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The Test Model Forced Turbulent Signals

Stochastic Models for TurbulenceMajda-Harlim Chapter 5

Mitch Bushuk & Themis Sapsis

February 16, 2011

The Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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)

The Test Model Forced Turbulent Signals

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 Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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)

The Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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)

The Test Model Forced Turbulent Signals

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)

The Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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

The Test Model Forced Turbulent Signals

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 A Mathematical Theory for Fourier Filter Reduction

6.1 A Mathematical Theory for Fourier Filter Reduction

6.1 A Mathematical Theory for Fourier Filter Reduction

6.1 A Mathematical Theory for Fourier Filter Reduction

6.1 A Mathematical Theory for Fourier Filter Reduction

6.1 A Mathematical Theory for Fourier Filter Reduction

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

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

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

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

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