A Fourier Integral approach to solving the avier-Stokes equations on
an unbounded space domain
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 continuous
families of ordinary differential and linear equations for the classical Fourier coefficients. Prove
theorems on existence, uniqueness and smoothness of solutions of the Navier Stokes equations.
Interpret these results for solutions of the Navier Stokes partial differential equations using the
r rˆˆ
ˆ
Fourier integral representation U = U , P = Pˆ .
Key Words: Fluid Mechanics, incompressible Navier-Stokes equations
1. Introduction
The main result in this paper can be stated as follows. If the data is smooth, spatially Schwartz,
and the body force and its higher order time derivatives satisfy generalized sector conditions,
r
then the (space) average of the kinetic energy of U (k ) is bounded for all forward time. A unique
r
physical solution (U , P) exists which is smooth in t ≥ 0 . The solution is the extension of the
unique regular (jointly smooth) short time solution determined by the data. Solutions are unique,
separately smooth and bounded in time and smooth in space 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 the author knows all frequency domain formulas but one which appear in this paper are
new. Cannone uses the Fourier transform to rewrite the variation of constants formula solving
the Navier-Stokes evolution equations in the formula just prior to (27) on page 15 of Harmonic
Analysis Tools for Solving the Incompressible avier-Stokes Equations.
The vector form of the Navier-Stokes equations for spatially Schwartz data is
1
r
r
r
r
r
r
U t + U ⋅ ∇U = η∆U − ∇P + F ,η > 0, x ∈ R 3 , t ≥ 0
r
r
U ⋅ ∇ = 0, x ∈ R 3 , t ≥ 0
r r
r r r
U (0, x ) = U 0 ( x ), x ∈ R 3 .
(1-1)
r r
The author seeks a unique solution of (1-1) given the data U 0, F which is jointly smooth in
space and time and bounded for all forward time.
r
r
In (1-1) U is the velocity vector field and ∇P is the pressure gradient to be determined, ∇ ⋅ U is
r
r
the divergence of the vector field and U ⋅ ∇ is the tensor matrix DxrU . Equation (1-1) is
equivalent to the formulation in [7] (7)-(11) Fefferman.
r
r
r
The body force F is smooth on [0, ∞) × R 3 F and the initial function U 0 are Schwartz (smooth)
on R 3 .
r
Since U t is the acceleration, the momentum equation (first equation of (1-1)) can also be
interpreted as Newton’s second law for incompressible fluids since the net force appears on the
left hand side if both sides are multiplied by the mass m. In fact the momentum equation is
Newton’s second law of motion for fluids combined with a dynamic version of Archimedes’ law
r
r r
r r
r
of hydrostatics. If U (t , x ) = 0 , the equation reduces to Archimedes’ law ∇P (t , x ) = F (t , x ) .
The second equation of (1-1) is called the equation of continuity. It is the reduction of the more
general Navier-Stokes equation for the density which follows from the incompressibility of the
fluid.
In this paper the following equations for the Fourier transform of the velocity are equivalent to
(1-1).
rˆ r
rˆ r r r r rˆ r r
rˆ r
r v r
r
r
dU (t , ω )
= −η | ω | 2 Uˆ (t , ω ) − i ∫ U (t , ω − φ )(ω − φ ) t U (t , φ ) dφ − iωPˆ (t , ω ) + F (t , ω ),
dt
Φ3
r
t ≥ 0, ω ∈ Ω 3
rˆ r r
r
U (t , ω ) ⋅ ω = 0, ω ∈ Ω 3 , t ≥ 0
rˆ
rˆ r r
r
U (0, ω ) = U 0 (ω ), ω ∈Ω 3
r r rˆ r
r
ω s U ∈L p (Ω 3 ), p = 1,2,3,.., ∞, s ∈W 3, t ≥ 0.
(1-2)
In (1-2) the Fourier transforms of the velocity, the initial velocity, the pressure and the body
rˆ rˆ rˆ
force are denoted by U ,U , F , Pˆ . Also i = − 1 and W 3 is the set of whole number triples.
0
The law governing the average mechanical energy of an incompressible fluid
2
Theorem 2-4 establishes the existence of a unique smooth solution defined for all forward time.
The following formulas provide a smooth generalization of Leray’s mechanical energy law ([12]
section 17 formula 3.4) for the Navier-Stokes equation with non zero body force.
t
t
r (k ) 2 r 1
r (k ) 2 r
r (k ) 2 r
r
r
r
1
|
|
U
d
y
−
|
U
|
d
y
=
−
|
∇
U
|
d
y
ds
+
U ( k ) ⋅ F ( k ) dyds,
η
0
∫
∫
∫
∫
∫
∫
2 R3
2 R3
0 R3
0 R3
(1-3)
η > 0, t ≥ 0, k = 0,1,2,...
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 ( k )
1
(2π )
−η
3
∫| U
r r
| 2 (t , ω )dω −
Ω3
1
(2π ) 3
t
∫
rˆ ( k )
1
(2π )
3
∫| U
Ω3
rˆ ( k ) r 2 r
∫ | ω | | U ( s, ω )| dωds +
r
r r
| 2 (0, ω )dω =
2
0 Ω3
1
(2π ) 3
t
∫
rˆ ( k ) r rˆ ( k ) r r
∫ U ( s, ω ) ⋅ F (s, ω )dωds, k = 0,1,2,...
(1-4)
0 Ω3
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
r
average kinetic energy of U (k ) .
The problem of finite time blow up
In theorem 2-3 the author shows that finite time blow up of solutions of (1-1) is impossible given
the conditions on the data, the equation of continuity, and the following conditions on the forcing
function
t
r (k ) r (k ) r
r (k ) 2 r
U
⋅
F
d
x
ds
<
η
|
∇
U
| dxds, t ≥ 0,η > 0, k = 0,1,2,...
∫∫
∫∫
t
0 R3
(1-5)
0 R3
In the frequency domain, inequalities (1-5) take the form
t
∫
t
rˆ ( k )
r rˆ ( k )
r r
r 2 rˆ ( k ) 2 r
U
(
s
,
ω
)
⋅
F
(
s
,
ω
)
d
ω
ds
<
η
|
ω
∫
∫ ∫ | | U | dωds, t ≥ 0,η > 0, k = 0,1,2,...
0 Ω3
(1-6)
0 Ω3
3
These conditions extend Leray’s result of 1934 in [9] that a solution continuous in the time
r r
variable and weakly differentiable in the space variables exists for all forward time when F = 0 .
r
Under the conditions (1-3) on F the solutions of (1-1) are bounded. Absolute
stability/boundedness extends the concept of Lyapunov stability/boundedness from
homogeneous nonlinear systems to nonlinear systems with a forcing function.
2. Existence and Extension
The notation Ω 3 = (−∞, ∞) 3 denotes the set of all continuous frequency triples; W denotes the set
of whole numbers.
r
Let S ( R 3 ) denote the space of spatially Schwartz functions. Here the initial function is in
r
S ( R 3 ) i.e.
r
r
r
sup xr∈R 3 | x p (1) y p ( 2) z p ( 3) | D k U 0 |≤ M k , k = 0,1,2,..., p ∈W 3, x = ( x, y, z ).
The notation
r
r
rˆ
r
U ∈ {C ∞ ([0, T ]), x ∈ R 3 } ∩ {S ( R 3 ), t ≥ 0}
denotes the space of functions which are separately smooth in time and space and Schwartz
r
r
over space. Thus U is separately smooth on (t , x ) ∈ [0, T ] × R 3 if it is smooth as a function of t
r
r
for each fixed x ∈ R 3 and smooth as a function of x for each fixed t ∈ [0, T ] .
The notation
r
rˆ
rˆ
r
U ∈ {C ∞ ([0, T ]), ω ∈ Ω 3 } ∩ {S (Ω 3 ), t ≥ 0}
used for functions of time and frequency in the Fourier transform domain is analogous.
Lemma 2-1. The Navier-Stokes ordinary differential equations for the Fourier transform of the
solution of the spatially Schwartz problem (1-1b) are
rˆ r
rˆ r r r rˆ r r r
rˆ r
r v r
r
dU (t , ω )
= −η | ω| 2 Uˆ (t , ω ) − i ∫ U (t , ω − φ )φ tU (t , φ )dφ −iωPˆ (t , ω ) + F (t , ω ),
dt
Φ3
r
t ≥ 0, ω ∈ Ω 3
r r r
r
U (t , ω ) ⋅ ω = 0, ω ∈ Ω 3 , t ≥ 0
rˆ
rˆ r r
r
U (0, ω ) = U 0 (ω ), ω ∈Ω 3
(2-1)
where the Fourier transform is defined by
4
rˆ r
r r
U (t , ω ) = ℑ{U (t , x )} =
r
1
∫ U (t , x)e
r r
− iω ⋅ x
(2π ) R 3
rˆ r
r
and similarly for F (t , ω ), Pˆ (t , ω ) .
3
r r
dx , ω ∈Ω 3 , t ≥ 0
(2-2)
PROOF
Note that the triple integral over R 3 in the definition of the Fourier transform (2-2) is well
r r r
defined for all forward time since, by the Schwartz property in x , F ,U 0 are continuous (in fact
r
smooth) and integrable over x ∈ R 3 and continuous and bounded on t ∈ [0, ∞ ) . Lemma 2-3
shows that the Fourier transform of the pressure, calculated below, is likewise well defined.
rˆ
dU r ˆ rˆ
The terms
, ω ⋅ P, F follow directly by the application of the Fourier transform to the terms
dt
of (1-1) and the differentiation property of Fourier transforms applied to the first partials in the
case of the pressure gradient term.
The transform of the Laplacian term can be calculated integrating by parts twice. Since the three
calculations are identical, it suffices to calculate the transform of the second partial derivative
with respect to the first component of the vector of space variables x
v −iωr ⋅ xr
U
∫ xx e dxdydz =
R3
∞ ∞
∫
r −iω (1) x ∞
| x = −∞ e −i[ω ( 2) y +ω (3) z ] dydz +
U
∫ xe
− ∞− ∞
(2-3)
r rr r
rˆ r
r r r
v
ω1 ∫ U x e −iω⋅ x dx = ω12 ∫ U e −iω⋅ x dx = −ω12U (t , ω ), t ≥ 0.
R3
R3
It follows that
v
r rˆ
ℑ{η∆U } = −η | ω | 2 U .
(2-4)
The Fourier transform of the Euler term is
r
r
r
r rr r
rˆ r r
rˆ r
rˆ r r r r rˆ r r
ℑ{U ⋅ ∇U } = ∫ U ⋅ ∇Ue −iω⋅ x dx = i[U (t , ω )ω t ] ∗ U (t , ω ) = i ∫ U (t , ω − φ )(ω − φ ) t U (t , φ )dφ
R3
rˆ r r rˆ r r r
= i ∫ U (t , φ )φ tU (t , ω − φ )dφ
Φ3
(2-5)
Φ3
The transform of the pressure gradient term is
r r
rˆ
− iω ⋅ x r
∇
Pe
d
x
=
i
ω
P.
∫
R
(2-6)
3
The transform of the equation of continuity is
5
r r
r r
r
− iω ⋅ x r
(
)
[
]
0
,
0
,
U
+
V
+
W
e
d
x
=
i
ω
U
+
ω
V
+
ω
W
=
i
ω
⋅
U
=
t
≥
ω
∈ Ω3.
x
y
z
1
2
3
∫
R
(2-7)
3
Divide both sides of (2-7) by i to obtain the third line of (2-1).
rˆ rˆ
rˆ rˆ
Since U ∈ S (Ω 3 ) ⇒ U ∈ L p (Ω 3 ), p = 1,2,3,..., ∞ the Schwartz conditions of (2-1a) permit the
r
r
Fourier integrals of U , P, F and the terms in (2-1) involving them to be well defined.
END PROOF
Remark 2-1. Equations (2-1) specify an infinite family of ordinary differential equations with a
continuous vector parameter whose solutions are the time dependent Fourier transforms of
r r r r r r
r
U (t , x ), F (t , x ),U 0( x ), P(t , x ) .
v r
r rˆ r r
Definition 2-1. Lωr Fˆ (t , ω ) = ω ⋅ F (t , ω ), ω ∈Ω 3 , t ≥ 0
(2-8)
r
r
Remark 2-2. For any ω , Lωr :C ∞ [0, ∞) → C ∞ [0, ∞) . The linear operator Lωr is a map from
vectors of infinitely continuous differentiable functions on [0, ∞ ) to a scalar continuous function
on [0, ∞ ) .
rˆ
d kU
Proposition 2-1. Any derivative of finite order of the velocity
, k = 0,1,2,... is in η ( Lωr ) .
dt k
PROOF
By (2-1) the assertion holds for the Fourier transform of the equation of continuity k = 0 . Take
the derivatives of order k = 1,2,3,... with respect to t of both sides of
rˆ r r
r
U (t , ω ) ⋅ ω = 0, ω ∈Ω 3 , t ≥ 0
(2-9)
to complete the proof.
END PROOF
Since differentiation of a function with respect to the space variables corresponds to frequency
multiplication of its transform, the following Banach space is the most economical one needed to
r
establish that solutions of the Navier-Stokes equations are smooth in (t , x ) ∈ [0, ∞) × R 3 .
rˆ
Definition 2-2. The Schwartz space S is the set of all three component vectors of functions
r
which are Schwartz in the frequency variable on ω ∈ Ω 3 . The solution space for the NavierStokes equations consists of functions which are smooth and bounded for all forward time,
6
r
uniformly in ω ∈ Ω 3 and Schwartz in the frequency for all t ≥ 0. Any finite order mixed partial
derivative with respect to frequency is bounded in the L p norm for any p = 1,2,..., ∞ with
r
respect to monomial frequency weights of any finite order on the space Sˆ (Ω 3 ) .
The following lemma shows that the family of differential equations simplifies to a linear
r r
ordinary vector differential equation at ω = 0 .
r r
Lemma 2-2. If ω = 0 the Navier-Stokes ordinary equations for the Fourier transform of the
velocity reduce to
rˆ
rˆ r rˆ r
rˆ r
dU r
(2-10)
(t ,0) = F (t ,0),U (0,0) = U 0 (0) .
dt
PROOF
Formula (2-10) follows immediately from equation (2-1).
END PROOF
The Fourier transform function is an auxiliary function for a family of equations which involves
it provided it does not appear in an equivalent form. The following lemma establishes that the
Fourier transform of the pressure function is an auxiliary function for the family of (ordinary
differential) equations (2-1). A closed formula for the Fourier transform of the pressure is
supplied.
Lemma 2-3. Equations (2-1) can be placed into the following equivalent form
 rr
1 0 0  
rˆ
rˆ rˆ
 rˆ r
r 2 rˆ  ωω t 
r
r r
U t = −η | ω | U +  r 2 −  0 1 0 [i (Uω t ) ∗ U − F ], t ≥ 0, ω ∈ Ω 3 , ω ≠ 0
|ω | 


 0 0 1
rˆ
rˆ r r
r
U (0, ω ) = U 0 (ω ), ω ∈ Ω 3
rˆ rˆ
U ∈ S ([0, T ) × Ω 3 ), T > 0.
(2-11)
The pressure coefficients satisfy the following equations
rˆ r rˆ r
r
r rˆ r r
r
r r
1
Pˆ (t , ω) = r 2 {−ω ⋅ i(U (t, ω)ω t ) *U + ω ⋅ F (t, ω)}, t ≥ 0, ω ∈ Ω3 , ω ≠ 0
|ω |
rˆ r r rˆ
r
r rˆ r r
r r
r r
1
Pˆ (0 + , ω) = r 2 {−ω ⋅ i(U 0 (ω)ω t ) *U 0 (ω) + ω ⋅ F (0 + , ω},ω ∈ Ω3 , ω ≠ 0.
|ω |
(2-12)
PROOF
7
Apply the linear operator Lωr of definition 2-1 to each side of equation (2-1) – where, by 2-5,
rˆ r
rˆ
rˆ r r r r rˆ r r
(Uω t ) ∗ U = ∫ (U (t , ω − φ )(ω − φ ) t U (t , φ )dφ -by forming the dot product of each term with the
Φ3
3
r
vector ω ∈ Ω to obtain
rˆ r
r dU (t , ω )
r r v r
ω⋅
= −ηω⋅ | ω| 2 Uˆ (t , ω )
dt
rˆ r r r r rˆ r r r r
r
r
r rˆ r r
− ω ⋅ ∫ iU (t , ω − φ )(ω − φ ) t U (t , φ dφ − ω ⋅ ωiPˆ (t , ω ) + ω ⋅ F (t , ω ), ω ∈ Ω 3 , t ≥ 0.
(2-13)
Φ3
rˆ r rˆ
r
r
By proposition 2-1, U (t , ω ),U t (t , ω ) ∈ η ( Lωr ), ω ∈ Ω 3 hence (2-13) reduces to
rˆ r r r r rˆ r r r r
r
r
r rˆ r r
0 = −ω ⋅ ∫ iU (t , ω − φ )(ω − φ ) t U (t , φ )dφ − ω ⋅ ωiPˆ (t , ω ) + ω ⋅ F (t , ω ), ω ∈ Ω 3 , t ≥ 0.
(2-14)
Φ3
r
Solve (2-14) for Pˆ (t , ω ) to obtain the first line of the pressure coefficient formulas. Insert the
first line of the pressure formula into (2-1) to obtain the first line of the velocity coefficient
formulas. The second line of the pressure transform formulas are obtained from the first line
rˆ r
from the initial transform U (t , ω ) .
s
The only remaining question is to show that Pˆ (t , ω ) is well defined. Equivalently, is it
r
bounded for ω ≠ 0 ?
By the Schwartz property in the frequency parameters,
rˆ r
rˆ r
∃M > 0 : max | F (t , ω} |, | Fω t |≤ M .
Thus
sup ωr∈Ω3
r rˆ
rˆ
r
(2-15)
r
| F (t , ω} |
ω ⋅ F (t , ω}
|
≤
r
r 2 |= sup ωr∈Ω
|ω |
|ω |
3
max{sup ωr∈Ω3
rˆ r
rˆ
rˆ
| Fω t |
| F |, sup ωr∈Ω3
r } = sup ωr∈Ω3 | F |, t ≥ 0.
|ω |
(2-16)
r
Since the convolution in the numerator is of quadratic order in ω ,
r
rˆ r r r r t rˆ r r
ω
r 2 ⋅ ∫ U (t , ω − φ )(ω − φ ) U (t , φ )dφ < ∞, t ≥ 0 .
|ω | Φ
(2-17)
3
8
r r
r r
Thus, the definition of Pˆ (t , ω ), ω ≠ 0 can be smoothly extended to Pˆ (t , ω ), ω = 0, t ≥ 0 by
calculating
r
lim ωr →0r Pˆ (t , ω ), t ≥ 0. .
(2-18)
END PROOF
Remark 2-3. From higher order derivatives of (2-1) and the projection of the kth order equation
r
of continuity, one can calculate formulas for Pˆ ( k ) (t , ω ) .
The equation that results from calculating any higher order derivative of (2-13a) is
 rr
1 0 0  
rˆ ( k +1)
rˆ r
rˆ
rˆ

r 2 rˆ ( k )  ωω t 
U
= −η | ω | U +  r 2 −  0 1 0 [(i (Uω t ) ∗ U ) ( k ) − F ( k ) ],
|ω | 


 0 0 1
r
t ≥ 0, ω ∈ Ω 3 , k ∈ r r
rˆ
r
U ( k ) (0, ω ) = 0, ω ∈ Ω 3 , k ∈ rˆ
r r
r
U ( k )(t , ω ) ⋅ ω = 0, t ≥ 0, ω ∈ Ω 3 , k ∈ .
(2-19a)
Remark 2-4. 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. The result is
that the Schwartz norm has two equivalent formulations
rˆ
rˆ
r
sup pr∈W 3 | ω1p (1)ω 2p ( 2)ω3p (3) || U 0 |= sup p∈W | ω | p | U 0 | .
(2-19b)
The first theorem establishes the existence of unique solutions for sufficiently short forward time
starting at time t =0 + . The following upper bound on the matrix operator of (2-19) is useful in
the proof of the first theorem
 r r t 1 0 0  


 ωω
0
1
0
−
r

 ≤
| ω |2



 0 0 1
r

 ω12 − | ω | 2

r 2 0 0
|ω |

 ω12 ω1ω 2 ω1ω 3  1 0 0  
r
r
r

 

ω 22 − | ω | 2  3 | ω | 2 − | ω | 2
1 
2

= 2.
r 2 0 |<
r 2
r |  ω1ω 2 ω 2 ω 2ω 3  −  0 1 0  |<| 0


| ω |2 
|
|
|
|
ω
ω



 ω1ω 3 ω 2ω 3 ω 32   0 0 1 
r 2
2


 0 0 ω3 − | ω | 
r

| ω | 2 

(2-20)
9
Theorem 2-1. Suppose
r
rˆ
rˆ v
r
F , Uˆ ∈ {C ∞ ([0, T ]), x ∈ R 3 } ∩ {S (Ω 3 ), t ≥ 0}
rˆ
rˆ
U 0 ∈ S (Ω 3 ).
rˆ
v r
There exists T > 0 such that Uˆ ∈ C ∞ ([0, T )) ∩ S (Ω 3 ), satisfies (2-11) and
rˆ
Pˆ ∈ C ∞ ([0, T )) ∩ S (Ω 3 ).
(2-21)
PROOF
The goal is to establish
rˆ
a. Continuity of U ( k ) in time
rˆ
rˆ
r
∃T > 0 :| U ( k ) (t 2 ) − U ( k ) (t1 ) |< M (k , ω ) | t 2 − t1 |< ∞, ∀t1 , t 2 ∈ [0, T ], M
rˆ
= sup 0≤t ≤T U ( k +1) (t ).
rˆ
b. The continuity of U ( k ) in time is uniform in k ∈ W
rˆ
rˆ
r
∃T > 0 :| U ( k ) (t 2 ) − U ( k ) (t1 ) |< M (ω ) | t 2 − t1 |< ∞,
rˆ
~
∀t1 , t 2 ∈ [0, T ], M = sup k∈W sup 0≤t ≤T U ( k +1) (t ).
rˆ
r
c. The continuity of U ( k ) is uniform in ω ∈Ω 3
~
rˆ
rˆ
~
∃T > 0 :| U ( k ) (t 2 ) − U ( k ) (t1 ) |< M | t 2 − t1 |< ∞, ∀t1 , t 2 ∈ [0, T ], M =
rˆ
sup ωr∈Ω 3 sup k∈W sup 0≤t ≤T U ( k +1) (t ).
(2-22)
(2-23)
(2-24)
Use the variation of constants formula to solve for the general Fourier transform of the
momentum
rˆ r
U (t , ω ) =
 r r t 1 0 0  
t
rˆ r t
rˆ
rˆ
rˆ
r


r
−η |ω |2 ( t − s )  ωω
3
e
U0 + ∫e
 | ωr | 2 −  0 1 0 [(i (Uω ) ∗ U ) − F ]ds, t ≥ 0, ω ∈ Ω .
0
 0 0 1



rˆ
Form the difference to establish continuity of the U in time
r
−η |ω |2 t
(2-25)
10
rˆ
rˆ
rˆ
rˆ
r2
r2
r
r
U (t 2 , ω ) − U (t1 , ω ) = e −η |ω| t ( 2 )U 0 − e −η |ω | t (1)U 0 +
t (2)
∫e
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 ωω
ω
0
1
0
[(
i
(
U
)
U
)
F
]ds −
∗
−
−
r



| ω |2
 0 0 1 



r
−η |ω |2 ( t ( 2 ) − s )
0
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 ωω
ω
e
0
1
0
[(
i
(
U
)
∗
U
)
−
F
]ds
−
r

∫0
| ω |2 



 0 0 1
r
t 2 > t1 ≥ 0, ω ∈ Ω 3
t (1)
(2-26)
r
−η |ω |2 ( t (1) − s )
Simplify the expression on the right side of (2-26)
rˆ
rˆ
r
r
U (t 2 , ω ) − U (t1 , ω ) =
rˆ
r2
r2
{e −η |ω | t ( 2) − e −η |ω| t (1) }U 0 +
t ( 2)
∫e
r
−η |ω |2 ( t ( 2 ) − s )
t (1)
t (1)
∫ [e
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 ωω
 | ωr | 2 −  0 1 0 [(i (Uω ) ∗ U ) − F ]ds −
 0 0 1 



r
−η |ω |2 ( t ( 2 ) − s )
−e
r
−η |ω |2 ( t (1) − s )
0
r
t 2 > t1 ≥ 0, ω ∈ Ω 3 .
(2-27)
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 ωω
] r 2 −  0 1 0 [(i (Uω t ) ∗ U ) − F ]ds
|ω | 


 0 0 1
Examine continuity near time 0 by setting t 2 = t , t1 = 0 in (2-27)
rˆ r
rˆ
rˆ
r2
r
U (t , ω ) − U (0, ω ) = {e −η |ω | t − 1}U 0 +
 r r t 1 0 0  
rˆ
rˆ
 rˆ r t

 ωω
∫0 e
 | ωr | 2 −  0 1 0 [((Uω ) ∗ U ) − F ]ds −
 0 0 1 



r
3
t ≥ 0, ω ∈ Ω .
t
r
−η |ω |2 ( t ( 2 ) − s )
(2-28)
By (2-20)
11
 r r t 1 0 0  
rˆ r
rˆ


 ωω
|  r 2 −  0 1 0  || ((Uω t ) ∗ U ) | ds ≤
∫0 e
|ω | 


 0 0 1
rˆ r t
rˆ r t −η |ωr |2 (t − s )
2 sup ωr∈Ω 3 sup s ≥0 | ((Uω ) ∗ U )ω || ∫ e
ds |=
t
r
−η |ω |2 ( t − s )
(2-29)
0
rˆ rˆ r
2 sup ωr∈Ω 3 sup s ≥0 | (U ∗ U ) | ω | 2
1
r
η |ω |
2
≤
2M 1
η
,t ≥ 0
Also
 r r t 1 0 0  
 rˆ

 ωω
sup ωr∈Ω3 ∫ e
|  r 2 −  0 1 0  || F | ds
|ω | 
0


 0 0 1
t
rˆ r
rˆ r t
r2
r2
2 sup ωr∈Ω 3 ∫ e −η |ω | ( t − s ) sup s ≥0 | F | | ω| 2 ds ≤ sup ωr∈Ω 3 sup s ≥0 | F || ω | 2 | ∫ e −η |ω | (t − s ) ds |
t
r
−η |ω |2 ( t − s )
0
(2-30)
0
rˆ r
= 2 sup ωr∈Ω3 sup s ≥0 | F || ω | 2
1
r
η |ω |
2
≤
1
η
rˆ
M
sup ωr∈Ω3 sup s ≥ 0 | F |< 2 2 , t ≥ 0.
η
It follows that
rˆ r
M
M
sup ωr∈Ω3 | U (t , ω ) |≤ Lt + 2[ 1 + 2 ], t ≥ 0,η > 0.
η
η
(2-31)
For derivatives of the momentum coefficients of any finite order k = 1,2,3,... the variation of
constants formula yields
rˆ
r2 r
r
r
ˆ
U ( k )(t , ω ) = (−η ) k | ω | 2 k e −η |ω | tU 0
 r r t 1 0 0  
rˆ ( k ) rˆ ( k )
 rˆ r t

 ωω
ω
0
1
0
[
i
(
U
)
∗
U
) − F ]ds,
+ ∫e
−
r

| ω |2 
0



 0 0 1
r
t ≥ 0, ω ∈ Ω 3 , k = 1,2,3,...
t
r
−η |ω |2 ( t − s )
(2-32)
Form the difference to prove continuity of the kth derivative of the momentum coefficient in
time,
12
rˆ
rˆ
rˆ
r2
r2
r rˆ
r
r
r
U ( k )(t 2 , ω ) −U ( k )(t1 , ω ) = (−η ) k | ω | 2 k e −η |ω| t ( 2 )U 0 − (−η ) k | ω | 2 k e −η |ω| t (1)U 0
t ( 2)
∫e
+
r
−η |ω |2 ( t ( 2 ) − s )
0
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 ωω
ω
0
1
0
[(
i
(
U
)
U
) − F ]ds −
∗
−
r


2

| ω |
 0 0 1 



 r r t 1 0 0  
t (1)
rˆ r
rˆ
rˆ
r2


 ωω
+ ∫ e −η |ω | (t (1) − s )  r 2 −  0 1 0 [(i (Uω t ) ∗ U ) ( k ) − F ( k ) ]ds
|ω | 
0


 0 0 1
r
t 2 > t1 ≥ 0, ω ∈ Ω 3 , k = 1,2,3,..
(2-33)
Simplify the difference of differentiated Fourier transforms at two distinct times
rˆ
rˆ
r2
r2
r rˆ
r
r
U ( k )(t 2 , ω ) −U ( k )(t1 , ω ) = (−η ) k | ω | 2 k [e −η |ω | t ( 2 ) − e −η |ω | t (1) ]U 0
t (2)
∫e
+
r
−η |ω |2 ( t ( 2 ) − s )
t (1)
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 ωω
i
U
U
0
1
0
[(
(
)
) − F ]ds −
ω
∗
−
r



| ω |2



 0 0 1
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 ωω
] r 2 −  0 1 0 [(i (Uω t ) ∗ U ) ( k ) − F ( k ) ]ds
−e
∫0 [e
|ω | 


 0 0 1
r
t 2 > t1 ≥ 0, ω ∈ Ω 3 , k = 1,2,3,...
t (1)
r
−η |ω |2 ( t ( 2 ) − s )
(2-34)
r
−η |ω |2 ( t (1) − s )
rˆ
r
To investigate continuity of U ( k )(t , ω ) near time 0, evaluate (2-34) at t1 = 0, t 2 = t
rˆ
rˆ
r2
r
r
U ( k )(t , ω ) = (−η ) k | ω | 2 k [e −η |ω | t − 1]U 0
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 ωω
ω
0
1
0
[(
i
(
U
)
∗
U
) − F ]ds
+ ∫e
−
r

| ω |2 
0



 0 0 1
r
t ≥ 0, ω ∈ Ω 3 , k = 1,2,3,...
t
r
−η |ω |2 ( t − s )
(2-35)
The upper bound on the momentum transform is
13
rˆ
rˆ
r2
r
r
|U ( k )(t , r ) |≤| η | k | ω | 2 k [1 − e −η |ω| t ] | U 0 |
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 ωω
|  r 2 −  0 1 0  | [| (i (Uω t ) ∗ U ) ( k ) | + | F ( k ) |]ds
+ ∫e
|ω | 
0


 0 0 1
r
t ≥ 0, ω ∈ Ω 3 , k = 1,2,3,...
t
r
−η |ω |2 ( t − s )
(2-36)
The upper bound of (2-36) can be simplified as follows.
rˆ
rˆ
r
r
r
|U ( k )(t , ω ) |≤| η | k | ω | 2 k η | ω | 2 | U 0 | t
 r r t 1 0 0  
rˆ r
rˆ
rˆ


 ωω
|  r 2 −  0 1 0  | [| (i (Uω t ) ∗ U ) ( k ) | + | F ( k ) |]ds
+ ∫e
|ω | 
0


 0 0 1
r
t ≥ 0, ω ∈ Ω 3 , k = 1,2,3,...
t
r
−η |ω |2 ( t − s )
By the hypothesis on the transforms of the data functions
rˆ r
sup t ≥0 | U 0 (ω ) |< L
rˆ
rˆ
r
r
r
r
r
r
sup s ≥0 | η | k | ω | 2 k | F ( k ) ( s, ω ) |<| ω |5k | F ( k ) ( s, ω ) |< M 2 (r ) < M 2 , | ω > η |
(2-37)
(2-38a)
(2-38b)
Since the variation of constants operator is defined on the smooth/ Schwartz transforms,
rˆ
r r
U ( s, ω )ω t is Schwartz and the convolution of Schwartz functions is a Schwartz function with
no effect on the smoothness in time,
rˆ r r
rˆ r
r
sup s ≥0 | ([(U ( s, ω )ω t ) ∗ U ( s, ω )]( k ) |< M 1 (ω ) < M 1 , k = 1,2,3,...
(2-39)
By (2-20),
t
∫e
r
−η |ω | 2 ( t − s )
0
 r r t 1 0 0  
rˆ r
rˆ


M (k )
 ωω
, k = 1,2,3,...
|  r 2 −  0 1 0  || ((Uω t ) ∗ U ) ( k ) | ds ≤ 2 1
η
|ω | 


 0 0 1
(2-40)
Also
14
 r r t 1 0 0  
 rˆ

 ωω
|  r 2 −  0 1 0  || F ( k ) | ds
∫0 e
|ω | 


 0 0 1
t
rˆ ( k ) r 2
rˆ ( k ) r 2 t −η |ωr |2 ( t − s )
r
−η |ω |2 ( t − s )
sup s ≥ 0 | F | | ω| ds ≤ sup ωr∈Ω3 sup s ≥ 0 | F || ω | | ∫ e
ds |
∫e
t
sup ωr∈Ω3
sup ωr∈Ω3
r
−η |ω |2 ( t − s )
0
rˆ
r
= sup ωr∈Ω3 sup s ≥ 0 | F ( k ) || ω | 2
(2-41)
0
1
r
η | ω |2
≤
1
η
rˆ
M (k )
sup ωr∈Ω3 sup s ≥0 | F ( k ) |< 2
, t ≥ 0, k = 1,2,3,...
η
It follows that
rˆ
r
M (k ) M 2 (k )
sup ωr∈Ω3 | U ( k ) (t , ω ) |≤ Lt + 2 1
+
, t ≥ 0,η > 0, k = 1,2,3,...
η
η
(2-42)
rˆ
r
Not only are the |U ( k )(t , ω ) | bounded for sufficiently short time but for all finite forward time.
rˆ
rˆ
Since U ( k ) , F ( k ) are Schwartz, Pˆ ( k ) is Schwartz in Ω 3 for all forward time follow automatically
from the given conditions on the boundary data.
END PROOF
The following proposition is used as a lemma for the next theorem.
rˆ
r
Proposition 2-2. The inner product of U ( k )(t , ω ) with the transformed Euler (convolution) term
satisfies
rˆ
rˆ r
r rˆ r
r
U ( k )(t , ω )[Uω t ∗ U (t , ω )] ( k ) = 0, k = 0,1,2,3,..., t ≥ 0, ω ∈ Ω 3 .
PROOF
The Fourier transform of the equation of continuity is
rˆ
r r
U ( k ) (t , ω ) ⋅ ω = 0, t ≥ 0.
(2-43)
By the Liebnitz rule, the kth order Euler term can be written
15
rˆ
r r r r r r
U ( k ) (t , ω ) ⋅ [U (t , ω )ω t ∗ U (t , ω )]( k ) =
rˆ
r
r k k  r r r
r
U ( k ) (t , ω ) ⋅ ∑  [U (t , ω )ω t ](l ) ∗ U ( k −l ) (t , ω ) =
l =0  l 
k
rˆ
r
r r r
k  r r r
r r
U ( k ) (t , ω ) ⋅ ∫ ∑  [U (t , ω − σ )](l ) (ω − σ ) t U ( k −l ) (t , σ )dσ
l
Ω3 l =0  
k
rˆ
k  r r r
r r r r
r r
= ∫ ∑  [U (t , ω − σ )]( l ) ⋅ U ( k ) (t , ω )(ω − σ ) ⋅ U ( k −l ) (t , σ )dσ .
l
Ω3 l =0  
(2-44)
The previous quantity is bounded above by
≤
k
k  r
r
r
∫ ∑  l  | U (t , ω − σ )]
(l )
r
rˆ
r
r r
r
r
| 2 | U ( k ) (t , ω ) | 2 | (ω − σ ) | 2 | U ( k −l ) (t , σ ) | 2 dσ
 
rˆ
rˆ r r
r
r r
 k  rˆ
r
r
≤ ∫ ∑   | U ( k ) (t , ω ) | 2 | U ( k −l ) (t , σ ) | 2 | U (t , ω − σ )](l ) | 2 | (ω − σ ) | 2 dσ
l
Ω3 l =0  
Ω3 l =0
k
=
(2-45)
 k  rˆ ( k ) r 2 rˆ ( k −l ) r 2 rˆ r r ( l ) r r 2
∫ ∑  l  | U (t, ω ) | | U (t ,σ ) | | U (t , ω − σ )] ⋅ (ω − σ ) | dσ = 0.
Ω3 l = 0  
k
The first equality follows by the Liebnitz rule and the fact that the time derivative distributes
over the convolution, the second by the definition of the convolution. The third line follows by
matrix vector multiplication, the fourth by the Schwartz inequality applied continuously to the
r
both families of vector dot products in R 3 parameterized by ω ∈ Ω 3 , the fifth by the definition
r r r
of the inner product of a vector with itself x ⋅ x =| x| 2 . The final inequality follows by applying
rˆ
r r
r
the higher order equation of continuity to each term (U ( l ) (t , ω ) ⋅ ω = 0, l = 0,1,2,..., k , t ≥ 0, ω ∈Ω 3 .
Similarly,
rˆ
r r r
r r
r
U ( k ) (t , ω ) ⋅ [U (t , ω )ω t ∗ U (t , ω )] ( k ) ≥
k
(2-46)
rˆ
rˆ r r
r
r r
 k  rˆ
r
− ∫ ∑   | U ( k ) (t , ω ) | 2 | U ( k −l ) (t , σ ) | 2 | U (t , ω − σ )]( l ) ⋅ (ω − σ ) | 2 dσ = 0, k = 0,1,2,...
l
Ω3 l =0  
Hence
rˆ
r r r
r r
r
r
U ( k ) (t , ω ) ⋅ [U (t , ω )ω t ∗ U (t , ω )] ( k ) = 0, t ≥ 0, ω ∈Ω 3 , k = 0,1,2,... .
(2-47)
END PROOF
The next theorem extends the domain of definition of solutions of (2-11) by showing that
solutions are bounded 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. It is equivalent to formula (1-4).
16
r
Theorem 2-2. Suppose ¬{U 0( k ) ≡ 0}, k = 0,1,2,... and
t
1
(2π ) 3
∫
rˆ ( k )
r rˆ ( k )
r r
ω
U
s
(
,
) ⋅ F ( s, ω )dωds < η
∫
0 Ω3
t
1
(2π ) 3
∫
r 2 rˆ ( k ) 2 r
ω
|
∫ | | U | dωds,
0 Ω3
(2-47)
t ≥ 0,η > 0, k = 0,1,2,...
a. Then the following formulas are well defined
rˆ ( k )
1
(2π )
−η
3
∫|U
r r
| 2 (t , ω )dω −
Ω3
t
1
(2π ) 3
∫
rˆ ( k )
1
(2π )
3
∫|U
0
Ω3
rˆ ( k )
r
r
ω
|
|
|
( s, ω )| 2 dωds +
U
∫
r
r r
| 2 (ω )dω =
2
0 Ω3
1
(2π ) 3
t
∫
rˆ ( k )
r rˆ ( k )
r r
ω
U
(
s
,
) ⋅ F ( s, ω )dωds,
∫
(2-48)
0 Ω3
k = 0,1,2,...
rˆ
r
b. The solution of (2-12) and every finite time derivative U ( k ) (t , ω ), k = 0,1,2,.. of it has a unique
extension (with respect to time) which is bounded and continuous in t for all forward time,
continuous (and asymptotically vanishing in Ω 3 ) and jointly continuous and bounded almost
rˆ
r
r
everywhere on [0, ∞ ) × Ω 3 such that U ∈ ( L1 ∩ L2 )(Ω 3 ), t ≥ 0.
PROOF
First note that
r
rˆ
r r
¬{U 0( k ) ≡ 0}, k = 0,1,2,... ⇒ ∫ | U 0( k ) | 2 ( x )dx > 0, k = 0,1,2,...
R
(2-49)
3
Thus
rˆ ( k )
1
(2π )
3
∫| U
Ω
0
r r
| 2 (ω )dω > 0, k = 0,1,2,...
(2-50)
3
By Plancherel’s theorem
1
rˆ ( k ) 2 r r
rˆ ( k ) 2 r r
|
U
|
(
ω
)
d
ω
<
∞
⇔
|
U
0
∫
∫ 0 | ( x )dx < ∞, k = 0,1,2,...
(2π ) 3 Ω3
R
(2-51)
3
rˆ
Next construct the formulas for the average energy of U ( k ) , k = 0,1,2,... Then eliminate the
modified Euler terms to simplify these formulas. Form the dot (Hermitian) product of each
17
rˆ ( k ) r
r
equation (2-19) with the complex vector U * (t , ω ), k = 0,1,2,... , integrate over ω ∈ Ω 3 and
integrate from 0 to t to obtain
1
( 2π )
+i
+
3
rˆ ( k ) r 2 r
∫ |U (t , ω )| dω −
Ω3
t
1
∫∫
( 2π ) 3
( 2π )
3
( 2π )
3
rˆ ( k )
r 2 r
∫ | U (0, ω )| dω = −η
Ω3
t
1
( 2π )
3
r
rˆ ( k )
∫ ∫| ω | | U
2
r
| 2 dωds
0 Ω3
rˆ ( k )
r r
r r rˆ ( k ) r rˆ * r r r
(
(
,
)
(
U
s
−
⋅
− φ ))U ( s, φ )U ( s, ω ) dφ dω
ω
φ
ω
∫
(2-52)
0 Ω3 Φ 3
t
1
1
rˆ * ( k ) rˆ ( k ) r
⋅ F dωds.
∫ ∫U
0 Ω3
Prior to integration from 0 to t, the transformed Euler term which would appear in (2-48) is
∫
rˆ
r r r r rˆ r rˆ * r r r
(
U
(
s
,
ω
− φ ) ⋅ (ω − φ ))U ( s, φ )U ( s, ω )dφ dω.
∫
(2-53)
Ω3 Φ 3
The higher order time derivatives of the transformed Euler term
rˆ
rˆ
r r r r rˆ r
r r r
i ∫ ∫ {(U ( s, ω − φ ) ⋅ (ω − φ ))U ( s, φ )}( k ) U ( k )* ( s, ω )dφ dω , k = 0,1,2,...
(2-54)
Ω3 Φ 3
vanish by proposition 2-2.
It follows immediately by (2-47), that
rˆ ( k
rˆ ( k )
r 2 r
r
r
|
U
(
t
,
ω
)|
d
ω
−
|
U
(t1 , ω )| 2 dω < 0,0 ≤t 1 < t 2 , k = 0,1,2,... .
2
∫
∫
Ω3
(2-55)
Ω3
In particular
1
rˆ ( k ) 2 r r
U
|
| (t , ω )dω ≤
∫
(2π ) 3 Ω3
1
rˆ ( k ) 2
r r
|
U
∫ 0 | (0, ω )dω = M k < ∞, t ≥ 0, k = 0,1,2,...
(2π ) 3 Ω3
(2-56)
By (2-56) and the strict inequality of (2-47) , it follows that the time average of the potential
energy is bounded for all forward time. Thus all terms appearing in formulas (2-48) are well
defined for all forward time.
rˆ
rˆ
r
Since || U || L2 < ∞ it follows that U ( k ) (t , ω ), k = 0,1,2,.... is bounded in t for all forward time
r
uniformly bounded with respect to ω ∈ Ω 3 . By the fundamental (extension) theory of ordinary
18
rˆ
r
differential equations U ( k ) (t , ω ), k = 0,1,2,... is continuous for all forward time uniformly with
r
respect to ω ∈ Ω 3 .
r
r
By the corresponding inequalities (1-5) it follows that U ( k ) (t , x ), k = 0,1,2,... is in ( L1 ∩ L2 )( R 3 )
rˆ
r
for all forward time, hence U ( k ) (t , ω ), k = 0,1,2,..., t ≥ 0 is continuous and asymptotically
r
rˆ
r
vanishing as a function of ω . Thus U ( k ) ∈ C 0 (Ω 3 ) . By the inequalities (2-47),
rˆ
r
r
U ( k ) ( L1 ∩ L2 )(Ω 3 ), k = 0,1,2,..., t ≥ 0. .
END PROOF
r
rˆ
r
r
r
Lemma 2-4. If U ( k ) ⋅ F ( k ) , ∇U ( k ) ∈ S ( R 3 ) ⊂ L 2 ( R 3 ), k = 0,1,2,..., t ≥ 0 such that U (k ) satisfies
the Navier-Stokes partial differential equation (1-1) and its finite time derivatives satisfy
t
t
r (k ) r (k ) r r
r (k ) r 2 r
U
⋅
F
(
s
,
x
)
d
x
ds
<
η
∫∫
∫ ∫ | ∇U (s, x ) | dxds, t ≥ 0, k = 0,1,2,...
0 R
3
0 R
(2-57a)
3
rˆ
rˆ
r rˆ
for all forward time if and only if U ( k ) ⋅ F ( k ) , | ω | 2 U ( k ) ∈L 2 (Ω 3 ), k = 0,1,2,...t ≥ 0 and the
solutions of the Navier-Stokes ordinary differential equations (2-1) and its finite time derivatives
(2-20) satisfy
t
∫
t
rˆ ( k )
r rˆ ( k )
r r
r 2 rˆ ( k )
r 2
U
(
s
,
ω
)
⋅
F
(
s
,
ω
)
d
ω
ds
<
η
∫
∫ ∫ | ω | | U (s, ω )| ds, t ≥ 0, k = 0,1,2,...
0 Ω3
(2-57b)
0 Ω3
for all forward time.
PROOF
By the strict inequality, the integral on the left of (2-57a) over R 3 must be finite. In order for the
Fourier transforms which appear in (2-57b) to be well defined
r
r
r
F ( k ) ⋅ U ( k ) , ∇U ( k ) ∈L 2 ( R 3 ), k = 0,1,2,..., t ≥ 0 if and only if
rˆ
rˆ
r rˆ
U ( k ) ⋅ F ( k ) , | ω | 2 U ( k ) ∈L 2 (Ω 3 ), k = 0,1,2,..., t ≥ 0
r
r
r
r
Note that, since F ( k ) , k = 0,1,2,... is Schwartz in x ∈ R 3 , U ( k ) ⋅ F ( k ) is integrable by Ho&&lder ' s
r
r
inequality if only U ( k ) ∈ L1 ( R 3 ) without assuming (2-57a).
Now suppose
19
t
∫
t
rˆ ( k )
r rˆ ( k )
r r
r 2 rˆ ( k )
r 2 r
U
(
s
,
ω
)
⋅
F
(
s
,
ω
)
d
ω
ds
<
η
|
ω
∫
∫ ∫ | | U ( s, ω ) | dωds, t ≥ 0, k = 0,1,2,...
0 Ω3
(2-58)
0 Ω3
By the definition of the Fourier transform, the previous inequalities hold if and only if
t
∫
t
r ( k ) r r ( k ) r −iωr ⋅ xr 2 r r
∫ ∫ U (s, x ) ⋅ F ( s, x ) | e | dxdωds < η ∫
0 Ω3 R3
r ( k ) r 2 −iωr ⋅ xr 2 r r
|
∇
U
( s, x ) | | e
| dx dωds,
∫∫
0 Ω3 R 3
(2-59)
k = 0,1,2,..., t ≥ 0.
But these inequalities hold if and only if
t
r (k ) r r (k ) r r
r (k ) r 2 r
U
(
s
,
x
)
⋅
F
(
s
,
x
)
d
x
ds
<
η
∫∫
∫ ∫ | ∇U (s, x ) | dxds, t ≥ 0, k = 0,1,2,...
t
0 R3
(2-60)
0 R3
r r
r
r
because | e −iω ⋅ x | 2 = 1, ω ∈ Ω 3 , x ∈ R 3 .
END PROOF
rˆ r
r
Theorem 2-3. If F (t , ω ) , and all time derivatives are Schwartz in ω ∈ Ω 3 for all forward time
rˆ r
rˆ r
r
and U 0 (ω ) is Schwartz in ω ∈ Ω 3 then any finite time derivative of U (t , ω ) ( the solution of the
r
equation of lemma 2-2) is Schwartz in ω ∈Ω 3 .
PROOF
r
a. For any fixed t , k the kth derivative is Schwartz in ω ∈ Ω 3
rˆ
r
supωr∈Ω3 | ω | p | U ( k ) |< ∞, ∀p ∈ W .
b.
(2-61)
The weighted upper bound is uniform in k
rˆ
r
sup k∈W sup ωr∈Ω3 | ω | p | U ( k ) |< ∞, ∀p ∈ W .
(2-62)
c. The weighted upper bound is uniform for all t ≥ 0 .
rˆ
r
sup t ≥0 sup k∈W sup ωr∈Ω3 | ω | p | U ( k ) |< ∞, ∀p ∈ W .
(2-63)
20
By the variation of constants formula, it suffices to show that the convolution integral is discrete
rˆ r
r2 r
r
ˆ
Schwartz since Φ(t , ω ) =e −η |ω | t U 0 are Schwartz in ω ∈ Ω 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
rˆ
r2
sup ωr∈Ω3 | ω1p (1)ω 2p ( 2 )ω 3p (3) || U |≤ sup ωr∈Ω 3 e −η |ω | t | U 0 || ω1p (1)ω 2p ( 2 )ω 3p (3) | +
t
sup ωr∈Ω3 | ω
p (1)
1
ω
p(2)
2
ω
p ( 3)
3
|| ∫ e
r
−η |ω |2 ( t − s )
0
r
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 ωω
 | ωr | 2 −  0 1 0 [i ((Uω ) ∗ U ) − F ]ds |
 0 0 1 



(2-64)
η > 0, p ∈ W 3 , t ≥ 0, k = 1,2,3,
Simplify the upper bound on the Schwartz weighted Fourier transform of the momentum of (264)
rˆ
r
sup ωr∈Ω3 | ω1p (1)ω 2p ( 2)ω 3p (3) ||U ( k )(t , r ) |≤
rˆ
r2
r
sup ωr∈Ω3 | ω1p (1)ω 2p ( 2)ω 3p (3) || (−4η ) k | π 2 k | r | 2 k e −η |ω| t | U 0 |
t
+ sup ωr∈Ω 3 | ω
p (1)
1
ω
p ( 2)
2
ω
p ( 3)
3
|| ∫ e
0
r
r
r
−η |ω |2 t ( t − s )
 r r t 1 0 0  
rˆ r t
rˆ ( k ) rˆ ( k )


 ωω
ω
0
1
0
[
i
((
U
)
∗
U
) − F ]ds |
−
r

| ω |2 



 0 0 1
η > 0, p ∈ W , t ≥ 0, ω ∈Ω 3 , k = 1,2,3,
rˆ
rˆ
Since U 0 ∈ S (Ω 3 )
r
∀t ≥ 0, p ∈W 3
rˆ
rˆ
r
sup ωr∈Ω e −η |ω| t | U 0 || ω1p (1)ω 2p ( 2 )ω 3p ( 3) |≤ sup ωr∈Ω | ω1p (1)ω 2p ( 2 )ω 3p (3) || U 0 |≤ M 2
2
3
(2-65)
(2-66)
3
Since
rˆ rˆ
rˆ r
rˆ rˆ
rˆ rˆ
U ∈ S (Ω 3 ) ⇒ (Uω t ) ∗ U ∈ S (Ω 3 ), F ∈ S (Ω 3 )
(2-67)
it follows by (2-20) (the norm on the matrix operator) that
21
rˆ
sup ωr∈Ω3 | ω1p (1)ω 2p ( 2 )ω 3p (3) || U ( k ) |≤
 r r t 1 0 0  
rˆ r t
rˆ
rˆ


 ωω
0
1
0
[
i
((
U
r
)
∗
U
)
−
F
]ds |
sup ωr∈Ω3 | ω1p (1)ω 2p ( 2 )ω 3p (3) || ∫ e
−
r



| ω |2
0



 0 0 1
t
rˆ r
rˆ
r2
≤ 2 ∫ e −η |ω | t (t − s ) sup rr∈ 3 sup s ≥0 | ω1p (1)ω 2p ( 2 )ω 3p (3) | ((Uω t ) ∗ U )ds |
t
r
−η |ω |2 ( t − s )
(2-68)
0
t
r2
+ 2 ∫ e −η |ω |
(t − s )
rˆ
sup s ≥0 | ω1p (1)ω 2p ( 2 )ω 3p (3) || F |ds |
0
Now switch to the vector norm Schwartz condition.
It suffices to consider the integral.
t
r
sup ωr∈Ω3 | ω| p + 2 | ∫ e
r
−η |ω |2 ( t − s )
0
rr
1 0 0 
rˆ
r
 rˆ r
ωω t 
[ r 2 −  0 1 0 ][(Uω t ∗ U ) ( k ) − F ( k ) ]ds |≤
|ω | 

 0 0 1
(2-69)
t
rˆ r
rˆ
r
r2
r
r
2(sup ωr∈Ω3 | ω| 2 | ∫ e −η |ω | (t − s ) ds | (sup ωr∈Ω3 sup t ≥ 0 | ω| p [| (Uω t ∗ U ) ( k ) | + | F ( k ) |])
0
The previous inequality follows by the product inequality, the inequality for the product of
suprema, the inequality relating suprema with respect to one vector parameter vs. one vector
r
r
parameter and one variable and the fact that | ω| p | . |≤ sup ωr∈Ω3 | ω| p | . | .
2(sup ωr∈Ω3
r
t
r
rˆ r
rˆ
r
| ω| 2 | ∫ e −η |ω | (t − s ) ds |) × (sup ωr∈Ω sup t ≥ 0 | ω| p [| (Uω t ∗ U ) ( k ) | + | F ( k ) |])
2
3
0
≤
2
η
(2-70)
{M 1 (k ) + M 2 (k )}.
The inequality above follows by integration for the first factor and the Schwartz bounds for each
term of the second factor.
The homogeneous term has a uniform upper bound
r
∀t ≥ 0, p ∈W 3
rˆ
rˆ
r2
sup ωr∈Ω3 e −η |ω| t | ω1p (1)ω 2p ( 2)ω 3p ( 3) || U 0 |≤ sup ωr∈Ω3 | ω1p (1)ω 2p ( 2)ω 3p (3) || U 0 |≤ M 2
The pressure function
r
r
r
1 r rˆ r t rˆ rˆ r
3
⋅
∗
−
≥
∈
Ω
−
Pˆ (t , r ) =
{
r
[
U
ω
U
F
(
t
,
ω
)]},
t
0
,
ω
0
| ω| 2
(2-71)
(2-72)
22
rˆ rˆ r
r
is Schwartz in ω since U , Uω t are Schwartz by hypothesis, the convolution of discrete
rˆ r
rˆ rˆ r
Schwartz functions is Schwartz and the difference of Schwartz functions [Uω t ∗ U − F (t , ω )] is
Schwartz.
r
By the same reasoning any finite order time derivative of the pressure transform Pˆ (t , ω ) is
Schwartz.
END PROOF
The following theorem is the main result of this paper.
r
r
r
r
r
Theorem 2-4. Suppose U ( k ) ⋅ F ( k ) , ∇U ( k )∈ S ( R 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,...
(2-73)
∫∫
∫∫
0 D
0 D
r
r
r
where F (k ) is jointly smooth in (t , x ) , Schwartz in x ∈ R 3 and bounded in t ∈ [0, ∞ ) . Then every
finite time derivative of the solution of the Navier-Stokes momentum equation is bounded,
r
continuous and uniquely determined in t. It is also smooth in x ∈ R 3 for all forward time.
PROOF
The conclusion follows directly by the properties of the inverse Fourier transform representation
rˆˆ
U of the velocity function, theorem 2-2 and theorem 2-3. In particular,
rˆ rˆ
rˆˆ r r
U ∈ S (Ω 3 ), t ≥ 0 ⇒ U = U ∈ C ∞ ( R 3 ), t ≥ 0
(2-74)
r
rˆ r ∞
r
r
ˆ
r
r
ˆ
U ∈ C ([0, ∞)), ω ∈ Ω 3 ⇒ U = U ∈ C ∞ ([0, ∞)), x ∈ R 3 .
By the formula in lemma 2-2, each component of the pressure gradient satisfies the same
r
properties as each component of the momentum vector (marginal smoothness in t , x , uniqueness,
and boundedness for all forward time).
END PROOF
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.
Selected Bibliography
23
1. Aris, R., Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Dover, NY,
1962 (Dover 1989).
2. Avrin, J., A One Point Attractor Theory for the avier-Stokes Equation on Thin Domains
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).
5. Fefferman C., Existence and Smoothness of the avier-Stokes Equation, 2000, http://
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.
8. Hirsch M. and Smale S., Differential Equations, Dynamical Systems, and Linear Algebra,
Academic Press, N.Y, 1974.
9. Leray J., Essai sur les mouvement d’un liquide visqueux Emplissant L’Espace,
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