Control of Chaotic Flows and Fluid Forces

Control of Chaotic Flows and Fluid Forces (Part 3)

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Control of Chaotic Flows and Fluid Forces (Previous Part)

Perturbation Expansion

The distribution function may be expanded as a multiscale perturbation expansion in terms of a small parameter \epsilon as follows:

    \[f_{i}=f^{eq}{i}+\epsilon f^{(1)}{i}+\epsilon^{2}f^{(2)}{i}+\cdots\]

The second component of the multiscale perturbation expansion of \frac{\partial}{\partial t} is not a second order derivative; it simply indicates the expansion of the time scale in the second position.

    \[\frac{\partial}{\partial t}=\epsilon \frac{\partial^{(0)}}{\partial t}+\epsilon^{2} \frac{\partial^{(1)}}{\partial t}+\cdots\]

and \frac{\partial}{\partial r^{*}}=\epsilon \frac{\partial}{\partial \alpha}.

Using the above expansions in Eq. (??), we obtained

(1)   \begin{equation*}\begin{split} \left( \vec \zeta{i\alpha} \epsilon \frac{\partial}{\partial \alpha}+\epsilon \frac{\partial^{(0)}}{\partial t}+\epsilon^{2} \frac{\partial^{(1)}}{\partial t}\right) \left(f^{eq}{i}+\epsilon f^{(1)}{i}+\epsilon^{2}f^{(2)}{i}\right)=-\frac{1}{\tau}\left( \epsilon f^{(1)}{i}+\epsilon^{2}f^{(2)}{i} \right)\\+\frac{\mathit{\Delta}t}{2\tau}\left( \vec \zeta{i\alpha} \epsilon \frac{\partial}{\partial \alpha}+\epsilon \frac{\partial^{(0)}}{\partial t}+\epsilon^{2} \frac{\partial^{(1)}}{\partial t}\right)\left( \epsilon f^{(1)}{i}+\epsilon^{2}f^{(2)}{i}+O(\epsilon^{3}) \right)\end{split}\end{equation*}

Control of Chaotic Flows and Fluid Forces Around Two Offset Cylinders in Presence of Control Plate (Part 2)

Equating coefficient of \epsilon

O(\epsilon):

(2)   \begin{equation*}\frac{\partial^{(0)} f^{eq}{i}}{\partial t}+\vec \zeta{i\alpha}\frac{\partial f^{eq}{i}}{\partial \alpha}=-\frac{1}{\tau}f^{(1)}{i} \end{equation*}

O(\epsilon^{2}):

(3)   \begin{equation*}\frac{\partial^{(1)} f^{eq}{i}}{\partial t}+\left( \frac{\partial^{(0)}}{\partial t}+\vec \zeta{i\alpha}\frac{\partial}{\partial \alpha} \right)f^{(1)}{i} -\frac{\mathit{\Delta}t}{2\tau}\left( \frac{\partial^{(0)}}{\partial t}+\vec \zeta{i\alpha}\frac{\partial}{\partial \alpha}\right)f^{(1)}{i}=-\frac{1}{\tau}f^{(2)}{i} \end{equation*}

Moments and Recombination

Zeroth Moment

To find zeroth moment of Eq. (2) of O(\epsilon) sum the equation over i

(4)   \begin{equation*}\sum_{i}\left( \frac{\partial^{(0)} f^{eq}{i}}{\partial t}+\vec \zeta{i\alpha}\frac{\partial f^{eq}{i}}{\partial \alpha}\right) = \sum{i}-\frac{1}{\tau}f^{(1)}{i} \end{equation*}

(5)   \begin{equation*} \frac{\partial^{(0)}}{\partial t}\sum{i}f^{eq}{i}+\frac{\partial}{\partial \alpha}\sum{i} \vec \zeta_{i\alpha}f^{eq}{i}=-\frac{1}{\tau}\sum{i}f^{(1)}{i} \end{equation*}

Distribution function is written as

    \[f_{i}=f^{neq}{i}+\epsilon f^{(1)}{i}+\epsilon^{2}f^{(2)}{i}+\cdots\]

By adding the above equation over i, we get

(6)   \begin{equation*} \sum_{i}f_{i}=\sum_{i}\left( f^{eq}{i}+\epsilon f^{(1)}{i}+\epsilon^{2}f^{(2)}{i}+\cdots\right) \end{equation*}

Since \rho= \sum_i f_{i}, and

    \[f^{eq}_{i}=\rho \omega_{i}\]

By adding the above equation over i, we get

    \[\sum_{i}f^{eq}_{i}=\rho \sum_{i}\omega_{i}\]

Since the sum of weight factor is always equal to one i.e. \sum_{i}\omega_{i}=1. Therefore from  above equation we get

    \[\sum_{i}f^{eq}_{i}=\rho\]

Using above equation in Eq. (6), we get

    \[0=\sum_{i}(\epsilon f^{(1)}_{i}+\epsilon^{2}f^{(2)}_{i}+\cdots)\]

From above equation, we get

    \[\sum_{i}f^{(n)}_{i}=0 \qquad and \quad \sum_{i}\vec \zeta_{i}f^{(n)}_{i}=0 \qquad \text{for all,} n\geqslant 1\]

Now since,

    \[\sum_{i}f^{(1)}_{i}=0, \quad \sum_{i}f^{eq}_{i}=\rho, \quad \sum_{i}\vec \zeta_{i\alpha}f^{eq}_{i}=\rho u_{\alpha}\]

So Eq. (5) becomes

(7)   \begin{equation*}\frac{\partial^{(0)}\rho}{\partial t}+\frac{\partial(\rho u_{\alpha})}{\partial \alpha}=0 \end{equation*}

First Moment

To find zeroth moment of Eq. (2) of O(\epsilon) we multiply this equation with \vec \zeta_{i\alpha} and sum over \textit{i} and index changing to \beta

    \[\sum_{i}\left(\vec \zeta_{i\alpha} \frac{\partial^{(0)} f^{eq}_{i}}{\partial t}+\vec \zeta_{i\beta}\vec \zeta_{i\alpha}\frac{\partial f^{eq}_{i}}{\partial \beta}\right) = \sum_{i}-\frac{1}{\tau}\vec \zeta_{i\alpha}f^{(1)}_{i}\]

(8)   \begin{equation*} \frac{\partial^{(0)}}{\partial t}\sum_{i}\vec \zeta_{i\alpha}f^{eq}_{i}+\frac{\partial}{\partial \beta}\sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{i\beta}f^{eq}_{i}=-\frac{1}{\tau}\sum_{i}\vec \zeta_{i\alpha}f^{(1)}_{i} \end{equation*}

Since

    \[\sum_{i}\vec \zeta_{i\alpha}f^{eq}_{i}=\rho u_{\alpha} \quad \sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{i\beta}f^{eq}_{i}=\Pi^{eq}_{\alpha \beta}, \quad \sum_{i}\vec \zeta_{i\alpha}f^{(1)}_{i}=0\]

So Eq. (8) becomes

(9)   \begin{equation*}\frac{\partial^{(0)}(\rho u_{\alpha})}{\partial t}+\frac{\partial (\Pi^{eq}_{\alpha \beta})}{\partial \beta}=0\end{equation*}

Second Moment

To find zeroth moment of Eq. (2) of O(\epsilon) we multiply this equation with \vec \zeta_{i\alpha}\vec \zeta_{b} and sum over \textit{i} and index changing to \gamma

    \[\sum_{i}\left(\vec \zeta_{i\alpha}\vec \zeta_{i\beta} \frac{\partial^{(0)} f^{eq}_{i}}{\partial t}+\vec \zeta_{i\gamma}\vec \zeta_{i\beta}\vec \zeta_{i\alpha}\frac{\partial f^{eq}_{i}}{\partial \gamma}\right) = \sum_{i}-\frac{1}{\tau}\vec \zeta_{i\alpha}\vec \zeta_{i\beta}f^{(1)}_{i}\]

(10)   \begin{equation*}\frac{\partial^{(0)}}{\partial t}\sum_{i}\vec \zeta_{i\alpha}\xi_{i\beta}f^{eq}_{i}+\frac{\partial}{\partial \gamma}\sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{i\beta}\vec \zeta_{i\gamma}f^{eq}_{i}=-\frac{1}{\tau}\sum_{i}\xi_{i\alpha}\vec \zeta_{i\beta}f^{(1)}_{i} \end{equation*}

Since

    \[\sum_{i}\vec \zeta_{i\alpha}\vec \zeta_{i\beta}f^{eq}_{i}=\Pi^{eq}_{\alpha \beta} \quad \sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{i\beta}\vec \zeta_{i\gamma}f^{eq}_{i}=\Pi^{eq}_{\alpha \beta \gamma}, \quad \sum_{i}\vec \zeta_{i\alpha}\vec \zeta_{i\beta}f^{(1)}_{i}=\Pi^{0}_{\alpha \beta}\]

So Eq. (10) becomes

(11)   \begin{equation*}\frac{\partial^{(0)}(\Pi^{eq}_{\alpha \beta})}{\partial t}+\frac{\partial (\Pi^{eq}_{\alpha \beta \gamma})}{\partial \gamma}=-\frac{1}{\tau}\Pi^{0}_{\alpha \beta}\end{equation*}

Now we calculate the zeroth, and first moment of O(\epsilon^{2}).

Zeroth Moment of O(\epsilon^{2})

To find zeroth moment of Eq. (3) of O(\epsilon^{2}) sum the equation over i

    \[\sum_{i}\left( \frac{\partial^{(1)} f^{eq}_{i}}{\partial t}+\left( \frac{\partial^{(0)}}{\partial t}+\vec \zeta_{i\alpha}\frac{\partial}{\partial \alpha} \right)f^{(1)}_{i} -\frac{\mathit{\Delta}t}{2\tau}\left( \frac{\partial^{(0)}}{\partial t}+\vec \zeta_{i\alpha}\frac{\partial}{\partial \alpha}\right)f^{(1)}_{i}\right) =-\sum_{i}\frac{1}{\tau}f^{(1)}_{i}\]

(12)   \begin{equation*}\begin{split}\frac{\partial^{(1)}}{\partial t}\sum_{i}f^{eq}_{i}+\frac{\partial^{(0)}}{\partial t}\sum_{i} f^{(1)}_{i} &+\frac{\partial}{\partial \alpha}\sum_{i} \vec \zeta_{i\alpha}f^{(1)}_{i} -\frac{\mathit{\Delta}t}{2\tau}\left( \frac{\partial^{(0)}}{\partial t}\sum_{i} f^{(1)}_{i}+\frac{\partial}{\partial \alpha}\sum_{i} \vec \zeta_{i\alpha}f^{(1)}_{i}\right)\\&=-\frac{1}{\tau}\sum_{i}f^{(2)}_{1}\end{split}\end{equation*}

Now since,

    \[\sum_{i}f^{1}_{i}=0, \qquad \sum_{i} \vec \zeta_{i\alpha}f^{(1)}_{i}=0, \qquad \sum_{i}f^{eq}_{i}=0, \qquad \sum_{i}f^{(2)}_{i}=0\]

So Eq. (12) becomes

(13)   \begin{equation*}\frac{\partial^{(1)} \rho}{\partial t}=0 \end{equation*}

First Moment of O(\epsilon^{2})

To find zeroth moment of Eq. (3) of O(\epsilon^{2}) we multiply this equation with \vec \zeta_{i\alpha} and sum over \textit{i} and index changing to \beta

(14)   \begin{equation*}\begin{split}\sum_{i}&\left( \frac{\partial^{(1)} f^{eq}_{i}}{\partial t} \vec \zeta_{i\alpha}+\left( \frac{\partial^{(0)}}{\partial t}+\vec \zeta_{i\beta}\frac{\partial}{\partial \beta} \right)\vec \zeta_{i\alpha}f^{(1)}_{i}-\frac{\mathit{\Delta}t}{2\tau}\left( \frac{\partial^{(0)}}{\partial t}+\vec \zeta_{i\beta}\frac{\partial}{\partial \beta}\right)\vec \zeta_{i\alpha}f^{(1)}_{i}\right) \\&=-\sum_{i}\frac{1}{\tau}\vec \zeta_{i\alpha}f^{(2)}_{i}\end{split} \end{equation*}

(15)   \begin{equation*} \begin{split}&\frac{\partial^{(1)}}{\partial t} \sum_{i} \vec \zeta_{i\alpha}f^{eq}_{i}+\frac{\partial^{(0)}}{\partial t}\sum_{i} \vec \zeta_{i\alpha}f^{(1)}_{i}+\frac{\partial}{\partial \beta} \sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{\beta}f^{(1)}_{i}\\&-\frac{\mathit{\Delta}t}{2\tau}\left( \frac{\partial^{(0)}}{\partial t}\sum_{i} \vec \zeta_{i\alpha}f^{(1)}_{i}+\frac{\partial}{\partial \beta}\sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{\beta}f^{(1)}_{i}\right)=-\frac{1}{\tau}\sum_{i}\vec \zeta_{i\alpha}f^{(2)}_{i}\end{split} \end{equation*}

Now since,

    \[\sum_{i} \vec \zeta_{i\alpha}f^{eq}_{i}=\rho u_{\alpha}, \quad \sum_{i}\vec \zeta_{i\alpha}f^{(1)}_{i}=0, \quad \sum_{i} \vec \zeta_{i\alpha}\vec \zeta_{i\beta}f^{(1)}_{i}=\Pi^{1}_{\alpha \beta}, \quad \sum_{i}\vec \zeta_{i\alpha}f^{(2)}_{i}=0\]

So Eq. (15) becomes

(16)   \begin{equation*}\frac{\partial^{(1)}}{\partial t}\left(\rho u_{\alpha} \right)+\frac{\partial}{\partial \beta}\left(1-\frac{\mathit{\Delta}t}{2\tau} \right)\Pi^{1}_{\alpha \beta}=0 \end{equation*}

Now by combining Eq. (7) and (13), we get

    \[\epsilon\left(\frac{\partial^{(0)}\rho}{\partial t}+\frac{\partial(\rho u_{\alpha})}{\partial \alpha}\right)+\epsilon^{2}\left( \frac{\partial^{(1)} \rho}{\partial t} \right)=0 \]

    \[\left( \epsilon\frac{\partial^{(0)}}{\partial t}+\epsilon^{2}\frac{\partial^{(1)}}{\partial t}\right)\rho+ \epsilon \frac{\partial(\rho u_{\alpha})}{\partial \alpha}=0\]

Control of Chaotic Flows and Fluid Forces (Previous Part)

Control of Chaotic Flows and Fluid Forces

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