A Fourier Series approach to solving the avier-Stokes equations with
spatially periodic data
William McLeod Rivera
College Park, MD 20740
Email address: WILLIAM_RIVERA35@comcast.net
rˆ
Abstract: The symbol U denotes the velocity or momentum (the mass multiplied by the
velocity). Transform the Navier Stokes momentum and density equations into infinite systems of
ordinary differential and linear equations for the classical Fourier coefficients. Prove theorems
on existence, uniqueness and smoothness of solutions of them. Interpret the results using the
r rˆˆ
ˆ
Fourier series representation U = U , P = Pˆ .
Key Words: Fluid Mechanics, incompressible Navier-Stokes equations, Fourier series
1. Introduction
The main result in this paper can be stated as follows. If the data is jointly smooth, spatially
periodic, and the body force and its higher order time derivatives satisfy the generalized sector
r
conditions described later a unique physical solution (U , P) exists which is separately smooth in
r
r
(t , x ) on t ≥ 0, x ∈ R 3 . This solution is the extension of the unique regular (smooth) short time
r
solution determined by the data. The solution (U , P ) is also bounded for all forward time.
In 1934 Leray (in [9]) formulated the regularity problem and related it to the smoothness
problem. In the year 2000, Fefferman formulated the problem (in [5]). In that same year, Bardos
wrote a monograph on the problem ([3]) which summarized the then literature. The author
interprets the remarks in [3] Bardos to indicate that the problem of regularity/smoothness can be
solved as formulated in (A) of [5] Fefferman.
What is new?
As far as this author knows the formulas for the Navier Stokes ordinary differential equations in
this paper are new. However for the problem defined on the aperiodic space domain Cannonein the formula immediately prior to (27) on page 15 of Harmonic Analysis Tools for Solving the
Incompressible avier-Stokes Equations. (2003)- uses the Fourier transform to rewrite the
variation of constants formula solution of the Navier-Stokes evolution equations.
The vector form of the avier-Stokes equations
The vector form of the Navier-Stokes equations for an incompressible fluid on the unit cube is
1
r
r
r
r
r
r
U t + U ⋅ ∇U = η∆U − ∇P + F ,η > 0, t ≥ 0, x ∈ [0,1]3
r
r
U ⋅ ∇ = 0, t ≥ 0, x ∈ [0,1]3
r
r
r
U (0, x, y, z ) = U 0 ( x, y, z ), x ∈ [0,1]3
r
r
r
U (t , x, y,0) = U (t , x, y,1) = h 1 ( x, y ), ( x, y ) ∈ [0,1]2 , t ≥ 0
r
r
r
U (t , x,0, z ) = U (t , x,1, z ) = h 2 ( x, z ), ( x, z ) ∈ [0,1] 2 , t ≥ 0
r
r
r
U (t ,0, y, z ) = U (t ,1, y, z )) = h 3 ( y, z ), ( y, z ) ∈ [0,1] 2 , t ≥ 0.
(1-1)
One seeks a unique solution of (1-1) on the unit cube which is jointly smooth in time and space
r r r
and bounded for all forward time given that the data U 0 , F , h i , i = 1,2,3 is smooth. Such solutions
extend to periodic solutions of period 1 in each space variable holding the others fixed.
r
r
In (1-1) U is the velocity vector field, ∇P is the pressure gradient to be determined, ∇ ⋅ U is the
 ∇U 
r 
r
r
 r
divergence of U , and U ⋅ ∇ is the matrix tensor DxrU =  ∇V ,U = (U ,V ,W ) .
 ∇W 


r
The first equation of (1-1) determines the momentum. The body force F is smooth on
r
r
[0, ∞) × [0,1]3 . The initial function U 0 is smooth on [0,1]3 . The boundary functions h i , i = 1,2,3
are smooth on [0,1] × [0,1] .
The momentum equation can be interpreted as Newton’s second law of motion for fluids
r
r r
combined with a dynamic version of Archimedes’ law of hydrostatics. If U (t , x ) = 0 , the
r r
r
equation reduces to Archimedes’ law ∇ P (t , x ) = F (t , x ) .
The second equation of (1-1) is the equation of continuity. It is the Navier-Stokes equation for
the density for an incompressible fluid.
The following equations for the classical Fourier series coefficients are equivalent to (1-1). The
conditions on the data functions are equivalent to the conditions that their Fourier coefficients
belong to discrete Schwartz frequency spaces
rˆ r
rˆ r r rˆ r
rˆ r
v r
r
r
r
dU (t , r )
= −4η | r | 2 π 2Uˆ (t , r ) − 2πU (t , r )r t ∗ U (t , r ) − 2πr Pˆ (t , r ) + F (t , r ),
dt
r
t ≥ 0, r ∈ 3,η > 0
rˆ v
rˆ r r
U (0, r ) = U 0 (r ), r ∈ 3
r r r
r
r ⋅ U (t , r ) = 0, r ∈ 3, t ≥ 0.
(1-2a)
2
The boundary conditions for the Fourier coefficents are
rˆ
rˆ
U (t ,0, r2 , r3 ) = h1 (r2 , r3 ), t ≥ 0, ¬(r2 ∧ r3 ) = 0
rˆ
rˆ
U (t , r1 ,0, r3 ) = h2 (r1 , r3 ), t ≥ 0, ¬(r1 ∧ r3 ) = 0
rˆ
rˆ
U (t , r1 , r2 ,0) = h3 (r1 , r2 ), t ≥ 0, ¬(r1 ∧ r2 ) = 0
r
t ≥ 0, r ∈ 3 .
(1-2b)
rˆ
rˆ rˆ
In (1-2), U , Pˆ , F ,h i , i = 1,2,3 denote the classical Fourier sine series coefficients of
r
r r
U , P , F ,h i , i = 1,2,3 and 3,W 3 denote the set of natural (respectively whole) number triples.
The law governing the average mechanical energy of an incompressible fluid
Theorem 2-4 establishes the existence of a unique solution defined (bounded) for all forward
r
r
time smooth in t uniformly in x ∈ [0,1]3 and smooth in x uniformly in t ≥ 0. The following
formulas provide a smooth generalization of Leray’s mechanical energy law ([12] section 17
formula 3.4) for the Navier-Stokes equation and all of its time derivatives with non zero body
force
t
t
r (k ) 2 r 1
r (k ) 2 r
r (k ) 2 r
r (k ) r (k ) r
1
U
d
y
−
|
U
|
d
y
=
−
|
∇
U
|
d
y
ds
+
U
|
|
η
0
∫0 ∫ 3
∫0 ∫ 3 ⋅ F dyds,η > 0, t ≥ 0, (1-3)
2 [0∫,1]3
2 [0∫,1]3
[ 0 ,1]
[ 0,1]
k = 01,2,...
Equations (1-3) state that the difference of the space average of the kinetic energy (at time
t > 0 ) minus that at time t = 0 + is equal to the viscosity times the potential energy minus the
average work done by the body force acting on the incompressible fluid where
r
r
| ∇U | 2 = ∇u ⋅ ∇u + ∇v ⋅ ∇v + ∇w ⋅ ∇w, U t = (u , v, w) . The energy formulas (1-3) are equivalent to
the formulas
rˆ
rˆ
r
r
∑r | U ( k ) (t , r )|2 −∑r | U ( k ) (0, r )| 2 =
r
r >0
r
r >0
t
t
rˆ
r 2 rˆ ( k ) 2
v rˆ r
− 4π η ∫ ∑r | r | | U | ds + ∫ ∑r U ( k ) ( s, r ) ⋅ F ( s, r )ds, t ≥ 0,η > 0, k = 0,1,2,...
(1-4)
2
r
0 r >0
r
0 r >0
established in theorem 2-2.
By Parseval’s theorem, the quantities on the left side of (1-3) and (1-4) are both equal to the
average kinetic energy.
The problem of finite time blow up
3
In theorem 2-3 the author shows that finite time blow up of solutions of (1-1) is impossible given
smooth data, the equation of continuity, and the following conditions on the forcing function
t
∫
t
r (k ) r (k ) r
U
⋅
F
d
x
ds
<
η
∫
∫
3
0 [ 0 ,1]
r (k )
∫ | ∇U
r
| 2 dx ds, t ≥ 0,η > 0, k = 0,1,2,...
(1-5)
3
0 [ 0 ,1]
In the frequency domain, inequalities (1-5) become
t
t
rˆ ( k ) v rˆ ( k ) r
r 2 rˆ ( k ) r 2
2
U
(
s
,
r
)
⋅
F
(
s
,
r
)
ds
<
4
π
η
|
r
∑
∑
∫ r
∫ r | | U (s, r ) | ds, t ≥ 0,η > 0, k = 0,1,2,... .
r
0 r >0
(1-6)
r
0 r >0
r
Under the conditions (1-5) on F the solutions of (1-1) are absolutely bounded. Absolute
stability/boundedness extends the concept of Lyapunov stability/boundedness from
homogeneous nonlinear systems to nonlinear systems with a forcing function.
2. Existence of a unique smooth short time solution and its forward time extension
The Navier-Stokes ordinary differential equations in the frequency domain comprise an infinite
rˆ r
system of ordinary differential equations for the time dependent Fourier coefficients U (t , r ) of
r r
the velocity U (t , x ) . It is possible to use the classical Fourier sine series coefficients because the
inhomogeneous Dirichlet boundary conditions match on opposite faces of the unit cube.
The notation
r
rˆ
rˆ
r
U ∈ {C ∞ ([0, T ]), r ∈ 3 } ∩ {S ( 3 ), t ≥ 0}
indicates the space of Fourier coefficients of momentum vectors which are separately smooth on
t ∈ [0, T ] and discrete Schwartz in the vector of frequency parameters. A Fourier coefficient is
discrete Schwartz in a parameter if it decays more rapidly than any fixed power of the norm of
rˆ r
r
r
the frequency | r | p , p ∈ W grows. Thus lim |rr|→∞ | r | p | U (t , r ) = 0, t ≥ 0, p ∈ W . Equivalently the
discrete Schwartz coefficients can be defined with upper bounds replacing limits (see definition
2-2 below).
Lemma 2-1. The Navier-Stokes ordinary differential equations for the Fourier coefficients of the
solution of the spatially periodic problem (1-1) are
rˆ r
rˆ r r rˆ r
rˆ r
v r
r
r
r
dU (t , r )
= −4η | r | 2 π 2Uˆ (t , r ) − 2πU (t , r )r t ∗ U (t , r ) − 2πr Pˆ (t , r ) + F (t , r ),
dt
r
t ≥ 0, r ∈ 3,η > 0
r rˆ r
r
r ⋅ U (t , r ) = 0, r ∈ 3, t ≥ 0
rˆ v
rˆ r r
U (0, r ) = U 0 (r ), r ∈ 3.
(2-1a)
4
where
rˆ
rˆ
U (t , r1 , r2 ,0) = h3 ( r1 , r2 ), t ≥ 0, ¬( r1 ∧ r2 ) = 0
rˆ
rˆ
U (t , r1 ,0, r3 ) = h2 ( r1 , r3 ), t ≥ 0, ¬( r1 ∧ r3 ) = 0
rˆ
rˆ
U (t ,0, r2 , r3 ) = h1 ( r2 , r3 ), t ≥ 0, ¬( r2 ∧ r3 ) = 0 3
(2-1b)
where ∗ in line 1 of (2-1a) denotes the convolution
rˆ r r rˆ r
rˆ r r r r rˆ r
U (t , r )r t *U (t , r ) = ∑r [U (t , r − q )(r − q ) t ]U (t , q ).
r
q >0
(2-2)
The discrete Fourier (integral) transform (the Fourier coefficient) of the velocity is
rˆ r
r r
U (t , r ) = ℑ{U (t , x )} =
r r
U
∫ (t , x ) sin 2πr1 x sin 2πr2 y sin 2πr3 zdxdydz ,
[ 0 ,1]3
(2-3)
r
r ∈ 3 , t ≥ 0.
rˆ r
r
and similarly for F (t , r ), Pˆ (t , r ) .
PROOF
Note that all triple integrals over [0,1]3 in the definition of the discrete Fourier transform defined
r r
r
by (2-3) are well defined for all forward time since F ,U 0 are smooth in x ∈ D and smooth and
bounded in t with all time derivatives continuous and bounded on t ∈ [0, ∞) . The boundary
r
functions h i , i = 1,2,3 are smooth on [0,1]2 .
rˆ
dU r ˆ rˆ
The terms
, r P , F are calculated from the discrete Fourier transform to the terms of (1-1) and
dt
the differentiation property of discrete Fourier transforms applied to the first partial derivatives
in the case of the pressure gradient term.
The Fourier series coefficient of the diffusion term can be calculated using integration by parts
twice. Since the three calculations are identical, it suffices to calculate the transform of the
r
second partial derivative of U with respect to the first component x,
5
v
∫U
sin 2πr1 x sin 2πr2 y sin 2πr3 zdxdydz =
xx
[ 0 ,1]3
1
1
r
∫ ∫U
x
sin 2πr1 x |1x =0 sin 2πr2 y sin 2πr3 zdydz =
y = 0 z =0
− 2πr1
(2-4)
v
∫ U x cos 2πr1 x sin 2πr2 y sin 2πr3 zdxdydz =
[ 0 ,1]3
− 4π 2r12
r
∫ U sin 2πr y sin 2πr zdxdydz = −4π
2
2
3
rˆ r
r
r12U (t , r ), t ≥ 0, r ∈ 3 .
[ 0 ,1]3
Thus the diffusion term is
r
r
r
r r
∂U (t ,1, x2 , x3 ) ∂U (t ,0, x2 , x3 )
2
2
2 ˆ
]+
−
ℑ{∆U } = (r1 + r2 + r3 )U (t , r ) + r1[
∂x1
∂x1
r
r
r
r
∂U (t , x1 ,1, x3 ) ∂U (t , x1 ,0, x3 )
∂U (t , x1 , x2 ,1) ∂U (t , x1 , x2 ,0)
] + r3 [
]+
r2 [
−
−
∂x3
∂x3
∂x2
∂x2
r
r
r
r
r
r
U (t ,1, x2 , x3 ) − U (t ,0, x2 , x3 ) + U (t, x1 ,1, x3 ) − U (t , x1 ,0, x3 ) + U (t , x1 , x2 ,1) − U (t , x1 , x2 ,0)
rˆ r
r
= (r12 + r22 + r32 )U (t , r ) ≠ 0, t ≥ 0, r ∈ 3 .
(2-5)
It follows that
v
r rˆ r
ℑ{η∆U } = −4ηπ 2 | r | 2 U , r ∈ 3 , t ≥ 0 .
(2-6)
The transform of the Euler term is
r
r
ℑ{U ⋅ ∇U } =
rˆ
r
r r
r
rˆ
r
r
∑ 2π [U (t , r − q)(r − q ) ]U (t, q), t ≥ 0, r ∈ t
3
.
(2-7)
r
q∈ 3
The transform of the pressure gradient term is
r
r
r
∫ ∇P sin 2πr x sin 2πr y sin 2πr zdx = 2πr Pˆ , t ≥ 0, r ∈ 1
[ 0,1]
2
3
3
.
(2-8)
3
The transform of the equation of continuity is
∫ (U
x
r
+ V y + W z ) sin 2πr1 x sin 2πr2 y sin 2πr3 zdx = 2π [r1U + r2V + r3W ] = 0,
[ 0 ,1]3
r
t ≥ 0, r ∈ 3 .
(2-9)
The boundary conditions (lines 3-6 of formula (2-1)) are derived by solving the homogeneous
Laplace’s equation constrained by the face matching boundary conditions using transform
(Fourier series) methods.
6
END PROOF
Remark 2-1. Equations (2-1) specify an infinite system of ordinary differential equations with
a discrete vector parameter whose solutions are the time dependent Fourier coefficients in the
r r r r r r
r
Fourier expansions of U (t , x ), F (t , x ),U 0( x ), P(t , x ) . In fact formulas (2-4) through (2-9) hold on
r
r ∈ W 3 . But, since the boundary conditions are Dirichlet and the Fourier sine series is used,
r r
there is no constant term corresponding to r = 0 while cases with one or two indices ri equal to
0 occur on the boundary of the frequency domain.
Definition 2-1. The family of linear operators defined by the transformed equation of continuity
is
v r
r rˆ r r
Lrr [ Fˆ (t , r )] = r ⋅ F (t , r ), r ∈ 3 , t ≥ 0.
(2-10)
r
r
Remark 2-2. For any r ∈ 3 , Lrr :C ∞ [0, ∞ ) →C ∞ [0, ∞ ) . The linear operator Lrr is a map from
vectors of smooth (infinitely continuous differentiable) functions on t ∈ [0, ∞) to smooth scalar
functions on [0, ∞ ) .
rˆ
d kU
Proposition 2-1. Any derivative of finite order of the velocity
, k = 0,1,2,... is in η ( Lrr ) (the
dt k
null space of Lrr ).
PROOF
By (1-1) this relation holds for the discrete Fourier transform of the equation of continuity (
k = 0 ). Take derivatives of order k = 1,2,3,... with respect to t of both sides of
rˆ r r
r r
U (t , r ) ⋅ r = 0, t ≥ 0, r > 0
(2-11)
to complete the proof.
END PROOF
Since differentiation of the momentum function with respect to the space variables corresponds
to multiplication of its transform by frequency variables, the following Banach space is the one
r
needed to establish that solutions of the Navier-Stokes equations are smooth in (t , x ) ∈ [0, ∞) × R 3 .
rˆ
Definition 2-2. The discrete Schwartz space (uniform in t ∈ [0, T ] ) for Fourier coefficients U
defined on [0, T ] × 3 is
rˆ
rˆ r
rˆ
rˆ
r r
r
S = {U ∈ C ∞ ([0, T ]) :|| r1s (1) r2s ( 2 ) r3s ( 3)U ||= sup rr∈ 3 | r1s (1) r2s ( 2 ) r3s ( 3)U |≤ M ( s ), r ∈ 3, s ∈ W 3 },
(2-12)
0 ≤ t ≤ T.
7
rˆ r
The following assertions follow from the hypotheses of [5] Fefferman or, in the case of U (t , r )
from the functional analytic conditions imposed in the last line of equations (2-1).
The following lemma establishes that P̂ is an auxiliary function for (2-1)- i.e. a function which
can be removed in an equivalent form of the equation.
Lemma 2-2. Equations (2-1) can be placed into the following equivalent form
 rr
1 0 0  
rˆ r
rˆ rˆ
rˆ

r
r 2 rˆ  r r t 
2
U t = −4π η | r | U +  r 2 −  0 1 0 [2π (Ur t ) ∗ U − F ], t ≥ 0, r ∈ 3
|r | 


 0 0 1
rˆ r r
r
U (t , r ) ⋅ r = 0, t ≥ 0, r ∈ 3
rˆ r
rˆ r r
U (0, r ) = U 0 (r ), r ∈ 3
rˆ
rˆ
rˆ rˆ
r
rˆ rˆ
r
U ∈ {U : U ∈ (C ∞ ∩ L∞ )[0, ∞), r ∈ 3 ,U ∈ S ( 3 ), t ≥ 0).
The boundary conditions for the velocity coefficients are
rˆ
rˆ
U (t , r1 , r2 ,0) = h3 ( r1 , r2 ), t ≥ 0, ¬( r1 ∧ r2 ) = 0
rˆ
rˆ
U (t , r1 ,0, r3 ) = h2 ( r1 , r3 ), t ≥ 0, ¬( r1 ∧ r3 ) = 0
rˆ
rˆ
U (t ,0, r2 , r3 ) = h1 ( r2 , r3 ), t ≥ 0, ¬( r2 ∧ r3 ) = 0.
(2-13a)
(2-13b)
The pressure coefficients satisfy the following equations
rˆ r t rˆ r rˆ r
r
r
1
−
r
⋅
U
r ) *U + r ⋅ F (t, r )}, t ≥ 0, r ∈ 3
{
2
(
π
r2
2π | r |
rˆ r r t rˆ r r rˆ + r r
r
r
1
3
Pˆ (0 + , r ) =
r 2 {−r ⋅ 2π (U 0 (r )r ) *U 0 (r ) + r ⋅ F (0 , r }, r ∈ .
2π | r |
r
Pˆ (t , r ) =
(2-14a)
The pressure coefficients satisfy the following boundary conditions
8
Pˆ (t, r1 , r2 ,0) =
rˆ
rˆ
1
t
r 2 {− r∑ 2π (r1 , r2 ,0) ⋅[U (t, q1 , q2 ,0)(q1 , q2 ,0) ][U (t, r1 − q1 , r2 − q2 ,0)]
2π | r | q∈ 2
rˆ
+ (r1 , r2 ,0) ⋅ F (t, r1 , r2 ,0)},(r1 , r2 ) > (0,0), t ≥ 0,
rˆ
rˆ
1
t
Pˆ (t, r1 ,0, r3 ) =
r 2 {−r∑ 2π (r1 ,0, r3 ) ⋅[U (t, q1 ,0, q3 )(q1 ,0, q3 ) ][U (t, r1 − q1 ,0, r3 − q3 )]
2π | r | q∈ 2
rˆ
+ (r1 ,0, r3 ) ⋅ F (t, r1 ,0, r3 )},(r1 , r3 ) > (0,0), t ≥ 0,
rˆ
rˆ
1
t
Pˆ (t,0, r2 , r3 ) =
r 2 {−r∑ 2π (0, r2 , r3 ) ⋅[U (t, q1 , q2 ,0)(0, q2 , q3 ) ][U (t,0, r2 − q2 , r3 − q3 )]
2π | r |
q∈ 2
rˆ
+ (0, r2 , r3 ) ⋅ F (t,0, r2 , r3 )},(r2 , r3 ) > (0,0), t ≥ 0.
(2-14b)
PROOF
The equations in the first line of (2-13a) form an infinite system of vector ordinary differential
equations-one for each Fourier coefficient as a function of time. The equations in the second line
of (2-13a) for an infinite set of homogeneous linear equations.
Apply the linear operator Lrr to each side of equation (2-1) by forming the dot product of each
r
term with the vector r to obtain
rˆ r
v r
r r
r dU (t , r )
r⋅
= −ηr ⋅ | r | 2 π 2Uˆ (t , r )
dt
rˆ r r r r rˆ r r r
r
r r rˆ r r
− r ⋅ ∑ [2πU (t , r − q )(r − q ) t ]U (t , q ) − r ⋅ r 2πPˆ (t , r ) + r ⋅ F (t , r ), r ∈ 3 , t ≥ 0.
(2-15)
r
q∈ 3
rˆ r rˆ
r
r
By proposition 2-1, U (t , r ),U t (t , r ) ∈ η ( Lrr ), r ∈ W 3 hence (2-15) reduces to
r
0 = −r ⋅
rˆ
r
r r
r
rˆ
r
r r
r
r rˆ
r
∑ 2π [U (t, r − q )(r − q ) ]U (t , q ) − r ⋅ r 2πPˆ (t , r ) + r ⋅ F (t , r ) .
t
(2-16)
r
q∈ 3
r
Solve (2-16) for Pˆ (t , r ) to obtain the first line of (2-14a). Insert the first line of the pressure
formula into (2-1) to obtain the first line of (2-13a). The second line of (2-14a) is obtained from
rˆ
rˆ r
the transform of the initial function U 0 (r ) . The coefficients h i , i = 1,2,3 on the boundary of the
rˆ
frequency domain can be inserted for U to further reduce the formulas (2-14b). The pressure
coefficients are independent of time on the boundary.
END PROOF
9
Remark 2-3. From higher order derivatives of (2-1) and the projection defined by the kth order
r
equation of continuity, one can calculate formulas for Pˆ ( k ) (t , r ) similar to (2-14).
The equation that results from calculating any finite order time derivative of equations (2-13a) is
 rr
1 0 0  
rˆ ( k +1)
rˆ r
rˆ
rˆ

r 2 rˆ ( k )  r r t 
2
U
= −4π η | r | U +  r 2 −  0 1 0 [(2π (Ur t ) ∗ U ) ( k ) − F ( k ) ],
|r | 


 0 0 1
r
t ≥ 0, r ∈ 3 , k ∈ rˆ
r r r
U ( k ) (0, r ) = 0, r ∈ 3 , k ∈ rˆ
r r
r
U ( k )(t , r ) ⋅ r = 0, t ≥ 0, r ∈ 3 , k ∈ .
(2-17)
The following theorem establishes the existence of smooth short time solutions of equation (213).
rˆ rˆ
rˆ
rˆ
rˆ v r
rˆ
Theorem 2-1. Suppose F , Uˆ ∈ C ∞ ([0, ∞) ∩ S ( 3 ),U 0 ∈ S ( 3 ), hi ∈ S ( 2 ), t ≥ 0, i = 1,2,3 ,then
rˆ
v r
there exists T > 0 such that Uˆ ∈ C ∞ ([0, T )) ∩ S ( 3 ), satisfies (2-11) and
rˆ
Pˆ ∈ C ∞ ([0, T )) ∩ S ( 3 ).
PROOF
The goal is to establish
rˆ
r
a. For any fixed r ∈ 3 , and any finite k, U ( k ) is continuous for sufficiently short time.
rˆ
rˆ
rˆ
r
∃T > 0 :| U ( k ) (t 2 ) − U ( k ) (t1 ) |< M (k , r ) | t2 − t1 |< ∞, ∀t1 , t2 ∈ [0, T ], M = sup0≤t ≤T U ( k +1) (t ) (2-18)
rˆ
b. The continuity of U ( k ) is uniform in k ∈ W
rˆ
rˆ
r
∃T > 0 :| U ( k ) (t 2 ) − U ( k ) (t1 ) |< M (r ) | t 2 − t1 |< ∞, ∀t1 , t 2 ∈ [0, T ],
rˆ
(2-19)
~
M = sup k∈W sup 0≤t ≤T U ( k +1) (t )
rˆ
r
c. The continuity of U ( k ) is uniform in r ∈ 3
~
rˆ
rˆ
~
∃T > 0 :| U ( k ) (t 2 ) − U ( k ) (t1 ) |< M | t 2 − t1 |< ∞, ∀t1 , t 2 ∈ [0, T ], M
rˆ
(2-20)
= sup rr∈ 3 sup k∈W sup 0≤t ≤T U ( k +1) (t )
10
Use the variation of constants formula to solve for the general Fourier coefficient of the
momentum
rˆ r
rˆ r
U (t , r ) = Ψ (r ) +
e
r
−η 4π 2 |r |2 t
 r r t 1 0 0  
t
rˆ
rˆ r t
rˆ
rˆ
r


r
−η 4π 2 |r |2 ( t − s )  r r
0
1
0
[(
2
π
(
U
r
)
∗
U
)
−
F
]ds, t ≥ 0, r ∈ 3 .
U0 + ∫e
−
r



| r |2
0
 0 0 1



(2-21)
Form the difference to establish continuity of the Fourier coefficients in time
rˆ
rˆ
r
r
U (t 2 , r ) − U (t1 , r ) =
rˆ
rˆ
2 r2
2 r2
e −η 4π |r | t ( 2 )U 0 − e −η 4π |r | t (1)U 0 +
t (2)
∫e
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 rr
 | rr | 2 −  0 1 0 [(2π (Ur ) ∗ U ) − F ]ds −
 0 0 1 



r
−η 4π 2 |r |2 ( t ( 2 ) − s )
0
(2-22)
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 rr
π
e
0
1
0
[(
2
(
U
r
)
∗
U
)
−
F
]ds
−
r


∫0

| r |2



 0 0 1
r
t 2 > t1 ≥ 0, r ∈ 3 .
t (1)
r
−η 4π 2 |r |2 ( t (1) − s )
Simplify the expression on the right side of (2-22)
rˆ
rˆ
r
r
U (t 2 , r ) − U (t1 , r ) =
{e −η 4π
t (2)
∫e
2
r
|r |2 t ( 2 )
− e −η 4π
r
−η 4π 2 |r |2 ( t ( 2 ) − s )
t (1)
t (1)
∫ [e
2
r
|r |2 t (1)
rˆ
}U 0 +
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 rr
π
U
r
U
F
0
1
0
[(
2
(
)
)
]ds} −
∗
−
−
r



| r |2
 0 0 1 



r
−η 4π 2 |r |2 ( t ( 2 ) − s )
−e
0
r
t 2 > t1 ≥ 0, r ∈ 3 .
r
−η 4π 2 |r |2 ( t (1) − s )
(2-23)
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 rr
] r 2 −  0 1 0 [(2π (Ur t ) ∗ U ) − F ]ds}
|r | 


 0 0 1
Examine continuity near time 0 by setting t2 = t , t1 = 0
11
rˆ r
rˆ r
U (t , r ) − U (0, r ) =
rˆ
2 r2
{e −η 4π |r | t − 1}U 0 +
t
r
−η 4π 2 |r |2 ( t ( 2 ) − s )
∫e
0
r
t ≥ 0, r ∈ 3 .
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 rr
π
0
1
0
[(
2
(
U
r
)
∗
U
)
−
F
]ds} −
−
r


2

| r |
 0 0 1 



(2-24)
It follows that
rˆ r
| U (t , r ) |≤ Lt
t
+ ∫ e −η 4π
2
r
|r | 2 ( t − s )
2{M 1 + M 2 }ds =
(2-25)
0
r
−η 4π 2 |r |2 t
1− e
Lt + (
r )2{M 1 + M 2 }
η 4π 2 | r | 2
r
t ≥ 0, r ∈ 3 .
For derivatives of the momentum coefficients of any finite order k = 1,2,3,... the variation of
constants formula yields
rˆ
2 r2 r
r
r
ˆ
U ( k )(t , r ) = (−4η ) k π 2 k | r |2 k e −η 4π |r | tU 0
 r r t 1 0 0  
rˆ r t rˆ ( k ) rˆ ( k )


r
 rr
0
1
0
[(
2
(
U
r
)
∗
U
)
−
F
]
ds
,.
t
≥
0
,
r
∈ 3,
+ ∫e
−
π
r


2

| r |
0
 0 0 1 



r
3
t ≥ 0, r ∈ , k = 1,2,3,...
t
r
−η 4π 2 |r |2 ( t − s )
(2-26)
To prove continuity of the kth derivative of the momentum coefficient in time, calculate
rˆ
rˆ
rˆ
2 r2
2 r2
r rˆ
r
r
r
U ( k )(t 2 , r ) −U ( k )(t1 , r ) = (−4η ) k π 2 k | r |2 k e −η 4π |r | t ( 2)U 0 − (−4η ) k π 2 k | r |2 k e −η 4π |r | t (1)U 0
t ( 2)
+
∫e
0
r
−η 4π 2 |r |2 ( t ( 2 ) − s )
 r r t 1 0 0  
rˆ r t rˆ ( k ) rˆ ( k )


 rr
U
r ) ∗ U ) − F ]ds −
0
1
0
[(
2
(
π
−
r



 | r |2



 0 0 1
 r r t 1 0 0  
t (1)
rˆ r t rˆ ( k ) rˆ ( k )
r


−η 4π 2 |r |2 ( t (1) − s )  r r
0
1
0
[(
2
(
U
r ) ∗ U ) − F ]ds
π
+ ∫e
−
r


2

| r |
0
 0 0 1 



r
3
t 2 > t1 ≥ 0, r ∈ , k = 1,2,3,..
(2-27)
12
rˆ
Simplify the difference of U ( k ) at two distinct times
rˆ
rˆ
2 r2
2 r2
r rˆ
r
r
U ( k )(t 2 , r ) −U ( k )(t1 , r ) = (−4η ) k π 2 k | r | 2 k [e −η 4π |r | t ( 2 ) − e −η 4π |r | t (1) ]U 0
t (2)
∫e
+
r
−η 4π 2 |r |2 ( t ( 2 ) − s )
t (1)
t (1)
∫ [e
r
−η 4π 2 |r |2 ( t ( 2 ) − s )
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 rr
π
U
r
U
0
1
0
[(
2
(
)
) − F ]ds −
∗
−
r



| r |2



 0 0 1
−e
r
−η 4π 2 |r |2 ( t (1) − s )
0
r
t 2 > t1 ≥ 0, r ∈ 3 , k = 1,2,3,...
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 rr
] r 2 −  0 1 0 [(2π (Ur t ) ∗ U ) ( k ) − F ( k ) ]ds
|r | 


 0 0 1
(2-28)
To investigate behavior near time 0, evaluate (2-27) at t1 = 0, t 2 = t
rˆ
rˆ
2 r2
r
r
U ( k )(t , r ) = (−4η ) k π 2 k | r |2 k [e −η 4π |r | t − 1]U 0
 r r t 1 0 0  
rˆ r t rˆ ( k ) rˆ ( k )


 rr
π
0
1
0
[(
2
(
U
r ) ∗ U ) − F ]ds
+ ∫e
−
r



 | r |2
0



 0 0 1
r
t ≥ 0, r ∈ 3 , k = 1,2,3,...
t
r
−η 4π 2 |r |2 ( t − s )
(2-29)
The upper bound on the momentum coefficient is
rˆ
rˆ
2 r2
r
r
|U ( k )(t , r ) |≤| 4η |k π 2 k | r |2 k [1 − e −η 4π |r | t ] | U 0 |
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 rr
|  r 2 −  0 1 0  | [| (2π (Ur t ) ∗ U ) ( k ) | + | F ( k ) |]ds
+ ∫e
|r | 
0


 0 0 1
r
t ≥ 0, r ∈ 3 , k = 1,2,3,...
t
r
−η 4π 2 |r |2 ( t − s )
(2-30)
The upper bound can be simplified as follows.
rˆ
r
r
r rˆ
|U ( k )(t , r ) |≤| 4η |k π 2 k | r |2 k 4ηπ 2 | r |2 | U 0 | t

 r12 r1r2 r1r3  1 0 0 
 
rˆ r
rˆ
rˆ
 1 
r

2
−η 4π 2 |r |2 ( t − s )

|  r 2 r1r2 r2 r2 r3  −  0 1 0  | [| (2π (Ur t ) ∗ U ) ( k ) | + | F ( k ) |]ds
+ ∫e

0
 | r | 
2 
0 0 1
r
r
r
r
r
1
3
2
3
3

 


r
3
t ≥ 0, r ∈ , k = 1,2,3,...
t
(2-31)
Use the trace norm to bound the matrix operator inside the integral above
13
r

 r12 − | r | 2

r 2 0 0
|r |

 r12 r1 r2 r1 r3  1 0 0  
r 2 
r 2
r 2

2




|
|
−
r
r
3
|
|
|
|
1
−
r
r
2
2
= 2.
r 2 |  r1 r2 r2 r2 r3  −  0 1 0  |<|  0
r 2 0  |<
r 2


|
|
|r | 
|
|
r
r

 r1 r3 r2 r3 r32   0 0 1 
r 2
2


 0 0 r3 − | r | 
r

| r | 2 

(2-32)
By the hypothesis on the data
rˆ r
sup t ≥0 | U 0 (r ) |< L(r ) < L
(2-32a)
rˆ
rˆ
r
r
r
r
r
r
sup s ≥0 | 4η |k π 2 k | r |2 k | F ( k ) ( s, r ) |<| r |5 k | F ( k ) ( s, r ) |< M 2 (r ) < M 2 , | r > max{4π ,η} |
(2-32b)
Since the variation of constants operator is defined on coefficients which are smooth time/
rˆ r r
discrete Schwartz frequency, U ( s, r )r t is discrete Schwartz in the frequency. The convolution
r
of discrete Schwartz functions is discrete Schwartz in the frequency r ∈ 3 which is also
smooth in the time variable,
rˆ r r
rˆ r
r
sup s ≥0 | (2π [(U ( s, r )r t ) ∗U ( s, r )]( k ) |< M 1 (r ) < M 1.
(2-33)
It follows that
rˆ
r
|U ( k )(t , r ) |≤ Lt
t
+ ∫ e −η 4π
2
r
|r | 2 ( t − s )
2{M 1 + M 2 }ds =
0
r
−η 4π 2 |r |2 t
(2-34)
1− e
Lt + (
r )2{M 1 + M 2 }
η 4π 2 | r | 2
r
t ≥ 0, r ∈ 3 , k = 1,2,3,...
rˆ
r
Not only are the coefficients |U ( k )(t , r ) | bounded for sufficiently short time but for all finite
forward time.
rˆ
The formulas (2-13b) and (2-14b) for U , Pˆ on the integer lattice “boundaries” of the frequency
domain and their discrete Schwartz property in 3 for all forward time follow automatically
from the given conditions on the boundary data.
END PROOF
14
rˆ
r
Proposition 2-2. The inner product of U ( k )(t , r ) with the transformed Euler (convolution) term
rˆ
rˆ
r rˆ r
r r
U ( k )(t , r )[Ur t ∗ U (t , r )] ( k ) = 0, k = 0,1,2,3,..., t ≥ 0, r > 0.
PROOF
By the Liebnitz rule, the kth order Euler term can be written
k
r ( k ) d k rˆ
r
rˆ
 k  rˆ
r rˆ
r
U rr ⋅ k [U rr (t )r t ∗ U rr (t )] = U rr( k ) ⋅ ∑  (U rr (t )r t ) ( l ) ∗ (U rr (t )) ( k −l )
dt
l =0  l 
k
r
rˆ
 k  rˆ
r r
= U rr( k ) (t ) ⋅ ∑∑  (U rr −qr (t )(r − q ) t ) (l ) (U qr (t )) ( k −l ) .
r r
r >q l =0  l 
(2-36)
The equality in the second line follows by the definition of convolution and the fact that the
r r
coefficients for the classical Fourier sign series are one sided r > 0 .
The next equality follows by matrix vector multiplication,
k
r
rˆ
rˆ
rˆ
k  r r
r rˆ
d k rˆ
U rr( k ) ⋅ k [U rr (t )r t ∗ U rr (t )] = ∑∑  (r − q ) ⋅ (U qr (t )) ( k −l ) (U rr −qr (t )) (l ) ⋅ (U qr (t )) ( k −l ) .
r r
dt
r >q l =0  l 
(2-37)
The next inequality follows by the Schwartz inequality for vectors in R 3 applied to each term
and both dot products,
r ( k ) d k rˆ
r rˆ
U rr ⋅ k [U rr (t )r t ∗ U rr (t )] ≤
dt
k
rˆ
rˆ
 k  r r 2 rˆ ( k −l )
  | r − q | | (U qr )
(t ) | 2 | (U rr −qr ) ( l ) (t ) | 2 | (U qr ) ( k −l ) (t ) | .
∑∑
r r
r >q l =0  l 
(2-38)
r r r
The next equality follows by the definition of the inner product of a vector with itself x ⋅ x =| x| 2 .
k
k  r
r
rˆ
r
∑∑  l  | r − q | | (U
2
q
rˆ
rˆ
) ( k −l ) (t ) |2 | (U rr −qr ) (l ) (t ) |2 | (U qr ) ( k −l ) (t ) =
 
k
rˆ ( k −l )
rˆ
 k  rˆ ( k −l )
r r 2
2
2
(l )
r)
r)
r r


|
(
U
(
t
)
|
|
(
U
(
t
)
|
|
(
U
∑∑
q
q
r − q ) (t ) ⋅ ( r − q ) | =


r r
r >q l =0  l 
r r
r >q l =0
(2-39)
0, k ∈ W .
The final inequality follows by applying the higher order equation of continuity to each term
rˆ
r
r r
(2-40)
U rr(l ) (t ) ⋅ r = 0, l = 0,1,2,..., k , t ≥ 0, r > 0 .
Similarly
15
rˆ
r rˆ
d k rˆ
U rr( k ) ⋅ k [U rr (t )r t ∗ U rr (t )] ≥
dt
k
rˆ
rˆ
 k  rˆ
r r
− ∑∑   | (U qr ) ( k −l ) (t ) | 2 | (U qr ) ( k −l ) (t ) | 2 | (U rr −qr ) ( l ) (t ) ⋅ (r − q ) | 2 = 0, k ∈ .
r r
r >q l =0  l 
(2-41)
Hence
rˆ
r rˆ
r r
d k rˆ
U rr( k ) ⋅ k [U rr (t )r t ∗ U rr (t )] = 0, t ≥ 0, r > 0, k ∈ .
dt
(2-42)
END PROOF
r r
In the next theorem notation r > 0 indicates that all discrete frequency indices are non negative
integers and not all indices are 0.
The next theorem extends the domain of definition of solutions of (2-13) by showing that they
and all of their finite time derivatives are bounded and continuous for all forward time. A
frequency domain formula for the total mechanical energy of the average velocity and any time
derivative of it also appears. This is the frequency domain analog of the extension of Leray’s
energy law augmented by the body force.
The key to proving the next theorem is to impose the generalized sector conditions for the
Fourier coefficients to control finite time blow up and to realize (from the variation of constants
formula) that higher order initial functions exist, and, if they are nontrivial -denoted by
rˆ r
r r
¬{U 0( k ) (r ) ≡ 0, r > 0}, k = 0,1,2,...
-then a useful inequality can be constructed.
rˆ
r
r r
Theorem 2-2. If ¬{U 0( k ) ( r ) ≡ 0, r > 0}, k = 0,1,2,... and
t
t
rˆ ( k ) v rˆ ( k )
r 2 rˆ ( k ) r 2
2
U
(
s
,
r
)
⋅
F
ds
<
4
π
η
∫ ∑r
∫ ∑r | r | | U (s, r ) | ds, t ≥ 0,η > 0, k = 0,1,2,...
r
0 r >0
(2-43)
r
0 r >0
then
a. the following formulas for the total mechanical energy are well defined
rˆ r
rˆ r
rˆ r
t
r 2 ∂ k U (t , r ) 2
∂ k U (t , r ) 2
∂ k U (0, r ) 2
2
| ds
| = −4π η ∫ ∑r | r | |
| −∑r |
|
∑
k
r r
r
r
∂t k
∂t k
∂
t
>
0
r >0
r >0
r
0
r
r
v
v
ˆ
ˆ
t
∂ k U t ( s, r ) ∂ k F t ( s, r )
+ ∫ ∑r
⋅
ds, k = 0,1,2,3,..., t ≥ 0.
r
∂s k
∂s k
0 r >0
(2-44)
16
rˆ
r
b. The solution of (2-1)/ (2-13a) and every finite time derivative U ( k ) (t , r ), k = 0,1,2,..., t ≥ 0 of
r
it is unique, continuous, and bounded in t for all forward time, square summable in r ∈ 3 for
r
all forward time and jointly bounded in (t , r ) ∈ [0, ∞) × 3 .
PROOF
r
r r
First note that, by continuity, the hypothesis implies ∫ | U 0( k ) | 2 ( x ) dx > 0, k = 0,1,2,... hence
D
rˆ
r
∑r | U 0( k ) (r ) |2 > 0, k = 0,1,2,...,
(2-45)
r
r >0
By Parseval’s theorem
r (k ) 2 r r
rˆ ( k ) r 2
∫ | U 0 | ( x )dx < ∞ ⇔ ∑r | U 0 (r )| < ∞, k = 0,1,2,...
(2-46)
r
r >0
D
rˆ
r
Construct the formulas for the average energy of U ( k ) (t , r ), k = 0,1,2,... Then apply proposition 2rˆ
r
2 to eliminate the U ( k ) (t , r ) dotted time derivative of the Euler terms to simplify these formulas.
rˆ
r
r
r
Form the dot product of each equation (2-13a) with U ( k ) (t , r ) , sum over r ∈ W 3 − {0} and
integrate from 0 to t to obtain
t
rˆ ( k ) r 2
rˆ ( k ) r 2
rˆ
r
2
∑r | U (t , r )| −∑r | U (0, r )| = −4π η ∫ ∑r | r |2 | U ( k ) | 2 ds +
r
r >0
r
r >0
t
2π ∫
∂
k
∑ ∑ ∂t
r r r r r
0 r ≥ q >0 q >0
k
0 r >0
rˆ
r r r r rˆ r v
r
{U ( k ) ( s, r − q ) ⋅ (r − q ))(U ( s, r ) ⋅ Uˆ ( k ) ( s, q )}ds +
(2-47)
t
rˆ ( k ) v rˆ ( k ) r
U
∫ ∑r (s, r ) ⋅ F ( s, r )ds, k = 0,1,2,...
r
0 r >0
rˆ
r r
By proposition 2-2 the U dotted transformed Euler term summed over r > 0 satisfies
rˆ r rˆ r r r v r
2π ∑r ∑r U (s, r ) ⋅ (U ( s, r − q ) ⋅ q )Uˆ ( s, q ) = 0.
r r
r
r ≥q >0 q >0
(2-48)
rˆ
By proposition 2-2 the U ( k ) dotted transformed Euler term of the kth order Navier Stokes
r
ordinary differential equations, likewise summed over r ∈W 3 satisfies
rˆ r
rˆ r r r r v r
2π ∑ r ∑r [U ( s, r )]( k ) ⋅ [(U ( s, r − q ) ⋅ r − q )Uˆ ( s, q )]( k ) = 0.
(2-49)
r r
r
r ≥ q > 0 q >0
17
a. By (2-39) the frequency average of the kinetic energy is a decreasing function of time
rˆ
rˆ
r
r
(2-50)
∑r | U (k ) (t 2 , r )| 2 < ∑r | U (k ) (t 1, r )|2 < ∞,0 ≤ t1 < t 2 , k = 0,1,2,... .
r
r >0
r
r >0
Since the frequency average of the kinetic energy is decreasing in time, the kinetic energy of
rˆ
U ( k ) is bounded for all forward time
rˆ
rˆ
rˆ
r
0 < ∑r |U ( k ) (t , r ) | 2 ≤ ∑r |U 0( k ) | 2 = k , t ≥ 0 ⇒ sup rr >0r | U ( k ) | (t ) < ˆ k ,
r
r >0
r
r >0
(2-51)
t ≥ 0, k = 0,1,2,3,...
It follows by the strict inequality of (2-43) and the boundedness of the frequency average that
the time average of the potential energy is bounded for all forward time. Since each term in the
total energy formula is finite on [0, ∞ ) the formulas (2-44) are well posed.
rˆ
rˆ
r
b. Since || U ( k ) || l2 < ∞ it follows that U ( k ) (t , r ) is bounded for all forward time uniformly with
r
respect to r ∈ 3 . By the fundamental (extension) theory of ordinary differential equations it
rˆ
r
r
follows that for each r ∈ 3 , U ( k ) (t , r ) is unique and continuous in t for all forward time
r
uniformly with respect to r ∈ 3 .
rˆ
r
r
Since U ( k ) (t , r ), k = 0,1,2,... is in L2 ( 3 ) for all forward time. Since
rˆ
rˆ
rˆ r
r
|| U ( k ) || l2 ≤|| U ( k ) || l∞ , k = 0,1,2,... , U (t , r ) is bounded everywhere in r ∈ 3 for all forward time t.
END PROOF
The following lemma establishes a link between the inequalities (2-18) in the hypothesis of
theorem 2-2 and inequality (5) and its higher order analogs.
r
r
r
r
Lemma 2-3. If U ( k ) ⋅ F ( k ) , ∇U ( k )∈ L2 [0,1]3 , k = 0,1,2,..., t ≥ 0 such that the U ( k ) , k = 0,1,2,...
satisfy the Navier-Stokes partial differential equations and its finite order time derivatives satisfy
t
∫
t
r (k ) r (k ) r
U
⋅
F
d
x
ds
<
η
∫
∫
0 [ 0 ,1]3
r (k ) 2 r
|
∇
U
| dxds, t ≥ 0, k = 0,1,2,...
∫
(2-52)
0 [ 0 ,1]3
rˆ
rˆ
rˆ
for all forward time if and only if U ( k ) ⋅ F ( k ) , ∇U ( k )∈ l 2 ( 3 ), k = 0,1,2,..., t ≥ 0 such that
rˆ
U ( k ) , k = 0,1,2,... satisfy the Navier-Stokes ordinary differential equations and its finite time
derivatives satisfy
18
t
t
rˆ ( k ) v rˆ ( k ) r
r 2 rˆ ( k ) r 2
2
|
U
(
s
,
r
)
⋅
F
(
s
,
r
)
|
ds
<
4
π
η
|
r
∫∑
∫ ∑ | |U ( s, r )| ds, t ≥ 0,η > 0, k = 0,1,2,...
r
0 r >0
r
0 r >0
(2-53)
PROOF
By the strict inequality, the integral on the left of (2-32a) over R 3 must be finite. In order for the
Fourier transforms which appear in (2-32b) to be well defined
r
r
r
r
F ( k ) ⋅ U ( k ) , ∇U ( k ) ∈L 2 ([0,1]3 ), k = 0,1,2,..., t ≥ 0 if and only if
r
rˆ
rˆ
r rˆ
U ( k ) ⋅ F ( k ) , | ω | 2 U ( k ) ∈l 2 (W 3), k = 0,1,2,..., t ≥ 0.
r
r
r
r
Note that, since F ( k ) , k = 0,1,2,... is smooth in x ∈ [0,1]3 , U ( k ) ⋅ F ( k ) is integrable by Ho&&lder ' s
r
r
inequality if only U ( k ) ∈ L1 ([0,1]3 ) without assuming (2-28a).
Now suppose
t
t
rˆ ( k ) v rˆ ( k ) r
r 2 rˆ ( k ) r 2
2
|
U
(
s
,
r
)
⋅
F
(
s
,
r
)
|
ds
<
4
π
η
|
r
∫∑
∫ ∑ | |U (s, r )| ds, t ≥ 0,η > 0, k = 0,1,2,...
r
0 r >0
r
0 r >0
(2-54)
By the definition of the discrete Fourier transform (i.e. the Fourier coefficient) the previous
inequality is equivalent to
t
r (k )
∫ ∑ ∫U
r
3
0 r ∈ [ 0 ,1]3
t
4ηπ
2
∫∑
r r
r
r
( s, x ) ⋅ F ( k ) ( s, x ) | × | sin 2πr1 x sin 2πr2 y sin 2πr3 z| 2 dxds <
r (k ) r
r
|
U
( s, x ) ⋅ ∇ | 2 | sin 2πr1 x sin 2πr2 y sin 2πr3 z| 2 dxds, t ≥ 0, k = 0,1,2,...
∫
(2-55)
r
3
0 r ∈ [ 0 ,1]3
r
Since 0 ≤| sin 2πr1 x sin 2πr2 y sin 2πr3 z| 2 ≤ 1, x ∈ R 3 , r ∈ 3 (2-55) holds if and only if
t
∫
t
r r r r
r
U
(
s
,
x
)
⋅
F
(
s
,
x
)
|
d
x
ds
<
η
∫
∫
0 [ 0 ,1]3
r
r
r
∫ | ∇U (s, x ) | dxds, t ≥ 0, k = 0,1,2,... .
2
(2-56)
0 [ 0 ,1]3
END PROOF
rˆ r rˆ
r
Theorem 2-3. If F (t , r ), hi , i = 1,2,3 , and all time derivatives are discrete Schwartz in r ∈ 3 for
rˆ r
r
all forward time and U 0 ( r ) is discrete Schwartz in r ∈ 3 then any finite time derivative of
rˆ r
r
U (t , r ) ( the solution of the equation of lemma 2-2) is Schwartz in r ∈ 3 .
PROOF
19
rˆ
r
a. For any fixed t, k U ( k ) is discrete Schwartz in r ∈ 3
rˆ
r
sup rr∈ 3 | r | p | U ( k ) |< ∞, ∀p ∈ W
rˆ
The upper bound for U ( k ) is uniform in k
rˆ
r
sup k∈W sup rr∈ 3 | r | p | U ( k ) |< ∞, ∀p ∈ W
rˆ
c. The upper bound for U ( k ) is uniform for all t ≥ 0 .
rˆ
r
sup t ≥ 0 sup k∈W sup rr∈ 3 | r | p | U ( k ) |< ∞, ∀p ∈ W
(2-57)
b.
(2-58)
(2-59)
By the variation of constants formula, it suffices to show that the convolution integral is discrete
rˆ r rˆ r
2 r2 r
r
ˆ
Schwartz since Ψ (r ), Φ(t , r ) =e −η 4π |r | t U 0 are discrete Schwartz in r ∈ 3 by inspection given
the hypotheses on the initial and boundary conditions.
Multiply the time convolution by any monomial formed by the product of any finite powers of
the frequency components
rˆ
sup rr∈ 3 | r1p (1) r2p ( 2 ) r3p ( 3) || U |≤
rˆ r
sup rr∈ 3 | r1p (1) r2p ( 2 ) r3p ( 3) || Ψ (r ) | +
rˆ
2 r2
sup rr∈ 3 e −η 4π |r | t | U 0 || r1 p (1) r2p ( 2 ) r3p (3) | +
t
sup rr∈ 3 | r1
p (1)
p ( 2 ) p ( 3)
2
3
r
r
|| ∫ e
r
−η 4π 2 |r |2 ( t − s )
0
r
r
(2-60)
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 rr
π
0
1
0
[
2
((
U
r
)
∗
U
)
−
F
]ds |
−

 | rr | 2 



 0 0 1
η > 0, p ∈ W , t ≥ 0, r ∈ 3 , k = 1,2,3,
Simplify the upper bound on the Schwartz weighted momentum coefficient of (2-60)
rˆ
r
sup rr∈ 3 | r1 p (1) r2p ( 2 ) r3p ( 3) ||U ( k )(t , r ) |≤
rˆ
2 r2
r
sup rr∈ 3 | r1 p (1) r2p ( 2 ) r3p ( 3) || (−4η ) k | π 2 k | r | 2 k e −η 4π |r | t | U 0 |
t
+ sup rr∈ 3 | r1
p (1)
p ( 2 ) p ( 3)
2
3
r
r
|| ∫ e
r
−η 4π 2 |r |2 ( t − s )
0
r
r
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 rr
π
0
1
0
[(
2
(
U
r
)
∗
U
) − F ]ds |
−
r



| r |2



 0 0 1
(2-61)
η > 0, p ∈ W , t ≥ 0, r ∈ 3 , k = 1,2,3,
20
The Fourier series coefficient of the solution of Laplace’s equation is discrete Schwartz
r r
rˆ
rˆ
rˆ r
h (r ) ∈ S ( 2 ) ⇒ Ψ (r ) ∈ S ( 3 ) ⇒
rˆ r
r
∀p ∈ W 3 , sup rr∈ 3 | r1 p (1) r2p ( 2 ) r3p (3) || Ψ (r ) |< M 1 .
(2-62)
rˆ
rˆ
Since U 0 ∈ S ( 3 )
r
∀t ≥ 0, p ∈W 3
sup rr∈ 3 e −η 4π
2
r
|r |2 t
rˆ
rˆ
| U 0 | r1p (1) r2p ( 2) r3p ( 3) ≤ sup rr∈ 3 | U 0 | r1p (1) r2p ( 2) r3p (3) ≤ M 2
(2-63)
Since
rˆ rˆ
rˆ r
rˆ rˆ
rˆ rˆ
U ∈ S ( 3 ) ⇒ (Ur t ) ∗ U ∈ S ( 3 ), F ∈ S ( 3 )
and
 r r t 1 0 0  


 rr
|  r 2 −  0 1 0  |< 2, r ∈ 3.
|r | 


 0 0 1
Use the norm on the matrix operator immediately above to get the following upper bound
rˆ
sup rr∈ 3 | r1p (1) r2p ( 2 ) r3p ( 3) || U ( k ) |≤
t
sup rr∈ 3 | r1p (1) r2p ( 2 ) r3p ( 3) || ∫ e
0
t
≤ 2 ∫ e −η 4π
2
r
|r |2 ( t − s )
2
r
|r |2 ( t − s )
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 rr
π
0
1
0
[
2
((
U
r
)
∗
U
)
−
F
]ds |
−
r

| r |2 



 0 0 1
rˆ r
rˆ
| r1p (1) r2p ( 2) r3p (3) | ((Ur t ) ∗ U )ds |
r
−η 4π |r |2 ( t − s )
2
2π sup rr∈ 3 sup s ≥0
(2-64)
0
t
+ 2 ∫ e −η 4π
rˆ
sup s ≥0 | r1p (1) r2p ( 2) r3p ( 3) || F |ds.
0
The convolution and body force terms in the integrand have upper bounds which are uniform
over all time and any order k of the derivative.
21
rˆ
sup rr∈ 3 | r1p (1) r2p ( 2) r3p ( 3) || U ( k ) |≤
t
2 ∫ e −η 4π
2
r
|r | 2 ( t − s )
[ 1 + 2 ]ds ≤
(2-65)
0
2[ 1 + 2 ]
1
r
1
, | r |≥ 3
r 2 ≤ 2[ 1 + 2 ]
2
η 4π | r |
η12π
2
For the kth derivative with respect to time of the classical Fourier coefficient
t
sup rr∈ 3 | r1
p (1)
p ( 2 ) p ( 3)
2
3
r
r
|| ∫ e
0
t
≤ 2 ∫ e −η 4π
2
r
|r |2 ( t − s )
2
r
|r |2 ( t − s )
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 rr
0
1
0
[(
2
(
)
) − F ]ds
U
r
∗
U
π
−

 | rr | 2 



 0 0 1
rˆ r
rˆ
| r1p (1) r2p ( 2 ) r3p (3) | (2π (Ur t ) ∗ U ) ( k ) ds |
r
−η 4π 2 |r |2 ( t − s )
2π sup rr∈ 3 sup s ≥0
0
t
+ 2 ∫ e −η 4π
rˆ
sup s ≥ 0 | r1p (1) r2p ( 2 ) r3p ( 3) || F ( k ) |ds ≤
(2-66)
0
t
2 ∫ e −η 4π
2
r
|r | 2 ( t − s )
[ P1 + P2 ]ds ≤
0
2[ P1 + P2 ]
1
r
1
, | r |≥ 3.
r 2 ≤ 2[ P1 + P2 ]
2
η 4π | r |
η12π
2
For coefficients in discrete Schwartz spaces the integer weights are unrestricted. Hence any sum
of squares can be exceeded by a product which is a single square. Thus the Schwartz norm has
two equivalent formulations
rˆ
rˆ
r
sup pr∈W 3 | U 0 | r1p (1) r2p ( 2) r3p ( 3) = sup p∈W | r | p | U 0 | .
(2-67)
The homogeneous term has a uniform upper bound
r
∀t ≥ 0, p ∈W 3
sup rr∈ 3 e −η 4π
2
r
|r |2 t
rˆ
rˆ
| U 0 | r1p (1) r2p ( 2 ) r3p (3) ≤ sup rr∈ 3 | U 0 | r1p (1) r2p ( 2) r3p (3) ≤ M 2
(2-68)
The pressure function
r
Pˆ (t , r ) =
rˆ r t rˆ rˆ r
r
r
1
3
r 2 {r ⋅ [ 2πUr ∗ U − F (t , r )]}, t ≥ 0, r ∈ 2π | r |
(2-69)
22
rˆ rˆ s
r
is discrete Schwartz in r since U , Ur t are discrete Schwartz by hypothesis, the convolution of
discrete Schwartz coefficient is discrete Schwartz and the sum of discrete Schwartz functions
rˆ r rˆ rˆ r
[Ur t ∗ U − F (t , r )] is discrete Schwartz.
r
By the same reasoning any finite order time derivative of the pressure transform Pˆ (t , r ) is
discrete Schwartz.
END PROOF
The following theorem is the main result of this paper. It interprets the results obtained for the
rˆ
Fourier coefficients U , Pˆ which solve the Navier-Stokes ODE to the Fourier series
rˆˆ ˆ
representation of the functions U , Pˆ which solve the Navier-Stokes PDE.
r
r
r
r
Theorem 2-4. Suppose U ( k ) ⋅ F ( k ) , ∇U ( k )∈ L2 [0,1]3 , k = 0,1,2,..., t ≥ 0 where the U ( k ) , k = 0,1,2,...
satisfy the Navier-Stokes partial differential equations and its finite order time derivatives such
that
t
t
r (k ) r (k ) r
r (k )
r
U
⋅
F
d
x
ds
<
η
||
U
⋅ ∇ || 2 dxds, t ≥ 0,η > 0, k = 0,1,2,...
∫∫
∫∫
0 D
(2-70)
0 D
r
r
where F ( k ) is smooth in (t , x ) and bounded in t ∈ [0, ∞) . Then every finite time derivative of the
solution of the Navier-Stokes momentum equation is bounded, continuous and uniquely
r
determined in t. It is also smooth in x ∈ [0,1]3 for all forward time.
PROOF
rˆˆ
The conclusion follows directly by the properties of the Fourier series representation U of the
velocity function and theorem 2-2. In particular,
rˆ rˆ
rˆˆ r r
U ∈ S ( 3 ), t ≥ 0 ⇒ U = U ∈ C ∞ ([0,1]3 ), t ≥ 0
.
r ∞ r∞
r ∞ r∞
rˆ
rˆˆ r
r
r
3
3
U ∈ (C ∩ L )([0, ∞ )), r ∈ ⇒ U = U ∈ (C ∩ L )([0, ∞ )), x ∈ [0,1]
(2-71)
By the projection formula (or by solving the Navier-Stokes momentum equation for ∇ P ) and
the first part of this theorem, the pressure gradient satisfies the same properties as each
r
component of the momentum vector (marginal smoothness in t , x , uniqueness, and boundedness
for all forward time).
END PROOF
23
Acknowledgments: The author studied the problem formulation in [7]. He received helpful
comments from an anonymous reviewer for the Proceedings of the Royal Society of Edinburgh,
Steve Krantz and the Journal of Mathematical Analysis and Applications, Gene Wayne, and
Charles Meneveau.
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1962 (Dover 1989).
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with o-Slip Boundary Conditions, Proceedings of the American Mathematical Society,
Volume 127, Number 3, March1999, pp. 725-735.
3. Bardos, C., A Basic Example of onlinear Equations, The avier-Stokes Equations,
2000, http://www.its.Caltech/~siam/tutorials/Bardos_The_NSE.pdf
4. Cafferelli, L. Kohn, R. Nirenberg, L. Partial Regularity of Suitable Weak Solutions of the
Navier-Stokes Equations, Comm. Pure. Appl. Math 35 (1982).
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www.claymath.org/millenium/NavierStokes_equations/
6. Goldberg, R., Methods of Real Analysis, 2nd edition, John Wiley &Sons, New York,
1976.
7. Hale J. , Ordinary Differential Equations, Krieger, Malabar, FLA , 1980.
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Academic Press, N.Y, 1974.
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J.Math.Pures Appl., se’rie 9, 13, 331-418,1934.
10. Powers, D.L., Boundary Value Problems, Second Edition, 1970.
11. Rivera, W.M., A ew Invariant Manifold with an application to smooth conjugacy at a
node, Dynamic Systems and Applications 2 (1993) 491-503.
12. Stein E. and Weiss G., Introduction to Fourier Analysis, Princeton University Press,
Princeton, NJ, 1971.
13. Sternberg S., Local contractions and a theorem of Poincare' , Amer. J. Math. 79, 1957,
809-824.
14. Strauss, W. A., Partial Differential Equations: An Introduction, John Wiley&Sons, New
York, 1992.
15. Vanderbauwhede, A. Centre Manifolds, ormal Forms and Elementary Bifurcations,
Dynamics Reported, Volume II, Kirchgraber and Walther,1989.
24