystyle y^{2}$ $\displaystyle\mapsto
2t+y^{2},\qquad y^{4}\mapsto 12t^{2}+12ty^{2}+y^{4}$

and get

$\displaystyle\mathfrak{p}_{\text{Motz}}(x,y,t)$
$\displaystyle=1-3(2t+x^{2})(2t+y^{2})+(12t^{2}+12tx^{2}+x^{4})(2t+y^{2})$
$\displaystyle\quad+(2t+x^{2})(12t^{2}+12ty^{2}+y^{4})$
$\displaystyle=1-12t^{2}+48t^{3}+6t(-1+6t)(x^{2}+y^{2})+(-3+24t)x^{2}y^{2}$
$\displaystyle\quad+2t(x^{4}+y^{4})+x^{4}y^{2}+x^{2}y^{4}$

for all $t\in\mathds{R}$. $\circ$

Since the heat kernel is a Schwartz function the convolution with polynomials
is well-defined and uniqueness of the solution of the heat equation shows that

$(\Theta_{t}*f_{0})(x)=\mathfrak{p}_{f_{0}}(x,t)$ (14)

holds for all $f_{0}\in\mathds{R}[x_{1},\dots,x_{n}]$. That the heat kernel
preserves polynomials holds for all convolution kernels which are integrable
with respect to polynomials. To make the paper self-contained, let us briefly
state and prove this known fact.

######  Theorem 3.12.

Let $d\in\mathds{N}_{0}$ and $\rho$ be a kernel such that
$\int_{\mathds{R}^{n}}y^{\alpha}\cdot\rho(y)\leavevmode\nobreak\ \mathrm{d}y$
is finite for all $\alpha\in\mathds{N}_{0}^{n}$ with $|\alpha|\leq d$, then

$\cdot\,*\rho:\mathds{R}[x_{1},\dots,x_{n}]_{\leq
d}\to\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}.$

###### Proof.

Let $p\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}$. Then from

$\displaystyle(p*\rho)(x)=\int_{\mathds{R}^{n}}p(x-y)\cdot\rho(y)\leavevmode\nobreak\
\mathrm{d}y$

and expanding $p(x-y)$ in the right side gives the assertion including the
degree bound $\deg(p*\rho)\leq d$. ∎

Since the heat kernel is non-negative, the convolution of the heat kernel with
a non-negative polynomial gives again a non-negative polynomial. Denote by
$\mathrm{Pos}(n,d)$ the set of all non-negative polynomials on
$\mathds{R}^{n}$ with degree at most $d\in\mathds{N}_{0}$:

$\mathrm{Pos}(n,d):=\\{p\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}\,|\,p(x)\geq
0\ \text{for all}\ x\in\mathds{R}^{n}\\}.$

Non-negativity and 3.12 gives the following.

######  Corollary 3.13.

Let $n\in\mathds{N}$, $d\in\mathds{N}_{0}$, and $f_{0}\in\mathrm{Pos}(n,d)$.
Then

$\mathfrak{p}_{f_{0}}(\,\cdot\,,t)\in\mathrm{Pos}(n,d)$

for all $t\geq 0$. Especially, if $f_{0}\neq 0$ then
$\mathfrak{p}_{f_{0}}(\,\cdot\,,t)>0$ on $\mathds{R}^{n}$ for all $t>0$.

######  Corollary 3.14.

Let $n\in\mathds{N}$ and $f_{0}\in\mathds{R}[x_{1},\dots,x_{n}]$. Assume there
exist $t>0$ and a point $\xi\in\mathds{R}^{n}$ such that
$\mathfrak{p}_{f_{0}}(\xi,t)<0$. Then $f_{0}\not\in\mathrm{Pos}(n,d)$.

Hence, recalling the Motzkin polynomial in 3.11 we find that
$\mathfrak{p}_{\text{Motz}}(\,\cdot\,,t)\in\mathrm{Pos}(2,6)$ for all $t\geq
0$. In the usual case (e.g. $f_{0}\in L^{2}(\mathds{R}^{n})$) the convolution
with the heat kernel has a smoothing effect on $f_{0}$, i.e., the regularity
increases and $\Theta_{t}*f_{0}$ is a $C^{\infty}$-function. But in the
polynomial case we already started with a $C^{\infty}$-function and this kind
of regularity does not change. We have even seen that for
$f_{0}\in\mathds{R}[x_{1},\dots,x_{n}]$ the function $\mathfrak{p}_{f_{0}}$
remains a polynomial for all $t\in\mathds{R}$. But from the Motzkin polynomial
in 3.11 we observe something additional. The Motzkin polynomial was of course
the first non-negative polynomial found that is not a sum of squares. However,
the heat kernel has another “smoothing effect” for this polynomial, as seen in
the following continuation. We denote by $\mathrm{SOS}(n,d)$ the set of all
sums of squares in $n\in\mathds{N}$ variables of degree less or equal to
$d\in\mathds{N}_{0}$:

$\mathrm{SOS}(n,d):=\\{p\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}\,|\,p\
\text{is a sum of squares}\\}.$

######  Example 3.15 (Motzkin polynomial, 3.11 continued).

We have

$\displaystyle\mathfrak{p}_{\text{Motz}}(x,y,1)$
$\displaystyle=37\cdot\left(1-\frac{11}{148}x^{2}-\frac{11}{148}y^{2}\right)^{2}+\frac{71}{2}\cdot\left(x-\frac{4}{71}xy^{2}\right)^{2}$
$\displaystyle\quad+\frac{71}{2}\cdot\left(y-\frac{4}{71}x^{2}y\right)^{2}+\frac{57}{2}x^{2}y^{2}+\frac{1063}{592}\cdot\left(x^{2}+\frac{27}{1063}y^{2}\right)^{2}$
$\displaystyle\quad+\frac{3815}{2126}\cdot y^{4}+\frac{63}{71}\cdot
x^{4}y^{2}+\frac{63}{71}x^{2}y^{4}\quad\in\mathrm{SOS}(2,6)$

i.e., $\mathfrak{p}_{\text{Motz}}(\,\cdot\,,1)$ is by 3.13 not just non-
negative, but in fact a sum of squares. This relation can easily be obtained
e.g. by the use of Macaulay2 [19] and the SumsOfSquares package [7]. In fact,
additional calculations indicate that

$\mathfrak{p}_{\text{Motz}}(\,\cdot\,,t)\in\begin{cases}\mathrm{Pos}(2,6)\setminus\mathrm{SOS}(2,6)&\text{for}\
t\in[0,T_{\mathrm{Motz}}),\ \text{and}\\\ \mathrm{SOS}(2,6)&\text{for}\
t\in[T_{\mathrm{Motz}},\infty),\end{cases}$

with

$\frac{31\,998}{1\,000\,000}\quad<\quad
T_{\mathrm{Motz}}\quad<\quad\frac{31\,999}{1\,000\,000}.$

The choice of the intervals $[0,T_{\mathrm{Motz}})$ and
$[T_{\mathrm{Motz}},\infty)$ is clear since $\mathrm{SOS}(2,6)$ is closed and
$\mathfrak{p}_{f_{0}}(\,\cdot\,,t)$ continuous in $t$, i.e.,
$\mathfrak{p}_{\text{Motz}}(\,\cdot\,,T_{\mathrm{Motz}})\in\mathrm{SOS}(2,6)$.
$\circ$

$\mathfrak{p}_{\text{Motz}}:[0,\infty)\ni
t\mapsto\mathfrak{p}_{\text{Motz}}(\,\cdot\,,t)\in\mathrm{Pos}(2,6)$ is a
continuous path through the cone of non-negative polynomials. The following
result shows that once $\mathfrak{p}_{f_{0}}$ enters $\mathrm{SOS}(n,d)$ e.g.
at time $t_{0}\geq 0$, then it stays in $\mathrm{SOS}(n,d)$ for all $t\geq
t_{0}$.

######  Theorem 3.16.

Let $\rho\geq 0$ be a kernel such that
$\int_{\mathds{R}^{n}}y^{\alpha}\cdot\rho(y)\leavevmode\nobreak\ \mathrm{d}y$
is finite for all $\alpha\in\mathds{N}_{0}^{n}$ with $|\alpha|\leq d$, then

$\cdot\,*\rho:\mathrm{SOS}(n,d)\to\mathrm{SOS}(n,d).$

###### Proof.

Let $p\in\mathrm{SOS}(n,d)$, i.e., there exists a symmetric
$Q\in\mathds{R}^{N\times N}$ with $N=\binom{n+d}{d}$ such that
$p(x)=(x^{\alpha})_{\alpha}^{T}\cdot Q\cdot(x^{\alpha})_{\alpha}$ where
$(x^{\alpha})_{\alpha}$ is the vector of all monomials $x^{\alpha}$ with
$|\alpha|\leq d$. We then have

$\displaystyle(p*\rho)(x)$
$\displaystyle=\int_{\mathds{R}^{n}}p(x-y)\cdot\rho(y)\leavevmode\nobreak\
\mathrm{d}y$
$\displaystyle=\int_{\mathds{R}^{n}}((x-y)^{\alpha})_{\alpha}^{T}\cdot
Q\cdot((x-y)^{\alpha})_{\alpha}\cdot\rho(y)\leavevmode\nobreak\ \mathrm{d}y$
and by Richter’s Theorem [45] we can replace $\rho(y)\leavevmode\nobreak\
\mathrm{d}y$ by a finitely atomic representing measure
$\mu=\sum_{i=1}^{k}c_{i}\cdot\delta_{y_{i}}$ with $c_{i}>0$ and get
$\displaystyle=\sum_{i=1}^{k}c_{i}\cdot((x-y_{i})^{\alpha})_{\alpha}^{T}\cdot
Q\cdot((x-y_{i})^{\alpha})_{\alpha}\in\mathrm{SOS}(n,d).\qed$

Besides sums of squares, other non-negative polynomials are linear
combinations of even powers of linear forms, i.e., they have a _Waring
decomposition_

$p(x)=\sum_{i=1}^{k}(a_{i}\cdot x)^{d}$

with even $d\in\mathds{N}$, $a_{i}\in\mathds{R}^{n+1}$, and $a_{i}\cdot
x:=a_{i,0}+a_{i,1}x_{1}+\dots+a_{i,n}x_{n}$ [44]. We denote by
$\mathrm{War}(n,d)$ the (Waring) cone of all these polynomials, i.e., we have
the proper inclusions

$\mathrm{War}(n,d)\quad\subsetneq\quad\mathrm{SOS}(n,d)\quad\subsetneq\quad\mathrm{Pos}(n,d).$

In 3.12 we have seen that convolution preserves being a polynomial, in 3.16 we
have seen that convolution preserves being a sum of squares, and the next
result shows that convolution also preserves being in the Waring cone.

######  Theorem 3.17.

Let $\rho\geq 0$ be a kernel such that
$\int_{\mathds{R}^{n}}y^{\alpha}\cdot\rho(y)\leavevmode\nobreak\ \mathrm{d}y$
is finite for all $\alpha\in\mathds{N}_{0}^{n}$, then
$\cdot\,*\rho:\mathrm{War}(n,d)\to\mathrm{War}(n,d)$.

###### Proof.

Let $p\in\mathrm{War}(n,d)$, i.e., $p(x)=\sum_{i=1}^{k}(a_{i}\cdot x)^{d}$.
Then

$\displaystyle(p*\rho)(x)$
$\displaystyle=\int_{\mathds{R}^{n}}p(x-y)\cdot\rho(y)\leavevmode\nobreak\
\mathrm{d}y$
$\displaystyle=\int_{\mathds{R}^{n}}\sum_{i=1}^{k}(a_{i}\cdot(x-y))^{d}\cdot\rho(y)\leavevmode\nobreak\
\mathrm{d}y$ and by Richter’s Theorem [45] we can replace
$\rho(y)\leavevmode\nobreak\ \mathrm{d}y$ by a finitely atomic representing
measure $\mu=\sum_{j=1}^{l}c_{j}\cdot\delta_{y_{j}}$ with $c_{j}>0$ and get
$\displaystyle=\sum_{j=1}^{l}\sum_{i=1}^{k}c_{j}\cdot(a_{i}\cdot(x-y_{j}))^{d}\in\mathrm{War}(n,d).\qed$

The Motzkin polynomial was the first polynomial found to be a non-negative
polynomial which is not a sum of squares. But others have been identified
[40]. We want to investigate some of these in chronological order.

######  Example 3.18 (Robinson polynomial [46]).

Let

$f_{\mathrm{Rob}}(x,y)=1-x^{2}-y^{2}-x^{4}+3x^{2}y^{2}-y^{4}+x^{6}-x^{4}y^{2}-x^{2}y^{4}+y^{6}$

be the Robinson polynomial, i.e.,
$f_{\mathrm{Rob}}\in\mathrm{Pos}(2,6)\setminus\mathrm{SOS}(2,6)$. Then by a
direct calculation using Macaulay2 with the SumsOfSquares package similar to
the Motzkin polynomial we find
$\mathfrak{p}_{\mathrm{Rob}}(\,\cdot\,,1)\in\mathrm{SOS}(2,6)$ and by 3.16 we
have

$\mathfrak{p}_{\mathrm{Rob}}(\,\cdot\,,t)\in\begin{cases}\mathrm{Pos}(2,6)\setminus\mathrm{SOS}(2,6)&\text{for}\
t\in[0,T_{\mathrm{Rob}}),\ \text{and}\\\ \mathrm{SOS}(2,6)&\text{for}\
t\in[T_{\mathrm{Rob}},\infty),\end{cases}$

with

$\frac{20\,946}{1\,000\,000}\quad<\quad
T_{\mathrm{Rob}}\quad<\quad\frac{20\,947}{1\,000\,000}.$

######  Example 3.19 (Choi–Lam polynomial [8]).

Let

$f_{\mathrm{CL}}(x,y,z)=1-4xyz+x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}$

be the Choi–Lam polynomial, i.e.,
$f_{\mathrm{CL}}\in\mathrm{Pos}(3,4)\setminus\mathrm{SOS}(3,4)$. We have

$\displaystyle\mathfrak{p}_{\mathrm{CL}}(x,y,z,t)$
$\displaystyle=1-4xyz+(2t+x^{2})(2t+y^{2})+(2t+x^{2})(2t+z^{2})$
$\displaystyle\quad+(2t+y^{2})(2t+z^{2})$
$\displaystyle=1+12t^{2}-4xyz+4t(x^{2}+y^{2}+z^{2})+x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}$
$\displaystyle=f_{\mathrm{CL}}(x,y,z)+12t^{2}+4t(x^{2}+y^{2}+z^{2})$

for all $t\in\mathds{R}$ and for $t=\frac{1}{9}$ we have

$\displaystyle\mathfrak{p}_{\mathrm{CL}}(x,y,z,1/9)$
$\displaystyle=\frac{31}{27}+\left(xy-\frac{2}{3}z\right)^{2}+\left(xz-\frac{2}{3}y\right)^{2}+\left(yz-\frac{2}{3}x\right)^{2}.$
$\circ$

Macaulay2 calculations with the SumsOfSquares package suggest that for

$T_{\mathrm{CL}}:=\min\\{t\geq
0\,|\,\mathfrak{p}_{\mathrm{CL}}(\,\cdot\,,t)\in\mathrm{SOS}(3,4)\\}$

we have

$\frac{1}{9}-7\cdot 10^{-9}\quad<\quad
T_{\mathrm{CL}}\quad<\quad\frac{1}{9}-6\cdot 10^{-9}.$

######  Example 3.20 (Schmüdgen polynomial [50]).

The polynomial

$\displaystyle f_{\text{Schm}}(x,y)$
$\displaystyle=(y^{2}-x^{2})x(x+2)[x(x-2)+2(y^{2}-4)]$
$\displaystyle\quad+200[(x^{3}-4x)^{2}+(y^{3}-4y)^{2}]\quad\in\mathrm{Pos}(2,6)\setminus\mathrm{SOS}(2,6)$

is the Schmüdgen polynomial and we find
$\mathfrak{p}_{\text{Schm}}(\,\cdot\,,1)\in\mathrm{SOS}(2,6)$. In fact,
Macaulay2 calculations with the SumsOfSquares package and 3.16 shows that
$\mathfrak{p}_{\text{Schm}}(\,\cdot\,,t)\in\mathrm{SOS}(2,6)$ for all $t\geq
2\cdot 10^{-4}$. $\circ$

######  Example 3.21 (Berg–Christensen–Jensen polynomial [3]).

The Berg–Christensen–Jensen polynomial

$f_{\text{BCJ}}(x,y)=1-x^{2}y^{2}+x^{4}y^{2}+x^{2}y^{4}\in\mathrm{Pos}(2,6)\setminus\mathrm{SOS}(2,6)$

is connected to the Motzkin polynomial $f_{\mathrm{Motz}}$ (3.11) by

$f_{\text{BCJ}}(x,y)=f_{\mathrm{Motz}}(x,y)+2x^{2}y^{2}$

and hence from 3.16 we see that

$\mathfrak{p}_{\text{BCJ}}(\,\cdot\,,t)\in\mathrm{SOS}(2,6)$

for all $t\geq\frac{1}{6}$. $\circ$

After all these historic examples let us have a look at a more modern example
from the vast literature of non-negative polynomials which are not sums of
squares.

######  Example 3.22 (Harris polynomial [20, $R_{2,0}$ in Lem. 5.1 and 6.8]).

Let

$\displaystyle f_{\text{Har}}(x,y)$
$\displaystyle=\phantom{+}\,16x^{10}-36x^{8}y^{2}+20x^{6}y^{4}+20x^{4}y^{6}-36x^{2}y^{8}+16y^{10}$
$\displaystyle\quad-36x^{8}\phantom{{}^{0}}+57x^{6}y^{2}-38x^{4}y^{4}+57x^{2}y^{6}-36y^{8}$
$\displaystyle\quad+20x^{6}\phantom{{}^{0}}-38x^{4}y^{2}-38x^{2}y^{4}+20y^{6}$
$\displaystyle\quad+20x^{4}\phantom{{}^{0}}+57x^{2}y^{2}+20y^{4}$
$\displaystyle\quad-36x^{2}\phantom{{}^{0}}-36y^{2}$ $\displaystyle\quad+16$

be the Harris polynomial, i.e.,
$f_{\text{Har}}=R_{2,0}\in\mathrm{Pos}(2,10)\setminus\mathrm{SOS}(2,10)$. With
3.9,

$\displaystyle\mathfrak{p}_{8}(x,t)$
$\displaystyle=1680t^{4}+3360t^{3}x^{2}+840t^{2}x^{4}+56tx^{6}+x^{8},$ and
$\displaystyle\mathfrak{p}_{10}(x,t)$
$\displaystyle=30240t^{5}+75600t^{4}x^{2}+25200t^{3}x^{4}+2520t^{2}x^{6}+90tx^{8}+x^{10}$

we calculate $\mathfrak{p}_{\text{Har}}$ and find
$\mathfrak{p}_{\text{Har}}(\,\cdot\,,1)\in\mathrm{SOS}(2,10)$. In fact,
Macaulay2 calculations and 3.16 show that
$\mathfrak{p}_{\text{Har}}(\,\cdot\,,t)\in\mathrm{SOS}(2,10)$ for all $t\geq
8\cdot 10^{-4}$. $\circ$

All examples so far become a sum of squares for large $t$. However, this is
not in general true.

######  Lemma 3.23.

Let $n,d\in\mathds{N}$ and

$f(x)=\sum_{|\alpha|\leq 2d}a_{\alpha}\cdot x^{\alpha}\in\mathrm{Pos}(n,2d)$

such that

$f_{2d}(x):=\sum_{|\alpha|=2d}a_{\alpha}\cdot
x^{\alpha}\not\in\mathrm{SOS}(n,2d),$

then
$\mathfrak{p}_{f}(\,\cdot\,,t)\in\mathrm{Pos}(n,2d)\setminus\mathrm{SOS}(n,2d)$
for all $t\geq 0$.

###### Proof.

Assume there is a $t\geq 0$ such that

$\mathfrak{p}_{f}(x,t)=\sum_{i=1}^{k}\left(\sum_{|\alpha|\leq
d}c_{i,\alpha}(t)\cdot x^{\alpha}\right)^{2}=\sum_{|\alpha|\leq
2d}a_{\alpha}(t)\cdot x^{\alpha}\in\mathrm{SOS}(n,2d).$

Since by 3.8 we have $a_{\alpha}(t)=a_{\alpha}(0)$ for all
$\alpha\in\mathds{N}_{0}^{n}$ with $|\alpha|=2d$ the sum of squares
decomposition of $\mathfrak{p}_{f}(\,\cdot\,,t)$ gives

$f_{2d}(x)=\sum_{i=1}^{k}\left(\sum_{|\alpha|=d}c_{i,\alpha}(t)\cdot
x^{\alpha}\right)^{2}\in\mathrm{SOS}(n,2d)$

which contradicts the assumption $f_{2d}\not\in\mathrm{SOS}(n,2d)$. ∎

The following example shows that the condition $f_{2d}\in\mathrm{SOS}(n,2d)$
in 3.23 is necessary, but not sufficient. Just take the polynomial
$f(w,x,y,z)=z^{6}-3x^{2}y^{2}z^{2}+x^{4}y^{2}+x^{2}y^{4}+w^{8}\in\mathrm{Pos}(3,8)\setminus\mathrm{SOS}(3,8)$.

######  Example 3.24.

Let
$f(x,y,z)=z^{6}-3x^{2}y^{2}z^{2}+x^{4}y^{2}+x^{2}y^{4}\in\mathrm{Pos}(3,6)\setminus\mathrm{SOS}(3,6)$
be the homogeneous Motzkin polynomial. Then
$\mathfrak{p}_{f}(\,\cdot\,,t)\in\mathrm{Pos}(3,6)\setminus\mathrm{SOS}(3,6)$
for all $t\geq 0$. $\circ$

######  Remark 3.25.

The result in 3.23 also holds for $f_{2d}\not\in\mathrm{War}(n,2d)$, i.e.,
$\mathfrak{p}_{f}(\,\cdot\,,t)\in\mathrm{Pos}(n,2d)\setminus\mathrm{War}(n,2d)$.
$\circ$

While we have seen in 3.23 and 3.24 that there are non-negative polynomials
which do not become sum of squares under the heat equation, the following
result shows that any non-negative polynomial becomes asymptotically close to
$\mathrm{SOS}(\mathds{R}^{n})$ under the heat equation, that is, the constant
polynomial becomes an attractor of the polynomial heat equation.

######  Lemma 3.26.

Let $n\in\mathds{N}$ and $f\in\mathrm{Pos}(\mathds{R}^{n})$ with $\deg f=2d$
for some $d\in\mathds{N}$. Then

$\lim_{t\to\infty}\mathfrak{p}_{f}(x,t)\cdot t^{-d}=c>0.$

###### Proof.

Let $k\in\mathds{N}$. It is easy to see that $\Delta^{k}g$ is constant on
$\mathds{R}^{n}$ for all $g\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq 2k}$ and
even equal to zero for all $g\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq 2k-1}$.

Since $f\in\mathrm{Pos}(\mathds{R}^{n})$ with $\deg f=2d$ we have that the
homogeneous part $f_{2d}$ of $f$ of degree $2d$ is non-zero and non-negative
on $\mathds{R}^{n}$. Let $S$ be the unit sphere in $\mathds{R}^{n}$. Since
$f_{2d}\in\mathrm{Pos}(\mathds{R}^{n})\setminus\\{0\\}$ we have

$\int_{S}f_{2d}(x)\leavevmode\nobreak\ \mathrm{d}x>0$

and by [27, Cor. 1] we have $\Delta^{m}f_{2d}\neq 0$ for all $m=1,\dots,d$.
Hence,
$\partial_{t}^{d}\mathfrak{p}_{f}(x,t)=\Delta^{d}\mathfrak{p}_{f}(x,t)=\Delta^{d}f_{2d}(x)=c>0$
which proves the statement. ∎

3.23 and 3.24 have only been proven here for $n\geq 3$, but for $n=2$ no
counterexample has been found. In fact, every time evolution of the
polynomials
$f\in\mathrm{Pos}(\mathds{R}^{2})\setminus\mathrm{SOS}(\mathds{R}^{2})$ in the
Examples 3.15, 3.18, 3.20, and 3.21 enters $\mathrm{SOS}(\mathds{R}^{2})$ and
by 3.16 never leaves $\mathrm{SOS}(\mathds{R}^{2})$ again. It is open if for
$n=2$ every time evolution of a non-negative polynomial becomes a sum of
squares; see 5.32.

##  4 Time-dependent set of atoms

###  4.1 Coefficients in $C_{b}^{\infty}$

Let $L:\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}\to\mathds{R}$ with
$d\in\mathds{N}_{0}\cup\\{\infty\\}$ be a (truncated) moment functional. When
a $p_{0}\in\mathds{R}[x_{1},\dots,x_{n}]_{\leq d}$ with $p_{0}\geq 0$ exists
with $L(p_{0})=0$, then $\mathrm{supp}\,\mu\subseteq\mathcal{Z}(p_{0})$ for
any representing measure $\mu$ of $L$.

In Section 2 we calculated the dual action of (3) on
$p_{0}\in\mathrm{Pol}(\mathds{R}^{n})$. It is easy to see that with $\nu>0$
and $p_{0}\geq 0$, then $p_{t}>0$ for all $t>0$. Therefore for a moment
functional $L$ we always have $L(p_{t})>0$ and no restriction of the support
of the representing measure is possible. Therefore, we can only study the
time-dependent set of atoms for $\nu=0$.

Let us remind the reader, that for any time-dependent vector field $g$ and
starting point $x_{0}\in\mathds{R}^{n}$ the system of ordinary differential
equations

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}G(x,t)&=g(x,t)\\\
G(x_{0},0)&=x_{0}\end{split}$ (15)

has by the Picard–Lindelöff Theorem a unique solution
$G:\mathds{R}^{n}\times\mathds{R}\to\mathds{R}^{n}$. That means a particle
located at $x_{0}\in\mathds{R}^{n}$ for time $t_{0}=0$ has the trajectory
$G(x_{0},t)$ when it experiences the force field $g$, i.e., at time
$t\in\mathds{R}$ is has the position $G(x_{0},t)\in\mathds{R}^{n}$.

The initial value problem (3) acts only on functions $f_{0}$. But by duality
in Section 2 also the action of (3) on a measure $\mu_{0}$ is well-defined.
The following solves the problem for a Dirac measure.

######  Lemma 4.27.

Let $g\in C(\mathds{R},C_{b}^{\infty}(\mathds{R}^{n}))^{n}$ be a time-
dependent vector field and $G:\mathds{R}^{n}\times[0,\infty)\to\mathds{R}^{n}$
be the unique solution of (15). Additionally, let $h\in
C(\mathds{R},C_{b}^{\infty}(\mathds{R}^{n}))$, $x_{0}\in\mathds{R}^{n}$, and
$c_{0}>0$. Then the measure valued differential equation

$\begin{split}\partial_{t}\mu_{t}(x)&=-g(x,t)\nabla\mu_{t}(x)+h(x,t)\cdot\mu_{t}(x)\\\
\mu_{0}(x)&=c_{0}\cdot\delta_{x_{0}}(x)\end{split}$ (16)

has the unique solution

$\mu_{t}(x)=c_{0}\cdot\exp\left(\int_{0}^{t}h(G(x_{0},s),s)\leavevmode\nobreak\
\mathrm{d}s\right)\cdot\delta_{G(x_{0},t)}(x)$

for all $t\in\mathds{R}$.

###### Proof.

Let $A(x,t)=-g(x,t)\nabla$ and $B(x,t)=h(x,t)$ be the two operators in (16),
i.e., (16) reads $\partial_{t}\mu_{t}=A\mu_{t}+B\mu_{t}$. Then by Trotters
approach we have

$\mu_{t}(x)=\lim_{N\to\infty}\left[\prod_{i=1}^{N}\exp\left(\int_{t_{i-1}}^{t_{i}}B(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)\exp\left(\int_{t_{i-1}}^{t_{i}}A(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)\right]\mu_{0}(x)$

for any decompositions $\mathcal{Z}_{N}=\\{t_{0}=0<t_{1}<\dots<t_{N}=t\\}$
with $\Delta\mathcal{Z}_{N}\to 0$ as $N\to\infty$.

But $A$ is the transport operator and hence when acting on
$c_{0}\cdot\delta_{x_{0}}(x)$ we can use the approximation

$\displaystyle\exp\left(\int_{t_{0}}^{t_{1}}A(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)c_{0}\cdot\delta_{x_{0}}(x)$ $\displaystyle\approx
c_{0}\cdot\delta_{x_{0}}\left(x-\int_{t_{0}}^{t_{1}}g(x,s)\leavevmode\nobreak\
\mathrm{d}s\right)$ $\displaystyle\approx
c_{0}\cdot\delta_{x_{0}+\int_{t_{0}}^{t_{1}}g(x_{0},s)\leavevmode\nobreak\
\mathrm{d}s}(x)$

on the interval $[t_{0},t_{1}]$ (and similarly by induction on all
$[t_{i-1},t_{i}]$ for $i=1,\dots,N$) which gives for $N\to\infty$ with
$\Delta\mathcal{Z}_{N}\to 0$ the same result $\mu_{t}(x)$. On the other hand
the operator $B$ acts only as multiplication and hence does not affect the
transport of the Dirac measure by $A$. The point $x_{0}$ is therefore
transported by $A$, which is seen by the approximation
$x_{i}:=x_{i-1}+\int_{t_{i-1}}^{t_{i}}g(x_{i-1},s)\leavevmode\nobreak\
\mathrm{d}s$. We therefore have $\mu_{t}(y)=c(t)\cdot\delta_{x(t)}(y)$ with
$x(t)=G(x_{0},t)$.

Since now the Dirac measure $\delta_{G(x_{0},t)}$ moves along the vector field
$g$ starting at $x_{0}$, it experiences at time $t$ the multiplication
operator $B(x(t),t)=h(G(x_{0},t),t)$ and therefore we have

$c(t)=c_{0}\cdot\exp\left(\int_{0}^{t}h(G(x_{0},s),s)\leavevmode\nobreak\
\mathrm{d}s\right)$

which proves the statement. Uniqueness follows either by the uniqueness of the
dual problem or directly by uniqueness of $G$ and ($*$). ∎

The solution of (16) for $\mathrm{d}\mu_{0}(x)=f_{0}(x)\leavevmode\nobreak\
\mathrm{d}x$ with $f_{0}\in\mathcal{S}(\mathds{R}^{n})$ can in general not be
written down explicitly. But for the case of
$\mu_{0}=c_{0}\cdot\delta_{x_{0}}$ 4.27 shows that (16) is simply solved since
the transport term $g(x,t)\cdot\nabla$ acts only on the point $x_{0}=x(0)$ to
get $x(t)$ and the multiplication $h(x,t)$ only acts on the coefficient
$c_{0}=c(0)$ to get $c(t)$. This simplifies the description of the solution
immensely and provides in the following result a way to trace the Carathéodory
number $\mathcal{C}(s)$.

######  Theorem 4.28.

Let $n\in\mathds{N}$,
$s(0)=(s_{\alpha}(0))_{\alpha\in\mathds{N}^{n}}\in\mathcal{S}_{n,\infty}$ be a
$n$-dimensional moment sequence with finite rank Hankel matrix
$\mathcal{H}(s(0))$, $g\in C(\mathds{R},C(\mathds{R}^{n},\mathds{R}^{n}))$,
and $h\in C(\mathds{R},C(\mathds{R},\mathds{R}))$. Then there is a unique time
evolution $s:\mathds{R}\to\mathcal{S}_{n,\infty}$ of $s(0)$ with respect to

$\partial_{t}f(x,t)=-g(x,t)\nabla f(x,t)+h(x,t)\cdot f(x,t).$

Additionally, for all $t\in\mathds{R}$ we have

$\mathrm{rank}\,\mathcal{H}(s(t))=\mathrm{rank}\,\mathcal{H}(s(0))$

and therefore the Carathéodory number is constant, i.e.,
$\mathcal{C}(s(t))=\mathcal{C}(s(0))$.

###### Proof.

Since $K:=\mathrm{rank}\,\mathcal{H}(s(0))$ is finite, there exists a unique
$K$-atomic representing measure
$\mu_{0}=\sum_{i=0}^{K}c(0)\cdot\delta_{x_{i}(0)}$ of $s(0)$. By linearity of
the time-evolution and 4.27 we have the unique time-evolution of
$\mu_{t}=\sum_{i=1}^{K}c_{i}(t)\cdot\delta_{x_{i}(t)}$ and therefore the
unique time-evolution of $s(t)$.

It remains to show that
$\mathcal{C}(s(t))=\mathrm{rank}\,\mathcal{H}(s(t))=\mathrm{rank}\,\mathcal{H}(s(0))=\mathcal{C}(s(0))$.
While the first and the third equality is clear from the rank, see e.g. [51,
Prop. 17.21], we have to show the second equality. For that it is sufficient
to show that the path of $x_{k}(t)$ never splits or two paths $x_{j}(t)$ and
$x_{k}(t)$ ($j\neq k$) intersect for any $t\in\mathds{R}$. But this follows
from 4.27 and the uniqueness of the solution $x_{j}(t)=G(x_{j}(0),t)$ of (15)
by Peano’s Theorem, i.e., when two integral curves $x_{j}(t)$ and $x_{k}(t)$
of (15) coincide for some $t^{\prime}\in\mathds{R}$, i.e.,
$x_{j}(t^{\prime})=x_{k}(t^{\prime})$, then $x_{j}(t)=x_{k}(t)$ for all
$t\in\mathds{R}$, especially for $t=0$ which contradicts the minimal choice of
$K=\mathrm{rank}\,\mathcal{H}(s(0))$. ∎

In the univariate case the previous result also holds for the truncated moment
problem.

######  Example 4.29.

Let $s(0)=(s_{k}(0))_{k=0}^{2d}$ with $d\in\mathds{N}$ be a moment sequence on
the boundary of the moment cone, i.e., $s$ is represented by
$\mu_{0}=\sum_{k=1}^{l\leq d}c_{k}(0)\cdot\delta_{x_{k}(0)}$ with
$c_{1}(0),\dots,c_{l}(0)>0$ and $x_{1}(0),\dots,x_{l}(0)$ pairwise different
and the non-negative polynomial
$p_{0}(x)=(x-x_{1}(0))^{2}\cdots(x-x_{l}(0))^{2}$ fulfills
$L_{s(0)}(p_{0})=0$. Let $s(t)$ evolve with respect to
$\partial_{t}f=-g\partial_{x}f+h\cdot f$. By 4.27 we have
$\mu_{t}=\sum_{k=1}^{l\leq d}c_{k}(t)\cdot\delta_{x_{k}(t)}$ and therefore
$L_{s(t)}(p_{t})=0$ for the non-negative polynomial
$p_{t}(x)=(x-x_{1}(t))^{2}\cdots(x-x_{l}(t))^{2}$. Hence, the boundary moment
sequence remains a boundary sequence for all times. $\circ$

######  Remark 4.30.

In the previous example we have seen that for univariate truncated moment
moment sequences on the boundary the time evolution with respect to (3) with
$\nu=0$ remains a boundary truncated moment sequence for all times. And
additionally, the time evolution is unique, depending only on the initial
moments. For multivariate moment sequences this no longer holds. Firstly,
multivariate (truncated) moment sequences on the boundary can be indeterminate
[13] and the time evolution then depends on the choice of the representing
measure $\mu_{0}$. And secondly, even if the truncated boundary moment
sequence is determinate, it can immediately enter the interior of the moment
cone. To see this, take an example from optimal design [44] where the isolated
zeros of a non-negative polynomial is maximal. For example take $10$
projective zeros of the Robinson polynomial [46] after rotation of the
projective space such that no zero lies at infinity (all $10$ zeros are then
in the affine space $\mathds{R}^{2}$). Since the zeros are isolated we find a
vector field $g:\mathds{R}^{2}\to\mathds{R}^{2}$ such that these points are in
general position for $t\neq 0$. Then by the Alexander–Hirschowitz theorem [1]
there is no non-negative polynomial of degree $6$ vanishing on all these $10$
points, i.e., the truncated moment sequence from the measure $\mu_{t}$ with
$t\neq 0$ belongs to an interior moment sequence. $\circ$

###  4.2 Extension to $C^{\infty}$

We have so far treated $g\in C_{b}^{\infty}(\mathds{R}^{n},\mathds{R}^{n})$.
We want to see what happens when we extend this class. In [6] we calculated
the explicit time-evolution of the moments for
$\partial_{t}f(x,t)=x\cdot\partial_{x}f(x,t)$ and we got

$s_{k}(t)=s_{k}(0)\cdot e^{-(k+1)\cdot t}.$

Also for $k=0$ the moments can be calculated only from the initial values
$s(0)$ by solving $\partial_{t}s_{0}(t)=0$ and $\partial_{t}s_{k}(t)=-k\cdot
s_{k-1}(t)$ for $k\in\mathds{N}$ by induction.

When we have $\partial_{t}f=x^{k}\partial_{x}f$ with $k\geq 2$, it is in
general not possible to calculate the time-dependent moments. For moment
sequences with finite rank we can at least ensure the existence of the time-
evolution $s(t)$ for small times and we observe a finite break down in time.

######  Example 4.31.

Let $k=2$ and $x_{0}=1$. Then to solve

$\displaystyle\partial_{t}\mu_{t}$
$\displaystyle=-x^{2}\cdot\partial_{x}\mu_{t}$ $\displaystyle\mu_{0}$
$\displaystyle=\delta_{1}$

we have by 4.27 to solve

$\displaystyle\partial_{t}x(t)$ $\displaystyle=a\cdot x(t)^{2}$ $\displaystyle
x(0)$ $\displaystyle=1,$

i.e., we have

$t=\int_{1}^{x}y^{-k}\leavevmode\nobreak\
\mathrm{d}y=(k-1)\cdot\left[1-x^{1-k}\right]$

and hence $x(t)=(1-t)^{-1}$. Since $\lim_{t\nearrow 1}x(t)=\infty$, we have
that the time-dependent moments $s_{l}(t)$ for
$\partial_{t}f=-x^{2}\cdot\partial_{x}f$ exist only for $t<1$ and
$s_{l}(1)=\infty$ for all $l\in\mathds{N}$. $\circ$

The previous example demonstrates why we have to restrict to $g\in
C_{b}^{\infty}$.

##  5 Summary and open questions

We investigated time-dependent moments from partial differential equations. At
first we gave the dual action on the polynomials (2.3). While for general
coefficients $g$ and $h$ we leave $\mathds{R}[x_{1},\dots,x_{n}]$ in general,
for coefficients independent on $x$ we remain in
$\mathds{R}[x_{1},\dots,x_{n}]$. We investigated the action of the heat
equation on non-negative polynomials more deeply. In Examples 3.15, 3.18, and
3.20 to 3.22 we have collected examples of non-negative polynomials on
$\mathds{R}^{2}$ which are not sums of squares. But under the action of the
heat equation after finite time they become sums of squares. The highest
degree part is a non-negative homogeneous polynomial in two variables and
hence a sum of squares [23]. If this holds for all bivariate non-negative
polynomials is still open.

######  Open Problem 5.32.

Let $f\in\mathrm{Pos}(\mathds{R}^{2})\setminus\mathrm{SOS}(\mathds{R}^{2})$.
Is it true that there always is a $T=T(f)>0$ such that
$\mathfrak{p}_{f}(x,t)\in\mathrm{SOS}(\mathds{R}^{2})$ for all $t\geq T$?

By 3.16 5.32 reduces to the questions that
$\mathfrak{p}_{f}(x,T)\in\mathrm{SOS}(\mathds{R}^{2})$ holds for at least one
$T>0$.

On $\mathds{R}^{n}$ with $n\geq 3$ this in general no longer holds and the
heat equation fails to produce sum of squares from non-negative polynomials in
three or more variables, see 3.23 and 3.24. Note, the Choi–Lam polynomial
(3.19) is a non-negative polynomial in $3$ variables, not a sum of squares,
but becomes a sum of squares under the heat equation. From the counterexamples
in 3.23 and 3.24 we see that it is necessary that the highest degree part is a
sum of squares. And from the calculations we get the impression that the SOS-
property is pushed from the highest degree part to the lower degree parts.
Supported by 3.19, Hilbert’s Theorem [23] then implies the second open
question.

######  Open Problem 5.33.

Let $f\in\mathrm{Pos}(\mathds{R}^{3})\setminus\mathrm{SOS}(\mathds{R}^{3})$
with $\deg f\leq 4$. Is it true that there always is a $T=T(f)>0$ such that
$\mathfrak{p}_{f}(x,t)\in\mathrm{SOS}(\mathds{R}^{3})$ for all $t\geq T$?

The third possibility in Hilbert’s Theorem [23] is $\deg f=2$. But then
$\mathfrak{p}_{f}(x,t)=f(x)+c\cdot t$ with $c>0$ [27], i.e., homogenization
shows already $f\in\mathrm{SOS}(n,2)$.

In 3.26 we show that under the heat equation any non-negative polynomial in
$\mathds{R}[x_{1},\dots,x_{n}]$ gets close to the constant polynomial

$\lim_{t\to\infty}\mathfrak{p}_{f}(x,t)\cdot t^{-d}=c>0,$

i.e., the positive constant polynomials are attractors of the polynomial heat
equation.

For the equation $\partial_{t}f=g\cdot\nabla f+h\cdot f$ we calculate the
time-dependent set of atoms, i.e., we solve the problem of how the atoms of a
finitely atomic representing measure
$\mu_{t}=\sum_{i=1}^{k}c_{i}(t)\cdot\cdot\delta_{x_{i}(t)}$ evolve in time. We
find that the atom positions $x_{i}(t)$ are governed by the transport term
$g\cdot\nabla$ but are unaffected by the scaling with $h$. The coefficients
(masses) $c_{i}(t)$ can then be analytically solved from the $x_{i}(t)$, see
4.27 and 4.28.

We also extended the treatment beyond smooth and bounded coefficients and
found that atoms under the transport term $-x^{2}\cdot\partial_{x}$ move to
infinity in finite time, i.e., a finite break down of the moments (and the
measure solution) appears even for linear partial differential equations, see
4.31.

## Acknowledgments

The second author and this project is financed by the Deutsche
Forschungsgemeinschaft DFG with the grant DI-2780/2-1 and his research
fellowship at the Zukunfskolleg of the University of Konstanz, funded as part
of the Excellence Strategy of the German Federal and State Government. This
work has also been supported by the European Union’s Horizon 2020 research and
innovation programme under the Marie Skłodowska-Curie Actions, grant agreement
813211 (POEMA), by the AI Interdisciplinary Institute ANITI funding, through
the French “Investing for the Future PIA3” program under the Grant agreement
n∘ ANR-19-PI3A-0004 as well as by the National Research Foundation, Prime
Minister’s Office, Singapore under its Campus for Research Excellence and
Technological Enterprise (CREATE) programme.

We thank the organizers of ICCOPT in Bethlehem, PA, in July 2022, and the
organizers of IWOTA in Krakow in September 2022 where [6] and the present work
have been presented. We also thank Tim Netzer for discussions on this work at
IWOTA 2022.

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