Gliklikh Y.E. Global and Stochastic Analysis with Applications to Mathematical Physics

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406 16 Hydrodynamics

∗

ξ(t)=

F (t, ξ(t)) (16.23)

u(t, x) on T

satisﬁes the Navier-Stokes equation (16.20).

The proof of Theorem 16.28 is reduced to the ﬁnite-dimensional arguments

used above.

The divergence-free vector ﬁeld u(t, x)onT

from Theorem 16.28, i.e., a

time-dependent vector in β

⊂ T

), can be obtained by right transla-

tion of the backward velocity

∗

ξ(t)ate, and so the Navier-Stokes equation

(16.20) plays the role of the Euler equation in the “algebra” T

)ac-

cording to the general approach to Euler equations described in Section 11.2.

The ﬂow of u(t, x)onT

,whichisacurveonD

) describing the motion

of a viscous incompressible ﬂuid, may be considered as the expectation of the

process ξ(t).

So, we need to construct a backward admissible process on D

)sat-

isfying (16.22) that also satisﬁes (16.23). It is a diﬃcult problem to ﬁnd a

process with given backward mean derivatives. That is why we shall try to

construct ξ(t) by solving ﬁrst a certain equation of type (16.21)andthen

changing the time direction in its solution.

Let a process η(t)onD

) be a solution of the stochastic diﬀerential

equation of type (16.21) with initial condition η(0) = e and suppose it exists

for t in a non-random time interval [0,T]. Consider the process with inverse

time direction ξ(t)=η(T −t). Our aim now is to construct an equation for η

such that (16.23) is fulﬁlled for ξ(t), and

∗

ξ(t)=¯u(t, ξ(t)) where ¯u(t, f )isan

admissible right-invariant vector ﬁeld with initial condition u(0,e)=u

∈ β

where u

= u

(x) is a divergence-free H

-vector ﬁeld on T

Since the backward mean derivative for ξ(t) is equal to the negative of

the forward mean derivative for η(T − t), we have

Dη(t)=−

∗

ξ(T − t)=

−u(T − t, η(t)). Hence, taking into account Theorem 8.7 and the fact that

TπS(X)=X and Tπ

= 0, we deduce that ξ(t) satisﬁes (16.23)ifη(t)

satisﬁes the equality

dη(t)=−˜u(T − t, η(t))dt + σ

A(η(t))dw(t) (16.24)

and the process ¯u(T − t, η(t)) in

β satisﬁes the equality

˜u(T − t, η(t)) = −S(˜u(T − t, η(t))) −

(T −t, ˜u(T − t, η(t))) (16.25)

where

(T −t, ˜u(T − t, η(t))) is the vertical lift of

F (T − t, ˜u(T −t, η(t))).

Denote by

the horizontal lift of the ﬁeld

A onto T D

). Introduce on

β the connection from Section 2.8. Recall that the projections of its geodesics

onto D

) are geodesics of the connection

H. Denote the exponential map

of this connection by exp

Theorem 16.29 Let the process u(T − t, η(t)) on

β satisfy the Itˆo equation

in Belopolskaya-Daletskii form

16.4 Viscous hydrodynamics 407

d¯u(T − t, η(t)) = exp

¯u(T −t,η(t))

(−S(¯u(T − t, η(t)))dt (16.26)

−F

(t, ¯u(T − t, η(t)))dt +

(˜u(T − t, η(t)))dw(t)).

Then the process η(t) and the right-invariant admissible vector ﬁeld ¯u on

) satisfy (16.24) and (16.25), respectively, and so ξ(t)=η(T −t) satis-

ﬁes (16.23) and (16.22). Hence the divergence-free vector ﬁeld u(t, x) on T

is a solution of (16.20).

Theorem 16.29 follows from the inﬁnite-dimensional version of Lemma 9.3.

The following ﬁnite-dimensional interpretation clariﬁes the construction.

The process η(t) with initial condition η(0) = e on D

)thatsatisﬁes

(16.24) is a random ﬂow on T

.Denotethisﬂowbyη(t, x) with η(0,x)=

x. The ﬂow η(t, x) is the general solution of the Itˆo stochastic diﬀerential

equation on T

dη(t, x)=−u(T − t, η(t, x))dt + σdw(t) (16.27)

with divu(t, x) = 0, the ﬁnite-dimensional version of (16.24). By direct cal-

culation of the forward mean derivatives for the ﬁnite dimensional process

η(t, x) we show that Dη(t, x)=−u(T − t, η(t, x)) and

PDDη(t, x)=

∂

∂t

u(T − t, η(t, x)) + (u(T − t, η(t, x)), ∇)u(T − t, η(t, x))

−

∇

u(T − t, η(t, x)) − gradp.

The latter equality is turned into (16.19) under the change of variables

η(t, x)=ξ(T − t). Thus equation (16.26) guarantees that for the process

η(t) satisfying (16.27) the relation PDDη(t, x)=F (t, x) holds.

For a stochastic diﬀerential equation with respect to a process ζ(t)on

) denote by ζ

(s) its solution with initial condition ζ

(t)=e. Consider

the following system on D

dη(t)=−˜u(T − t, η(t))dt + σ

A(η(t))dw(t)

u(t)=PEQ

◦ η

T −t

(T )) +



F (s)ds (16.28)

where Q

is introduced in Deﬁnition 5.16 and u

= u(0) ∈

is the initial

value for u(t) (see above). Notice that the ﬁrst equation of (16.28)is(16.24).

Theorem 16.30 If the process η(t) and the vector u(t) satisfy (16.28), then

u(t), considered as a divergence-free vector ﬁeld on T

, satisﬁes (16.20).

Indeed, taking into account the routine stochastic presentation of solu-

tions of PDEs one can easily derive from the second equation of (16.28)that

PDDη(t, x)=F (t, x).

408 16 Hydrodynamics

The system (16.28) is similar to the system considered by Ya. Belopol-

skaya in [22](seealso[24]). Equation (16.27) is a part of another system of

stochastic diﬀerential equations, connected with the Navier-Stokes equation,

that was also studied by B. Busnello [36].

We should also mention [42] where the theory of forward-backward stochas-

tic diﬀerential equations on the diﬀeomorphism group of a torus is used to

describe viscous incompressible hydrodynamics, and [39] where the inﬁnite-

dimensional Wiener process in the stochastic diﬀerential equations on that

group is used.

16.4.2 Solutions of Burgers, Reynolds and

Navier-Stokes equations via stochastic

perturbations of inviscid ﬂows

If the backward mean derivative of a process satisfying the stochastic Newton

law of the previous section is not generated by a right-invariant vector ﬁeld,

in the “algebra”, after passing to the Euler approach, some other types of

hydrodynamical equations may arise.

In this section we introduce a special stochastic perturbation of a ﬂow of

diﬀuse matter such that the perturbed ﬂow satisﬁes the stochastic Newton

law of type (16.15), and show that the corresponding curve in the tangent

space at the unit satisﬁes Burgers’ equation. The same perturbation of a ﬂow

of a perfect incompressible ﬂow without external force satisﬁes the stochastic

Newton law (16.23) with

F = 0, but yields a curve in the tangent space at

the unit that is a solution of a Reynolds type equation. Nevertheless, under

the action of a certain special external force on the ﬂow, this curve becomes

a solution of a Navier-Stokes equation without external force. As above, we

consider a ﬂuid motion on the ﬂat n-dimensional torus T

In this section we take s>

+ 2 so that the diﬀeomorphisms from

)andD

)areC

-smooth and T

) consists of C

-smooth

vector ﬁelds. We often use the operator Q introduced in Deﬁnition 5.16(iii).

Everywhere below we use the same process W

(σ)

(t) constructed from a

Wiener process w(t)inR

by formula (10.5) (see also Theorem 10.5). If, in

the formula, several random elements appear with subscript ω, this means

that they all are taken at “the same” ω ∈ Ω, i.e., sometimes the formula may

be considered as a description of a non-random element depending on the

parameter ω ∈ Ω.

Let g(t) be a solution of (16.1) with initial conditions g(0) = e and ˙g(0) =

∈ T

). Recall (see Proposition 16.2) that such a solution exists on

some time interval t ∈ [0,T] (for the sake of convenience we take a closed

interval inside the domain of g(t)). Recall also that g(t) is a ﬂow of diﬀuse

matter without external forces. Consider v(t)=R

−1

g(t)

˙g(t)= ˙g(t) ◦ g

−1

(t) ∈

16.4 Viscous hydrodynamics 409

). This inﬁnite-dimensional vector, considered as a vector ﬁeld on

, will also be denoted by v(t, m). Recall that this vector ﬁeld satisﬁes the

Hopf equation (16.4).

Consider a process on D

)oftheformη(t)=W

(σ)

(t) ◦g(t), t ∈ [0,T],

where W

(σ)

(t) is deﬁned by (10.5). In the ﬁnite-dimensional terminology,

η(t) is a random diﬀeomorphism of T

of the form η(t, m)=g(t, m)+σw(t)

modulo the factorization with respect to the integral lattice. Introduce the

process ξ(t)=η(T − t), i.e., in the ﬁnite-dimensional notation, ξ(t, m)=

g(T − t, m)+σw(T − t).

Since w(t) is a martingale with respect to its own “past”, one can easily

derive from the properties of conditional expectation that D

∗

ξ(t)= ˙g(T −

t, m)=v(T − t, g(T − t, m)) and so

∗

ξ(t)=

˙g(s)

|s=T −t

=0.

Consider the random process

(s)=ξ(s) ◦ξ

−1

(t)=W

(σ)

(T −s) ◦g(T −s) ◦g

−1

(T −t) ◦(W

(σ)

(T −t))

−1

Notice that the random diﬀeomorphism (W

(σ)

(t))

−1

acts by the rule

(σ)

(t))

−1

(m)=m −σw(t).

Obviously ξ

(t)=e. A ﬁnite-dimensional description of this process can be

given as follows.

By construction m = ξ(t, ξ

−1

(t, m)) = g(T − t, ξ

−1

(t, m)) + σw(T − t).

Then g(T −t, ξ

−1

(t, m)) = m −σw(T −t)andsoξ

−1

(t, m)=g

−1

(T −t, m −

σw(T − t)). Thus,

(s, m)=ξ(s, g

−1

(T −t, m −σw(T − t))

= g



T −s, g

−1

(T −t, m −σw(T − t))



+ σw(T − s).

We have ξ

(t, m)=m − σw(T − t)+σw(T − t)=m, indeed, ξ

(t)=e on

Since ξ

(t)=e,the“present”σ-algebra N

is trivial and so the con-

ditional expectation with respect to it coincides with the ordinary mathe-

matical expectation. Hence, using the relation between v(t)andg(t)andthe

deﬁnition of D

∗

, one can easily derive that

∗

(s)

|s=t

= E (v(T − t, m − σw(T − t)))

= E



−1

(σ)

(T −t)

v(T − t)



(16.29)

(here t is ﬁxed and the derivative is taken with respect to s).

Introduce on T

the vector ﬁeld V (t, m)=E(v(t, m − σw(t))). As an

inﬁnite dimensional vector, we denote it also by V (t)=E(Q

−1

(σ)

(t)

v(t)).

Formula (16.29)meansthatD

∗

(s)

|s=t

= V (T − t).

Theorem 16.31 The vector ﬁeld V (T −t, m) satisﬁes the Burgers equation

410 16 Hydrodynamics

V (T −t, m)+(V (T −t, m)·∇)V (T −t, m)−

∇

V (T −t, m)=0, (16.30)

where ∇

is the Laplace-Beltrami operator which on the ﬂat torus coincides

with the ordinary Laplacian.

Proof. For t ∈ [0,T]andω ∈ Ω, deﬁne the curve ζ

t,ω

(s)ins ∈ [0,T] depend-

ing on the parameter ω by the formula

t,ω

(s)=R

−1

(σ)

(T −t)

g(T −s, g

−1

(T −t)) = g(T −s, g

−1

(T −t, m −σw(T ))).

Note the only diﬀerence between ζ

t,ω

(s)andξ

t,ω

(s): the stochastic summand

σw(T − s) appears in the expression for ξ

t,ω

(s) while it is absent in ζ

t,ω

(s).

This means that ζ

t,ω

(s) is an a.s. smooth curve with random initial condition

t,ω

(t)=(W

(σ)

(T − t))

−1

We have

t,ω

(s)

|s=t

= −TR

−1

(σ)

(T −t)

v(T − s). Since g(T − s)isa

geodesic, from Proposition 5.7 it follows that “for almost all ω” (i.e., a.s.

for ω ∈ Ω) the curve ζ

t,ω

(s) is also a geodesic, i.e.,

t,ω

(s) = 0. From the

construction of the operator L

in Section 5.2 it follows that the action of the

diﬀeomorphism W

(σ)

(t) coincides with that of L

σw

(t)

. Hence, by Theorem

5.19 the curve (W

(σ)

(T − t))ζ

t,ω

(s)=L

(σw

(T −t))

t,ω

(s) is a.s. geodesic as

well, i.e.,

(σw

(T −t))

t,ω

(s) = 0. Note that

(σw

(T −t))

t,ω

(s)

|s=t

= Q

t,ω

(s)

|s=t

= −Q

−1

(σ)

(T −t)

v(T − t).

Recall that EQ

−1

(σ)

(T −t)

v(T −t)=V (T −t)andD

∗

(s)

|s=t

= V (T −t)

(see above). Then from the above arguments and constructions we derive

that

∗

(s)

|s=t

= D

∗

V (T −t, ξ

(s))

|s=t

= −E



(σw

(t))

t,ω

(s)

|s=t



=0.

But since D

∗

(s)

|s=t

= V (T −t), by formula (8.25) the backward derivative

∗

V (T −t, ξ

(s)) coincides with the left-hand side of (16.30). Hence (16.30)

is satisﬁed. 

Now, let us turn to the case of viscous incompressible ﬂuids. Let g(t)bea

solution of (16.9)onD

) with initial conditions g(0) = e and ˙g(0) = u

∈

). Theorem 16.11 tells us that such a solution exists in some time

interval t ∈ [0,T] (for the sake of convenience we again take a closed interval

inside the domain of g(t)). Recall that g(t) is a ﬂow of a perfect incompressible

ﬂuid without external forces. Consider u(t)= ˙g(t) ◦g

−1

(t) ∈ T

). This

inﬁnite-dimensional vector, considered as a divergence-free vector ﬁeld on

, will be denoted u(t, m). Recall that this vector ﬁeld satisﬁes the Euler

equation (16.11) without external forces (see Section 16.2).

16.4 Viscous hydrodynamics 411

Since W

(σ)

(t) takes values in D

)(seeTheorem10.5), we can repeat on

) the above constructions for D

), i.e., deﬁne η(t)=W

(σ)

(t)◦g(t),

where t ∈ [0,T], and ξ(t)=η(T − t) (i.e., in the ﬁnite-dimensional notation

ξ(t, m)=g(T −t, m)+σw(T −t)). It is easy to see that D

∗

ξ(t)= ˙g(T −t, m)=

u(T − t, g(T −t, m)) and so

∗

ξ(t)=

˙g(s)

|s=T −t

=0onD

As above, the process ξ

(s)=ξ(s) ◦ ξ

−1

(t) has the property ξ

(t)=e.Its

ﬁnite-dimensional description is quite analogous to the case of D

Introduce on T

the vector ﬁeld U (t, m)=E(u(t, m − σw(t))) (a direct

analog of V (t, m)). We also denote this ﬁeld as an inﬁnite dimensional vector

by U(t)=E(Q

−1

(σ)

(t)

u(t)).

Lemma 16.32 The vector ﬁeld U(t, m) is divergence-free.

Proof. By construction, for an elementary event ω ∈ Ω, the diﬀeomorphism

(σ)

(t)

)

−1

is a shift of the entire torus by a constant vector. Hence, Q

applied to TR

−1

(σ)

(t)

u(t) is the parallel translation on the torus of the entire

divergence-free vector ﬁeld u(t) by the same constant vector into the opposite

direction. Thus Q

−1

(σ)

(t)

u(t) is a random divergence-free vector ﬁeld on

the torus. Hence its expectation is divergence-free. 

So, U(t) ∈ T

). In particular, we have proved above that

∗

(s)

|s=t

= U(T − t). (16.31)

Since nothing like Theorem 5.19 holds on D

), we have PD

∗

(s)

|s=t

∗

U(T −t, ξ

(s))

|s=t

= 0 (the value of this mean derivative is calculated in

Remark 16.34 below). Hence there is no analog of Theorem 16.31.Wecan

prove only the following:

Theorem 16.33 The vector ﬁeld U(t, m) satisﬁes the following Reynolds

type equation (see, e.g., [207]):

∂

∂t

U + E [((u ·∇)u)(t, m −σw(t))] −

∇

U − grad p =0. (16.32)

Proof. It follows from the Itˆo formula that

du(t, m − σw(t))

∂u

∂t

(t, m − σw(t))dt +

∇

u(t, m − σw(t))dt − σu



dw(t),

where ∇

, as above, is the Laplace-Beltrami operator and u



is the linear

operator of the derivative of u in m ∈T

Recall that u(t, m) satisﬁes the Euler equation without external force, i.e.,

∂u

∂t

= −P ((u ·∇)u). Since



u(t, m − σw(t))



∂

∂t

Eu(t, m − σw(t)) =

∂

∂t

U(t)

412 16 Hydrodynamics

and E(σ(∇u)dw(t))=0,wederivethat

∂

∂t

U = E



u(t, m − σw(t))



= E



−P ((u ·∇)u)(t, m − σw(t)) +

∇

u(t, m − σw(t))



= −E [((u ·∇)u)(t, m − σw(t))] +

∇

U +gradp.



There are standard methods for transforming (16.32) into the standard

Reynolds form (see [207]). For a divergence-free vector ﬁeld X(m)onT

(i.e., for a vector X ∈ T

)) introduce the random divergence-free

vector ﬁeld

(t, m)=X(m − σw(t)) − E(X(m − σw(t)) (i.e., the vector

(t)=Q

−1

(σ)

(t)

X − E(Q

−1

(σ)

(t)

X)inT

)). For X = u(t),

we obtain

u(t)

(t, m)=u(t, m − σw(t)) − U(t, m)andsou(t, m − σw(t)) =

U(t, m)+

u(t)

(t, m)andE

u(t)

(t, m) = 0. Then one can easily see that

E([(u ·∇)u](t, m − σw(t))) = (U ·∇)U + E[(

u(t)

·∇)

u(t)

]. Thus, (16.32)

transforms into

∂

∂t

U +(U ·∇)U −

∇

U − grad p = −E[(

u(t)

·∇)

u(t)

], (16.33)

which is the standard form of the Reynolds equation. It diﬀers from the

Navier-Stokes type relation with viscosity

by the external force −E[(

u(t)

∇)

u(t)

] which depends on u(t, m), not on U(t, m). Recall that (16.33)de-

scribes the turbulent motion of a ﬂuid if the dependence of E[(

u(t)

·∇)

u(t)

]

on U is given (say, derived from experimental data, see [207]).

Remark 16.34. For ξ

(s) as introduced above, formula (16.31) tells us that

∗

(s)

|s=t

= U(T − t). Then, taking into account formula (8.25), one can

easily derive that

∗

(s)

s=t

∗

U(T −s, ξ

(s))

s=t

∂

∂t

U +(U ·∇)U −

∇

U −grad p.

Thus, (16.33) implies that

∗

(s)

s=t

= −PE[(

u(T −t)

·∇)

u(T −t)

Our next aim is to show that a slight modiﬁcation of the above argument

allows us to annihilate the external force in (16.33) by introducing a special

random force ﬁeld on D

)into(16.12).

For a random divergence-free a.s. H

s+1

-vector ﬁeld X

(m)onT

(i.e.,

forarandomvectorX

∈ T

s+1

) ⊂ T

)), construct the random

vector ﬁeld

(t, m) which, for any ω ∈ Ω, is given by the formula

16.4 Viscous hydrodynamics 413

(t, m)=X

(m −σw

(t)) −E(X

(m −σw

(t)).

Introduce the non-random H

vector ﬁeld PE[(

·∇)

] and then con-

struct the random vector F

(t, X

)inT

) by the formula

(t, X

)=Q

(σ)

(t)

PE[(

·∇)

Note that PE[(

·∇)

] and hence F

(t, X

) lose their derivatives, i.e.,

they are H

-vector ﬁelds only since X

(and so

)isH

s+1

.ThusF

(t, X

)

is well-deﬁned only on an everywhere dense subset T

s+1

)ofT

Now introduce the right-invariant vector force ﬁeld

(t, g, Y

)onD

where Y

∈ T

), which for g ∈D

s+1

)andω ∈ Ω is determined by

the formula

(t, g, Y

)=TR

(t, T R

−1

where TR

−1

is a divergence-free a.s. H

s+1

-vector ﬁeld.

Consider the equation

˙g

(t)=

(t, g

(t), ˙g

(t)) (16.34)

on D

) whose right-hand side is well-deﬁned on the everywhere dense

subset D

s+1

)inD

). Equation (16.34) has no diﬀusion term and

so it is an ordinary diﬀerential equation with parameter ω ∈ Ω. Here we

do not investigate the solvability of (16.34) but suppose that for the initial

conditions g

(0) = e and ˙g

(0) = u

∈ T

s+1

) it a.s. has a unique

s+1

-solution g

(t) which is a.s. well-deﬁned on a non-random time interval

t ∈ [0,T]forsomeT>0. Consider the divergence-free a.s. H

s+1

-vector ﬁeld

(t, m)onT

given by the relation ˙g

(t)=u

(t, g

(t)). The analog of the

above-mentioned vector U now takes the form

U(t, m)=E



(t, m − σw

(t))



= EQ

−1

(σ)

(t)

(t). (16.35)

As in Lemma 16.32 it is easy to see that the vector ﬁeld (16.35) is divergence-

free.

Theorem 16.35 The divergence-free vector ﬁeld U given by (16.35) satisﬁes

the Navier-Stokes equation without external force and with viscosity

∂

∂t

U +(U ·∇)U −

∇

U −grad p

=0. (16.36)

Proof. For the random ﬁeld u



(t, m) of linear operators and the random

ﬁeld u



(t, m) of bilinear operators (here the primes denote derivatives of u

in m ∈T

) the stochastic integrals





(t, m)dw

(t)and

414 16 Hydrodynamics





(t, m)(dw

(t), dw

(t)) =



tr u



dt =



∇

are well-deﬁned. Then by applying standard arguments to the Taylor series

expansion of u

, one can easily see that the Itˆo formula is well-deﬁned for

(t, m − σw

(t)) and so

E (du

(t, m − σw

(t)))

= E



∂

∂t

(t, m − σw

(t))dt +

∇

(t, m − σw

(t))dt



From (16.34) it follows (see (16.12)and(16.13)) that

∂

∂t

= −P [(u

·∇)u

(t, u

(t)). Thus, in the same manner as in the proof of Theorem 16.33 and

the derivation of (16.33), we obtain

∂

∂t

U(t, m)=E



(t, m − σw

(t))



= −E



·∇)u



(t, m − σw

(t))



∇

U +gradp + EQ

−1

(σ)

(t)

(t, u

(t))

= −(U ·∇)U +

∇

U +gradp − E[(

(t)

·∇)

(t)

]

+EQ

−1

(σ)

(t)

(t, u

(t)). (16.37)

But by construction and by formulae (5.13)and(5.14)weget

−1

(σ)

(t)

(t, u

(t))

= EQ

−1

(σ)

(t)

(σ)

(t)



(

(t)

·∇)

(t)



= EQ

−1

(σ)

(t)

(σ)

(t)

(σ)

(t)

−1



(

(t)

·∇)

(t)



= PE



(

(t)

·∇)

(t)



. (16.38)

Since U is divergence-free, the vector ﬁelds

∂

∂t

U and ∇

U are divergence-

free as well. Hence, grad p in (16.37) is taken from relation (5.7)forY =

E([(u

·∇)u

](t, m − σw

(t))), i.e.,

PE([(u

·∇)u

](t, m − σw

(t))) = E ([(u

·∇)u

](t, m − σw

(t)))−grad p.

Deﬁne grad p

and grad p

by the relations P (U ·∇)U =(U ·∇)U − grad p

and PE[(

(t)

·∇)

(t)

]=E[(

(t)

·∇)

(t)

]−grad p

. Clearly, grad p =

grad p

+gradp

(i.e., to within an additive constant p = p

+ p

). Thus

(16.36) follows from (16.37)and(16.38) in the natural form

∂

∂t

U+(U·∇)U−

∇

U −grad p

=0. 

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415