Exact linear representations of the nonlinear

Cauchy problems and their smooth solutions

Yu. N. Kosovtsov

Lviv, Ukraine

email<EMAIL_ADDRESS>
###### Abstract

The paper establishes conditions under which there are exact linear
representations of nonlinear partial differential equations (Cauchy problems).
By introducing a certain linear operator $A$, it is shown that under these
conditions there are three equivalent equations (one linear and two
nonlinear), while the formal operator solution is common for all of them and
is the expansion of the desired function $v(t,x)$ into a formal Taylor series.

Using the Borel-Whitney lemma, which states that any smooth function $v(t,x)$
in a neighborhood of a point is defined by its formal Taylor series, we obtain
the following statement. If all parameters of the operator $A$ are smooth
functions, then there exists a smooth function $\tilde{v}(t,x)$, which for
$t=0$ has the same power expansion as $v(t,x)$ and this function solves all
equivalent equations.

The Navier-Stokes and Euler equations are considered as a non-trivial examples
of this approach.

##  1 Equivalence of equations

One of the motivations for this paper is the formal scheme given in [1] for
obtaining solutions to nonlinear equations by the operator method.

The purpose of this section is to obtain _formal_ solutions for $v=v(t,x)$ to
the nonlinear equations of type ($t,x\in\mathbb{R}\times\mathbb{R}^{n}$)

$\frac{\partial}{\partial t}v=F(x,v,\textbf{D}_{j}^{\alpha_{j}}v),\qquad
v(t,x)|_{t=0}=u_{i}(x),$ (1)

where $\textbf{D}_{j}^{\alpha_{j}}v$ \- some given finite sequence of
derivatives of the function $v$ with respect to components of
$x\in\mathbb{R}^{n}$.

Assuming that the functions $u$ and $F$ are smooth (of class $C^{\infty}$) in
all their variables, we define in the spirit of [1] a differential operator

$A=\int_{\mathbb{R}^{n}}d\zeta
F(\zeta,u(\zeta),\textbf{D}_{j}^{\alpha_{j}}u(\zeta))\frac{\delta}{\delta
u(\zeta)},\qquad\zeta\\\ \in\\\ \mathbb{R}^{n},$ (2)

where the function

$F(x,u(x),\textbf{D}_{j}^{\alpha_{j}}u(x))=F(x,u_{0},u_{1},...,u_{m})$

is _the right-hand side of the equation_ (1) _under_ $t=0$ and it is assumed
that $u$ is an _arbitrary but fixed_ function, we can consider $F$ as the
linear multiplication operator, $\frac{\delta}{\delta u(\zeta)}$ is the
functional derivative. We have specially transferred the symbol $d\zeta$ at
the beginning of the formula to emphasize that the right operator is executed
first, and then the integration is performed.

Here by functional derivative we mean the linear mapping with the following
property

$\frac{\delta u(x)}{\delta u(\zeta)}=Dirac(x-\zeta),$

the chain rule is also valid in this context.

We emphasize that here this operator _do not depend on the ”time” $\,\,t$_.

Entered operator is a linear operation, i.e., operator $A$ obeys the following
properties:

$AcG(u(x))=cAG(u(x)),\qquad\mathrm{for\;any\;scalar}\;c\in\\\ \mathbb{R},$
$A[G(u(x))+H(u(x))]=AG(u(x))+AH(u(x)),\qquad\mathrm{for\;any}\;G,H\in
C^{\infty}.$

and operator $A$ is a _derivation_ , since

$A[G(u(x))\cdot H(u(x)]=[AG(u(x))]\cdot H(u(x))+G(u(x))\cdot AH(u(x))$

and in a particular case

$Au(x)=F(x,u(x),\textbf{D}_{j}^{\alpha_{j}}u(x)).$

Note also that it is obvious that $A^{j}u\in C^{\infty}$ for any
$j\in\mathbb{N}$.

First of all consider the following add-on _linear operator_ differential
equation

$\frac{\partial}{\partial t}v(t,x)=Av(t,x)),\qquad v(x,t)|_{t=0}=u(x),$ (3)

transform it to equivalent integral equation

$v(t,x)=u(x)+\int_{0}^{t}Av(\tau,x))d\tau$ (4)

and try to solve (4) by Picard iteration method starting from
$v_{0}(t,x)=u(x)$. We have

$v_{1}(t,x)=u(x)+\int_{0}^{t}Au(x)d\tau=u(x)+tAu(x);$
$v_{2}(t,x)=u(x)+\int_{0}^{t}A[(u(x)+\tau
Au(x)]d\tau=u(x)+tAu(x)+\frac{t^{2}A^{2}}{2}u(x);$ $...$
$v(t,x)=\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}u(x)=e^{tA}u(x).$ (5)

Thus, we obtain a solution to the equation (3) in the form of a formal
functional series

$v(t,x)=\sum_{n=0}^{\infty}\frac{t^{n}a_{n}(x)}{n!},$ (6)

where $a_{n}(x)=A^{n}u(x)$ and it is natural to assume that (5) _is a formal
expansion of the function $v(t,x)$_ (6) _to the Taylor functional series at
the point $t=0$_ and that

$a_{n}(x)=A^{n}u(x)=\frac{\partial^{n}}{\partial t^{n}}v(t,x))|_{t=0}.$ (7)

This is true (with some reservations), because using the equation (3), we get
the following chain

$\frac{\partial}{\partial t}v(t,x)|_{t=0}=Av(t,x)|_{t=0}=Au(x),$
$\frac{\partial^{2}}{\partial t^{2}}v(t,x)|_{t=0}=A\frac{\partial}{\partial
t}v(t,x)|_{t=0}=A^{2}u(x),$ $\cdot\cdot\cdot$ $\frac{\partial^{n}}{\partial
t^{n}}v(t,x)|_{t=0}=A^{n}u(x).$

We note here another important circumstance for what follows. If we
differentiate the operator solution (5) with respect to $t$ and take into
account that the operator $A$ commutes with itself, we have

$\frac{\partial}{\partial t}v(t,x)=Ae^{tA}u(x)=e^{tA}Au(x)$ (8)

and as a result we get _three formally equivalent_ equations. The first is the
original equation (3)

$\frac{\partial}{\partial t}v(t,x)=A(e^{tA}u(x))=Av(t,x).$

The second equation

$\frac{\partial}{\partial t}v(t,x)=e^{tA}Au(x).$

Under certain conditions, this equation is transformed into a nonlinear
equation (1):

$\frac{\partial}{\partial t}v(t,x)=F(x,v,\textbf{D}_{j}^{\alpha_{j}}v),\qquad
v(t,x)|_{t=0}=u(x).$

To show this we need to consider the following properties of the $e^{tA}$
operator.

Note that the operator $e^{-tA}$ is formally the inverse to $e^{tA}$ since

$e^{-tA}e^{tA}=1.$

Let $g(u(x))$ be a _function_ that is the operator of multiplication on the
$g(u(x))$. Then, as can be seen from the definition of the derivation operator
$A$, the commutator $[g(u(x)),A]$ is a function too.

Now consider the following construction of operators

$B=e^{tA}\,g(u(x))\,e^{-tA}.$

As far as the commutator $[g(u(x)),A]$ is a function, then all repeated
commutators are functions too. So with help of the well known Baker-Campbell-
Hausdorff formula (see (26)-(28) in [2]) we conclude that operator $B$ is _a
function_ and

$B=(e^{tA}g(u(x))),$ (9)

where the outward brackets in the right-hand side stress that action of
operator is restricted by the brackets area. An analogous conclusion holds for
$e^{-tA}\,g(u(x))\,e^{tA}$.

Let us consider further the following obvious chain

$\displaystyle e^{tA}\,g_{1}\,g_{2}\,\dots\,g_{p}$
$\displaystyle=e^{tA}\,g_{1}\,e^{-tA}\,e^{tA}\,g_{2}\,e^{-tA}\,\dots\,e^{tA}\,g_{p}$
$\displaystyle=(e^{tA}\,g_{1})\,(e^{tA}\,g_{2})\,\dots\,(\,e^{tA}\,g_{p}).$

With what is observed here and the fact that the operator $A$ is linear, leads
to conclusion that for any smooth function $F(g_{1},\,g_{2},\,\dots,\,g_{n})$,
which can be expanded in formal power series with respect to
$g_{1},\,g_{2},\,\dots,\,g_{n}$, we obtain the following nice property:

$e^{tA}\,\,F(g_{1},\,g_{2},\,\dots,\,g_{p})=F((e^{tA},\,(e^{tA}\,g_{2}),\,\dots,\,(e^{tA}\,g_{p})).$
(10)

This property of the operator $e^{tA}$ coincides with familiar property of the
shift operator $\exp\\{t\frac{\partial}{\partial s}\\}$, as shift operator is
the particular case of the operator exponent.

It be assumed henceforth that $F$ (or $F_{i}$ when considering equations or
systems involving $i$ functions) is smooth function on all their arguments.

Then, as a consequence of this, assuming that $F(x,u_{0},u_{1},...,u_{m})$ is
a smooth function in the variables $u_{i}$ (i.e., expandable into a formal
Taylor series) and taking into account (5), we have

$e^{tA}F(x,u(x),\textbf{D}_{j}^{\alpha_{j}}u(x))=F(x,e^{tA}u(x),\textbf{D}_{j}^{\alpha_{j}}e^{tA}u(x))=F(x,v,\textbf{D}_{j}^{\alpha_{j}}v).$
(11)

This implies (assuming that $F(x,u_{0},u_{1},...,u_{m})$ is a smooth function
in the variables $u_{i}$), that the second _linear equation_ (3) is
_equivalent to the nonlinear equation_ (1).

From a combination of these equations, one can obtain the third equivalent
equation

$Av=F(x,v,\textbf{D}_{j}^{\alpha_{j}}v),\qquad v(t,x)|_{t=0}=u(x)$

It follows from the equivalence of these equations that _all three equations_
have the same formal solution (5)

$v(t,x)=\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}u(x)=e^{tA}u(x)$

and wherein

$\frac{\partial^{n}}{\partial t^{n}}v(t,x)|_{t=0}=A^{n}u(x).$

The last equality can be verified directly (albeit by rather cumbersome
calculations) on concrete examples of nonlinear equations of the type (1) with
a concrete operator $A$ defined in accordance with (2). Both sides of the
partial differential equation can be differentiated $k$ times by $t$ and give
recurrence relations for the coefficients of the formal power series for
$v(t,x)$ that determine the coefficients $\frac{\partial^{n}}{\partial
t^{n}}v(t,x)|_{t=0}$. Comparing the obtained coefficients with $A^{k}u(u)$, we
find their complete coincidence.

Finally, noting that the formal Taylor series expansion following Lagrange can
be represented using the differential shift operator
$e^{t\frac{\partial}{\partial s}}$ (operator $\frac{\partial}{\partial s}$ is
a _derivation_) as

$v(t,x)=\sum_{n=0}^{\infty}\frac{t^{n}a_{n}(x)}{n!}=[e^{t\frac{\partial}{\partial
s}}v(s,x)]|_{s=0},$

then we have

$[e^{t\frac{\partial}{\partial s}}v(s,x)]|_{s=0}=e^{tA}u(x).$ (12)

The shift operator under consideration has the following well-known property.
For any at least differentiable function $f(s)$111The shift operator we use,
in particular, is related to the solution of the simplest linear partial
differential equation $\frac{\partial}{\partial
t}g(t,s)=\frac{\partial}{\partial s}g(t,s),\,\,g(t,s)|_{t=0}=f(s)$ by Picard
iteration method, with simplest differential operator
$A=\frac{\partial}{\partial s}$. As above, starting from $g_{0}(t,s)=f(s)$, we
obtain the solution $g(t,s)=e^{t\frac{\partial}{\partial s}}f(s)$. In this
case, it must be assumed that the function $f(s)$ is infinitely
differentiable. But it is well known that this equation has a solution
$g(t,s)=f(t+s)$, where $f(s)$ is subject to the simpler requirement of only
one-time differentiability. It is the property $e^{t\frac{\partial}{\partial
s}}f(s)=f(t+s)$ that determines the name of the operator. Often the shift
operator is defined by this property.

$e^{t\frac{\partial}{\partial s}}f(s)=f(e^{t\frac{\partial}{\partial
s}}s)=f(s+t),$

that is, the shift operator $e^{t\frac{\partial}{\partial s}}$ can be
_introduced ”inside”_ a differentiable function.

Obviously, the same is true for a function $f(s,x)$ of several variables
differentiable with respect to $s$.

Let’s suppose that

$v(t,x)=[e^{t\frac{\partial}{\partial s}}v(s,x)]|_{s=0}=e^{tA}u(x),$ (13)

then

$v(t+s,x)=e^{t\frac{\partial}{\partial s}}v(s,x)=e^{(t+s)A}u(x),$

differentiating the last expressions, we obtain

$\frac{\partial}{\partial t}v(t+s,x)=\frac{\partial}{\partial
s}v(t+s,x)=\frac{\partial}{\partial s}e^{t\frac{\partial}{\partial
s}}v(s,x)=\frac{\partial}{\partial s}e^{(t+s)A}u(x)=Ae^{(t+s)A}u(x),$

From this chain, further we need only the equality

$\frac{\partial}{\partial s}e^{t\frac{\partial}{\partial
s}}v(s,x)=Ae^{(t+s)A}u(x).$ (14)

We now start by expanding both sides of the _nonlinear_ equation (1) into a
Taylor series by applying the shift operator $e^{t\frac{\partial}{\partial
s}}$, we have

$[e^{t\frac{\partial}{\partial s}}\frac{\partial}{\partial
t}v(s,x))]|_{s=0}=[e^{t\frac{\partial}{\partial
s}}F(x,v(s,x),\textbf{D}_{j}^{\alpha_{j}}v(s,x))|_{s=0}$

or

$\frac{\partial}{\partial t}e^{t\frac{\partial}{\partial
s}}v(s,x))|_{s=0}=F(x,e^{t\frac{\partial}{\partial
s}}v(s,x)|_{s=0},\textbf{D}_{j}^{\alpha_{j}}e^{t\frac{\partial}{\partial
s}}v(s,x)|_{s=0}),$

$\frac{\partial}{\partial s}e^{t\frac{\partial}{\partial
s}}v(s,x))|_{s=0}=F(x,e^{t\frac{\partial}{\partial
s}}v(s,x)|_{s=0},\textbf{D}_{j}^{\alpha_{j}}e^{t\frac{\partial}{\partial
s}}v(s,x)|_{s=0}),$

further, using (12) and (14), we arrive at

$Ae^{tA}u(x)=F(x,e^{tA}u(x),\textbf{D}_{j}^{\alpha_{j}}e^{tA}u(x)),$

and denoting $v(t,x)=e^{tA}u(x)$ as a result we get the equivalent (1)
equation

$Av(t,x)=F(x,v(t,x),\textbf{D}_{j}^{\alpha_{j}}v(t,x)),\qquad
v(t,x)|_{t=0}=u(x).$ (15)

Combining now (1) and (15) we get one more equivalent equation

$\frac{\partial}{\partial t}v(t,x)=Av(t,x),\qquad v(t,x)|_{t=0}=u(x),$

which is nothing else than the linear equation (3) we have considered above,
for which we have found a formal solution (5) that satisfies our assumption
(13).

Thus we can conclude:

Proposition 1. _Let the function_ $F(x,v_{0},v_{1},...,v_{m})$ _in
equation_(1) _be smooth (of class_ $C^{\infty}$) _in all its variables and the
operator $A$ is defined according to the recipe specified in_ (2). _Then the
equations_ (1), (3), _and_ (15) _are equivalent and have the same solution_
(5).

Note that Proposition 1 deals with nonlinear PDE (1), linear (3), and
nonlinear (15) equations with _variational_ derivatives. All these equations
are united by the operator $A$ constructed on the basis of the equation (1).
This result has parallels with the well known exact representations
_nonlinear_ ODEs by _linear_ PDEs.

For nonlinear PDEs in a broader sense, apparently for the first time, rather
as a hypothesis, this result is given in [1] (see also [3]).

_Note._ The expression (7) unambiguously relates the action of the known
(maybe rather complicated) operator $A$ on the known (fixed) function $u(x)$
with the action of the known differentiation operator
$\frac{\partial}{\partial t}$ to the desired (i.e., unknown) function
$v(t,x))$ at the point $t=0$. From a practical point of view, (7) means that
from the assumption of the convergence of the Taylor series for $v(t,x))$, one
could justify the convergence of the series
$\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}u(x)$, which is not available due to
the unknown function $v(t,x))$.

Or, conversely, only known functions are involved in the formal _operator_
solution: an arbitrary but fixed function $u(x)$ and functions of type
$F(x,u_{0},u_{1},...,u_{m})$, generating the set of functions
$a_{n}=A^{n}u(x)$. Therefore, using some version of the ”inverse” Taylor
theorem, based on the properties of the known functions $u$ and $F$ and,
accordingly, $a_{n}$, justify the properties of $v(t,x))$, which may be more
realistic.

Moreover, recovery of the function $v(t,x))$ from the coefficients
$\frac{\partial^{k}}{\partial t^{k}}v|_{t=0}$ is possible only up to a flat
function, i.e. _ambiguously_. But the recovery of the function $v(t,x))$ from
the coefficients $A^{k}u$ is _unique_.

Here is a quote from [4]: ”Normally, in applied mathematics or physics, the
Taylor series has been used to approximate $f$, and it is not surprising that
analyticity has been emphasized. But in fact analyticity is neither necessary
nor sufficient for such approximations to be valid, in terms of the use to
which they are put. Analyticity has been over-emphasized, and the vanishing of
the remainder term $R_{k}=f(x_{0},x)-(a_{0}+a_{1}x+...+a_{k}x^{k})$ as $k$
tends to $\infty$ has figured too prominently, thus - in an elegant phrase of
Zeeman’s - allowing the tail of the Taylor series to wag the dog.”

##  2 The Borel - Whitney lemma

The Borel-Whitney lemma shows that any smooth function in a neighborhood of a
point is given by its formal Taylor series up to a term (flat function) whose
Taylor series at that point consists of all zeros. This fact is very
convenient to use to prove various assertions about local properties of
functions, mappings, PDE’s solutions, and so on.

This line of work, started by E. Borel [5] and continued by H. Whitney [6],
[7], is being developed by many authors up to the present. The current state
of the problem can be found in [8] and in a series of papers by Ch. Fefferman
(see, for example, [9]).

Consider a formal power series in variables $t=(t_{1},...,t_{p})$ whose
coefficients are smooth functions of the variables $x=(x_{1},...,x_{m})$

$\sum_{|\alpha|=0}^{\infty}b_{n_{1}...n_{p}}(x)t^{\alpha}.$ (16)

A natural question arises: does every formal series (16) correspond to a
smooth function $V(t,x)$ whose Taylor series in $t$ at zero coincides with the
given series? The answer to this question is positive.

For the purposes of this paper, we only need the elementary lemma due to Emile
Borel and Hassler Whitney [5] \- [7], [10], [11].

Lemma (Borel - Whitney). _For any formal series_(16) _and an open bounded
domain_ $\Omega\subset\mathbb{R}^{m}$, _there exists a smooth function_
$V(t,x):\mathbb{R}\times\Omega\rightarrow\mathbb{R}$ _whose Taylor series in
the variable $t$ at the point_ $t=0$ _coincides with_ (16).

_Proof._ Here, for simplicity, we assume that $p=1$ and $m=1$ (in the general
case, the proof is the same, but the formulas are more cumbersome).

Let $\psi:\mathbb{R}\rightarrow\mathbb{R}$ be a smooth function (cutoff
function) such that

$\psi(t)={\Huge\left\\{{}_{0,\qquad|t|\geq
1.}^{1,\qquad|t|\leq\frac{1}{2},}\right.}$

Let us show that for any formal series

$\sum_{n=0}^{\infty}b_{n}(x)t^{n},\qquad t,x\\\ \in\\\ \mathbb{R}$ (17)

with smooth coefficients $b_{n}(x)$ and any open bounded domain
$\Omega\subset\mathbb{R}$ there exists a smooth function
$V(t,x):\mathbb{R}\times\Omega\rightarrow\mathbb{R}$ whose Taylor series in
the variable $t$ at the point $t=0$ coincides with (17).

We will look for the function $V(t,x)$ in the form of a series

$V(t,x)=\sum_{n=0}^{\infty}b_{n}(x)t^{n}\psi(t/r_{n}),\qquad\qquad$ (18)

where $b_{n}(x)$ are the same as in (17) and the numbers $r_{n}>0$ are defined
by the formula

$r_{n}=\frac{1}{n!(1+\beta_{n})},\,\,\mathrm{where}\,\,\beta_{n}=\max_{0<i<n}\,\max_{x\in\overline{\Omega}}|b_{n}^{(i)}(x)|,$

where $\overline{\Omega}$ means the closure of the region $\Omega$,
$b_{n}^{(i)}(x)$ is the $i$-times derivative with respect to the variable $x$.

It is not difficult to verify that for any integer $i\geq 0$ the following
estimate is obtained:

$\sum_{n=i+1}^{\infty}|b_{n}^{(i)}(x)t^{n}\psi(t/r_{n})|\leq\sum_{n=i+1}^{\infty}\beta_{n}r_{n}^{n}=\sum_{n=i+1}^{\infty}(\frac{1}{n!})^{n}\frac{\beta_{n}}{(1+\beta_{n})^{n}}\leq\sum_{n=i+1}^{\infty}\frac{1}{n!}<\infty,$

which shows that the series obtained by $i$-times differentiation with respect
to the variable $x$ of the series on the right side of formula (18) converges
absolutely and uniformly in $\mathbb{R}\times\Omega$.

Here we first used that $|t^{n}\psi(t/r_{n})|\leq|t|^{n}$ for $|t|\leq r_{n}$
and $|t^{n}\psi(t/r_{n})|=0$ for $|t|>r_{n}$ (both statements follow from the
definition of the function $\psi$). Then we used the inequality
$(1+|b_{n}|)^{n}\geq|b_{n}|)$, which is valid for all $n\geq 1$ (which is why
we started the summation from one and not from zero, but discarding a finite
number terms of the series does not affect its convergence).

This implies that series (18) defines a function $V(t,x)$ that has continuous
partial derivatives $\partial^{i}V/\partial x^{i}$ of all orders in
$\mathbb{R}\times\Omega$, $i\geq 0$. Similarly, for the series obtained from
(18) by single differentiation with respect to the variable $t$ and $i$ -times
differentiation with respect to the variable $x$, we obtain the following
estimate

$\sum_{n=i+2}^{\infty}|b_{n}^{(i)}(x)(t^{n}\psi(t/r_{n}))^{\prime}|\leq\sum_{n=i+2}^{\infty}\beta_{n}|nt^{n-1}\psi(t/r_{n})+t^{n}/r_{n}\psi^{\prime}(t/r_{n})|$
$\leq\sum_{n=i+2}^{\infty}\beta_{n}r_{n}^{n-1}(n+M_{1})=\sum_{n=i+2}^{\infty}\frac{\beta_{n}(n+M_{1})}{(n!(1+\beta_{n}))^{n-1}}\leq\sum_{n=i+2}^{\infty}\frac{n+M_{1}}{n!}<\infty,$

where number $M_{1}=\max_{[-1,1]}\psi^{\prime}(t)|$.

Using similar reasoning, it is easy to obtain a similar estimate (meaning
absolute and uniform convergence) for the series obtained from (18) by $k$ \-
times differentiation with respect to the variable $t$ and $i$ \- times
differentiation with respect to the variable $x$ . Therefore, the series (18)
converges to some function $V(t,x)$ defined and smooth in
$\mathbb{R}\times\Omega$. Verification that the Taylor series of this function
$V(t,x)$ with respect to the variable $t$ at the point $t=0$ coincides with
(17) is not difficult. The proof of the assertion in the case of $p>1$ and
$m>1$ is completely analogous to the case we have considered. $\Box$

The formal solution of the Cauchy problems (1), (3) and (15) in the form (5)
that we obtained earlier does not have to be a smooth function. However, due
to theorems due to E. Borel and H. Whitney there exists a smooth function
$\tilde{v}$ of the form

$\tilde{v}(t,x)=\sum_{n=0}^{\infty}\frac{t^{n}A^{n}u(x)}{n!}\psi(t/r_{n}),$
(19)

which for $t=0$ has the same power expansion as $v(t,x)$. Here the $\psi(t)$
is a cutoff function.

Let us make one important remark for what follows. The formulation of the
Borel-Whitney Lemma is based on the expansion of the _desired_ function
$v(t,x)$ into a formal Taylor series. In this case, there is a remark about
the ambiguity of the function $\tilde{v}(t,x)$ up to a flat function whose
Taylor series at this point consists of all zeros.

When using the expression (19), the expansion coefficients are determined
uniquely by _known_ functions and there is no reason for the appearance of an
additional flat function. This is also confirmed by the widely used proof of
the uniqueness of the solution of linear equations by the Picard iteration
method.

Thus, we can formulate the following

Proposition 2. _Let an operator $A$ of type_ (2) _with functions_ $u,F\in
C^{\infty}$ _for_ $x\in\mathbb{R}^{n}$. _Then there exists a smooth function_
$\tilde{v}(t,x)$ (19), _which for_ $t=0$ _has the same power expansion as_
$v(t,x)$ and this function solves equations (1), (3) _and_ (15).

##  3 Some generalizations

Above, we considered only single equations whose right-hand side does not
explicitly depend on $t$ and, accordingly, the introduced operator $A$ also
does not depend on $t$. Here we briefly consider the cases of systems of
equations and the possibility of using the proposed approach to equations of
the same type, but whose right-hand side _depends_ explicitly on $t$. In doing
so, we rely on the ideas presented in [1].

For the system of $m$ equations for the $m$ desired functions $v_{i}(t,x)$
($t,x\in\mathbb{R}\times\mathbb{R}^{n}$)

$\frac{\partial}{\partial
t}v_{i}=F_{i}(x,\textbf{v},\textbf{D}_{j}^{\alpha_{j}}\textbf{v}),\qquad
i=1...m,\,j\in\mathbb{N},\qquad v_{i}(t,x)|_{t=0}=u_{i}(x),$ (20)

where $v_{i}=v_{i}(t,x)$, $\textbf{v}=v_{1},...,v_{m}$,
$\textbf{D}_{j}^{\alpha_{j}}\textbf{v}$ \- some given finite sequence of
derivatives of the functions $v_{i}$ with respect to components
$x\in\mathbb{R}^{n}$.

Now let’s introduce the (_single_) operator $A$

$A=\int_{\mathbb{R}^{n}}d\zeta\sum_{k=1}^{m}F_{k}(\zeta,\textbf{u}(\zeta),\textbf{D}_{j}^{\alpha_{j}}\textbf{u}(\zeta))\frac{\delta}{\delta
u_{k}(\zeta)},\qquad\zeta\\\ \in\\\ \mathbb{R}^{n},$ (21)

where $u_{i}=u_{i}(x)$ $i=1,...,m$ is a set of initial conditions.

The operator $A$ introduced here is also independent of $t$. Therefore,
following, almost without changes, the same scheme as in Section 2, we obtain
a formal solution of the system (20)

$v_{i}(t,x)=\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}u_{i}(x)=e^{tA}u_{i}(x),\qquad(i=1,...,m).$
(22)

and as a consequence of Proposition 1 and Proposition 2 with the functions
$u_{i},F_{i}\in C^{\infty}$ for $x\in\mathbb{R}^{n}$ there exists a smooth
functions $\tilde{v}_{i}(t,x)$ (19), having for $t=0$ the same power
expansions as $v_{i}(t,x)$.

If the right-hand sides of the equations explicitly depend on $t$

$\frac{\partial}{\partial
t}v_{i}=F_{i}(t,x,\textbf{v},\textbf{D}_{j}^{\alpha_{j}}\textbf{v}),\qquad
i=1...m,\,j\in\mathbb{N},\qquad v_{i}(t,x)|_{t=0}=u_{i}(x),$ (23)

then, respectively, the introduced operator $A$ also depends on $t$

$A=\int_{\mathbb{R}^{n}}d\zeta\sum_{k=1}^{m}F_{k}(t,\zeta,\textbf{u}(\zeta),\textbf{D}_{j}^{\alpha_{j}}\textbf{u}(\zeta))\frac{\delta}{\delta
u_{k}(\zeta)},\qquad\zeta\\\ \in\\\ \mathbb{R}^{n}.$ (24)

In this case, the operators $A(t_{1})$ and $A(t_{2})$ do not commute with
$t_{1}\neq t_{2}$, which leads to significant complications in the scheme we
are describing, starting with the interpretation of the Picard iteration
method. To represent the expansion obtained in this case in the form of an
operator exponent, one usually introduces the proposed by F. Dyson time
ordered (chronological) exponent, which differs significantly in its
properties from the usual operator exponent [2]. The most important thing for
this paper is that it does not directly solve the equations in the form of a
Taylor series expansion in $t$.

In [1], [2] a trick was proposed to _exactly_ reduce the _time ordered_
exponent to _usual_ one by simply modifying the operator $A$ (24) by replacing
$t\rightarrow s$ and adding the term $\frac{\partial}{\partial s}$:

$\tilde{A}=\int_{\mathbb{R}^{n}}d\zeta\sum_{k=1}^{m}F_{k}(s,\zeta,\textbf{u}(\zeta),\textbf{D}_{j}^{\alpha_{j}}\textbf{u}(\zeta))\frac{\delta}{\delta
u_{k}(\zeta)}+\frac{\partial}{\partial s},\qquad\zeta\\\ \in\\\
\mathbb{R}^{n}$ (25)

and now $\tilde{A}$ _does not depend_ on $t$ and therefore we obtain a formal
solution of the system (23)

$v_{i}(t,x)=[\sum_{n=0}^{\infty}\frac{t^{n}\tilde{A}^{n}}{n!}u_{i}(x)]|_{s=0}=[e^{t\tilde{A}}u_{i}(x)]|_{s=0},\qquad(i=1,...,m),$
(26)

namely in the form of a Taylor series expansions of $v_{i}(t,x)$ in $t$ for
$t=0$. With the same consequences from Proposition 1 and Proposition 2.

This list of possible generalizations is far from be complete.

##  4 Example. Smoothness of solutions of the Navier-Stokes equations

Incompressible flows of homogeneous fluids in space $\mathbb{R}^{3}$ are
solutions of the system of equations [12]-[14]

$\frac{\partial{v}}{\partial t}=\nu\Delta
v-\sum_{j=1}^{3}v_{j}\frac{\partial{v}}{\partial x_{j}}-\nabla p+f,$ (27)

$\mathrm{div}\;v\equiv\sum_{j=1}^{3}\frac{\partial{v_{j}}}{\partial
x_{j}}=0,\qquad(x,t)\\\ \in\\\ \mathbb{R}^{3}\times[0,\infty),$ (28)

$v|_{t=0}=u,\qquad x\\\ \in\\\ \mathbb{R}^{3},$ (29)

where $v(x,t)\equiv(v_{1},v_{2},v_{3})$ is the fluid velocity, $p(x,t)$ is the
scalar pressure, the gradient operator is $\nabla=(\frac{\partial}{\partial
x_{1}},\frac{\partial}{\partial x_{2}},\frac{\partial}{\partial x_{3}})$ and
the Laplace operator $\Delta\equiv\sum_{j=1}^{3}\frac{\partial^{2}}{\partial
x_{j}^{2}}$, a kinematic constant viscosity $\nu\geq 0$, initial conditions
$v|_{t=0}=u(x)\equiv(u_{1},u_{2},u_{3})$, $f=f(x,t)$ is given, externally
applied force. Further we will consider Navier-Stokes equations in the absence
of external forces, i.e. with $f=0$.

We restrict our attention to Navier-Stokes equations and correspondingly to
operator $A$ on all of $\mathbb{R}^{3}$, thus avoiding the delicate discussion
of boundary conditions.

It is well known [15],[16] that we can find an equation $p=p(v)$ to eliminate
the pressure from (27). If we take the divergence at each side of the momentum
equation (27), which yields, using (28) that

$\Delta\emph{p}=-\sum_{i,j=1}^{3}\frac{\partial{v_{j}}}{\partial
x_{i}}\frac{\partial{v_{i}}}{\partial x_{j}}.$ (30)

The equation (30) with respect to the function $p$ is the well known Poisson
equation, whose solution in $\mathbb{R}^{3}$ [15] has the form (for brevity,
we will sometimes denote this solution as $p_{v}(x,t)\equiv p_{v}$)

$p(x,t)=\frac{1}{4\pi}\int_{\mathbb{R}^{3}}d\xi\sum_{i,j=1}^{3}\frac{\partial{v_{j}(t,\xi)}}{\partial\xi_{i}}\frac{\partial{v_{i}(t,\xi)}}{\partial\xi_{j}}\frac{1}{|x-\xi|},\qquad(\xi\\\
\in\\\ \mathbb{R}^{3}).$ (31)

Substituting the solution (31) into (27), we obtain (Leray’s Formulation of
the Navier-Stokes Equation [15]) _non-linear system of integro-differential
equations_ containing _only_ unknown functions $v_{i}$.

$\frac{\partial{v}}{\partial t}=\nu\Delta
v-\sum_{j=1}^{3}v_{j}\frac{\partial{v}}{\partial x_{j}}-\nabla
p_{v},\qquad(x\\\ \in\\\ \mathbb{R}^{3},t\geq 0).$ (32)

It is easy to see [15] that the systems of the Navier-Stokes equations
(27)-(29) are equivalent to the system of the equations (31), (32), provided
that functions $u_{i}$ in initial conditions are smooth and

$\sum_{i=1}^{3}\frac{\partial{u_{i}}}{\partial x_{i}}=0,\qquad(x\\\ \in\\\
\mathbb{R}^{3}).$ (33)

Although the Leray’s formulation of the Navier-Stokes equation has been known
for a long time, it has not been widely used to analyze its solutions. The
main theses of this can be traced to the following [15].

The equation (32) is quadratically nonlinear and contains a nonlocal,
quadratically nonlinear operator. These facts make the Navier-Stokes equation
hard to study analytically.

Although (32) constitutes a closed system for $v$, this formulation turns out
to be not very useful for further analysis except by rather crude methods
based on the energy principle. The main reason for this is that this
formulation hides all the properties of vorticity stretching and interaction
between rotation and deformation.

We will consider the vector function $u(x)\equiv(u_{1}(x),u_{2}(x),u_{3}(x))$,
where arbitrary functions $u_{i}(x)\in C^{\infty}$, $x\in\mathbb{R}^{3}$
which, as we discussed in the previous section, are infinitely many times
differentiable. In accordance with the trick described above let us introduce
the linear differential operator

$\displaystyle
A=\int_{\mathbb{R}^{3}}d\zeta\,\sum_{i=1}^{3}{\Large\\{}\nu\Delta_{\zeta}u_{i}(\zeta)-\sum_{j=1}^{3}u_{j}(\zeta)\frac{\partial{u_{i}(\zeta)}}{\partial\zeta_{j}}$
$\displaystyle-\frac{1}{4\pi}\frac{\partial}{\partial\zeta_{i}}{\Large[}\int_{\mathbb{R}^{3}}d\xi\sum_{i,j=1}^{3}\frac{\partial{u_{j}(\xi)}}{\partial\xi_{i}}\frac{\partial{u_{i}(\xi)}}{\partial\xi_{j}}\frac{1}{|\zeta-\xi|}{\Large]}{\Large\\}}\frac{\delta}{\delta
u_{i}(\zeta)}.$ (34)

In short notation

$A=\int_{\mathbb{R}^{3}}d\zeta\,\sum_{i=1}^{3}{\Large\\{}\nu\Delta_{\zeta}u_{i}(\zeta)-\sum_{j=1}^{3}u_{j}(\zeta)\frac{\partial{u_{i}(\zeta)}}{\partial\zeta_{j}}-\nabla_{\zeta}\;p_{u}(\zeta){\Large\\}}\frac{\delta}{\delta
u_{i}(\zeta)},$ (35)

where

$p_{u}(\zeta)\equiv\frac{1}{4\pi}\int_{\mathbb{R}^{3}}d\xi\sum_{i,j=1}^{3}\frac{\partial{u_{j}(\xi)}}{\partial\xi_{i}}\frac{\partial{u_{i}(\xi)}}{\partial\xi_{j}}\frac{1}{|\zeta-\xi|}.$
(36)

It is well known that if $u_{i}\in C^{\infty}$, then $p_{u}\in C^{\infty}$
too. Therefore, the operator $A$ satisfies the conditions of Proposition 1 and
Proposition 2.

Then there exists a smooth functions $\tilde{v}_{i}(t,x)$ (19) having the same
power expansions for $t=0$ as $v_{i}(t,x)$ and this functions solves equations
(32) and, accordingly, (27), provided that functions $u_{i}$ in initial
conditions are smooth and satisfy condition (33).

Note that the scheme of proof given here is also applicable in the case of
$x\in\mathbb{R}^{n}$ for any $n\in\mathbb{N}$ and in the _presence of smooth_
external forces.

And nothing prevents us from putting a kinematic constant viscosity $\nu=0$
and getting the same results for the Euler equations.

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