Multi variable Langevin equation

Question

I need to solve the following system

$\frac{\partial f(t)}{\partial t}=a_1 f(t)+a_1 g(t)+s_1(t) \\ \frac{\partial g(t)}{\partial t}=a_3 f(t)+a_4 g(t)+s_2(t) $

with $s_i$ being a noise, delta-correlated : $<s_i(t_1)s_j(t_2)>=2B ~~\delta_{i,j}~~ \delta(t_1-t_2)$

In particular I want the mean squares $<f^2>_{t \to \infty}$ and $<g^2>_{t \to \infty}$

Could you please help to solve this, or recommend a comprehensible document/book?

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So I recognise here the Langevin equation, which I only know how to solve in 1 dimension. When I try this to solve I quickly fail :

From the second equation : $ f(t)=\frac{1}{a_3} \left( \frac{\partial g(t)}{\partial t} - a_4 g(t) - s_2(t) \right) $

Substituting in equation 1 : $ \frac{1}{a_3}\left( \frac{\partial^2 g(t)}{\partial t^2}-a_4 \frac{\partial g(t)}{\partial t} - \frac{\partial}{\partial t} s_2(t) \right)= a_1 f(t) + ... $

Which is already problematic because of the term $\frac{\partial}{\partial t} s_2(t)$: a stochastic process can't be differentiated.

This is treated for the general n-dimensional case by Zwanzig but I find the book a bit dry and poorly detailed.

Thank you

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align} \totald{{\sf F}\pars{t}}{t}={\sf W}{\sf F}\pars{t} + {\sf s}\pars{t}\,, \qquad\qquad \left\{\begin{array}{rcl} {\sf F}\pars{t} & \equiv & {\fermi\pars{t} \choose {\rm g}\pars{t}} \\[3mm] {\sf W} & \equiv & \pars{% \begin{array}{cc} a_{1} & a_{1} \\ a_{3} & a_{4} \end{array}} \\[3mm] {\rm s}\pars{t} & = & {{\rm s}_{1}\pars{t} \choose {\rm s}_{2}\pars{t}} \end{array}\right. \end{align}

$$ \expo{-{\sf W}t}\totald{{\sf F}\pars{t}}{t} -\expo{-{\sf W}t}{\sf W}{\sf F}\pars{t} =\expo{-{\sf W}t}{\sf s}\pars{t}\quad\imp\quad \totald{\bracks{\expo{-{\sf W}t}{\sf F}\pars{t}}}{t} =\expo{-{\sf W}t}{\sf s}\pars{t} $$

$$ \expo{-{\sf W}t}{\sf F}\pars{t} - \expo{-{\sf W}t_{0}}{\sf F}\pars{t_{0}} =\int_{t_{0}}^{t}\expo{-{\sf W}\xi}{\sf s}\pars{\xi}\,\dd \xi $$

\begin{align} {\sf F}\pars{t} &=\expo{{\sf W}\pars{t - t_{0}}}{\sf F}\pars{t_{0}} +\int_{t_{0}}^{t}\expo{-{\sf W}\pars{\xi - t}}{\sf s}\pars{\xi}\,\dd \xi \\[3mm]&=\expo{{\sf W}\pars{t - t_{0}}}{\sf F}\pars{t_{0}} +\int_{t_{0} - t}^{0}\expo{-{\sf W}\xi}{\sf s}\pars{\xi + t}\,\dd \xi \\[3mm]&=\expo{{\sf W}\pars{t - t_{0}}}{\sf F}\pars{t_{0}} -\int_{-t_{0} + t}^{0}\expo{{\sf W}\xi}{\sf s}\pars{-\xi + t}\,\dd \xi \end{align}

$$\color{#66f}{\large% {\sf F}\pars{t} =\expo{{\sf W}\pars{t - t_{0}}}{\sf F}\pars{t_{0}} +\int_{0}^{t - t_{0}}\expo{{\sf W}\xi}{\sf s}\pars{t - \xi}\,\dd \xi} $$

In order to continue you have to know some details about $\ds{{\sf W}}$. You can go ahead as the usual Langevin $\ds{1D}$ equation.