Stochastic Models for Turbulenceqidi/filtering18/Lecture5.pdf · The Test Model Forced Turbulent Signals Motivation • Turbulent ﬂows are highly irregular and need to be analyzed

The Test Model Forced Turbulent Signals

Stochastic Models for TurbulenceMajda-Harlim Chapter 5

Mitch Bushuk & Themis Sapsis

February 16, 2011

Di Qi


Outline

1. Motivation and a Test Model for Turbulence

2. Turbulent signals with Forcing and Dissipation

3. Statistics of Turbulent Solutions in Physical Space

4. Turbulent Rossby Waves


Motivation

• Turbulent flows are highly irregular and need to be analyzed in

a statistical sense. This lends itself to the idea of modeling

turbulence as a stochastic process

• Coarse grained models are unable to resolve small scale

turbulence

• Large physical systems often have small scale turbulent

processes which are governed by unknown dynamics

We choose to parametrize resolved and unresolvedturbulence with spatially correlated white noise forcing


A Stochastic Test Model for Turbulent Signals

Consider the initial value problem for the following scalar

stochastic PDE:

@u(x ,t)@t = P( @

@x )u(x , t)� �( @@x )u(x , t) + F (x , t) + �(x)W (t)

u(x , 0) = u0(x)

where,

• P( @@x ) is an operator constructed from odd derivatives

• �( @@x ) is an operator constructed from even derivatives

• F (x , t) is a deterministic forcing

• �(x)W (t) is spatially correlated white-noise forcing

• u0(x) ⇠ N(x ,�2)


Test Model Operators

The operators satisfy

P( @@x )e

ikx= p(ik)e ikx

�( @@x )e

ikx= �(ik)e ikx

Assume p(ik) is wave-like:

p(ik) = i!k

where �!k is the dispersion relation.

�(ik) is the damping operator and satisfies

�(ik) > 0 8 k 6= 0


The Stochastically Forced Dissipative Advection Equation

Taking P( @@x ) = �c @

@x and �( @@x ) = �d + µ @2

@x2 , we have,

@u(x ,t)@t = �c @u(x ,t)

@x � du(x , t) + µ@2u(x ,t)@x2 + F (x , t) + �(x)W (t)

• p(ik) = i!k = �ick

• �(ik) = d + µk2

We seek a Fourier series solution:

u(x , t) =1X

k=�1uk(t)e

ikx

where u�k = u⇤k


Fourier Series Solution

@u(x ,t)@t = P( @

@x )u(x , t)� �( @@x )u(x , t) + F (x , t) + �(x)W (t)

Each uk(t) satisfies the SDE:

duk(t) = [p(ik)� �(ik)]uk(t)dt + Fk(t)dt + �kdWk(t)

Multiplying by the integrating factor e(�(ik)�p(ik))t, we have,

d(e(�(ik)�p(ik))t uk(t)) = e(�(ik)�p(ik))t(Fk(t)dt + �kdWk(t))

e(�(ik)�p(ik))t uk(t)� uk(0) =R t0 e(�(ik)�p(ik))s Fk(s)ds + �k

R t0 e(�(ik)�p(ik))sdWk(s)

Thus,

uk(t) = uk(0)e(p(ik)��(ik))t+R t0 e(�(ik)�p(ik))(s�t)Fk(s)ds +

�kR t0 e(�(ik)�p(ik))(s�t)dWk(s)


Large Time Behaviour

Taking F (x , t) = 0, we have

uk(t) = uk(0)e(p(ik)��(ik))t+ �k

R t0 e(�(ik)�p(ik))(s�t)dWk(s)

Note:Wk =W1+iW2p

2

E[uk(t)] = uk(0)e(p(ik)��(ik))t ! 0

E[uk(t)uk(t)⇤] = uk(0)uk(0)⇤e�2�(ik)t+

�k�k⇤2�(ik)(1� e�2�(ik)t

) !�2k

2�(ik)

We define the energy spectrum,

Ek =�2k

2�(ik) 1 k < 1


Autocorrelation Function

Let �(ik) = �(ik)� p(ik)

Note that: �⇤(ik) = �(ik) + p(ik)

Rk(t, t + ⌧) = E[(uk(t)� ¯uk)(uk(t + ⌧)� ¯uk)]

= E[(�kR t0 e(�(ik)(s�t)dWk(s))(�k

R t+⌧0 e(�(ik)(s

0�t�⌧)dWk(s 0))⇤]

=

�2ke

��(ik)(2t+⌧)�p(ik)⌧R t0

R t+⌧0 e�(ik)s+�⇤(ik)s0 1

2E[dW1(s)dW1(s 0) +dW2(s)dW2(s 0)]

= �2ke

��(ik)(2t+⌧)�p(ik)⌧R t0

R t+⌧0 e�(ik)s+�⇤(ik)s0�(s � s 0)dsds 0

= �2ke

��(ik)(2t+⌧)�p(ik)⌧R t0 e�(ik)s+�⇤(ik)sds

= �2ke

��(ik)(2t+⌧)�p(ik)⌧ 12�(ik)(e

2�(ik)t � 1)

=�2k

2�(ik)e��(ik)(2t+⌧)�p(ik)⌧

(e2�(ik)t � 1)

=�2k

2�(ik)e��(ik)⌧�p(ik)⌧

(1� e�2�(ik)t)


Autocorrelation Function

Finally, we have,

R(t, t + ⌧) =�2k

2�(ik)e��(ik)⌧�p(ik)⌧

(1� e�2�(ik)t)

In the large t limit,

R(t, t + ⌧) =�2k

2�(ik)e��(ik)⌧�p(ik)⌧

Thus,

Real(R(t, t + ⌧)) =�2k

2�(ik)e��(ik)⌧

cos(!k⌧) = Eke��(ik)⌧cos(!k⌧)

Decorrelation Time:1

�(ik)


Calibrating the Noise Level

From observations, we can roughly determine the energy spectrum

and decorrelation time at each wavenumber.

Recall that Ek =�2k

2�(ik) . Thus, we can produce an estimate of the

noise level using

�k =p2�(ik)Ek

A typical turbulent energy spectra has the power law form:

Ek = E0|k |��

For example, if �(ik) = d + µk2,

�k = E 1/20 |k |��/2

(d + µk2)1/2

• � > 2 ) decreasing noise at small spatial scales

• � < 2 ) increasing noise at small spatial scales


Damped Forced Solutions

Consider a forcing of the following form:

Fk(t) =

⇢Ae i!0(k)t k M0 k > M

The mean dynamics satisfy:

¯uk(t) = ¯uk(0)e��(ik)t+R t0 e�(ik)(s�t)Fk(s)ds

=

(¯uk(0)e��(ik)t

+Aei!0(k)t

�(ik)+i(!0(k)�!k )(1� e(��(ik)�!0(k))t) k M

¯uk(0)e��(ik)t k > M


Test Problem with Resonant Forcing

We choose a resonant forcing: !0(k) = !k 8 k such that |k | M

Numerical integrations were performed on:

@u(x ,t)@t = �c @u(x ,t)

@x � du(x , t) + µ@2u(x ,t)@x2 + F (x , t) + �(x)W (t)

with parameters:

• c = 1

• d = 0

• µ = 0.01

• A = 0.1

• M = 20

• kmax = 61

• Ek = 1, Ek = k�5/3

• �k =p0.02k , �k =

p0.02k1/6


Test Problem with Resonant Forcing

Sta$s$cs'of'Turbulent'solu$ons'in'physical'space'

We'have'the'solu$on'

Sta$s$cal'behavior'7'Mean'

'''''

'''''

Turbulent'Rossby'waves'Barotropic'Rossby'waves'with'phase'varying'only'in'the'north7south'direc$on'

known'from'observa$ons'that'on'scales'of'order'of'thousands'of'kilometers'these'waves'have'a'k−3'energy'spectrum'

Chapter 6:Filtering Turbulent Signals: Plentiful Observations

In particular, in this simplified context, we address the basic issues outlined in 1.a)-1.d) of Chapter 1.

6.1 A Mathematical Theory for Fourier Filter Reduction







6.1 The Number of Observation Points Equals the Number of Discrete MeshPoints:MathematicalTheory





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Stochastic Models for Turbulenceqidi/filtering18/Lecture5.pdf · The Test Model Forced Turbulent Signals Motivation • Turbulent ﬂows are highly irregular and need to be analyzed