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Discretizing a Fluctuation Dissipation Theorem

  1. Dec 3, 2015 #1
    Hey!

    I want to discretize a fluctuation dissipation theorem for the white noise ζ of a stochastic differential equation on a 2D domain (sphere). For that I integrate over "Finite Volume" elements with area A and A' (see below).

    [itex]
    \begin{eqnarray*} \int_{A} d A \int_{A'} d A' \langle\zeta(\mathbf{r},t)\zeta(\mathbf{r}',t')\rangle &=& -2\int_{A} d A \int_{A'} d A'(\nabla_s^{\mathbf{r}})^2\delta(\mathbf{r}-\mathbf{r}')\delta(t-t')\\ &=& 2\int_{A} d A \int_{A'} d A'\nabla_s^{\,\mathbf{r}'}\cdot\nabla_s^{\,\mathbf{r}}\delta(\mathbf{r}-\mathbf{r}')\delta(t-t')\\ &=& 2\int_{A} d A \int_{\partial A'} d S'\left[\nabla_s^{\,\mathbf{r}}\delta(\mathbf{r}-\mathbf{r}')\cdot\mathbf{n}'\right]\delta(t-t')\\ &=& 2\int_{A} d A \nabla_s^{\,\mathbf{r}}\cdot\underbrace{\int_{\partial A'} d S'\left[\delta(\mathbf{r}-\mathbf{r}')\mathbf{n}'\right]}_{\mathbf{g}}\delta(t-t') \end{eqnarray*}
    [/itex]

    Note that ∇s is the 2D surface gradient (thus only w.r.t. angles on the sphere, not radius). The superscript on ∇s indicates whether differentiation is w.r.t to r or r'. All quantities are assumed to be nondimensionalised to the same length-scale (sphere radius R): dA and dA' each have dimension 'area', each ∇s has dimension '1/length', δ(r-r') has dimension '1/area', thus the whole thing has dimension '1'.

    1.) In the first step I rewrite one of the two ∇s w.r.t. r', which gives a negative sign due to the δ(r-r') function.

    2.) In the second step I use the divergence theorem for surfaces to transform the surface integral dA' into an integral over the boundary.

    3.) Next I write the remaining ∇s operator in front of the integral g, because the operator doesn't depend on r'.

    4.) The integral g depends on the postition of the two elements. Assuming the elements are different and not neighbors, the delta function δ(r-r') in g vanishes, because r and r' do never 'meet'. Now, when the elements are identical, A=A', I would assume that g=n, which allows me to use the divergence theorem again for the remaining integral. This gives me 2∫dS n⋅n δ(t-t')=2Lδ(t-t'), where L is the perimeter of the element, measured in lengthscale R. The case of neighboring elements is more complicated.

    Am I doing this correct? I'm especially not sure about evaluating integral g: the integral g has dimension '1/lenght' but my assumed result in the case of A=A' is n, which has dimension 1. Actually I was expecting something like 2L2δ(t-t') in the end, which would be consistent with the way I discretize δ(t-t')...

    Maybe somebody has an idea...
     
  2. jcsd
  3. Dec 8, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 26, 2016 #3
    Just in case someone has a similar problem. I ended up with this approach (taken from a preprint):

    Here, we discretize the noise ζ in a field equation into element noise vector z, with components zi of the finite volume (FVM) elements i.
    We define the correlation function between zi and zj by integrating
    \begin{equation}
    \langle\zeta(\mathbf{r},t)\zeta(\mathbf{r}',t')\rangle=-2 \nabla_s^2\delta(\mathbf{r}-\mathbf{r}')\delta(t-t')\;,
    \end{equation}
    over element areas Ai and Aj:
    \begin{eqnarray}
    &&\langle z_i(t)z_j(t')\rangle\equiv\iint\limits_{A_i}\! d A_i \iint\limits_{ A_j}\! d A_j \langle\zeta(\mathbf{r}_i,t)\zeta(\mathbf{r}_j,t')\rangle\quad\\
    %
    &&=2\int_{\partial A_i}\limits\! d S_i\, \mathbf{n}_i\cdot\!\int_{\partial A_j}\limits\! d S_j\mathbf{n}_j\,\delta(\mathbf{r}_i-\mathbf{r}_j)\delta(t-t')\label{eq:element_noise_vector2}\\
    %
    &&=2 \sum_{q} l_{iq}\sum_{p} l_{jp}\delta_{q,p}\mathbf{n}_{iq} \cdot \mathbf{n}_{jp} \delta(t-t')\;.\label{eq:element_noise_vector3}
    \end{eqnarray}
    In (\ref{eq:element_noise_vector2}) we used the divergence theorem and in (\ref{eq:element_noise_vector3}) we converted the line integrals into sums over the element boundaries.

    e8t8s4ch.gif

    Here we discretize δ(ri-rj) by partitioning the surface into rhombi of area A (see figure above) and defining:
    \begin{equation*}
    \delta_{q,p}= \begin{cases}
    1/A_{\Diamond} & \mathrm{for} \enspace q=p\;, \\
    0 & \mathrm{for} \enspace q\neq p\;,
    \end{cases}
    \end{equation*}
    where q and p are the indices of the boundaries of element i and j respectively.
    Three cases have to be considered: Supposed the elements i and j are neither identical nor neighbors, δq,p vanishes in (\ref{eq:element_noise_vector3}) for all q and p. For i=j, however,

    $$\delta_{q,p}=1/A_{\Diamond}\; and\; \mathbf{n}_{iq} \cdot \mathbf{n}_{jp}=1$$

    for all q and p. Finally, for neighboring elements, there is one common boundary where

    $$\delta_{q,p}=1/A_{\Diamond}\; and\;\mathbf{n}_{iq} \cdot \mathbf{n}_{jp}=-1$$.

    Thus, one finds:
    \begin{eqnarray}
    \langle\underline{z}(t)\otimes\underline{z}(t')\rangle&=&\frac {2Nl^2}{A_{\Diamond}} \left(\underline{\underline{1}}-\frac 1 N \underline{\underline{Q}}\right) \delta(t-t')\;,
    \label{eq:fluc_diss_element_conti}
    \end{eqnarray}
    where N is the number of element boundaries. Here, Qij=1 if elements i and j are neighbors and zero otherwise. Note that in (\ref{eq:fluc_diss_element_conti}), we assumed constant boundary length l and number of boundaries N. This is reasonable for a refined icosahedron with 642 FVM elements. The form of (\ref{eq:fluc_diss_element_conti}) acknowledges the conservation law for the noise.
     
    Last edited: Jan 26, 2016
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