Turbulent Rivers
Björn Birnir
Center for Complex and Nonlinear Science
and
Department of Mathematics
University of California, Santa Barbara
May 1, 2006
Abstract
The existence of solutions describing the turbulent flow in rivers is proven.
The existence of an associated invariant measure describing the statistical
properties of this one dimensional turbulence is established. The turbulent
solutions are not smooth but Hölder continuous with exponent 3/4. The
scaling of the solutions’ second structure (or width) function gives rise to
Hack’s law [14]; stating that the length of the main river, in mature river
basins, scales with the area of the basin l ∼ Ah , h = 0.58 being Hack’s exponent.
1
Introduction
The flow of water in streams and rivers is a fascinating problem with many application that has intrigued scientists and laymen for many centuries, see Levi [19].
Surprisingly it is still not completely understood even in one or two-dimensional
approximation of the full three-dimensional flow. Erosion by water seems to determine the features of the surface of the earth, up to very large scales where the
influence of earthquakes and tectonics is felt, see [26, 27, 25, 4, 3, 28]. Thus water flow and the subsequent erosion gives rise to the various scaling laws know for
river networks and river basins, see [10, 6, 7, 8, 9].
1
One of the best known scaling laws of river basins is Hack’s law [14] that
states that the area of the basin scales with the length of the main river to an exponent that is called Hack’s exponent. Careful studies of Hack’s exponent, see [9]
show that it actually has three ranges, depending on the age and size of the basin,
apart from very small and very large scales where it is close to one. The first range
corresponds to a spatial roughness coefficient of one half for small channelizing
(very young) landsurfaces. This has been explained, see [3] and [11], as Brownian motion of water and sediment over the channelizing surface. The second
range with a roughness coefficient of 2/3 corresponds to the evolution of a young
surface forming a convex (geomorphically concave) surface, with young rivers,
that evolve by shock formation in the water flow. These shocks are called bores
(in front) and hydraulic jumps (in rear), see Welsh, Birnir and Bertozzi [28]. Between them sediment is deposited. Finally there is a third range with a roughness
coefficient 3/4. This range that is the largest by far and is associated with what
is called the mature landscape, or simply the landscape because it persists for a
long time, is what this paper is about. This range is associated with turbulent flow
in rivers and we will develop the statistical theory of turbulent flow in rivers that
leads to Hack’s exponent.
Starting with the three basic assumption on river networks: that the their structure is self-similar, that the individual streams are self-affine and the drainage
density is uniform, see [6], river networks possess several scalings laws that are
well documented, see [24]. These are self-affinity of single channels, which we
will call the meandering law, Hack’s law, Horton’s laws [15] and their refinement
Tokunaga’s law, the law for the scaling of the probability of exceedance for basin
areas and stream lengths and Langbein’s law. The first two laws are expressed
in terms of the meandering exponent m, or fractal dimension of a river, and the
Hack’s exponent h. Horton’s laws are expressed in terms of Horton’s ratio’s of
link numbers and link lengths in a Strahler ordered river network, Tokunaga’s
law is expressed in term of the Tokunaga’s ratio’s, the probability of exceedance
is expressed by decay exponents and Langbein’s law is given by the Langbein’s
exponents, [6].
In a series of paper’s Dodds and Rothman [10, 6, 7, 8, 9] showed that all
the above ratios and exponents are determined by m and h, the meandering and
Hack’s exponents, see [14], [10]. The origin of the meandering exponent m is still
a mystery, but in this paper we show how Hack’s exponent is determined by the
scaling exponent of turbulent one-dimensional flow. Specifically, it is determined
by the scaling exponent of the second structure function, see [12], in the statistical
theory of the turbulent flow.
2
Two dimensionless numbers the Reynolds number and the Froude number are
used to characterize turbulent flow in rivers and streams. If we model the river as
an open channel with x parameterizing the downstream direction, y the horizontal
dept and U is the mean velocity in the downstream direction, then the Reynolds
number
fturbulent Uy
R=
=
fviscous
ν
is the ratio of the turbulent and viscous forces whereas the Froude number
F=
fturbulent
fgravitational
=
U
(gy)1/2
is the ratio of the turbulent and gravitational forces. ν is the viscosity and g is the
gravitational acceleration. Other forces such as surface tension, the centrifugal
force and the Coriolis force are insignificant in streams and rivers.
The Reynolds number indicates whether the flow is laminar or turbulent with
the transition to turbulence starting at R = 500 and the flow usually being fully
turbulent at R = 2000. The Froude number measures whether gravity waves, with
speed c = (gy)1/2 in shallow water, can overcome the flow velocity and travel
upstream. Such flow are called rapid or shooting flows, c > U, in distinction to
tranquil flows, c < U, they correspond to the Froude numbers
1. F > 1, supercritical, c > U
2. F = 1, critical, c = U
3. F < 1, subcritical, c < U
Now for streams and rivers the Reynolds number is typically large O = 105 −
106 , whereas the Froude numbers is small typically O = 10−1 − 10−2 , see [17].
Thus the flows are highly turbulent and ought to be tranquil but this is not the
whole story as we will now explain.
In practice streams and rivers have varied boundaries which are topologically
equivalent to a half-pipe. These boundaries are rough and resist the flow and this
had lead to formulas involving channel resistance. The most popular of these are
Chézy’s law
1/2
V = ucCr1/2 so , uc = 0.552m/s
and Manning’s law
1
1/2
V = um r2/3 so , um = 1.0m/s
n
3
where so is the slope of the channel and r is the hydraulic radius. C is called
Chézy’s constant and measures inverse channel resistance. n is Manning’s roughness coefficient, see [17]. We get new effective Reynolds and Froude numbers
with these new averaged velocities V ,
g
R = 2 2 R, F ∗ =
3uc C
∗
g
2
uc C2 so
1/2
F
It turns out that in real rivers the effective Froude number is approximately one
and the effective Reynolds number is also one, when R = 500 for typical channel
roughness C = 73.3. Thus the transition to turbulence typically occurs in rivers
when the effective turbulent forces are equal to the viscous forces.
In this paper we will ignore the boundaries of the river. The point is that
in a straight segment of a reasonably deep and wide river the boundaries do not
influence the details of the river current in the center, except as a source of flow
disturbances. We will simply assume that these disturbances exist, in the flow
in the center of the river and not be concerned with how they got there. For
theoretical purposes we will conduct a thought experiment where we start with an
unstable uniform flow and then put the disturbances in as small white noise. Then
the mathematical problem is to determine the statistical theory of the resulting
turbulent flow.
The details of the flow close to the boundary are obviously important and
give rise to the Prantl-von Karman universal velocity distribution law for smooth
boundaries and the Chézy’s and Manning’s roughness coefficients for rough boundaries. However, these properties of the flow at the boundaries are a separate problem that will not be addressed in this paper.
The outline of the paper is as follow. In Section 2 we pose the problem. It turns
out to be a stochastic initial value problem. In Section 3 we derive a priori estimates necessary for the existence theory and pose the one-dimensional problem
solved in this paper. The existence of unique solutions that are Hölder continuous
stochastic processes, with Hölder exponent 3/4, is proven in Section 4. A good
and useful estimate of these solutions is derived in Section 5 and the existence
of a unique invariant measure on the space L2 (T 1 ), where the solutions reside,
is proven in Section 6, following McKean [20] and Da Prato and Zabczyk [23].
The scaling of the second structure function (3/2) is found in Section 7 and the
derivation of Hack’s law recalled from [4] and [3].
4
2
The Initial Value Problem
Consider the Navier-Stokes equation
(1)
vt + v · ∇v = ∆v + ∇p
v(x, 0) = vo
with the incompressibility conditions
∇·v = 0
(2)
Eliminating the pressure p using (2) gives the equation
vt + v · ∇v = ∆v + ∇{∆−1 [trace(∇v)2 ]}
(3)
We want to consider turbulent flow in the center of a wide and deep river and to do
that we consider the flow to be in a box and impose periodic boundary conditions
on the box. Since we are mostly interested in what happens in the direction along
the river we take our x axis to be in that direction.
We will assume that the river flows fast and pick an initial condition of the
form
v(0) = Uo e1
(4)
where Uo is a large constant and e1 is a unit vector in the x direction. Clearly
this initial condition is not sufficient because the fast flow will be unstable and
the white noise ubiquitous in nature will grow into small velocity and pressure
oscillations, see for example [1]. But we perform a thought experiment where
white noise is introduced into the fast flow at t = 0. This experiment may be hard
to perform in nature but it is easily done numerically. It means that we should
look for a solution of the form
v(x,t) = Uo + u(x,t)
(5)
where u(x,t) is smaller that Uo but not necessarily small. However, in a small
initial interval [0,to ] u is small and satisfies the equation (3) linearized about the
fast flow Uo
(6)
ut +Uo ∂x u = ∆u + f
u(x, 0) = 0
5
driven by the noise
f=
∑ hk
1/2
dβtk ek
k6=0
The ek = e2πik·x are (three-dimensional) Fourier components and each comes with
its own independent Brownian motion βtk . None of the coefficients of the vectors
1/2
1/2 1/2 1/2
hk = (h1 , h2 , h3 ) vanish because the instabilities are seeded by truly white
noise (white both is space and in time). f is not white in space because the coef1/2
ficients hk must have some decay in k so that the noise term in (6) makes sense.
1/2
However to determine the decay of the hk s will now be part of the problem.
The form of the noise expresses the fact that in turbulent flow there is a continuous sources of small white noise that grows and saturates into turbulent noise
1/2
that drives the fluid flow. The decay of the coefficients hk expresses the spatial
coloring of this larger noise in turbulent flow.
The justification for considering the initial value problem (6) is that for a short
time interval [0,to ] we can ignore the nonlinear terms
−u · ∇u + ∇{∆−1 [trace(∇v)2 ]}
in the equation (3). But this is only true for a short time to , after this time we have
to start with the solution of (6)
uo (x,t) =
∑
1/2
hk
k6=0
Z t
2 |k|2 +2πiU k )(t−s)
o 1
e(−4π
0
dβks ek (x)
(7)
as the first iterate in the integral equation
Z t
u(x,t) = uo (x,t) +
to
K(t − s) ∗ [−u · ∇u + ∇∆−1 (trace(∇u)2 )]ds
(8)
where K is the (oscillatory heat) kernal in (7). In other words to get the turbulent
solution we must take the solution of the linear equation (6) and use it as the first
term in (8). It will also be the first guess in Picard iteration.The solution of (6) can
be written in the form
1/2
uo (x,t) = ∑ hk Atk ek (x)
k6=0
where the
Atk =
Z t
2 |k|2 +2πiU k )(t−s)
o 1
e(−4π
0
6
dβks
(9)
are independent Orstein-Uhlenbeck processes with mean zero, see for example
[23].
Now it is easy to see that the solution of the integral equation (8) u(x,t) satisfies the driven Navier-Stokes equation
ut +Uo ∂x u = ∆u − u · ∇u + ∇∆−1 (trace(∇u)2 ) + ∑ hk dβtk ek , t > t0
1/2
k6=0
(10)
ut +Uo ∂x u = ∆u + ∑ hk dβtk ek , u(x, 0) = 0, t ≤ t0
1/2
k6=0
and the above argument is the justification for studying the initial value problem
(10). We will do so from here on. The solution u of (10) still satisfies the periodic
boundary conditions and the incompressibility condition
∇·u = 0
(11)
The mean of the solution uo of the linear equation (6) is zero by the formula (7)
and this implies that the solution u of (10) also has mean zero
Z
ū(t) =
3
T3
u(x,t)dx = 0
(12)
A Priori Estimates
In this section we will explain the probabilistic setting and prove the a priori estimates necessary for the existence proof. The problem to be solved will be also be
posed.
We let (Ω, F , P), Ω is a set (of events) and F a σ algebra on Ω, denote a
probability space with P the probability measure of Brownian motion and Ft a filtration generated by all the Brownian motions βtk on [t, ∞). We denote by V [0, T ]
the space having the following properties, see Oxendal [22].
Definition 3.1 V [0, T ] is the space of functions satisfying the conditions:
1. f (ω,t) is measurable with respect to F × B where B is the σ-algebra of the
Borel sets on [0, ∞), ω ∈ Ω,
2. f (ω,t) is adapted to the filtration Ft ,
7
3.
Z T
f 2 (ω,t)dt) < ∞.
E(
0
We denote by W(p,m) [0, T ] the space of functions, whose W (m,p) Sobolev norm
2
kuk(m,p) (ω,t) lies in ∈ V [0, T ] and we let L(m,p)
denote the space of functions in
(2,2)
W (p,m) whose Sobolev norm lies in L2 (Ω, P). The norm in the space L(m,p) [0, T ] =
2
L2 ([0, T ]; L(m,p)
) is
1/2
T
Z
E(
kuk2(m,p) dt)
0
In particular for p = 2, we get the norm
Z T
kuk2 (2,2) = E(
L(m,2)
0
|(1 − ∆2 )m/2 u|22 dt)
(2,2)
Here E denotes the expectation in L2 (Ω, P). The space L(m,2) [0, T ] is clearly the
product space L2 (Ω, P) ×W (m,2) (T 1 ) × L2 [0, T ] and we will proved the existence
2
of solutions in the subspace of continuous functions C([0, T ]; L(m,2)
)
Let < ·, · > denote the inner product in L2 (R3 ) The following a priori estimates
provide the foundation of the probabilistic version of LeRay’s theory.
Lemma 3.1 The L2 norms |u|2 (ω,t) and |∇u|2 (ω,t) satisfy the identity
d|u|22 + 2|∇u|22 dt = 2 ∑ hk
1/2
k6=0
< u, ek > dβtk + ∑ hk dt
k6=0
and the bounds
|u|22 (ω,t) ≤ 2
(14)
∑
1/2
hk
k6=0
Z t
0
e−2λ1 (t−s) < u, ek > dβks
+
Z t
(15)
0
|∇u|22 (ω, s)ds ≤
∑
k6=0
1/2
hk
Z t
0
8
1 − e−2λ1t
2λ1
< u, ek > dβks +
∑ hk
k6=0
t
hk
2 k6∑
=0
(13)
where λ1 is the smallest eigenvalue of −∆ with vanishing boundary conditions on
the box [0, 1]3 . The expectations of these norms are also bounded
1 − e−2λ1t
2λ1
E(|u|22 )(t) ≤
(16)
Z t
(17)
0
∑ hk
k6=0
t
hk
2 k6∑
=0
E(|∇u|22 (s))ds ≤
Proof: The identity (13) follows from Le Ray’s theory and Ito’s Lemma. We
apply Ito’s Lemma to the L2 norm of u squared,
Z
d
T3
|u|2 dx = 2
∂u
1/2
· udxdt + 2 ∑ hk
3
T ∂t
k6=0
Z
1/2
where k ∈ Z3 and hk
2
T3
u · ek dxdβtk + ∑ hk
k6=0
Z
T3
dxdt (18)
∈ R3 . Now by use of the Navier-Stokes equation (10)
Z
d|u| = 2
Z
T3
∆u · u + (−u · ∇u + ∇∆−1 (trace(∇u)2 ) · udx
+2 ∑ hk
Z
= −2|∇u|22 dt + 2 ∑ hk
Z
k6=0
k6=0
T3
T3
u · ek dxdβtk + ∑ hk dt
k6=0
u · ek dxdβtk + ∑ hk dt
k6=0
since the divergent-free vector u is orthogonal both to the gradient ∇∆−1 (trace(∇u)2
and u · ∇u by the divergence theorem. The first term in the last expression is also
obtained by integration by parts. This is the identity (13). The inequality (14) is
obtained by Poincaré’s inequality
λ1 |u|22 ≤ |∇u|22
(19)
where λ1 is the smallest eigenvalue of −∆ with vanishing boundary conditions on
the cube [0, 1]3 . By Poincaré’s inequality
d|u|22 + 2λ1 |u|22 dt ≤ d|u|22 + 2|∇u|22 dt
= 2 ∑ h1/2 < u, ek > dβtk + ∑ hk dt
k6=0
k6=0
9
Solving the inequality gives (14). (15) is obtained by integrating (13)
|u|22 (t) + 2
Z t
0
|∇u|22 (ω, s)ds = 2
∑h
k6=0
1/2
Z t
0
< u, ek > dβks +
t
hk
2 k6∑
=0
since |u|2 (0) = 0 and dropping |u|22 (t) > 0.
Finally we take the expectations of (14) and (15) to obtain respectively (16)
and (17), using that the function < u, ek > (ω,t) is adapted to the filtration Ft and
Fubini’s theorem for (17).
QED
In a deep and wide river it is reasonable to think that the directions transverse
to the main flow, y the direction across the river, and z the horizontal direction,
play a secondary role in the generation of turbulence. As a first approximation to
the flow in the center of a deep and wide, fast-flowing river we will now drop these
directions. Of course y and z play a role in the motion of the large eddies in the
river but their motion is relatively slow compared to the smaller scale turbulence.
Thus our initial value problem (10) becomes
(20)
2
k
ut +Uo ux = uxx − uux + ∂−1
x ((ux ) ) + ∑ hk dβt ek , t > t0
1/2
k6=0
(21)
ut +Uo ux = uxx + ∑ hk dβtk ek , u(x, 0) = 0, t ≤ t0
1/2
k6=0
We still have periodic boundary condition on the unit interval but the incompressibility condition can be dropped. Recall from Equation (12) that the mean ū is still
zero. This equation now describes the turbulent flow in the center of a relatively
straight section of a fast river. The full three-dimensional flow will be treated in a
subsequent publication [2].
The one-dimensional version of Lemma 3.1 is
Corollary 3.1 In one dimension the L2 norms of u and ux satisfy
d|u|22 + 2|ux |22 dt = 2 ∑ h1/2 < u, ek > dβtk + ∑ hk dt
k6=0
k6=0
10
(22)
and the bounds
|u|22 (ω,t) ≤ 2 ∑ h1/2
(23)
k6=0
Z t
e−2λ1 (t−s) < u, ek > dβks
0
+
Z t
(24)
0
|ux |22 (ω, s)ds ≤
∑h
1/2
k6=0
Z t
0
1 − e−2λ1t
2λ1
< u, ek > dβks +
∑ hk
k6=0
t
hk
2 k6∑
=0
where λ1 is the smallest eigenvalue of −∂xx with vanishing boundary conditions
on the interval [0, 1]. The expectations of these norms are also bounded
Z t
(26)
0
4
1 − e−2λ1t
2λ1
E(|u|22 )(t) ≤
(25)
∑ hk
k6=0
t
hk
2 k6∑
=0
E(|ux |22 (s))ds ≤
Existence of Turbulent Solutions
In this section we prove the existence of the turbulent solutions of the initial value
problem (20). The following theorem states the existence of turbulent solutions
in one dimension. First we write the initial value problem (20) as an integral
equation
Z t
u(x,t) = uo (x,t) +
0
1
2
K(t + to − s) ∗ [− (u2 )x + ∂−1
x (ux ) )]ds
2
(27)
Here K is the oscillatory heat kernal (7) in one dimension and
uo (x,t) =
∑ hk
1/2 k
At ek (x)
k6=0
the Atk s being the Orstein-Uhlenbeck processes from Equation (9).
We start by proving a Gronwall estimate that will be used in later sections and
+
to given an a priori bound on the solution. If qp is a rational number let qp denote
any real number s > qp .
11
Lemma 4.1 Let u1 and u2 be two solutions of (27) with initial functions, i.e. solutions to (21), uo1 (x,t) = ∑k6=0 (h1k )1/2 Atk ek and uo2 (x,t) = ∑k6=0 (h2k )1/2 Atk ek , then
+
E(ku1 − u2 k25 + ) ≤
( 4 ,2)
(1 + (2πk)(5/2) ) 1 1/2
((hk ) − (h2k )1/2 )2
∑
2
2(2πk)
k6=0
(h1 + h2k )
×(1 + t ∑ k
4λ1
k6=0
(28)
Z t
(h1 +h2 )
∑k6=0 k4λ k (t 2 −s2 )
1
e
ds)
0
Proof: We consider the difference of two solutions to the integral equation (27)
u1 (ω, x,t) − u2 (ω, x,t)
Z
h
1 t −(4π2 k2 −2πikUo )(t+to −s)
1 1/2
2 1/2 k
= ∑ ((hk ) − (hk ) )At +
e
2 0
k6=0
×
\ 2
−1\ 2
\2
\2
(2[∂−1
x ((u1 )x ) − ∂x ((u2 )x ) ] − [ ∂x ((u1 ) ) − ∂x ((u2 ) )])ds
=
∑
h
i
ek (x)
((h1k )1/2 − (h2k )1/2 )Atk
k6=0
1
+
2
Z t
e−(4π
2 k2 −2πikU )(t+t −s)
o
o
2(
0
d
d
d
d
d
d
(u
((u
1 )x ∗ ((u1 )x − (u2 )x )
1 )x − (u2 )x ) ∗ (u2 )x
+
)
iπk
iπk
i
− 2πik (b
u1 ∗ (b
u1 − ub2 ) + (b
u1 − ub2 ) ∗ ub2 ))ds ek (x)
Thus
|u1 (ω,t) − u2 (ω,t)|22
h
≤ 2 ∑ ((h1k )1/2 − (h2k )1/2 )2 |Atk |2
k6=0
+
Z t
1
|
4
2 k2 −2πikU )(t+t −s)
o
o
e−(4π
2(
0
d
d
d
d
d
d
(u
((u
1 )x ∗ ((u1 )x − (u2 )x )
1 )x − (u2 )x ) ∗ (u2 )x
+
)
iπk
iπk
i
− 2πik (û1 ∗ (û1 − û2 ) − (û1 − û2 ) ∗ û2 )]ds|2
≤ 2 ∑ ((h1k )1/2 − (h2k )1/2 )2 |Atk |2
k6=0
Z t
+ C(
0
(|u1 |2 + |u2 |2 + |(u1 )x |2 + |(u2 )x |2 )(|u1 − u2 |2 + |(u1 )x − (u2 )x |2 ds)2
by the Cauchy-Schwarz inequality and putting a derivative on the oscillatory heat
kernal and overestimating the result by a constant C. Another application of
12
Cauchy-Schwarz gives
|u1 (ω,t) − u2 (ω,t)|22 ≤ 2 ∑ ((h1k )1/2 − (h2k )1/2 )2 |Atk |2
k6=0
Z t
+ C
0
(|u1 |22 + |u2 |22 + |(u1 )x |22 + |(u2 )x |22 )ds
Z t
0
(|u1 − u2 |22 + |(u1 )x − (u2 )x |22 )ds
and an application of Corollary 3.1 followed by taking the expectation gives
E(|u1 (ω,t) − u2 (ω,t)|22 ) ≤
(29)
(h1 + h2k )
+ tC ∑ k
2λ1
k6=0
Z t
0
((h1k )1/2 − (h2k )1/2 )2
∑
8π2 k2
k6=0
E((|u1 − u2 |22 + |(u1 )x − (u2 )x |22 ))ds
5+
Now acting on the difference u1 − u2 above by ∂x4 gives
(30)
+
5+
4
E(|∂x (u1 (ω,t) − u2 (ω,t))|22 ) ≤
(h1k + h2k )
+ tC ∑
2λ1
k6=0
Z t
0
(2πk)(5/2) ((h1k )1/2 − (h2k )1/2 )2
∑
8π2 k2
k6=0
E((|u1 − u2 |22 + |(u1 )x − (u2 )x |22 ))ds
with a different constant from above since the derivatives act only on the oscillatory heat kernal. Then adding the two inequalties (29) and (30) gives the inequality
+
(31)
E(ku1 (ω,t) − u2 (ω,t)k25 + ) ≤
( 4 ,2)
(h1k + h2k )
2λ1
k6=0
+ tC ∑
Z t
0
(1 + (2πk)(5/2) )((h1k )1/2 − (h2k )1/2 )2
∑
8π2 k2
k6=0
E(ku1 (ω,t) − u2 (ω,t)k25 + )ds
( 4 ,2)
Solving the last inequality in a similar manner as the proof of Gronwall’s inequality gives (28).
QED
Theorem 4.1 If the solution of the linear equation (6) satisfies the condition
(32)
E(kuo k25 + ) =
( 4 ,2)
∑ (1 + (2πk)(5/2)
+
)hk E(|Atk |2 )
k6=0
+
1
(1 + (2πik)(5/2) )
≤ ∑
hk < ∞
2 k6=0
(2πk)2
13
then the integral equation (27) has a unique solution in the space C([0, ∞); L 25 + )
( 4 ,2)
of stochastic processes with
kuk
Z T
2
L (2,2)
5+
(4
= E(
0
,2)
kuk25 + dt) < ∞
( 4 ,2)
for any finite T .
Proof: Setting u2 and h2k equal to zero in Equation (31) gives
+
sup E(ku(ω,t)k25 + ) ≤
( 4 ,2)
t
(33)
(1 + (2πk)(5/2) )hk
∑
8π2 k2
k6=0
hk
sup E(kuk25 + )
( 4 ,2)
2λ1 t
k6=0
+ t 2C ∑
For t small this inequality shows that the integral equation (27) maps a bounded set
in C([0,t); L 25 + ) into itself. Similarly the Equation (31) shows that the integral
( 4 ,2)
equation (27) defines a contraction on C([0,t); L 25 + )
( 4 ,2)
E(ku1 (ω,t) − u2 (ω,t)k25 + )
( 4 ,2)
≤ t 2C ∑
k6=0
(h1k + h2k )
2λ1
sup E(ku1 (ω,t) − u2 (ω,t)k25 + )
( 4 ,2)
t
for t small. The proof is finished by setting u2 and h2k equal to zero in Equation
(28), this gives the a priori estimate of u
+
E(kuk25 + ) ≤
( 4 ,2)
(1 + (2πk)(5/2) )
hk
hk (1 + t ∑
∑
2
2(2πk)
4λ1
k6=0
k6=0
Z t
h
∑k6=0 4λk (t 2 −s2 )
1
e
ds) (34)
0
This estimate allows us to extend the time interval of local existence to the whole
half-axis [0, ∞).
QED
Corollary 4.1 The solution of the linearized equation (6) uniquely determines the
solution of the integral equation (27).
Proof: When the initial functions u1o and u2o in Lemma 4.1 are the same then
u1 = u2 .
QED
14
5
An Estimate of the Turbulent Solutions
The estimate of the turbulent solution (34) is exponentially increasing (it increases
2
as c1tec2t ) and is not terribly useful. In this section we prove a much more realistic
estimate on a finite time interval.
The mechanism of the turbulence production are fast oscillations seeded by
white noise and driven by the fast constant flow Uo . The bigger Uo is the more efficient this mechanism becomes. The following lemma plays a key role in the proof
of the useful estimate of the turbulent solution. It is a version of the RiemannLebsgue Lemma which captures the averaging effect of the oscillations.
Lemma 5.1 Let
Z T
ŵ =
w(s)e−2πikUo s ds
0
then
c
ŵ(k) = ∆w(k)
where
1
1
∆w = (w(s) − w(s +
))
2
2kUo
and ∆w satisfies the estimate
1
|∆w| ≤
2(2kUo )1/2
Z
0
T
∂w
( )2 dt
∂t
1/2
(35)
Proof: The proof is similar to the proof of the Riemann-Lebesgue lemma for the
Fourier transform in time
Z T
ŵ(k) =
w(s)e−2πikUo s ds
0
=−
Z T
1
w(s)e−2πikUo (s− 2kUo ) ds
0
=−
Z T
w(s +
0
1
)e−2πikUo sds
2kUo
Taking the average of the first and the last expression we get
1
ŵ =
2
Z T
0
(w(s) − w(s +
1
c
))e−2πikUo s ds = ∆w
2kUo
15
Now
|∆w| =
1
1
|(w(s) − w(s +
))|
2
2kUo
!1/2
Z s+ 1
2kUo ∂w 2
( ) dt
1
≤
2(2kUo )1/2
1
≤
2(2kUo )1/2
∂t
s
T
Z
0
∂w
( )2 dt
∂t
1/2
by the Cauchy-Schwarz inequality.
QED
The lemma allows us to estimate the sup norm (in t) of w in terms of the L2 norm
(in t) with a gain of (kUo )−1/2 .
Theorem 5.1 If the velocity Uo of the uniform flow is sufficiently large and
(36)
E(kuo k25 + ) =
( 4 ,2)
∑ (1 + (2πk)(5/2)
+
)hk E(|Atk |2 )
k6=0
+
1
(1 + (2πik)(5/2)
≤ ∑
hk < ∞
2 k6=0
(2πk)2
and if
T≤
Uo
2(C ∑k6=0
+
(1+(2πk)(5/2) )
2(2πk)2
hk (1 + 12 ∑k6=0 hk ))1/2
(37)
then the solution of the integral equation (27) satisfies the estimates
+

1/2
(1+(2πk)(5/2) )
hk 
∑
k6
=
0
2(2πk)2

kuk25 + dt) ≤ Uo 

( 4 ,2)
C(1 + 12 ∑k6=0 hk )
0
Z T
E(
and
(38)
+
sup E(kuk25 + )(t) ≤
( 4 ,2)
t
1
(1 + (2πk)(5/2) )
hk + DUo2
∑
2
2 k6=0
(2πk)
on the time interval [0, T ], C and D being constants.
16
(39)
Corollary 5.1 Onsager’s Conjecture
The solutions of the integral equation (27) are Hölder continuous with exponent
3/4.
Remark 5.1 The hypothesis (36) is the answer to the question we posed in Section
1/2
2 how fast the coefficients hk had to decay in Fourier space. They have to decay
5+
5+
sufficiently fast for the supremum in t of the expectation of the H 4 = W ( 4 ,2)
Sobolev norm of the initial function uo , to be finite. In other words the sup in t of
the L 25 + norm has to be finite.
( 4 ,2)
We now prove the theorem with the help of one lemma.
Proof: We write the integral equation (27) in the form
u(ω, x,t)
=
∑
1/2
[hk Atk
k6=0
=
∑ [hk
1/2 k
At
k6=0
1
+
2
+
Z t
2 k2 −2πikU )(t+t −s)
o
o
e−(4π
2 k2 −2πikU )(t+t −s)
o
o
0
Z
1 t
2
e−(4π
0
\2 \
2
(2∂−1
x (ux ) − ∂x (u ))(k, s)ds]ek (x)
(
ûx ∗ ûx
− 2πikû ∗ û)(k, s)ds]ek (x)
iπk
We first estimate the second term in the last integral above
Z t
2 k2 (t+t
e−4π
o −s)
(2πik
0
l6=0,k
=
−e−4π
∑ (û(k − l)û(l))(s))e2πikUo(t+to−s))ds
2 k2 (t+t −(s+ 1 ))
o
2kUo
(
Z
1 t
2
2 k2 (t+t
2πik[e−4π
0
o −s)
∑ (û(k − l)û(l))(s)
l6=0,k
1
∑ (û(k − l)û(l))(s + 2kUo ))]e2πikUo(t+to−s)ds
l6=0,k
17
by the same computation as in Lemma 5.1,
Z
2 2
1 t
=
[ ∑ 2πike−4π k (t+to −s) û(k − l, s)
2 0 l6=0,k
− 2πke
1
−4π2 k2 (t+to −(s+ 2kU
))
o
1
û(k − l, s +
) û(l, s)
2kUo
2 2
1 + ∑ û(k − l, s +
) 2πike−4π k (t+to −s) û(l, s)
2kUo
l6=0,k
1
1
−4π2 k2 (t+to −(s+ 2kU
))
o
− 2πke
û(l, s +
) ]e2πikUo (t+to −s) ds
2kUo
Z t
1
=
[ ∑ (∆(e0 û(k − l))û(l))(s) + û(k − l, s +
)∆(e0 û(l))(s)]e2πikUo (t+to −s) ds
2kUo
0 l6=0,k
2 2
using the notation from Lemma 5.1, where e0 denotes 2πke−4π k (t+to −s) and w =
e0 û(k − l), etc. The sum in the discrete convolution of the Fourier coefficients excludes l = k, 0 because û(0) = 0, since the mean vanishes, ū = 0. By the CauchySchwarz inequality we get the inequality
"
Z
t
≤
(
0
∑ (∆(e0û(k − l)))2)1/2( ∑ (û(l))2)1/2
l6=0,k
l6=0,k
∑ (û(k − l))2)1/2( ∑ (∆(e0û(l)))2)1/2
+(
l6=0,k
#
ds
l6=0,k
for the last expression above. Now by Lemma 5.1
s+ 2kUo
∂e0 û(k − l) 2
1
(∆(e
û(k
−
l))(s))
≤
(
) dr
∑
∑
2kUo s
∂r
l6=0,k
l6=0,k
!
∂e0 û(k − l) 2
∂K 0 ∗ u
1
1
= 2 2 ∑(
) (s + c) = 2 2 | o |22 (s + c)
4k Uo l6=0,k
∂s
4k Uo
∂s
0
Z
2
1
1
by the mean-value theorem, 0 ≤ c ≤ 2kU
, and Plancherel’s identity where Ko0 is
o
the the derivative of the oscillatory heat kernal from (7), started at time to , and ∗
denotes the convolution in x, so
Z t
0
1
2
∂d
x u (k, s)ds ≤
kUo
Z t
∂Ko0 ∗ u
|
0
∂s
18
|2 (s + c)|u|2 (s +
1
)ds
2kUo
Thus the second (inertial) term in the integral in the integral equation (27) can be
estimated as
1
∑ 4|
k6=0
Z t
Z t
∂Ko0 ∗ u
1
)ds)2
∂s
2kUo
0
0
(40)
The first (pressure) term in the integral in the integral equation (27) is estimated in much the same way and we get the estimate
Z t
1
1
2
2
∂d
(
x u (k, s)ds| ≤
∑
4Uo2 k6=0 k2
|
Z t
∂Ko ∗ u
1
)ds)2
∂s
2kU
0
o
k6=0 0
(41)
We make a small error when we make one u and and one ux term depend on
0
1
c ∂Ko ∗u
instead
of
s.
But
this
error
is
bounded
by
|
s + 2kU
U
∂s |2 and is easily absorbed
o
o
into the terms below.
Collecting the estimates (40) and (41) we estimate of the solution u the integral
equation (27)
∑|
\2
2
∂−1
x (ux ) (k, s)ds| ≤
|u|22 (ω, x,t)
≤
∑
1
1
(
∑
Uo2 k6=0 k2
|
π2 1
[ (
3Uo2 4
(|
|2 (s + c)|u|2 (s +
hk |Atk |2 +
k6=0
(42)
Z 2
∂Ko0 ∗ u
|
+ (
0
∂s
|2 (s + c)|ux |2 (s +
Z t
∂Ko0 ∗ u
0
∂s
|2 (s + c)|ux |2 (s +
|2 (s + c)|u|2 (s +
1
)ds)2
2kUo
1
)ds)2 ]
2kUo
∂K 0 ∗u
Next we estimate the term | ∂so |2 by use of the Navier-Stokes equation (20).
We convolve Ko0 with the equation (20) and utilizing that Ko0 is a solution of the
heat equation with the convection term (Uo ux ), we write the resulting equation as
1
2
(Ko0 ∗ u)t = 2Ko000 ∗ u − Ko0 ∗ (u2 )x + Ko0 ∗ ∂−1
x (ux )
2
Thus by Minkowski’s inequality
1
2
|(Ko0 ∗ u)t |2 = 2|Ko000 ∗ u|2 + |Ko0 ∗ (u2 )x |2 + |Ko0 ∗ ∂−1
x (ux ) |2
2
1
≤ |Ko00 ∗ u2 |2 + |Ko ∗ (ux )2 |2 ≤ C(|u|2 + |u|24 + |ux |24 )
2
by integration by parts and the fact that Ko and all its derivatives are bounded for
to > 0. Similarly, by Poincaré’s inequality we get
|(Ko ∗ u)t |2 ≤ C(|ux |2 + |u|24 + |ux |24 )
19
Now by Sobolev’s inequality
5+
|ux |4 ≤ C(|∂x4 u|2 + |u|2 )
and
1+
4
|u|4 ≤ C(|∂x u|2 + |u|2 )
5+
1+
4
Interpolating |∂x u|2 between |∂x4 u|2 and |u|2 , we get that
|u|24 + |ux |24 ≤ Ckuk25 +
( 4 ,2)
This finally gives the estimate
|(Ko0 ∗ u)t |2 ≤ C(kuk25 +
+ |u|2 )
(43)
|(Ko ∗ u)t |2 ≤ C(kuk25 +
+ |ux |2 )
(44)
( 4 ,2)
and similarly
( 4 ,2)
Now we use the estimates above and (43) to substitute into (42) to get an estimate
of u
|u|22 (ω, x,t) ≤
∑
hk |Atk |2 +
k6=0
π2 1
[ (
3Uo2 4
Z t
∂Ko0 ∗ u
(|
∂s
0
Z t
∂Ko0 ∗ u
|
+(
(45)
∂s
0
≤
Z
C 1 t
∑ hk |Atk |2 + Uo2 [ 4 (
0
k6=0
|2 (s + c)|u|2 (s +
1
)ds)2
2kUo
|2 (s + c)|ux |2 (s +
1
)ds)2 ]
2kUo
(|u|22 + kuk25 + |u|2 )ds)2
( 4 ,2)
Z t
+(
0
(|ux |22 + kuk25 + |ux |2 )ds)2 ]
( 4 ,2)
by (43) and (44)
≤
∑
k6=0
hk |Atk |2 +
Z t
C
[(
Uo2
0
|u|22 ds)2
Z t
1
+
(
2
0
Z t
+ (
0
Z t
C
≤ ∑ hk |Atk |2 + 2 [(
Uo
k6=0
0
|ux |22 ds)2
Z t
+
0
20
|ux |22 ds)2
1
kuk( 5 + ,2) (|ux |2 + |u|2 )ds)2 ]
2
4
kuk( 5 + ,2) ds
4
Z t
0
kuk( 5 + ,2) |ux |22 ds]
4
by Poincaré’s inequality and since by the Cauchy-Swartz inequality
Z t
(
kuk25 + |ux |2 ds)2
( 4 ,2)
0
≤
Z t
0
Z t
kuk( 5 + ,2) ds
0
4
kuk( 5 + ,2) |ux |22 ds
4
5+
Then by use of Corollary 3.1 and since the H 4 (T 1 ) norm dominates the H 1 (T 1 )
norm, we get that
|u|22 (ω, x,t) ≤
∑ hk |Atk |2 +
k6=0
Z t
1
C
(1
+
hk )(
Uo2
2 k6∑
=0
×( ∑ hk
1/2
Z t
0
k6=0
0
kuk25 + ds)2 +
( 4 ,2)
C
Uo2
kuk25 + (s + c) < u, ek > (s +
( 4 ,2)
Z t
0
kuk25 + ds
( 4 ,2)
1
)dβs+ 1 )
2kUo
2kUo
Differentiating the integral equation (27) with respect to x gives a similar esti5+
4
mated for |∂x u| and adding the two gives
kuk25 + (ω,t) ≤
( 4 ,2)
∑ (1
(5/2)+
+ (2πk)
)hk |Atk |2 +
k6=0
+
C
1/2
+ 2 ∑ (1 + (2πk)(5/2) )hk
Uo k6=0
×
Z t
0
Z t
0
Z t
C
1
(1 + ∑ hk )(
2
Uo
2 k6=0
0
kuk25 + dt)2
( 4 ,2)
kuk25 + dt
( 4 ,2)
kuk25 + (s + c) < u, ek > (s +
( 4 ,2)
1
)dβs+ 1
2kUo
2kUo
Integrating with respect to t then gives the inequality
Z t
0
≤
kuk25 + (ω,t)dt −
( 4 ,2)
∑ (1 + (2πk)(5/2) )hk
+
k6=0
Z t
1
C
(1 + ∑ hk )t(
2
Uo
2 k6=0
Z t
0
0
kuk25 + dt)2
( 4 ,2)
|Atk |2 dt +
+
C
1/2
(1 + (2πk)(5/2) )hk
∑
2
Uo k6=0
Z tZ s
0
0
ku|25 + (r + c) < u, ek > (r +
( 4 ,2)
Applying Lemma (5.2) to the last inequality gives the bound
v
u
+
(1+(2πk)(5/2) )
Z t
u∑
hk
2
2
t k6=0
8π k
E( kuk25 + ds) ≤ Uo
1
( 4 ,2)
C(1 + 2 ∑k6=0 hk )
0
21
1
)dβr+ 1
2kUo
2kUo
(46)
Then we use the inequality (31) with u2 = 0 and h2k = 0. This gives
+
E(kuk25 + ) ≤
( 4 ,2)
(1 + (2πk)(5/2) )
hk
hk + tC0 ∑
∑
2
2
8π k
2λ1
k6=0
k6=0
Z t
0
E(kuk25 + )ds
( 4 ,2)
+
(1 + (2πk)(5/2) )
≤∑
hk + DUo2
2 k2
8π
k6=0
where D is a constant independent of the hk s. This inequality finishes the proof of
the theorem.
QED
Lemma 5.2 Consider the inequality
x(t) − bx2 (t) ≤ a + d,
x(0) = 0,
where E(d) = 0 and b is deterministic. If E(a) <
1
Proof: x − bx2 obtains it maximum 4b
at x =
1
then E(x(t)) must stay below its maximum 2b
.
1
4b
1
2b
then E(x) ≤
1
2b
1
thus if E(a) = E(a + d) < 4b
QED
Remark 5.2 Corollary 5.1 is the resolution of a famous question in turbulence: Is
turbulence always caused by the blow-up of the velocity u or its gradient ux ? The
answer according to Theorem 5.1 is no; the solutions are not singular. However,
they are not smooth either, contrary to the belief, stemming from LeRay’s theory
[18], that if solutions are not singular then they are smooth. By Corollary 5.1
the solutions are Hölder continuous with exponent 3/4 in one dimension. This
confirms a conjecture made by Onsager [21] in 1949.
6
The Existence of the Invariant Measure
If φ is a bounded function on L2 (T 1 ) then the invariant measure dµ for the SPDEs
(20 and 21) is given by the limit
Z
lim E(φ(u(ω,t))) =
t→∞
L2 (T 1 )
φ(u)dµ(u)
(47)
In this section we proof that this limit exists and is unique.
Theorem 6.1 The integral equation (27) possesses a unique invariant measure.
22
Corollary 6.1 The invariant measure dµ is ergodic and strongly mixing.
The corollary follows immediately from Doob’s Theorem, see for example [23].
We prove the theorem in three lemmas. First we define a transition probability
Pt (v, Γ) = L (u(v,t))(Γ), Γ ⊂ E ,
where L is the law of u(t) and v = uo is the initial condition and E is the σ algebra
generated by the Borel subsets of L2 (T 1 ). Then
RT (v, ·) =
1
T
Z T
P(v, ·)dt
0
is a probability measure on L2 (T 1 ). By the Krylov-Bogoliubov theorem, see [23],
if the sequence of measures RT are tight then the invariant measure dµ is the weak
limit
Z
1 T
dµ(·) = lim
P(v, ·)dt
T →∞ T 0
Namely,
Z
R∗T dν(Γ) =
and
< R∗T ν, φ >=
Z
L2 (T 1 )
L2 (T 1 )
RT (v, Γ)dν(v)
φ(v)RT (v, Γ)dν(v) →
Z
L2 (T 1 )
φ(v)dµ(v)
as T → ∞.
Lemma 6.1 The sequence of measures
1
T
Z T
P(v, ·)dt
0
is tight.
Proof: By the inequalities (25) and (26)
1
T
Z T
0
E(kuk2(1,2) )(t)dt ≤
(1 + λ1 )
hk
2λ1 k6∑
=0
The space H 1 (T 1 ) = W (1,2) is relatively compact in L2 (T 1 ) so it suffices to show
that u(t) lies in a bounded set in H 1 (T 1 ) almost surely, or for all ε > 0 there exists
an R such that,
Z
1 T
P(ku(t)k(1,2) < R)dt > 1 − ε
T 0
23
for T ≥ 1. But this follows from Chebychev’s inequality. Namely,
!
Z
1 T
1 (1 + λ1 )
P(ku(t)k(1,2) ≥ R)dt ≤
hk < ε
T 0
R
2λ1 k6∑
=0
for R sufficiently large. It is no restriction that we are taking the initial data to lie
in H 1 (T 1 ) because this is a dense subset of L2 (T 1 ). This proofs that the sequence
of measures is tight.
QED
Next we proof the strong Feller property, see [23].
Lemma 6.2 Suppose that
1
C ∑ hk < √
8 2
k6=0
(48)
where C is a Sobolev coefficient, then the Markovian semigroup Pt generated by
the integral equation (27) is strongly Feller.
Proof: We compare the operators
T = ∂2x + 2∂−1
x ux ∂x
and
S = 2∂2x
on L2 (T 1 ). The operator
A = S − T = ∂2x − 2∂−1
x ux ∂x
is S bounded, see Kato [16],
kAvk ≤ akvk + bkSvk
with coefficients a = 8π2C|ux |22 and b = 3/4,
1/2
2
kAvk = k∂2x vk + 2k∂−1
x ux vx k ≤ |∂x v|2 + 2λ1 |vx |∞ |ux |2
2
by Poincaré’s inequality, since ∂−1
x ux vx = 0 at x = 0, where λ1 = 4π . The last
expression is estimated as
3
3
≤ |2∂2x v|2 + 4Cλ1 |ux |22 |v|2 = kSk + 4Cλ1 |ux |22 |v|2
4
4
24
by interpolation, where C is a Sobolev’s constant. This means that for every (ω,t)
the distance between the corresponding eigenspaces of T and S is
δ(T, S) ≤
9
(a2 + b2 )1/2
= 4( + 16λ21C2 |ux |42 )1/2
(1 − b)
16
see Kato [16]. Thus
√
3
δ(T, S) ≤ 4 2 sup( , 4Cλ1 |ux |22 )
t 4
√
and since the first eigenvalue of −S is 2λ1 = 8π2 ≥ 3 2 is suffices that
Z
√
1 T
16 2Cλ1
E(|ux |22 )dt < 8π2
T 0
for all T . But using Equation (24) from Corollary 3.1 this condition becomes
1
C ∑ hk < √
8 2
k6=0
We will now use this hypothesis to solve the equation
vt = T1 v
where
T1 = ∂2x + 2∂−1
x ux ∂x − u∂x − ux
and
S1 = 2∂2x − u∂x − ux
Notice that
T1 − S1 = T − S = A
is the same operator as above.
vt = T1 v = S1 v + (T1 − S1 )v
gives
vt ≤ S1 v + kT1 − S1 kv
for positive solutions v. The last inequality may be solved to give
Rt
v(t) ≤ V (t, S1 )e
25
0 kAkdt
v0
using the commutativity of S1 and kAk where V (t, S1 ) is the solution operator generated by S1 . Now V is computed below and is an expectation so the expectation
of v satisfies
Rt
1 Rt
E(v) ≤ V E(e 0 kAkdt )v0 ≤ Ve( t 0 E(kAk)dt)t v0 .
We see below that
kV k ≤ e−8π
1 Rt
t 0 E(kAk)dt
and thus the hypothesis
2t
< 8π2 gives
E(v) ≤ v0
with a density ∂y E ≤ 1.
Now the initial value problem
vt = S1 v = vxx + uvx − ux v
v(0) = δ(x)dx
can be solved by a combination of Feynmann-Kac and Cameron-Martin, see McKean [20]. Namely,
v(x,t) = ∂y E(e−
Rt
1 Rt 2
0 u(t−s,xs )dxs − 2 0 u (t−s,xs )ds− 0 ux (t−s,xs )ds
Rt
= ∂y E(e
Rt
1 Rt 2
o u(s,xs )dxs − 2 0 u (s,xs )ds
; xt ≤ y)
; xt ≤ x)
where ∂y E denotes the density, upon time-reversal of the Brownian path x(t −s) →
x(s), s ≤ t. Evidently
Rt
|∂y E(e
1 Rt 2
0 u(s,xs )dxs − 2 0 u (s,xs )ds
; xt ≤ x)| ≤ e−8π
2t
as claimed above, and thus following McKean [20],
φt (u) − φt (w) =
Z 1Z 1
Z 1
BM
0
0
0
∇φ(h) · (u − w)(y,t)v(x,t)dxd(δu)dy
where v = ∂y E, h = w + (u − w)δu. δu being the variation
δu(x, y,t) =
∂u(x,t)
∂u(y, 0)
with y fixed, BM denotes the Brownian mean over the individual motions and
φt (u) = E(φ(u(t))). Thus
kφt (u) − φt (w)k ≤ |∇φ|∞
Z 1
0
|∂E|dxku − wk ≤ |∇φ|∞ ku − wk
26
since ∂E ≤ 1. The proof of the strong Feller property is finished by noting that
any ψ ∈ L∞ (L2 (T 1 )), the space of bounded functions on L2 , can be approximated
by a function φ ∈ Cb1 (L2 ), the space of differentiable bounded functions on L2 , and
that for such functions |∇φ|∞ ≤ C|ψ|∞ so that
kψt (u) − ψt (w)k ≤ C|ψ|∞ ku − wk
QED
Remark 6.1 The hypothesis (48) is an additional restriction on the coefficients
hk , in addition to (36). It says that the energy in the largest eddies (first few hk s)
cannot be to big. If (48) is not satisfied these eddies drive the flow in addition
to the uniform flow Uo and their effect on the flow must be computed. This can
be done but it means that these lowest Fourier modes are then excluded from the
stationary state. This is commonly referred to as a large scale cut-off.
Finally we prove irreducibility, see [23], of Pt . The proof of this lemma is an
application of stochastic control theory.
Lemma 6.3 The Markovian semigroup Pt generated by the integral equation (27)
is irreducible.
Proof: We first consider the linear deterministic equation
(49)
zt +Uo zx = zxx + w(x,t), t ≥ to
z(x, T ) = b(x), T > to
zt +Uo zx = zxx + Qv(x,t), t < to
z(x, 0) = 0
and the deterministic equation
(50)
2
yt +Uo yx = yxx − yyx + ∂−1
x (yx ) + Qv(x,t), t ≥ to
y(x, T ) = b(x), T > to
yt +Uo yx = yxx + Qv(x,t), t < to
y(x, 0) = 0
where Q : H −1 → L2 and kernel Q is empty. We can pick a function w ∈ C([0, T ]; H −1 (T 1 ))
such that z(x, T ) = b(x) and a corresponding function v ∈ L2 ([0, T ]; H −1 (T 1 )).
2
Namely, Qv = zzx − ∂−1
x (zx ) + w, since the kernal of Q is empty; then y = z is a
27
solution of the deterministic Navier-Stokes equation (50) above. This means that
(50) is exactly controllable, see Curtain and Zwartz [5].
Now we compare y and the solution u of the integral eqution (27). By the same
estimates as in the proof of Lemma 4.1 we get the inequality
Z t
E(ku − yk25 + ) ≤ E(kuo (x,t) − yo (x,t)k25 + ) + tC
( 4 ,2)
( 4 ,2)
0
E(ku − yk25 + )(s)ds
( 4 ,2)
Here y satisfies the integral equation
Z t
y = yo +
0
2
Ko ∗ (−yyx + ∂−1
x (yx ) )ds
where Ko is the oscillatory heat kernal started at to = 0 and yo =
the inequality (28) implies that
Rt
0 K ∗ wds.
Now
(51)
E(ku − yk( 5 + ,2) ) ≤ E(kuo (x,t) − z(x,t)k( 5 + ,2) )(1 + tC
4
Z t
4
eC(t
2 −s2 )
ds) <
0
εδ
2
since we can pick v such that
E(kuo (x,t) − yo (x,t)k( 5 + ,2) ) ≤
4
εδ
(1 + TC
2
Z T
eC(T
2 −s2 )
ds)−1
0
for t ≤ T . This implies that the probability
P(ku(T ) − bk( 5 + ,2) ≤ ε) ≥
4
ε
P ku(T ) − y(T )k( 5 + ,2) ≤
and
2
4
ky(T ) − b(T )k( 5 + ,2) ≤
4
ε
≥ 1−δ
2
by (51) and Chebychev’s inequality and since (50) is exactly controllable, where
δ is arbitrarily small.
QED
Proof of Theorem 6.1 and Corollary 6.1
Proof: Theorem 6.1 and Corollary 6.1 are now easily proven in the following
manner. If the Markovian semigroup Pt is strongly Feller and invariant, as it is by
Lemmas 6.2 and 6.3, it is also t-regular and then by Doob’s Theorem, see [23],
the invariant measure is unique and strongly mixing.
QED
28
7
Hack’s Law
We now use the invariant measure in Section 6 to derive the scaling of the second
structure function of the turbulent flow that solves Equation (27).
Lemma 7.1 The second structure function of the turbulent flow that solves the
integral equation (27) scales as
Z
s2 (x) =
L2 (T 1 )
3+
|u(y + x) − u(y)|2 dµ(u) ≤ C|x| 2
Proof: By Corollary 5.1 the solutions of (27) are Hölder continuous with exponent 3/4. Thus
3+
|u(y + x) − u(y)| ≤ Ckuk( 5 + ,2) |x| 4
4
Moreover, by Theorem 5.1 u satisfies the bound
+
E(kuk25 + ) ≤
( 4 ,2)
1
(1 + (2πk)(5/2) )
hk + DUo2
2
2 k6∑
(2πk)
=0
by Equation (39). Substituting these bounds into the integral and using that
Z
E(kuk25 + ) =
kuk25 + dµ
( 4 ,2)
( 4 ,2)
2
L
gives the estimate.
QED
Now Hack’s law is proven in the following manner. In [3] the equations describing the sediment flow are linearized about convex (concave in the terminology of geomorphology) surface profiles describing mature surfaces. Then the
colored noise generated by the turbulent flow (during big rainstorms) drives the
linearized equations and the solutions obtain the same color (scaling), see Theorem 5.3 in [3]. The resulting variogram (second structure function) of the surfaces
scales with the roughness exponent χ = 34 , see Theorem 5.4 in [3]. This determines the roughness coefficient χ of mature landsurfaces.
The final step is the following derivation of Hack’s law is copied from [4].
29
7.1
The Origin of Hack’s Law
The preceding results allow us to derive some of the fundamental scaling results
that are known to characterize fluvial landsurfaces. In particular, the avalanche
dimension computed in [4] and derived in [3], given the roughness coefficient χ,
allows us to derive Hack’s Law relating the length of a river l to the area A of
the basin that it drains. This is the area of the river network that is given by the
avalanche dimensions
A ∼ lD
and the avalanche dimensions is D = 1 + χ. This relation says that if the length
of the main river is l then the width of the basin in the direction, perpendicular
to the main river, is l χ . Stable scalings for the surface emerge together with the
emergence of the separable solutions describing the mature surfaces, see [4]. We
note that in this case χ = 34 , hence we obtain
(52)
1
l ∼ A 1+χ
≈ A0.57
a number that is in excellent agreement with observed values of the exponent of
Hack’s law of 0.58, see [13].
Acknowledgments This research for this paper was started during the author’s
sabbatical at the University of Granada, Spain. It was also finished in Granada two
years later. The author was supported by grants number DMS-0352563 from the
National Science Foundation whose support is gratefully acknowledged. Some
simulations are being done on a cluster of workstations, funded by a National Science Foundation SCREMS grant number DMS-0112388. The author wished to
thank Professor Juan Soler of the Applied Mathematics Department at the University of Granada, for his support and the whole research group Delia Jiroveanu,
José Luis López, Juanjo Nieto, Oscar Sánchez and Marı́a José Cáceres for their
help and inspiration. His special thanks go to Professor José Martı́nez Aroza who
cheerfully shared his office and the wonders of the fractal universe.
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