# Existence, Smoothness and Uniqueness (in smooth space) of the Navier-Stokes
equation by using a new Boundary Integral representation.

Edmund Chadwick<EMAIL_ADDRESS>
(May 25 2023)

###### Abstract

Consider an exterior space-time domain where the incompressible Navier-Stokes
equation and continuity equation hold with no bodies or force fields present,
and smooth velocity at initial time. This is equivalent to the velocity being
impulsively instantaneously started into motion and further assume that this
force impulse is bounded. A smooth solution with a Stokeslet far-field decay
for all subsequent time is sought and found, demonstrating existence and
smoothness. This is given by a space-time boundary integral velocity
representation by a single layer potential linear distribution of Navier-
Stokes fundamental solutions called NSlets. This is obtained by extending the
theory of hydrodynamic potentials to also include a non-linear potential that
subsequently drops out of the formulation. Zero initial velocity gives the
null solution and so there can be only one smooth solution demonstrating
uniqueness in smooth space, but this is not to say that there aren’t other
possible solutions in the wider class of non-smooth spaces.

##  1 Introduction

A general solution to the Navier-Stokes equation is one of the millennium
problem challenges [19]. This challenge seeks to prove (or otherwise) that
there exists a smooth solution satisfying the Navier-Stokes equation for the
particular problem of a smooth divergence free velocity at initial time and an
incompressible fluid filling all of space. So, non-smooth solutions such as
jumps and bifurcations are not considered because a solution is sought that is
restricted to $C^{\infty}$ only. Existence and smoothness have been proven for
the two-dimensional case, and is demonstrated by Ladyzhenskaya [24] for the
weak solution by using generalised functions. However, this approach has not
successfully been extended to the three-dimensional unsteady problem. This is
because of the question of blow-up of the solution due to a finite-time
singularity. At small scale the inertial (transport) term dominates and the
Navier-Stokes equation becomes supercritical and is demonstrated to blow up
when considering averaged quantities [28]. Ramm [27] gives a velocity
representation by a domain distribution of unsteady Stokeslets, with the non-
linear velocity term within the domain integral itself. By using Fourier
Transforms, he then is able to give bounds for the velocity and velocity
derivative as well as demonstrate the solution is unique. However, it is
unclear how this approach deals with blow-up and there has been a challenge to
the accuracy of the analysis in the review [25] that suggests one of the
integrals used in the analysis is non-integrable (unbounded). For the three-
dimensional steady Stokes problem, Ladyzhenskaya also gives a strong solution
by a different approach, using the theory of hydrodynamic potentials which
uses Green’s boundary integral method. Oseen [26] similarly gives solutions
for the three dimensional steady Oseen flow, and also unsteady Stokes flow
using Oseenlet and Stokeslet unsteady force impulse fundamental solutions.
Chadwick [8] extends this approach giving a boundary integral representation
for Navier-Stokes flow by introducing its fundamental solutions called NSlets.
In this paper [8] the near field of the NSlet is assumed to be the Stokeslet.
However, this is not the case and so the same difficulty as in [27] arises as
the expected behaviour at small scale given in [28] is not replicated,
particularly for turbulent flows associated with high Reynolds number.

Instead, in the present paper the NSlets are shown to reduce to Eulerlets near
to their origin. This is a crucial difference that also indicates the
difficulties in getting existence and smoothness proofs for the Navier-Stokes
equation, since the Eulerlets contain unstable domain singularities [10]
consequently giving the possibility of blow up [28] as well as turbulence. So
by using standard approaches, it is difficult to see how this singular small
scale nature can be treated mathematically to produce bounds for the solution,
explaining why the Navier-Stokes equation is so difficult to solve.

However, in the present paper it is found that the integral flux of the
Eulerlet in the boundary integral formulation is finite and can be evaluated
removing this difficulty, and so for smooth initial conditions with no bodies
or other force fields, the boundary integral representation in Chadwick [8] is
able to be used in the present paper despite the singular and unstable nature
of the Eulerlet. This follows the methodology of Oseen [26] that uses the
Green’s integral representation theory of hydrodynamic potentials for the
linear problems of Oseen and Stokes flow. In Oseen’s work, a potential by an
integral along the radius variable (from its origin) is defined. Similarly,
this approach is followed here for the non-linear Navier-Stokes equation which
gives an additional non-linear potential, but it is shown that this non-linear
potential drops out across the integral boundary resulting in a velocity
representation by a linear distribution of Navier-Stokes fundamental solutions
(NSlets). Once in this linear form, existence, smoothness and uniqueness are
obtained by following an approach similar to Ladyzhenskaya’s strong solution
for the three-dimensional steady Stokes problem by extending and developing it
for a space-time domain. The space-time boundary can be moved near to the
fluid point, where the fundamental solution approximates to the Eulerlet, and
so the representation reduces to a linear distribution of Eulerlets with
strength given by the force impulse distribution. Since the force impulse
generating the flow is assumed bounded and the integral flux of the Eulerlet
is bounded even though the Eulerlet itself is singular at its spatial origin,
it is then shown that the velocity is bounded as well as its derivatives, so
the solution exists and is smooth. Choosing the space-time boundary to be the
space domain at initial time, the velocity is represented by a boundary
integral distribution of NSlets such that the force impulse strength is given
by the initial velocity. So, considering zero initial velocity leads to a zero
velocity flow field at all subsequent time, demonstrating uniqueness in
$C^{\infty}$ only.

This boundary integral approach builds upon previous work on providing
boundary integral formulations in incompressible fluid dynamics by the author
which has incrementally developed Oseen’s mathematical analysis for boundary
integral representations and is outlined next. The work started by considering
the far-field Oseen representation for large Reynolds number [1], which
resulted in a far-field velocity representation by an integral distribution of
Oseenlets [2] [20]. From this, a slender body theory in Oseen flow was
developed [3], and verified experimentally in low-speed wind tunnels [6]. In
the large Reynolds number limit, this then gives insight into the form of the
near-field potential flow, and a slender wing theory in potential flow was
developed [4] as well as an understanding of the underlying vortex flow
structure [5] [11]. The work was then recast within a matched asymptotic
framework but using a novel approach with the boundary integrals of the
representations being matched. This was achieved at low Reynolds number for an
outer Oseen flow and an inner Stokes flow [12] and also later in [17] [18]
[16]. The method was also applied at high Reynolds number to matching an outer
Oseen flow to an inner Euler flow [21] [14] [13] [9] [10]. This leads to a
velocity representation in Euler flow by an integral distribution of Euler
flow fundamental solutions called Eulerlets. As well, it is shown that the
representation reduces to all the standard classic steady Euler flow models
(potential aerofoil theory, vortex-lattice theory, thin-wing theory, slender-
wing theory, slender-body theory) under the appropriate approximations [10].
Furthermore, this model was able to model bluff body flows by an Euler slip
wake expected from the theory [10]. Work was then directed to producing a
similar velocity representation in Navier-Stokes flow by an integral
distribution of fundamental solutions for the Navier-Stokes equation called
NSlets [7] [8]. This was then successfully applied to flow past a circular
cylinder at low Reynolds number, intermediate Reynolds number, and high
Reynolds number, as well as high Reynolds number flow past a slender body with
elliptical cross-section [8]. This was also shown to have at least the same
accuracy as Kusukawa [22] for the problem of two-dimensional steady flow past
a semi-infinite flat plate (Blasius solution). The Blasius solution requires
solving a non-linear partial differential equation, but instead with this
method the solution is given by a velocity represented by a Green’s integral
linear distribution of BL-lets [8].

##  2 Problem statement

This paper seeks to prove the following theorem.

###### Theorem 1

Consider the motion of a fluid in $\mathbb{R}^{3}$ described by the
incompressible Navier-Stokes equation and continuity equation, such that body
forces are taken to be zero, for the unknown velocity field
$\in\mathbb{R}^{3}$ and pressure $\in\mathbb{R}$ defined for time $t\geq 0$,
with initial condition given for the velocity as a $C^{\infty}$ divergence
free vector field that decays like a Stokeslet in the far-field. Then, the
pressure and velocity exist, are smooth and unique at all subsequent time with
bounded energy.

This theorem requires the following two lemmas.

###### Lemma 1

The problem can be restated as a boundary value problem in a space-time domain
$\Sigma_{+}$ enclosed by a hyper-surface $\partial\Sigma$ between one
spherinder (spherical cylinder) surface $\partial\Sigma_{X}$ given by:
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=X^{2}$, $t^{2}=X^{2}$;
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\leq X^{2}$, $t=X$; and
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\leq X^{2}$, $t=-X$; and another spherinder
surface $V$ of vanishing (instantaneous) time interval given by:
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\leq X^{2}$, $t=0$; where ${\bf x}$ is co-
ordinate position and $t$ is time, and $X\rightarrow\infty$, see figure 1.

Figure 1: The space-time domain $\Sigma^{+}$

The boundary conditions are: The spherinder hyper-surface $V$ is the volume at
initial time over which an instantaneous smooth force impulse distribution
giving a total finite force impulse is applied; and over the far-field
spherinder hyper-surface $\partial\Sigma_{X}$ the flow field behaves like a
Stokeslet.

The spherinder hyper-surfaces have defined normals (except in the limit
approaching the edges), are smooth surfaces piece-wise connected, and so
amenable to Green’s boundary integral analysis. Green’s analysis is performed
in the space-time domain enclosed by the initial time and far-field space-time
boundary integrals, and the vanishingly small space-time boundary enclosing
the space-time point. A proof by ansatz is considered. So, in this domain it
is assumed that the velocity and pressure (and therefore the NSlet velocity
and pressure also) are $C^{\infty}$ smooth. This enables a solution for the
velocity to be found by a Green’s integral representation. From this solution
it is then shown that a velocity and pressure which are $C^{\infty}$ smooth
exist, completing the proof by ansatz. (Of course, this proof by ansatz says
nothing about whether or not non-smooth solutions exist as well and so this
remains unanswered.) It is then shown that there can be only one solution,
uniqueness, in the $C^{\infty}$ smooth class. In particular, the Green’s
integral representation for the velocity $u_{k}({\bf x},t)$ is given by a
boundary integral distribution of NSlet fundamental solutions with strength
given by (minus) the initial velocity, distributed over the all of space
volume $V$ at the initial time $t=0$ so

$u_{k}({\bf x},t)=-\int_{V}u_{i}({\bf x^{\prime}},0)u_{ki}({\bf x}-{\bf
x^{\prime}},t)dV^{\prime},$ (1)

where $u_{ki}$ is the $i$th velocity component of the $k$th NSlet,
$dV^{\prime}$ is an element of the volume $V$, and the force impulse ${\bf J}$
is given by

$J_{i}=-\int_{V}u_{i}({\bf x^{\prime}},0)dV^{\prime},$ (2)

where $-u_{i}({\bf x},0)$ is also the force impulse distribution. (It is noted
that this automatically reduces to boundary integral formulations for the
linearised Oseen and Stokes formulations as for this problem there is no flux
through the surface.)

Proof: Given in Sections 3-8.

###### Lemma 2

The velocity representation at the space-time point $({\bf x},t)$ can
equivalently be given by a distribution of fundamental solutions over a small
space-time spherinder surface $\partial\Sigma_{\epsilon}$ enclosing the space-
time point. The surface is parameterised by $({\bf x^{\prime}},t^{\prime})$
where $\lvert x_{i}-x_{i}^{\prime}\rvert=\epsilon$, $t-\Delta t\leq
x_{0}^{\prime}\leq t$; and $0\leq\lvert
x_{i}-x_{i}^{\prime}\rvert\leq\epsilon$, $x_{0}^{\prime}=t-\Delta t$ and
$x_{0}^{\prime}=t$. The constants $\epsilon$ and $\Delta t$ are let be as
small as required to enable the approximation of the NSlet by the Eulerlet.

The velocity is given by

$u_{k}({\bf x},t)=\int_{\partial\Sigma_{\epsilon}}F_{i}({\bf
x^{\prime}},t^{\prime})u_{ki}({\bf x}-{\bf
x^{\prime}},t-t^{\prime})d\sigma^{\prime},$ (3)

where $d\sigma^{\prime}$ is an element of the hyper-surface and the force
impulse ${\bf J}$ is given by

$J_{i}=\int_{\partial\Sigma_{\epsilon}}F_{i}({\bf
x^{\prime}},t^{\prime})d\sigma^{\prime},$ (4)

where $F_{i}$ is a force impulse distribution. Taking $\epsilon$ and $\Delta
t$ sufficiently small, the NSlet tends towards the Eulerlet and the velocity
is represented by an integral over the curved part only of the spherinder
surface $\partial\Sigma_{\epsilon*}$ given by $\lvert
x_{i}-x_{i}^{\prime}\rvert=\epsilon$, $t-\Delta t\leq x_{0}^{\prime}\leq t$
only. This is

$u_{k}({\bf x},t)=u_{k}({\bf x},t-\Delta
t)+\int_{\partial\Sigma_{\epsilon*}}F_{i}({\bf
x^{\prime}},t^{\prime})u_{ki}^{E}({\bf x}-{\bf
x^{\prime}},t-t^{\prime})d\sigma^{\prime}(1+O(f(\epsilon))),$ (5)

for the velocity Eulerlet $u_{ki}^{E}$ and bounded function $f(\epsilon)$.
From this, it is shown that if the initial velocity $u_{k}({\bf x},0)$ is
smooth, then the velocity $u_{k}({\bf x},t)$ is bounded and smooth also, and
so also the pressure from the Helmholtz theorem. Similarly, the NSlet velocity
and pressure are shown to be bounded and smooth for any point away from its
origin where it is singular.

Proof: Given in Section 9.

##  3 Governing Equations

Consider the incompressible Navier-Stokes equation and continuity equation
given by

$\rho f_{i}=\rho u_{i;0}+\rho u_{j}u_{i;j}+p^{\dagger}_{;i}-\mu
u_{i;jj}=0\;,\;\;u_{i;i}=0,$ (6)

where $f_{i}({\bf x^{\prime}})$ is the body force taken to be zero,
$u_{i}({\bf x^{\prime}})$ is the velocity, $p^{\dagger}({\bf x^{\prime}})$ is
the pressure, $\rho$ is the fluid density and $\mu$ is the dynamic viscosity.
The capital index refers to space-time co-ordinates $x^{\prime}_{I}$, $0\leq
I\leq 3$ with $x^{\prime}_{0}$ being the time variable, and a lower index
refers to spatial Cartesian co-ordinates $x^{\prime}_{i}$, $1\leq i\leq 3$.
The analysis in this paper uses lengthy complex terms including differentials
that intricately mix the physical variables and fundamental solutions. In
order to make these expressions sufficiently concise and readable, the
Einstein summation index convention is used, and so
$a_{I}b_{I}=a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$. We also borrow the
General Relativity derivative representation by a semi-colon and by a comma.
However, here the semi-colon represents a derivative with respect to the
primed Cartesian co-ordinate $x^{\prime}_{I}$ such that $\frac{\partial g({\bf
x^{\prime}})}{\partial x^{\prime}_{I}}=g_{;I}$, for some function $g$. The
primed co-ordinate will refer to an integrated value over the boundary
integral given later in the Green’s Boundary Integral Method. To reduce the
number of parameters in the problem ($\rho$ and $\mu$) from two to one, it is
usual to express the quantities in dimensionless form, giving a single
parameter which is the Reynolds number. However, for this problem there is no
body and therefore no characteristic length, and so this is not possible. To
overcome this, Fefferman [19] considers dividing by $\rho$ to get

$f_{i}=u_{i;0}+u_{j}u_{i;j}+p_{;i}-\nu u_{i;jj}=0\;,\;\;u_{i;i}=0,$ (7)

where $p({\bf x^{\prime}})$ is the pressure for a density of one (or pressure
divided by the density constant), and now $\nu=\mu/\rho$ which is the
kinematic viscosity is the only parameter in the equation. (Comparing to the
dimensionless Navier-Stokes equation, then $\nu$ is in the position of the
reciprocal of the Reynolds number with Bernoulli pressure (ignoring the half)
used as the characteristic pressure, which provides an easy way to transform
into dimensionless form for other problems where there is a characteristic
body length.)

This way of representing the Navier-Stokes equation is used widely by those
working on the Navier-Stokes millennium problem, and so it shall also be
followed here. For ease of use, like Fefferman and others, we call $p$ the
pressure from now on, but to obtain the standard boundary integral formulation
the actual pressure is easily inserted back into the expression from
$p^{\dagger}=\rho p$.

The millennium problem is for the time dependent problem, and so to solve this
problem we must consider the space-time domain and the three fundamental
solutions to equation (7) $1\leq k\leq 3$ which we call NSlets, that are
generated by applying unit impulse forces in the $k$ direction at the point
$x_{I}=x^{\prime}_{I}$, and such that the solutions decay and vanish in the
far-field. Using the Dirac delta function notation, this is described by

$f_{ki}=u_{ki,0}+u_{{\bf k}j}u_{{\bf k}i,j}+p_{k,i}-\nu
u_{ki,jj}=-\delta(x_{I}-x_{I}^{\prime})\delta_{ki}\;,\;\;\;u_{ki,i}=0,$ (8)

where the first index $k$, denotes the $k$th solution. The comma represents a
derivative with respect to the Cartesian co-ordinate $x_{I}$ such that
$\frac{\partial g({\bf x-x^{\prime}})}{\partial x_{I}}=g_{,I}$.

A bold index does not use the Einstein summation index convention, and so
$a_{\bf k}b_{\bf k}=a_{1}b_{1}$ for $k=1$, $a_{\bf k}b_{\bf k}=a_{2}b_{2}$ for
$k=2$ and $a_{\bf k}b_{\bf k}=a_{3}b_{3}$ for $k=3$. The term $\delta_{ki}$ is
Kronecker delta, such that $\delta_{ki}=1$ for $i=k$, and $\delta_{ki}=0$ for
$i\neq k$. The Dirac delta function $\delta(x_{I}-x_{I}^{\prime})$ is given in
space-time, and so
$\delta(x_{I}-x_{I}^{\prime})=\delta(x_{0}-x_{0}^{\prime})\delta(x_{1}-x_{1}^{\prime})\delta(x_{2}-x_{2}^{\prime})\delta(x_{3}-x_{3}^{\prime})$,
where the Dirac delta function $\delta(x_{0}-x_{0}^{\prime})$ (for example) is
defined as being zero everywhere except at $x_{0}=x_{0}^{\prime}$ such that
$\int_{a}^{b}\delta(x_{0}-x_{0}^{\prime})dx_{0}=1$ if $a<x_{0}^{\prime}$ and
$b>x_{0}^{\prime}$, and is zero otherwise. Hence, the Dirac delta function is
a notation and the description (8) should be read as the equation

$\int_{\partial\Sigma_{c}}f_{kiJ}n_{J}d\sigma=-\delta_{ki}$ (9)

where: $\partial\Sigma_{c}$ is a closed space-time hyper-surface boundary
enclosing the point $x^{\prime}_{I}$ having an outward pointing normal
$n_{J}$, and $d\sigma$ is an element of the boundary; and the function
$f_{kiJ}$ is given by

$f_{kiJ}=u_{ki}\delta_{0J}+u_{{\bf k}j}u_{{\bf k}i}+p_{k}\delta_{ij}-\nu
u_{ki,j},$ (10)

where the vector product with mixed space-time and spatial indices is defined
as $a_{J}b_{j}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$. We note that (9) can also be
rewritten in terms of an integral of the primed co-ordinate. In this case, the
unit normal at the point $x_{I}$ for the surface integral over elements
$d\sigma$ enclosing the primed co-coordinate $x^{\prime}_{I}$ is minus the
unit normal at the primed co-ordinate point $x^{\prime}_{I}$ for the surface
integral over elements $d\sigma^{\prime}$ enclosing the co-coordinate $x_{I}$,
see figure 2.

Figure 2: The normal direction between integrations for the primed and
unprimed co-ordinate

So

$n_{J}d\sigma^{\prime}=-n_{J}d\sigma,$ (11)

and

$\int_{\partial\Sigma^{\prime}_{c}}f_{kiJ}n_{J}d\sigma^{\prime}=\delta_{ki},$
(12)

where $\partial\Sigma^{\prime}_{c}$ is the reflection about all four space-
time axes of the boundary $\partial\Sigma_{c}$, see figure 2.

##  4 NSlet asymptotic approximations

Existence, smoothness and uniqueness is established from the velocity
representation by NSlets. This representation is obtained from the theory of
hydrodynamic potentials, see for example [24]. This requires the evaluation of
boundary integrals near to and far from a fluid point, and so the near-field
and far-field asymptotic approximations of the NSlet. So in this section,
consider approximations to the NSlet near to and far from its origin. It shall
be shown that near to its origin, the NSlet approximates to the Eulerlet, and
far from its origin, the NSlet approximates to the Stokeslet.

###  4.1 Near to the origin: the Eulerlet

First, consider the NSlet close to its origin $x_{I}=x^{\prime}_{I}$. We shall
find that this approximates to the Eulerlet, and this is demonstrated next.
From the NSlet equation (8), it is seen that if the time dependent term
$u_{ki,0}$ doesn’t contribute to the integral flux represented by the term
$\delta\delta_{ki}$, then the velocity would be multiplied by the factor
$\delta(x_{0})$ since all the other terms involve spatial derivatives only.
However, such a velocity has no solution for $x_{0}^{\prime}\neq x_{0}$ and so
is not a meaningful solution to (8). This means that the time-dependent term
in the NSlet must contribute to the integral flux. (This does not mean that
steady state problems cannot be tackled, as the fundamental solution for
steady state problems is for unit forces and not unit force impulses, and the
steady state problem is obtained by integrating through all time the time-
dependent problem.) Consider the dimensional size of the velocity, time and
distance close to the NSlet to be $U$, $T$ and $L$ respectively. Then this
time dependent term has dimensions $UT^{-1}$ and the flux has dimensions
$L^{-3}T^{-1}$, so the velocity goes as $U\sim L^{-3}$. It is now possible to
compare the orders of the other terms, so $u_{{\bf k}j}u_{{\bf k}i,j}\sim
L^{-7}$ and $u_{ki,jj}\sim L^{-5}$. This means that the velocity becomes
increasingly singular such that the local Reynolds number defined as $UL/\nu$
tends to infinity, and to leading order the viscous term drops out, and so the
velocity tends toward the Eulerlet velocity $u_{ki}\sim u^{E}_{ki}$.
Similarly, let the pressure tend to an Euler pressure given by $p_{k}\sim
p^{E}_{k}+p^{B}_{k}$. Then substituting into (8) the Eulerlet velocity and
pressure satisfy $u^{E}_{ki,0}+u^{E}_{{\bf k}j}u^{E}_{{\bf
k}i,j}+p^{E}_{k,i}+p^{B}_{k,i}=-\delta\delta_{ki}$ where $\delta$ is a
function of $x_{I}-x_{I}^{\prime}$, and so $\delta(x_{I}-x_{I}^{\prime})$ as
given in (8). The Eulerlet has been given in [10] and is found by letting

$u^{E}_{{\bf k}j}u^{E}_{{\bf k}i,j}+p^{B}_{k,i}=0$ (13)

and

$u^{E}_{ki,0}+p^{E}_{k,i}=-\delta\delta_{ki}.$ (14)

Although the equation (13) has no time dependency, the equation (14) does,
leading to a solution dependent on time. In particular, it is noted that the
Dirac delta function $\delta$ is for four-dimensional space-time and not
three-dimensional space. We can now determine the Eulerlet velocity and
pressure $u^{E}_{ki}$ and $p^{E}_{k}$ by solving the equation (14), and give
the resulting Bernoulli pressure term $p^{B}_{k}$ from (13).

The pressure term $p_{k}^{E}$ is determined by taking the divergence of (14)
to get $p^{E}_{k,ii}=-\delta_{,k}$. Letting
$p^{E}_{k}=\delta(x_{0}-x_{0}^{\prime})p^{*}_{,k}$ gives
$p^{*}_{,ii}=-\delta(x_{i}-x_{i}^{\prime})$ and so $p^{*}=\frac{1}{4\pi R}$
where $R$ is the radius from the NSlet origin such that
$R^{2}=(x_{i}-x_{i}^{\prime})(x_{i}-x_{i}^{\prime})$, which gives

$p^{E}_{k}=\frac{\delta(x_{0}-x_{0}^{\prime})}{4\pi}\left[\frac{1}{R}\right]_{,k}.$
(15)

Substituting the pressure (15) into (14) then gives

$u^{E}_{ki,0}=-\delta\delta_{ki}-\frac{\delta(x_{0}-x_{0}^{\prime})}{4\pi}\left[\frac{1}{R}\right]_{,ki}$
(16)

which integrates up to give

$u^{E}_{ki}=-H(x_{0}-x_{0}^{\prime})\delta(x_{i}-x_{i}^{\prime})\delta_{ki}-\frac{1}{4\pi}H(x_{0}-x_{0}^{\prime})\left[\frac{1}{R}\right]_{,ki}$
(17)

where $H(x_{0}-x_{0}^{\prime})$ is the Heaviside function
$H(x_{0}-x_{0}^{\prime})=1$ for $x_{0}>x_{0}^{\prime}$ and zero otherwise. So,
the Eulerlet is dependent on time through the Heaviside function being zero
before time $x_{0}=x_{0}^{\prime}$ and generally non-zero afterwards. It is
also clear that the Eulerlet is not simply a source potential, in particular
the first term in (17) has a singular delta function which in the steady case
gives a singular wake line in the flow field for the steady Eulerlet [10], and
so is not based on a simple potential flow. For $R>0$, then
$u^{E}_{ki}=\phi^{E}_{k,i}=-\frac{1}{4\pi}\left[\frac{1}{R}\right]_{,ki}$, and

$\displaystyle p^{B}_{k,i}$ $\displaystyle=$ $\displaystyle-\phi^{E}_{{\bf
k},j}\phi^{E}_{{\bf k},ji}=\left[-(1/2)\phi^{E}_{{\bf k},j}\phi^{E}_{{\bf
k},j}\right]_{,i}$ $\displaystyle p^{B}_{k}$ $\displaystyle=$
$\displaystyle-(1/2)\phi^{E}_{{\bf k},j}\phi^{E}_{{\bf k},j}=-(1/2)u^{E}_{{\bf
k}j}u^{E}_{{\bf k}j},$ (18)

which is the Bernoulli pressure.

We can check the Eulerlet contribution from (14), which is

$\displaystyle\int_{\partial\Sigma_{\delta}}(u^{E}_{ki}n_{0}+p^{E}_{k}n_{i})d\sigma$
$\displaystyle=$
$\displaystyle\int_{V_{R}}\left(-\delta(x_{i}-x_{i}^{\prime})\delta_{ki}-\frac{1}{4\pi}\left[\frac{1}{R}\right]_{,ki}\right)dV$
(19)
$\displaystyle+\int_{T}\int_{S_{R}}\frac{\delta(x_{0}-x_{0}^{\prime})}{4\pi}\left[\frac{1}{R}\right]_{,k}n_{i}dSdt$
$\displaystyle=$
$\displaystyle-\delta_{ki}-\frac{1}{4\pi}\int_{V_{R}}\left[\frac{1}{R}\right]_{,ki}dV$
$\displaystyle-\frac{1}{4\pi}\int_{S_{R}}\frac{(x_{k}-x_{k}^{\prime})}{R^{3}}\frac{(x_{i}-x_{i}^{\prime})}{R}dS$
$\displaystyle=$
$\displaystyle-\delta_{ki}+\frac{1}{3}\delta_{ki}-\frac{1}{3}\delta_{ki}$
$\displaystyle=$ $\displaystyle-\delta_{ki},$

where the boundary $\partial\Sigma_{\delta}$ is defined as the boundary
consisting of the set of points $x_{I}$ centred on $x^{\prime}_{I}$ including:
those such that $(x_{j}-x_{j}^{\prime})(x_{j}-x_{j}^{\prime})=R^{2}$ and
$-T\leq x_{0}-x_{0}^{\prime}\leq T$ with outward pointing normal
$n_{j}=(x_{j}-x_{j}^{\prime})/R$, those such that
$0\leq(x_{j}-x_{j}^{\prime})(x_{j}-x_{j}^{\prime})\leq R^{2}$ and
$x_{0}-x_{0}^{\prime}=T$ with outward pointing normal $n_{0}=1$, and finally
those such that $0\leq(x_{j}-x_{j}^{\prime})(x_{j}-x_{j}^{\prime})\leq R^{2}$
and $x_{0}-x_{0}^{\prime}=-T$ which give no contribution because
$H(x_{0}-x_{0}^{\prime})=0$, see figure 3. So, $\partial\Sigma_{\delta}$ and
all the boundary surfaces considered are piecewise continuous Lyapunov
surfaces so have defined normals up to the piecewise corners, and so the
Green’s boundary integral and flux conditions hold. The values of $T$ and $R$
are considered small, the volume $V_{R}$ is the volume of the sphere of radius
$R$, and the surface $S_{R}$ is the surface of the sphere of radius $R$. So,
when a space-time infinitesimal element $d\sigma$ is over the same time, then
$d\sigma=dV$, the volume infinitesimal element, and when a space-time
infinitesimal element $d\sigma$ doesn’t move over time, then $d\sigma=dsdt$,
where $ds$ is an infinitesimal surface and $dt$ is an infinitesimal change in
time. Additionally, it is noted that the viscous contribution in (8) is zero
since

$\displaystyle\int_{\partial\Sigma_{\delta}}\nu u^{E}_{ki,j}n_{j}d\sigma$ (20)
$\displaystyle=$
$\displaystyle-\frac{\nu}{4\pi}\int_{0}^{T}\int_{S_{R}}\left[\frac{1}{R}\right]_{,kij}n_{j}dsdt$
$\displaystyle=$ $\displaystyle-\frac{\nu
T}{4\pi}\int_{S_{R}}\left[-\frac{(x_{k}-x_{k}^{\prime})}{R^{3}}\right]_{,ij}n_{j}ds$
$\displaystyle=$ $\displaystyle-\frac{\nu
T}{4\pi}\int_{S_{R}}\left[\frac{R^{3}\delta_{ki}-3R(x_{i}-x_{i}^{\prime})(x_{k}-x_{k}^{\prime})}{R^{6}}\right]_{,j}n_{j}ds$
$\displaystyle=$ $\displaystyle-\frac{\nu
T}{4\pi}\int_{S_{R}}\left[\frac{-3R^{2}\delta_{ki}+9(x_{i}-x_{i}^{\prime})(x_{k}-x_{k}^{\prime})}{R^{6}}\right]ds$
$\displaystyle=$ $\displaystyle-\frac{\nu
T}{4\pi}\int_{S_{R}}\left[\frac{-3R^{2}\delta_{ki}+9\delta_{ki}(x_{1}-x_{1}^{\prime})(x_{1}-x_{1}^{\prime})}{R^{6}}\right]ds$
$\displaystyle=$ $\displaystyle-\frac{\nu
T}{4\pi}\int_{S_{R}}\left[\frac{-3R^{2}\delta_{ki}+3(x_{j}-x_{j}^{\prime})(x_{j}-x_{j}^{\prime})\delta_{ki}}{R^{6}}\right]ds$
$\displaystyle=$ $\displaystyle 0,$

where we have used $x_{k,i}=\delta_{ki}$,
$(x_{j}-x_{j}^{\prime})(x_{j}-x_{j}^{\prime})=R^{2}$, and we have made some
use of the symmetry properties of the integral. So, the Eulerlet also
satisfies the equation for the NSlet (8). However, the Eulerlet is not the
NSlet even though it satisfies the NSlet equation because it does not decay in
the far-field.

Similarly, we can show equation (13) holds since

$\displaystyle\int_{\partial\Sigma_{\delta}}u^{E}_{{\bf k}j}u^{E}_{{\bf
k}i}n_{j}+p^{B}_{k}n_{i}d\sigma$ (21) $\displaystyle=$
$\displaystyle\int_{\partial\Sigma_{\delta}}\phi^{E}_{{\bf k},j}\phi^{E}_{{\bf
k},i}n_{j}-(1/2)\phi^{E}_{{\bf k},j}\phi^{E}_{{\bf k},j}n_{i}d\sigma$
$\displaystyle=$ $\displaystyle 0$

from the symmetry of the surface ${\partial\Sigma_{\delta}}$.

###  4.2 Far from the origin: the Stokeslet

Far from the origin, the velocity decays to zero, and so the quadratic term
disappears giving the time-dependent Stokeslet defined by

$u^{S}_{ki,0}+p^{S}_{k,i}-\nu
u^{S}_{ki,jj}=-\delta\delta_{ki}\;,\;\;u^{S}_{ki,i}=0.$ (22)

We note that this is also the time dependent Oseenlet in a reference frame
moving with velocity $U$, more usually attributed to far-field
representations. The steady Stokeslet solution cannot be used here, as it does
not satisfy the unit force impulse but instead the unit force, as the term
$\delta$ is the Dirac delta function in four-dimensional space time and not in
three-dimensional space. This fundamental solution for the Stokes equation
appears to have been given first by Oseen [26] and later named the Stokeslet.
The Stokeslet solution is given in the appendix from Fourier Transforms
following the approach of Chan and Chwang [15] as

$u^{S}_{ki}=\frac{1}{4\pi\nu}H(x_{0}-x_{0}^{\prime})\left(\frac{{\rm
erf}\eta}{R}\right)_{,0}\delta_{ki}-\frac{1}{4\pi}H(x_{0}-x_{0}^{\prime})\left(\frac{{\rm
erf}\eta}{R}\right)_{,ki}$ (23)

where $\eta=\frac{R}{\sqrt{4\nu(x_{0}-x_{0}^{\prime})}}$, and

$p^{S}_{k}=\frac{\delta(x_{0}-x_{0}^{\prime})}{4\pi}\left[\frac{1}{R}\right]_{,k}$
(24)

for the pressure.

##  5 Formulation of the Boundary Integral

We now follow Green’s Boundary Integral formulation in [8], as used by Oseen
[26]. Oseen starts with the integral given in the theory of hydrodynamic
potentials (described, for example, in [24])

$\int_{\Sigma}\left[f_{i}u_{ki}-f_{ki}u_{i}\right]d\Sigma^{\prime}=0$ (25)

integrated over the primed Cartesian co-ordinate $x^{\prime}_{I}$ such that in
the integrand the terms are functions of the following variables:
$f_{i}(x^{\prime}_{I})$ given by equation (7), $u_{i}(x^{\prime}_{I})$ which
is the fluid velocity, $u_{ki}(x_{I}-x^{\prime}_{I})$ which is the $ith$
component of velocity for the $kth$ NSlet Green’s function centred at
$x_{I}=x^{\prime}_{I}$, and $f_{ki}(x_{I}-x^{\prime}_{I})$ which is given by
equation (8).

Since the NSlet is undefined at $x_{I}=x^{\prime}_{I}$, then this point is
omitted from $\Sigma$. Define $\Sigma$ as consisting of two domains, one
exterior to a surface boundary $S(x^{\prime}_{I})$ between the boundaries
$\partial\Sigma_{\delta}$ and $\partial\Sigma_{X}$, and the other interior to
the surface boundary $S(x^{\prime}_{I})$. Here, the boundary
$\partial\Sigma_{\delta}$ is defined in section 4.1, but now over an
integration of the $x_{I}^{\prime}$ co-ordinates, centred on $x_{I}$.
Similarly, the boundary $\partial\Sigma_{X}$ is defined such that it consists
of the set of points $x^{\prime}_{I}$ including: those such that
$(x^{\prime}_{j}-x_{j})(x^{\prime}_{j}-x_{j})=X^{2}$ and $-X\leq
x^{\prime}_{0}<0$ or $0<x^{\prime}_{0}\leq X$, those such that
$0\leq(x^{\prime}_{j}-x_{j})(x^{\prime}_{j}-x_{j})\leq X^{2}$ and
$x^{\prime}_{0}=X$, and finally those such that
$0\leq(x^{\prime}_{j}-x_{j})(x^{\prime}_{j}-x_{j})\leq X^{2}$ and
$x^{\prime}_{0}=-X$, for some values of $X$ large. The domain $\Sigma$ is
represented by the figure 3.

Figure 3: The space-time domain $\Sigma$

Following [8], using the fact that
$[g(x_{I}-x^{\prime}_{I})]_{;I}=-[g(x_{I}-x^{\prime}_{I})]_{,I}$, for a
function $g$, the integrand is rewritten as

$\displaystyle f_{i}u_{ki}-f_{ki}u_{i}$ $\displaystyle=$
$\displaystyle\left\\{u_{i;0}+u_{j}u_{i;j}+p_{;i}-\nu u_{i;jj}\right\\}u_{ki}$
(26) $\displaystyle-\left\\{u_{ki,0}+u_{{\bf k}j}u_{{\bf k}i,j}+p_{k,i}-\nu
u_{ki,jj}\right\\}u_{i}$ $\displaystyle=$
$\displaystyle\left\\{u_{i;0}+u_{j}u_{i;j}+p_{;i}-\nu u_{i;jj}\right\\}u_{ki}$
$\displaystyle-\left\\{-u_{ki;0}-u_{{\bf k}j}u_{{\bf k}i;j}-p_{k,i}-\nu
u_{ki;jj}\right\\}u_{i}$ $\displaystyle=$
$\displaystyle(u_{i}u_{ki})_{;0}+(pu_{ki})_{;i}+(p_{k}u_{i})_{;i}$
$\displaystyle+(\nu u_{ki;j}u_{i}-\nu u_{i;j}u_{ki})_{;j}$
$\displaystyle+u_{j}u_{i;j}u_{ki}+(u_{{\bf k}j}u_{{\bf
k}i}u_{i})_{;j}-u_{i;j}u_{{\bf k}i}u_{{\bf k}j}$ $\displaystyle=$
$\displaystyle\left[u_{i}u_{ki}\delta_{0J}+pu_{kj}+p_{k}u_{j}\right.$
$\displaystyle\left.+\nu u_{ki;j}u_{i}-\nu u_{i;j}u_{ki}+u_{{\bf k}j}u_{{\bf
k}i}u_{i}\right]_{;J}$ $\displaystyle+u_{j}u_{i;j}u_{ki}-u_{i;j}u_{{\bf
k}i}u_{{\bf k}j}$ $\displaystyle=$
$\displaystyle\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}\right]_{;J}+Q_{k},$

where $\tau_{ij}=-p\delta_{ij}+\nu u_{i;j}$ is the stress tensor, and $Q_{k}$
is the quadratic term contribution (in the sense that it originates from the
quadratic term present in the differential equations (7) and (8)) defined by

$\displaystyle Q_{k}$ $\displaystyle=$ $\displaystyle
u_{j}u_{i;j}u_{ki}-u_{i;j}u_{{\bf k}i}u_{{\bf k}j}$ (27) $\displaystyle=$
$\displaystyle u_{j}u_{i;j}u_{ki}-[u_{i}u_{{\bf k}j}u_{{\bf
k}i}]_{;j}+u_{i}u_{{\bf k}j}u_{{\bf k}i;j}$ $\displaystyle=$ $\displaystyle
u_{j}u_{i;j}u_{ki}+u_{i}u_{{\bf k}j}(u_{{\bf k}i;j}-u_{{\bf k}j;i})$
$\displaystyle-[u_{i}u_{{\bf k}j}u_{{\bf k}i}]_{;j}+[(1/2)u_{i}u_{{\bf
k}j}u_{{\bf k}j}]_{;i}$ $\displaystyle=$ $\displaystyle
u_{j}u_{i;j}u_{ki}+u_{i}u_{{\bf k}j}(u_{{\bf k}i;j}-u_{{\bf k}j;i})$
$\displaystyle-[u_{i}u_{{\bf k}j}u_{{\bf k}i}-(1/2)u_{j}u_{{\bf k}i}u_{{\bf
k}i}]_{;j},$

###  5.1 The non-linear potential

The quadratic term $Q_{k}$ is then given in terms of the divergence of a
vector field $q_{kj}$ directed on a radial spoke where

$Q_{k}=q_{kj;j}$ (28)

and

$q_{kj}=W_{k}\hat{R}_{j}+q^{*}_{kj}$ (29)

such that $\hat{R}_{j}=\frac{x_{j}^{\prime}-x_{j}}{R}$ where $R=\lvert
x_{j}^{\prime}-x_{j}\rvert$ and $q^{*}_{kj}=(1/2)u_{j}u_{{\bf k}i}u_{{\bf
k}i}-u_{i}u_{{\bf k}j}u_{{\bf k}i}$. (It is noted that defining the quadratic
term is this way doesn’t imply any loss of generality and can be applied to
any differential equation problem, but it just happens to work out in this
particular case because the quadratic term disappears across boundaries and
because the evaluation of the integrals involving the quadratic term do not
contribute asymptotically close to and far from the point, see sections 6 and
7. This may not necessarily work out for other non-linear partial differential
equations.) Therefore the tensor $W_{k}\hat{R}_{j}$ has value $W_{k}$ directed
along the radial spoke $\hat{R}_{j}$, see figure 4,

Figure 4: The integral path of the non-linear potential

and so

$\displaystyle[W_{k}\hat{R}_{j}]_{;j}$ $\displaystyle=$ $\displaystyle
u_{j}u_{i;j}u_{ki}+u_{i}u_{{\bf k}j}(u_{{\bf k}i;j}-u_{{\bf k}j;i})$ (30)
$\displaystyle=$ $\displaystyle g_{k}(x_{I},x^{\prime}_{I}).$

This is the divergence of a four dimensional vector in space-time, and so
expressing in the co-ordinate system $(x_{0},R,\theta,\alpha)$ where
$(R,\theta,\alpha)$ are spherical co-ordinates in space, gives

$[W_{k}\hat{R}_{j}]_{;j}=\frac{1}{R^{2}}\frac{\partial}{\partial
R}\left\\{R^{2}W_{k}\right\\}=g_{k}.$ (31)

It is noted that in the linear (Stokes and Oseen) approach given by Oseen
[26], a potential is defined which is a divergence of physical variables of
the fundamental solution. Oseen then integrates along the radial variable of
integration to give an integral expression for this potential. Here, we use a
similar approach but for this non-linear potential $W_{k}$. So, integrating
gives

$W_{k}({\bf x},{\bf
x^{\prime}})=-\frac{1}{R^{2}}\int_{R_{X}}^{R}R^{*2}g_{k}({\bf x},{\bf
x^{*}})dR^{*},$ (32)

where $R_{X}$ is the radial distance to the outer boundary
$\partial\Sigma_{X}$ of the domain $\Sigma$, the variable of integration is
$R^{*}=\lvert x^{*}_{i}-x_{i}\rvert$, such that
$x_{i}^{*}-x_{i}=R^{*}\hat{R}_{i}$, $x_{0}^{*}=x_{0}^{\prime}$, and when
$R^{*}=R$ then ${\bf x^{*}}={\bf x}^{\prime}$, see figure 4.

So, by putting the results from (26) and (27) into (25), we can apply the
divergence theorem to get a boundary integral formulation

$\displaystyle\int_{\Sigma}\left[f_{i}u_{ki}-f_{ki}u_{i}\right]d\Sigma^{\prime}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]_{;J}d\Sigma^{\prime}$
(33) $\displaystyle=$
$\displaystyle\int_{\partial\Sigma_{Coll}}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]n_{J}^{\Sigma}d\sigma^{\prime},$
$\displaystyle=$ $\displaystyle 0$

where $n_{J}^{\Sigma}$ is the outward pointing normal to $\Sigma$, and
$\partial\Sigma_{Coll}$ is the collection of boundaries enclosing both the
exterior and interior domains that constitute $\Sigma$. The first term in the
integrand yields an integral associated with the double layer potential, the
second term in the integrand with the single layer potential, see for example
the theory of hydrodynamic potentials given in Ladyzhenskaya [24]. The third
term in the integrand is given by the non-linear potential that depends on
integration of quantities within the whole domain. The ansatz proof assumes
that the physical variables are all smooth, and so the divergence theorem
applied to the non-linear term to get the boundary representation is possible
since the term is assumed to be smooth. There may be other solutions that are
not $C^{\infty}$ smooth due to the non-elliptic nature of the Navier-Stokes
equation, but these other possible solutions are not considered here and
instead it is only investigated whether a smooth solution exists. This
representation also has similarities with the dual-reciprocity method used for
boundary integral representations for non-linear partial differential
equations, except that the dual-reciprocity method approximates the non-linear
domain term by a series expansion of known functions, such as radial basis
functions, whose divergence is known and so can then be represented over the
boundary.

The whole concept of the boundary integral framework requires linearity, and
the non-linear third term in (33) appears to preclude this. However, in the
subsequent analysis we shall see that a representation is found such that the
non-linear term (as well as the double-layer potential) drops out thus
justifying its use.

##  6 Evaluation of the Boundary Integral

###  6.1 Boundary integral evaluation in the vicinity of the point

A similar approach to Chadwick [8] is given, but here in greater detail. We
calculate part of the integral in (33) which is around the integral path
${\partial\Sigma_{\delta}}$ in the vicinity of the point, since here the NSlet
approximates to the Eulerlet. From (33), the contribution to this integral
from the integral path ${\partial\Sigma_{\delta}}$ is

$\displaystyle I_{\delta}$ $\displaystyle=$
$\displaystyle\int_{\partial\Sigma_{\delta}}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]n^{\Sigma}_{J}d\sigma^{\prime}$
$\displaystyle=$
$\displaystyle-\int_{\partial\Sigma_{\delta}}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]n_{J}d\sigma^{\prime}$
$\displaystyle=$
$\displaystyle-\int_{\partial\Sigma_{\delta}}\left[u_{i}(u_{ki}\delta_{0J}+u_{{\bf
k}j}u_{{\bf k}i}+p_{k}\delta_{ij}-\nu u_{ki,j})\right.$
$\displaystyle\left.-u_{ki}\tau_{ij}+(1/2)u_{j}u_{{\bf k}i}u_{{\bf
k}i}-u_{i}u_{{\bf k}j}u_{{\bf
k}i}+W_{k}\hat{R}_{j}\right]n_{J}d\sigma^{\prime}$ $\displaystyle=$
$\displaystyle-\int_{\partial\Sigma_{\delta}}\left[u_{i}u_{ki}\delta_{0J}+u_{j}p_{k}-\nu
u_{i}u_{ki,j}\right.$ $\displaystyle\left.-u_{ki}\tau_{ij}+(1/2)u_{j}u_{{\bf
k}i}u_{{\bf k}i}+W_{k}\hat{R}_{j}\right]n_{J}d\sigma^{\prime}.$

Only the first term in the integrand involves the normal component $n_{0}$,
all others have the factor $n_{j}$. For $n_{j}$, then $R>0$ over the boundary
and so

$\displaystyle p_{k}$ $\displaystyle=$ $\displaystyle p_{k}^{B}+p_{k}^{E}$
(35) $\displaystyle=$ $\displaystyle-(1/2)u^{E}_{{\bf k}j}u^{E}_{{\bf
k}j}+\frac{\delta(x_{0})}{4\pi}\left[\frac{1}{R}\right]_{,k}$

where $u_{ki}$ tends towards $u^{E}_{ki}$. So, the integral evaluation becomes

$\displaystyle I_{\delta}$ $\displaystyle=$
$\displaystyle-\int_{\partial\Sigma_{\delta}}\left[u_{i}u^{E}_{ki}\delta_{0J}+u_{j}p^{E}_{k}-\nu
u_{i}u^{E}_{ki,j}\right.$
$\displaystyle\left.-u^{E}_{ki}\tau_{ij}+W_{k}\hat{R}_{j}\right]n_{J}d\sigma^{\prime},$

Evaluate the integral of each term in the integrand of (LABEL:ER:integral) in
turn. We see that from (19) and (11), the first two terms
$u_{i}u^{E}_{ki}\delta_{0J}+u_{j}p^{E}_{k}$ give an integral contribution
$-u_{k}$. Similarly, from (20), the third term $-\nu u_{i}u^{E}_{ki,j}$ gives
a zero leading order integral contribution; That is, the leading order in the
Taylor series expansion of the velocity

$u_{i}({\bf x^{\prime}})=u_{i}({\bf x})+u_{i,q}({\bf
x})(x^{\prime}_{q}-x_{q})+u_{i,ql}({\bf
x})\frac{(x^{\prime}_{q}-x_{q})(x^{\prime}_{l}-x_{l})}{2!}+..$ (37)

The Taylor series expansion can be used because we are looking for a solution
for the class of variables in $C^{\infty}$ for which the Taylor series can be
employed. The next order in the expansion gives a contribution

$\displaystyle\nu
u_{i,q}\int_{\partial\Sigma_{\delta}}u_{ki,j}x_{q}n_{j}d\sigma$ (38)
$\displaystyle=$
$\displaystyle-\nu\frac{u_{i,q}T}{4\pi}\int_{S}\left\\{-\frac{3\delta_{ki}}{R^{4}}+9\frac{(x_{i}-x_{i}^{\prime})(x_{k}-x_{k}^{\prime})}{R^{6}}\right\\}(x_{q}-x_{q}^{\prime})ds$
$\displaystyle=$ $\displaystyle 0$

from symmetry. The subsequent orders in the Taylor series expansion are all at
most of order O$(T)\rightarrow 0$.

The next term in the integrand of (LABEL:ER:integral) is
$-u^{E}_{ki}\tau_{ij}$ and gives an integral contribution

$\displaystyle\int_{\partial\Sigma_{\delta}}u^{E}_{ki}\tau_{ij}n_{J}d\sigma^{\prime}$
(39) $\displaystyle=$
$\displaystyle\tau_{ij}\int_{\partial\Sigma_{\delta}}u^{E}_{ki}n_{J}d\sigma^{\prime}$
$\displaystyle=$
$\displaystyle-\tau_{ij}\int_{\partial\Sigma_{\delta}}u^{E}_{ki}n_{J}d\sigma$
$\displaystyle=$
$\displaystyle-\tau_{ij}\int_{0}^{T}\int_{S}u^{E}_{ki}n_{J}dsdt$
$\displaystyle=$
$\displaystyle\frac{\tau_{ij}T}{4\pi}\int_{S}\left[\frac{1}{R}\right]_{,ki}n_{J}ds$
$\displaystyle=$
$\displaystyle\frac{\tau_{ij}T}{4\pi}\int_{S}\left[\frac{R^{3}\delta_{ki}-3R(x_{i}-x_{i}^{\prime})(x_{k}-x_{k}^{\prime})}{R^{6}}\right]\frac{(x_{j}-x_{j}^{\prime})}{R}ds$
$\displaystyle=$ $\displaystyle 0$

since subsequent terms in the Taylor series expansion are of order
O$(T)\rightarrow 0$.

Finally, the last term is

$-\int_{\partial\Sigma_{\delta}}W_{k}\hat{R}_{j}n_{J}d\sigma^{\prime}=-\int_{T}\int_{S}W_{k}\hat{R}_{j}n_{J}d\sigma^{\prime},$
(40)

and $R>0$ on the surface $S$, so $u^{E}_{{\bf k}i,j}=u^{E}_{{\bf k}j,i}$ and
$q^{E}_{k}=\frac{1}{R^{2}}\int_{R_{X}}^{R}R^{*2}g_{k}^{E}dR^{*}$. This
integral is split into three parts as follows

$W_{k}=-\frac{1}{R^{2}}\left\\{\int_{R_{X}}^{R_{L}}+\int_{R_{L}}^{R_{\epsilon}}+\int_{R_{\epsilon}}^{R}\right\\}R^{*2}g_{k}dR^{*}=I_{1}+I_{2}+I_{3},$
(41)

where $R_{L}$ is a sufficiently large constant such that for $R^{*}>R_{L}$
then the NSlet approximates to the Stokeslet $u_{ki}\sim u_{ki}^{S}$, and
where $R_{\epsilon}$ is a sufficiently small constant such that for
$R^{*}<R_{\epsilon}$ then the NSlet approximates to the Eulerlet $u_{ki}\sim
u_{ki}^{E}$. Each of the three integrals are calculated in turn.

First, it is seen that for $R_{X}$ sufficiently large then $u_{ki}\sim
u_{ki}^{S}\sim\frac{1}{R^{*3}}$. This means that from (30)
$g_{k}\sim\frac{1}{R^{*3}}$ and so, from (41), $I_{1}\sim 1/R^{2}\ln R_{X}$.

Second, it is seen that between the two constant limits $R_{L}$ and
$R_{\epsilon}$ then $u_{ki}\sim constant$ (bounded). This means that from
(41), $I_{2}\sim 1/R^{2}$.

Third, it is seen that for $R$ sufficiently small then $u_{ki}\sim
u_{ki}^{E}\sim\frac{1}{R^{*3}}$ over the spherical surface radius $R$. (There
is only a contribution from the spherical surface because there is no
contribution when $J=0$, since from (40) $\hat{R}_{j}n_{J}=\hat{R}_{j}n_{j}$.)
This means that from (30) $g_{k}\sim\frac{1}{R^{*3}}$ and so, from (41),
$I_{3}\sim 1/R^{2}\ln R$.

So, from (41), $I_{1}\sim 1/R^{2}\ln R_{X}$, $I_{2}\sim 1/R^{2}$ and
$I_{3}\sim 1/R^{2}\ln R$, and so the integral evaluation of (40) where
boundary $\partial\Sigma_{\delta}$ has time interval $T$ and spherical radius
$R$ small is

$\int_{\partial\Sigma_{\delta}}W_{k}\hat{R}_{j}n_{J}d\sigma^{\prime}\sim
TMaximum(\lvert\ln R_{X}\rvert,\lvert\ln R\rvert).$ (42)

Choosing $T$ to be sufficiently small, of smaller order than
$Maximum(\lvert\ln R_{X}\rvert,\lvert\ln R\rvert)$ ensures that the integral
contribution disappears in the limit. For example, we can choose $R_{X}\sim
1/R$, $T\sim R$, and as $R\rightarrow 0$, then the integral contribution goes
as $R\ln R\rightarrow 0$.

So, bringing all the terms together gives the integral contribution

$I_{\delta}=-u_{k}.$ (43)

###  6.2 Boundary integral evaluation far from the point

In the far-field around $\partial\Sigma_{X}$, the time interval $T$ and radius
$R$ are large, the NSlet tends towards the Stokeslet (which is the Oseenlet in
a moving reference frame), and this contribution is shown to be zero next. In
the far-field, we have the contribution $u_{ki}\sim
u_{ki}^{S}\sim\frac{1}{R^{*3}}$, and $u_{i}\sim
u_{i}^{S}\sim\frac{1}{R^{*3}}$. So $g_{k}\sim\frac{1}{R^{*10}}$, and therefore
$W_{k}\sim\frac{1}{R^{2}}\frac{1}{R^{7}}=\frac{1}{R^{9}}$. So, the far-field
integral is

$\displaystyle\lvert I_{X}\rvert$ $\displaystyle=$
$\displaystyle\lvert\int_{\partial\Sigma_{X}}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]n_{J}^{\Sigma}d\sigma^{\prime}\rvert$
$\displaystyle\sim$
$\displaystyle\lvert\int_{\partial\Sigma_{X}}\left[u_{i}f^{S}_{kiJ}-u^{S}_{ki}\tau_{ij}+q^{S}_{kj}\right]n_{J}^{\Sigma}d\sigma^{\prime}\rvert$

However,
$\int_{\partial\Sigma_{X}}u_{i}f^{S}_{kiJ}n^{\Sigma}_{J}d\sigma^{\prime}\sim
R^{3}(1/R^{3})(1/R^{2})$ for a leading order pressure term decay $1/R^{2}$ and
the time integral $T$ being chosen of order $R$ large. Similarly,
$\int_{\partial\Sigma_{X}}u_{ki}^{S}\tau_{ij}n^{\Sigma}_{j}d\sigma^{\prime}\sim
R^{3}(1/R^{3})(1/R^{2})$. The integral contribution
$\int_{\partial\Sigma_{X}}q_{ki}^{S}n^{\Sigma}_{j}d\sigma^{\prime}\sim
R^{3}(1/R^{9})$. (The contribution to $q_{kj}$ from the term
$W_{k}\hat{R}_{j}$ in (29) is identically zero because $R=R_{X}$ and so
$W_{k}$ is identically zero from (32), which means that $g_{k}=0$ from (30).)
So

$I_{X}\rightarrow 0.$ (45)

##  7 Velocity representation

We now give an NSlet representation for the boundary integral (33). In the
near-field around $\partial\Sigma_{\delta}$, the NSlet tends towards the
Eulerlet and gives a contribution $-u_{k}$, see (43). In the far-field around
$\partial\Sigma_{X}$, the NSlet tends towards the Stokeslet and gives zero
contribution. Therefore, putting these results into (33) then gives the
velocity by the boundary integral representation

$u_{k}=\int_{\partial\Sigma}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]n_{J}^{\Sigma}d\sigma^{\prime}$
(46)

for some boundary $\partial\Sigma$. The equation (46) demonstrates why a
representation for the velocity is valid here, because the calculation in the
vicinity of the point (43) gives the velocity, and far from the point (45)
gives zero. It is noted that the first term in the integrand has a factor
$f_{kiJ}$ which contains the fundamental solution stress term giving the
integral for the double layer potential, the second term in the integrand has
a factor $u_{ki}$ which is the fundamental solution velocity term giving the
integral for the single layer potential.

###  7.1 Initial Value Problem boundary

Consider the variables stipulated at time $x_{0}=0$, and zero before this
time. Let the flow be instantaneously set into motion at $x_{0}=0$ initial
time by applying force impulses throughout the compact spherical volume
$V_{0}$ of radius $X_{0}$. So, every variable may be discontinuous across this
space-time boundary. The outward pointing normal is
$n_{I}^{\Sigma}=-\hat{x}_{0}$, where $\hat{x}_{0}$ is the unit vector in the
time direction. So, from (46)

$u_{k}=\int_{V}u_{i}f_{ki0}n_{0}^{\Sigma}dV^{\prime}=-\int_{V}u_{i}u_{ki}dV^{\prime},$
(47)

where the space-time boundary $\partial\Sigma$ is chosen as the volume
$V=\lim_{X_{0}\rightarrow\infty}V_{0}$ at time $x_{0}=0$. It is noted that the
representation (47) contains a single layer potential only. We note that
throughout the volume $V$ at time $x_{0}=0$, the velocity $u_{i}$ has been
stipulated as an initial condition, and so is known. Therefore, if this
representation exists then it will be unique in $C^{\infty}$, as substituting
the homogeneous boundary condition $u_{i}=0$ in the integrand gives the
velocity $u_{k}=0$ for all subsequent time. To ascertain whether the
representation exists, we choose a different boundary for $\partial\Sigma$ in
the domain $\Sigma$.

###  7.2 A general force impulse boundary $\partial\Sigma$

Consider the velocity representation (46) for any general closed force impulse
hyper-surface $\partial\Sigma$ in the domain $\Sigma$. The hyper-surface
divides the fluid domain into two, an external flow field exterior to
$\partial\Sigma$ and an internal flow field interior to $\partial\Sigma$, see
figure 5.

Figure 5: The flow field either side of the boundary $\partial\Sigma$

Let the fluid velocity $u_{i}$ for the exterior problem (exterior to
$\partial\Sigma$) and the fluid velocity $u_{i}^{-}$ for the interior problem
(interior to $\partial\Sigma$) be the same over $\partial\Sigma$. So, this is
equivalent to a body boundary condition for a moving exterior problem, with
the same body boundary condition for a moving interior problem. Thus, $u_{i}$
is continuous across (moving from one side to the other) the boundary, since
on either side of the boundary the fluid has the same velocity. Furthermore,
the Green’s functions become singular at their origin, but are continuous
across the rest of the flow field and so are continuous across this boundary.
However, the fluid stresses may be discontinuous, and this implies that forces
are applied on the fluid over this boundary. This means that $u_{i}$ and
$q_{kj}$ are continuous across the boundary, but $\tau_{ij}$ is discontinuous.
Therefore, over the external surface the velocity is given by

$u_{k}=\int_{\partial\Sigma}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}+q_{kj}\right]n_{J}^{\Sigma}d\sigma^{\prime}.$
(48)

Over the internal surface with interior values denoted by the superscript -,
the boundary integral is given by

$0=\int_{\partial\Sigma}\left[u_{i}f_{kiJ}-u_{ki}\tau_{ij}^{-}+q_{kj}\right]n_{J}^{\Sigma}d\sigma^{\prime}.$
(49)

The space-time boundary can be rewritten in terms of a time integral over a
surface integration in the following way. The outward pointing normal to the
boundary $n_{I}$ is given by $n_{I}=-n_{I}^{\Sigma}$. Let the angle $\alpha$
be the angle this normal makes with the outward pointing normal to the surface
$S$ at time $x_{0}$ denoted by $n_{i}^{S}$, see figure 6.

Figure 6: An element of the boundary

Then, $n_{j}=n_{j}^{S}\cos\alpha$ and
$d\sigma^{\prime}\cos\alpha=dx_{0}^{\prime}ds^{\prime}$ where $ds^{\prime}$ is
a surface element at time $x_{0}$. Combining these, we get
$n_{j}d\sigma^{\prime}=n_{j}^{S}dx_{0}^{\prime}ds^{\prime}$. So, putting these
into the combined equations (48) and (49), gives

$\displaystyle u_{k}$ $\displaystyle=$
$\displaystyle-\int_{\partial\Sigma}u_{ki}(\tau_{ij}-\tau_{ij}^{-})n_{J}^{\Sigma}d\sigma^{\prime}$
(50) $\displaystyle=$
$\displaystyle\int_{\partial\Sigma}u_{ki}(\tau_{ij}-\tau_{ij}^{-})n_{j}d\sigma^{\prime}$
$\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{S}u_{ki}(\tau_{ij}-\tau_{ij}^{-})n_{j}^{S}ds^{\prime}dx_{0}^{\prime}$

for the velocity $u_{k}$ at position $x_{j}$ and time $t$. Note that in the
same way that the double layer potential drops out because there is no flux
across the surface which is well-known from Stokes and Oseen boundary integral
formulations, the quadratic inertia term $q_{kj}$ also cancels and so drops
out. Again, the velocity representation (50) is given in terms of the single
layer potential only. The distribution strength can be expressed in terms of
the force impulse distribution, given next.

##  8 Force impulse

The force impulse $J_{i}$ exerted over a boundary $\partial\Sigma$ due to the
fluid motion is defined as

$J_{i}=\int_{\partial\Sigma}F_{i}d\sigma^{\prime}\;,$ (51)

where $F_{i}=(f_{iJ}-f^{-}_{iJ})n_{J}^{\Sigma}$, and
$f_{iJ}^{-}=u_{i}^{-}\delta_{0J}+u_{j}^{-}u_{i}^{-}+p^{-}\delta_{ij}-\nu
u_{i;j}^{-}$ is called the complement to the flow on the other side of the
boundary $\partial\Sigma$ to the flow field with velocity $u_{i}$ and pressure
$p$. The exterior domain $\Sigma$ is bounded by $\partial\Sigma$ and
$\partial\Sigma_{\infty}$, a boundary infinitely distant, and the interior
domain $\Sigma^{-}$ is bounded by $\partial\Sigma$, see figure 5. Then

$J_{i}=\int_{\partial\Sigma}F_{i}d\sigma^{\prime}=\int_{\partial\Sigma}(f_{iJ}-f_{iJ}^{-})n_{J}^{\Sigma}d\sigma^{\prime}$
(52)

since
$\int_{\partial\Sigma}f_{iJ}^{-}n_{J}^{\Sigma}d\sigma^{\prime}=-\int_{\Sigma^{-}}f_{iJ;J}^{-}d\Sigma^{\prime}=0$.
Similarly,
$\int_{\partial\Sigma+\partial\Sigma_{\infty}}f_{iJ}n_{J}^{\Sigma}d\sigma^{\prime}=\int_{\Sigma}f_{iJ;J}d\Sigma^{\prime}=0$
and so

$J_{i}=-\int_{\partial\Sigma_{\infty}}f_{iJ}n_{J}^{\Sigma}d\sigma^{\prime}.$
(53)

Now consider moving the boundary of domain $\partial\Sigma$ into the domain
$\Sigma$ to $\partial\Sigma^{*}$, so that $u_{i}$, $p$ are given in a new
domain $\Sigma^{*}$, only and that force impulses are positioned over
$\partial\Sigma^{*}$ creating a new complementary flow field $u_{i}^{*-}$,
$p^{*-}$ within domain $\Sigma^{*-}$, see figure 7.

Figure 7: The flow field either side of the boundary $\partial\Sigma^{*}$

Then in the exterior flow field we have
$J_{i}=-\int_{\partial\Sigma_{\infty}}f_{iJ}n_{J}^{\Sigma^{*}}d\sigma^{\prime}=\int_{\partial\Sigma^{*}}f_{iJ}n_{J}^{\Sigma^{*}}d\sigma^{\prime}$.
Furthermore, inside the interior flow field
$\int_{\partial\Sigma^{*}}f_{iJ}^{-}n_{J}^{\Sigma^{*}}d\sigma^{\prime}=0$, and
so

$J_{i}=\int_{\partial\Sigma^{*}}(f_{iJ}-f_{iJ}^{-})n_{J}^{\Sigma^{*}}d\sigma^{\prime}=\int_{\partial\Sigma^{*}}F_{i}d\sigma^{\prime}.$
(54)

So, $J_{i}$ is independent of the boundary $\partial\Sigma$ and any boundary
within $\Sigma$ may be taken, and (51) still holds.

###  8.1 The force impulse over the surface

Over the general surface $\partial\Sigma$, the velocity is continuous, and so
$u_{i}=u_{i}^{-}$ on $\partial\Sigma$. So

$\displaystyle f_{iJ}-f_{iJ}^{-}$ $\displaystyle=$
$\displaystyle(p-p^{-})\delta_{ij}-\nu(u_{i;j}-u_{i;j}^{-})$ (55)
$\displaystyle=$ $\displaystyle-(\tau_{ij}-\tau_{ij}^{-}),$

where the stresses either side of $\partial\Sigma$ are defined as
$\tau_{ij}=-p\delta_{ij}+\nu u_{i;j}$ and $\tau_{ij}^{-}=-p^{-}\delta_{ij}+\nu
u_{i;j}^{-}$. So, over the surface, the force impulse is

$\displaystyle J_{i}$ $\displaystyle=$
$\displaystyle\int_{\partial\Sigma}F_{i}d\sigma^{\prime}=-\int_{\partial\Sigma}(\tau_{ij}-\tau_{ij}^{-})n_{j}^{\Sigma}d\sigma^{\prime}$
(56) $\displaystyle=$
$\displaystyle\int_{\partial\Sigma}(\tau_{ij}-\tau_{ij}^{-})n_{j}d\sigma^{\prime}$

where $n_{j}=-n_{j}^{\Sigma}$ is the outward pointing normal from the surface.
So, over the surface $J_{i}$ is the force impulse generated by the fluid (both
internal and external) acting over the surface over time.

###  8.2 The force impulse over an initial time

For an initial value problem, suppose at an initial time the flow is
impulsively started from rest. In this case $u_{i}^{-}=p^{-}=0$ and so
$f_{iJ}^{-}=0$, and also $n_{J}^{\Sigma}=-\delta_{0J}$. So
$f_{iJ}n_{J}^{\Sigma}=-u_{i}^{o}$, where $u_{i}^{o}$ is the velocity $u_{i}$
at initial time $x_{0}=0$. The initial value space-time boundary is a volume
at initial time, and so the force impulse is

$J_{i}=\int_{\partial\Sigma}F_{i}d\sigma^{\prime}=-\int_{V}u_{i}^{o}dV^{\prime}.$
(57)

So, the fluid has been set into motion by a force impulse $-J_{i}$ acting over
the whole volume $V$ at initial time $x_{0}=0$.

##  9 Existence, Smoothness and Uniqueness

So, over any boundary $\partial\Sigma$ within the domain the velocity can be
expressed as

$u_{k}=\int_{\partial\Sigma}F_{i}u_{ki}d\sigma^{\prime}$ (58)

from (50) since $F_{i}=-(\tau_{ij}-\tau_{ij}^{-})n_{J}^{\Sigma}$, where the
force impulse is $J_{i}=\int_{\partial\Sigma}F_{i}d\sigma^{\prime}$ from (51)
and is bounded. (So, if the force impulse is scaled, for example doubled, then
the velocity is scaled in the same way. Although this appears to give a
contradiction with observed motions, it doesn’t. To understand this, consider
the simplified case of steady flow $U$ past a fixed body. The velocity
boundary condition must also be satisfied which is also proportional to $U$
giving a further doubling, and so for classical inviscid flows the force is
proportional to $U^{2}$ as expected, but not for general Navier-Stokes flows.
This is because the steady NSlet is a complex function of $U$ as well, and so
doubling $U$ doubles the Reynolds number and consequently the relative ratio
of the inertia and viscous terms in the NSlet equation which is a complex
change not given by any simple proportion, as expected from observed motions.)

Through appropriate topological deformation let the boundary
$\partial\Sigma=\partial\Sigma_{\epsilon}+\partial\Sigma_{\infty}$, where the
fluid variables $u_{i}$ and $p$ are defined inside $\partial\Sigma_{\epsilon}$
which surrounds the fluid point $x_{I}$, and outside the far-field boundary
$\partial\Sigma_{\infty}$. In particular $\partial\Sigma_{\epsilon}$ is
parameterised by $x_{I}^{\prime}$ which lies on the spherical cylinder hyper-
surface given by $\lvert x_{i}-x_{i}^{\prime}\rvert=\epsilon$, $t-\Delta t\leq
x_{0}^{\prime}\leq t$, and $0\leq\lvert
x_{i}-x_{i}^{\prime}\rvert\leq\epsilon$, $x_{0}^{\prime}=t-\Delta t$ and
$x_{0}^{\prime}=t$, see figure 8.

Figure 8: The boundary force impulse for a point enclosed by
$\partial\Sigma_{\delta}$

###  9.1 Existence

In the far-field, the NSlet fundamental solution tends to the Stokeslet whose
velocity and stresses decay to zero at least as $1/R^{3}$. So, the fluid
velocity and stresses decay in the same way and the far-field contribution in
(58) is zero. So, (58) becomes

$\displaystyle u_{k}({\bf x},t)$ $\displaystyle=$
$\displaystyle\int_{\partial\Sigma_{\epsilon}}F_{i}u_{ki}d\sigma^{\prime}$
(59) $\displaystyle=$ $\displaystyle u_{k}({\bf x},t-\Delta
t)+\int_{\partial\Sigma_{\epsilon*}}F_{i}u_{ki}d\sigma^{\prime}$
$\displaystyle=$ $\displaystyle u_{k}({\bf x},t-\Delta
t)+\int_{\partial\Sigma_{\epsilon*}}F_{i}u_{ki}^{E}d\sigma^{\prime}(1+{\rm
O}(f(\epsilon)).$

The boundary $\partial\Sigma_{\epsilon*}$ is defined by the part of the
spherical cylinder hyper-surface $\lvert x_{i}-x_{i}^{\prime}\rvert=\epsilon$,
$t-\Delta t\leq x_{0}^{\prime}\leq t$ only. It is close to the point and so
the (integral flux of the) NSlet $u_{ki}$ approximates to the (integral flux
of the) Eulerlet $u_{ki}^{E}$ for small $\epsilon,\Delta t>0$. So the function
$f(\epsilon)$ is finite and small for small $\epsilon>0$.

Using the Cauchy-Schwarz inequality

$\displaystyle\lvert u_{k}({\bf x},t)\rvert$ $\displaystyle\leq$
$\displaystyle\lvert u_{k}({\bf x},t-\Delta t)\rvert$
$\displaystyle+\left\\{\int_{\partial\Sigma_{\epsilon*}}\lvert F_{i}({\bf
x^{\prime}},t^{\prime})\rvert d\sigma^{\prime}\right\\}\lvert u_{ki}^{E}({\bf
x-x^{\prime}},t-t^{\prime})\rvert^{\rm max}(1+{\rm O}(f(\epsilon))).$

The integration $\int_{\partial\Sigma_{\epsilon*}}F_{i}d\sigma^{\prime}$ can
be evaluated over patches of this boundary $\partial\Sigma$ independently one
from the other. The integration
$\int_{\partial\Sigma_{\epsilon*}}F_{i}d\sigma^{\prime}$ is the finite force
impulse $J_{i}$, and so we expect that integrations over a patch of this
boundary where $F_{i}$ is positive will be finite, and similarly where it is
negative. Therefore, $\int_{\partial\Sigma_{\epsilon*}}\lvert F_{i}\rvert
d\sigma^{\prime}$ is finite. As well, $\lvert u_{ki}^{E}({\bf
x^{\prime}-x},t-t^{\prime})\rvert^{\rm max}$ is finite. So, if $u_{k}({\bf
x},t-\Delta t)$ is finite then $u_{k}({\bf x},t)$ is as well. It is noted that
a small finite fixed $\epsilon$ and $\Delta t$ give a fixed value for $\lvert
u_{ki}^{E}({\bf x^{\prime}-x},t-t^{\prime})\rvert^{\rm max}$ and $f(\epsilon)$
that approximates the NSlet by the Eulerlet, and this is independent of fluid
properties at $x_{I}$. So from the initial condition, $u_{k}({\bf x},0)$ is
finite and this provides a continuation to show $u_{k}({\bf x},\Delta t)$ is
finite, and repeating with the same $\epsilon$ and $\Delta t$ provides a
continuation to show $u_{k}({\bf x},2\Delta t)$ is finite. Continuing
repeating in this way enables us to show that $u_{k}({\bf x},t)$ is finite
everywhere in the domain.

###  9.2 Smoothness

Consider the first derivative $u_{k,j}$. Similarly, we get

$\displaystyle u_{k,j}({\bf x},t)$ $\displaystyle=$ $\displaystyle
u_{k,j}({\bf x},t-\Delta t)+\int_{\partial\Sigma_{\epsilon*}}F_{i}({\bf
x^{\prime}},t^{\prime})u_{ki,j}^{E}({\bf
x-x^{\prime}},t-t^{\prime})d\sigma^{\prime}(1+{\rm O}(g(\epsilon)))$
$\displaystyle\lvert u_{k,j}({\bf x},t)\rvert$ $\displaystyle\leq$
$\displaystyle\lvert u_{k,j}({\bf x},t-\Delta t)\rvert$
$\displaystyle+\left\\{\int_{\partial\Sigma_{\epsilon*}}\lvert F_{i}({\bf
x^{\prime}},t^{\prime})\rvert d\sigma^{\prime}\right\\}\lvert
u_{ki,j}^{E}({\bf x-x^{\prime}},t-t^{\prime})\rvert^{\rm max}(1+{\rm
O}(g(\epsilon))),$

for some finite function $g(\epsilon)$ where $\epsilon>0$ is small. In the
same way, from the initial condition, $u_{k,j}({\bf x},0)$ is finite and this
provides a continuation demonstrating that $u_{k,j}({\bf x},t)$ is finite as
well. Repeating for all the other derivatives demonstrates that $u_{k}$ is
smooth. If the velocity $u_{k}$ is smooth, then so is the pressure (see
appendix B).

###  9.3 Uniqueness

From (47), the homogeneous boundary condition over the initial value problem
boundary with stipulated initial velocity is $F_{i}=-u_{i}=0$. This gives
$F_{i}=0$ over the initial value boundary $V$, which then gives the velocity
at general point $x_{I}$ for time $t>0$ to be $u_{k}=0$. Hence, the solution
is unique. It is noted that Ladyzhenskaya states that uniqueness has only been
proved for a certain time interval [24], page 141. However Ladyzhenskaya
considers a wider class of solutions than $C^{\infty}$, and conversely it is
not stated that uniqueness cannot be extended beyond this interval. For a
wider class of solutions with different initial conditions then the Navier-
Stokes equation could be non-unique. Therefore, by uniqueness we mean the
$C^{\infty}$ smooth solution is unique for the problem of smooth initial
conditions and make no claim about the wider class of possible solutions with
different initial conditions.

##  10 Properties of the NSlet

Consider the velocity $u_{k}^{(q)}$ generated by a $q$ unit force impulse at
the origin gives the velocity $u_{k}^{(q)}$ to be the NSlet velocity $u_{qk}$
itself. Then likewise

$\displaystyle u_{k}^{(q)}({\bf x},t)$ $\displaystyle=$ $\displaystyle
u_{k}^{(q)}({\bf x},t-\Delta t)$
$\displaystyle+\int_{\partial\Sigma_{\epsilon*}}F_{i}^{(q)}{\bf
x^{\prime}},t^{\prime})u_{ki}^{E}({\bf
x-x^{\prime}},t-t^{\prime})d\sigma^{\prime}(1+{\rm O}(f(\epsilon)))$
$\displaystyle\lvert u_{k}^{(q)}({\bf x},t-\Delta t)\rvert$
$\displaystyle\leq$ $\displaystyle\lvert u_{k}^{(q)}({\bf x},t\rvert$
$\displaystyle+\left\\{\int_{\partial\Sigma_{\epsilon*}}\lvert
F_{i}^{(q)}({\bf x^{\prime}},t^{\prime})\rvert d\sigma^{\prime}\right\\}\lvert
u_{ki}^{E}({\bf x-x^{\prime}},t-t^{\prime})\rvert^{\rm max}(1+{\rm
O}(f(\epsilon))),$

where $F_{i}^{(q)}$ is the force impulse distribution $F_{i}$ for the $q$ unit
force impulse. Far from its origin the NSlet $u_{ki}$ tends toward the
Stokeslet $u_{ki}^{S}$ which decays to zero. So, in the far-field the NSlet
tends towards a smooth solution. In this case, the continuation is from the
far-field $t$ large to the time co-ordinate $x_{0}=t>0$. So, the unit force
impulse $u_{k}^{(q)}$, which is the NSlet velocity $u_{qk}$, is bounded and so
exists at any point $x_{0}=t>0$ (although it becomes singular at its origin as
$x_{I}\rightarrow 0$). Following the same approach as in Chapter 9, it is seen
the NSlet exists, is smooth and is unique.

###  10.1 Series summation

Far from the point the NSlet $u_{ki}$ tends toward the Stokeslet $u_{ki}^{S}$.
Also, near to the point the NSlet tend towards the Eulerlet $u_{ki}^{E}$,
which is a Stokeslet in the limit as $x_{0}\rightarrow 0$, so
$u_{ki}^{E}=\lim_{x_{0}\rightarrow 0}u_{ki}^{S}$. So, in the far-field and the
near-field the following asymptotic expansion is valid:

$\displaystyle u_{ki}$ $\displaystyle=$ $\displaystyle u_{ki}^{\rm
0}+u_{ki}^{\rm I}++u_{ki}^{\rm II}+...$ $\displaystyle u_{ki}$
$\displaystyle=$ $\displaystyle\sum_{{\rm N}=0}^{\rm N=\infty}u_{ki}^{\rm N},$
(63)

where $u_{ki}^{\rm 0}=u_{ki}^{S}$. In the above equation, the Stokeslets
$u_{ki}^{S}$ are known, see appendix A, and satisfy the Stokeslet equation

$u_{ki,0}^{S}+p_{k,i}^{S}-\nu u_{ki,jj}^{S}=-\delta\delta_{ki}.$ (64)

The next term in the expansion then satisfies

$u_{ki,0}^{\rm I}+p_{k,i}^{\rm I}-\nu u_{ki,jj}^{\rm I}=-u_{{\bf
k}j}^{S}u_{{\bf k}i,j}^{S},$ (65)

and so on. The expansion tends to the first term in the near- and far-fields,
and so is convergent. Since the NSlet velocity is smooth, the expansion can be
continued into the remainder of the domain. To recap, the NSlet tends to the
Stokeslet in the far-field which in turn tends to the Eulerlet as
$x_{0}\rightarrow x_{0}^{\prime}$, and the NSlet tends to the Eulerlet in the
near-field which has a singularity on $R=0$ as $x_{0}\rightarrow
x_{0}^{\prime}$, see figure 9.

Figure 9: A figurative representation of the NSlet

##  11 Conclusion

A Green’s integral representation for the velocity is given over a four-
dimensional space-time hyper-surface with a Green’s function fundamental
solution kernel, called an NSlet, having strength given by a force impulse
distribution over the hyper-surface. The force impulse for the problem is
assumed to be bounded. Choosing the integral representation for the velocity
to be close to the fluid point, then the NSlet approximates to the Eulerlet
which is a known smooth function (away from its spatial origin). From this, it
is shown that if the velocity is smooth at an initial time then it is smooth
for all subsequent time as well. If the velocity exists and is smooth then
from the Helmholtz theorem so is the pressure. The force impulse distribution
is given by the initial velocity, and if this is taken to be zero then the
velocity at all later time is zero, demonstrating uniqueness for the smooth
class of solutions $C^{\infty}$.

##  Appendix A Stokeslet

We follow the approach of Chan and Chwang [15] to give the unsteady Stokeslet.
The Stokeslet equation is

$u^{S}_{ki,0}+p^{S}_{k,i}-\nu
u^{S}_{ki,jj}=-\delta\delta_{ki}\;,\;\;u^{S}_{ki,i}=0.$ (66)

Taking the divergence of the first equation in (66) gives
$p^{S}_{k,ii}=-\delta_{,k}$. Letting $p^{S}_{k}=\delta(x_{0})p^{*}_{,k}$ gives
$p^{*}_{,ii}=-\delta(x_{i})$ and so $p^{*}=\frac{1}{4\pi R}$, which gives

$p^{S}_{k}=\frac{\delta(x_{0})}{4\pi}\left[\frac{1}{R}\right]_{,k}.$ (67)

Substituting this back into the first equation in (66) gives

$u^{S}_{ki,0}-\nu u^{S}_{ki,jj}=-\delta\delta_{ki}-\left(\frac{1}{4\pi
R}\right)_{,ki}\delta(x_{0}).$ (68)

We can split the velocity into two parts, and determine each separately, such
that

$u^{S}_{ki}=u^{A}\delta_{ki}+u^{B}_{,ki}$ (69)

where

$u^{A}_{,0}-\nu u^{A}_{,jj}=-\delta$ (70)

and

$u^{B}_{,0}-\nu u^{B}_{,jj}=-\frac{1}{4\pi R}\delta(x_{0}).$ (71)

###  A.1 Evaluation of $u^{A}$

Apply Fourier Transforms to (70) to get

$\left(\frac{1}{\sqrt{2\pi}}\right)^{4}\int_{\Sigma_{\infty}}\left(u^{A}_{,0}-\nu
u^{A}_{,jj}\right)e^{-\alpha_{J}x_{J}}d\Sigma=-\left(\frac{1}{\sqrt{2\pi}}\right)^{4},$
(72)

where $\alpha_{I}$ are the Fourier Transform variables, and
$\int_{\Sigma_{\infty}}$ is the integration across all space-time. So

$\left(\frac{1}{\sqrt{2\pi}}\right)^{4}\left(i\alpha_{0}+\nu\alpha_{R}^{2}\right)\int_{\Sigma_{\infty}}u^{A}e^{-\alpha_{J}x_{J}}d\Sigma=-\left(\frac{1}{\sqrt{2\pi}}\right)^{4}.$
(73)

Let
${\bar{\bar{u}}}^{A}=\left(\frac{1}{\sqrt{2\pi}}\right)^{4}\int_{\Sigma_{\infty}}u^{A}e^{-\alpha_{J}x_{J}}d\Sigma$,
and let
${\bar{u}}^{A}=\left(\frac{1}{\sqrt{2\pi}}\right)^{3}\int_{V_{\infty}}u^{A}e^{-\alpha_{j}x_{j}}dV$,
where $dV$ is an element of volume $V_{\infty}$ which is an integration across
all of space. This gives

$\displaystyle{\bar{\bar{u}}}^{A}$ $\displaystyle=$
$\displaystyle-\left(\frac{1}{\sqrt{2\pi}}\right)^{4}\left(\frac{1}{i}\right)\left(\frac{1}{\alpha-i\nu\alpha_{R}^{2}}\right)$
$\displaystyle{\bar{u}}^{A}$ $\displaystyle=$
$\displaystyle-\left(\frac{1}{\sqrt{2\pi}}\right)^{3}H(x_{0})e^{\nu\alpha_{R}^{2}x_{0}}$

using the result for the inverse Fourier Transform
$F^{-1}\\{1/(\alpha-ia)\\}=H(x_{0})\sqrt{2\pi}ie^{-\alpha x}$ given in
standard transform tables, for the real variable $x$, transform variable
$\alpha$ and constant $a$. Taking the inverse transforms in the spatial
directions by using the inverse Fourier Transform
$F^{-1}\\{e^{-a\alpha^{2}}\\}=\left(e^{-x^{2}/(4a)}\right)/\sqrt{2a}$ given in
standard transform tables for the Gaussian, we then get

$\displaystyle u^{A}$ $\displaystyle=$
$\displaystyle-\left(\frac{1}{\sqrt{2\pi}}\right)^{3}H(x_{0})e^{-R^{2}/(4\nu
x_{0})}\left(\sqrt{\frac{1}{2x_{0}}}\right)^{3}$ (75) $\displaystyle=$
$\displaystyle\frac{1}{4\pi\nu}H(x_{0})\left(\frac{{\rm
erf}\eta}{R}\right)_{,0}$

where $\eta=\frac{R}{\sqrt{4\nu x_{0}}}$ and ${\rm
erf}\eta=(2/\sqrt{\pi})\int_{0}^{\eta}e^{-\eta^{\prime 2}}d\eta^{\prime}$,
which means

$\displaystyle({\rm erf}\eta)_{,0}$ $\displaystyle=$
$\displaystyle\eta_{,0}({\rm erf}\eta)_{,\eta}$ (76) $\displaystyle=$
$\displaystyle-\frac{R}{4}\frac{1}{\sqrt{\nu}}\frac{1}{\sqrt{x_{0}}^{3}}\frac{2}{\sqrt{\pi}}e^{-\eta^{2}}.$

###  A.2 Evaluation of $u^{B}$

We then have

$[u^{B}_{,0}-\nu u^{B}_{jj}]_{,qq}=-\left(\frac{1}{4\pi
R}\right)_{,qq}\delta(x_{0})=\delta.$ (77)

So $u^{B}_{,jj}=-u^{A}$ which gives $u^{B}_{,0}=-\nu u^{A}$ and so

$u^{B}=-\frac{1}{4\pi}H(x_{0})\frac{{\rm erf}\eta}{R}.$ (78)

This then gives the Stokeslet velocity to be

$u^{S}_{ki}=\frac{1}{4\pi\nu}H(x_{0})\left(\frac{{\rm
erf}\eta}{R}\right)_{,0}\delta_{ki}-\frac{1}{4\pi}H(x_{0})\left(\frac{{\rm
erf}\eta}{R}\right)_{,ki}.$ (79)

##  Appendix B Pressure smooth

By applying the Helmholtz decomposition (fundamental theorem of vector
calculus) it is shown that if the velocity is smooth then the pressure is
smooth as well. The Navier-Stokes equation is rewritten in terms of the head
function $h$ such that

$u_{i,0}+h_{i}+p_{,i}-\nu u_{i,jj}=0\;,\;\;h_{i}=u_{j}u_{i,j}.$ (80)

Applying the Helmholtz decomposition to the velocity and head functions

$\displaystyle u_{i}$ $\displaystyle=$
$\displaystyle\phi_{,i}+w_{i}\;,\;\;w_{i}=\epsilon_{ijk}A_{k,j}$
$\displaystyle h_{i}$ $\displaystyle=$ $\displaystyle
h_{,i}+h_{i}^{w}\;,\;\;h_{i}^{w}=\epsilon_{ijk}B_{k,j}$

for curl free scalar potentials $\phi$ and $h$, and divergence free vectors
$w_{i}$ and $h_{i}^{w}$, represented by the vector potentials $A_{k}$ and
$B_{k}$ respectively, and $\epsilon_{ijk}$ is the Levi-Civita symbol. (This
symbol is such that $\epsilon_{ijk}=1$ for permutations $(i,j,k)=(1,2,3)$,
$(2,3,1)$ and $(3,1,2)$, $\epsilon_{ijk}=-1$ for permutations
$(i,j,k)=(1,3,2)$, $(2,1,3)$ and $(3,2,1)$, and zero otherwise.) The wake
velocity divergence is $w_{i,i}=\epsilon_{ijk}A_{k,ji}=0$, and so
$u_{i,i}=\phi_{,ii}=0$. So, the curl free component of the Navier-Stokes
equation is $\phi_{,i0}+h_{,i}+p_{,i}=0$ since $\nu\phi_{,ijj}=0$. Integrating
this then gives the pressure as

$p=-\phi_{,0}-h.$ (82)

From the Helmholtz theorem [23], the potentials are given by

$\displaystyle\phi$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi}\int_{S}\frac{u_{i}n_{i}}{\lvert
R-R^{\prime}\rvert}ds^{\prime}$ $\displaystyle h$ $\displaystyle=$
$\displaystyle-\frac{1}{4\pi}\int_{V}\frac{u_{j;i}u_{i;j}}{\lvert
R-R^{\prime}\rvert}dV^{\prime}+\frac{1}{4\pi}\int_{S}\frac{u_{j}u_{i;j}n_{i}}{\lvert
R-R^{\prime}\rvert}ds^{\prime},$

for some volume $V$ enclosed by a surface $S$ with outward pointing normal
$n_{i}$. So, if the velocity $u_{i}$ is smooth then the potentials $\phi$ and
$h$ are smooth, and the pressure $p$ is smooth.

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