Detailed balance condition for coupled Langevin equation

Question

Suppose $a$ and $m$ are real variables and they satisfy the following two coupled Langevin equations: $$ \dot{a}=F_a(a,m)+\eta_a(t);\quad\dot{m}=F_m(a,m)+\eta_m(t); $$ where $\eta_a$ and $\eta_m$ are white noises with strength $\Delta_m$ and $\Delta_a$. According to this, the detailed balance condition is $$\Delta_m\partial_mF_a=\Delta_a\partial_aF_m,$$ but I have no idea how this is derived, supposing it is right.

Edit: in response to the request for more references.

I am well aware that if $i,j$ are two arbitrary discrete states of a system, the detailed balance condition will be $$P(i)W(i\rightarrow j)=P(j)W(j\rightarrow i),$$ where $W(i\rightarrow j)$ is the transition probability from state $i$ to $j$. The continuous version can be derived in the same spirit. Once we know the probability of finding the system at a particular state and transition probabilities, the detailed balance condition should be fairly easy to derive. However, it does not seem trivial to derive the detailed balance condition from a general Langevin equation, and this is what I ask for.

Edit 2: the two noise terms $\eta_a$ and $\eta_m$ are uncorrelated.

The equation comes from here.

Answer 1

Definitions

Define $W(2|1)$ as the transition probability per unit time from $1$ to $2$. This gets us the Master equation

$$\partial_t p(a, m) = \int\!\!\!\!\int\! \left(W(a,m|a',m')p(a',m') - W(a',m'|a,m)p(a,m)\right)\mathrm{d}a'\mathrm{d}m'$$

Define further, for fun, $\mathbb{W}(a,m|a',m') = W(a,m|a',m') - \delta(a-a')\delta(m-m')\int\!\!\int W(a'',m''|a,m)\mathrm{d}a''\mathrm{d}m''$. With direct substitution we have $\partial_t p = \int\!\!\int\mathbb{W}(a,m|a'm')p(a',m')\mathrm{d}a'\mathrm{d}m'$, an operation we denote by

$$\partial_t p = \mathbb{W}p$$

Now the continuous detailed balance condition is: $$W(a,m|a',m')p^\text{eq}(a',m') = W(a',m'|a,m)p^\text{eq}(a,m)$$

where $p^\text{eq}(a,m)$ is the equilibrium probability distribution, defined by the equation $\mathbb{W}p^\text{eq} = 0$. The detailed balance can be written in the form

$$\frac{W(a,m|a',m')}{p^\text{eq}(a,m)} = \frac{W(a',m'|a,m)}{p^\text{eq}(a',m')}$$

We may now multiply this by some function $f(a,m)$ and integrate over $a$ and $m$. If we require the relation to hold for all $f$, we have the equivalent form

$$\int\!\!\!\!\int\frac{W(a,m|a',m')}{p^\text{eq}(a,m)}f(a,m)\mathrm{d}a\mathrm{d}m = \int\!\!\!\!\int\frac{W(a',m'|a,m)}{p^\text{eq}(a',m')}f(a,m)\mathrm{d}a\mathrm{d}m, \quad \forall f$$

For symmetry's sake, let's do the same with another function $g(a', m')$:

$$\int\!\!\!\!\int\!\!\int\!\!\!\!\int\frac{W(a,m|a',m')}{p^\text{eq}(a,m)}f(a,m)g(a',m')\mathrm{d}a\mathrm{d}m\mathrm{d}a'\mathrm{d}m' = \int\!\!\!\!\int\!\!\int\!\!\!\!\int\frac{W(a',m'|a,m)}{p^\text{eq}(a',m')}f(a,m)g(a',m')\mathrm{d}a\mathrm{d}m\mathrm{d}a'\mathrm{d}m', \quad \forall f, g$$

Rearranging and throwing $\mathbb{W}$ into the mix, we have

$$\int\!\!\!\!\int\frac{f(a,m)\ \int\!\!\int \mathbb{W}(a,m|a',m')g(a',m')\mathrm{d}a'\mathrm{d}m'}{p^\text{eq}(a,m)}\mathrm{d}a\mathrm{d}m = \int\!\!\!\!\int\frac{\int\!\!\int \mathbb{W}(a',m'|a,m)f(a,m)\mathrm{d}a\mathrm{d}m\ g(a',m')}{p^\text{eq}(a',m')}\mathrm{d}a'\mathrm{d}m', \quad \forall f, g$$

Now using the definitions from above, we conclude that the detailed balance is satisfied if and only if

$$(\mathbb{W}g, f) = (g, \mathbb{W}f), \quad \forall f,g$$

where the inner product is defined as

$$(g, f) = \int\!\!\!\!\int\frac{g(a,m)f(a,m)}{p^\text{eq}(a,m)}\mathrm{d}a\mathrm{d}m$$

Derivation

Phew. Finally, time to do the hard part. The Fokker-Planck equation corresponding to your pair of stochastic differential equations is (assuming there is no correlation between the noise sources):

$$\partial_t p = -\partial_a(F_a p) -\partial_m(F_m p) + \frac{1}{2}\sigma_a^2\partial_a^2p + \frac{1}{2}\sigma_m^2\partial_m^2p$$

This also conveniently defines the operator $\mathbb{W}$.

The strategy now is to take $(\mathbb{W}f, g)$ and integrate it by parts (only once), then do the same with $(f, \mathbb{W}g)$. Finally we're going to equate the expressions. Let's do these steps one at a time:

$$(\mathbb{W}f, g) = \int\!\!\!\!\int \left((F_a f - \frac{\sigma_a^2}{2}\partial_a f) \partial_a\!\!\left(\frac{g}{p^\text{eq}}\right) + (F_m f - \frac{\sigma_m^2}{2}\partial_m f) \partial_m\!\!\left(\frac{g}{p^\text{eq}}\right)\right)\mathrm{d}a\mathrm{d}m$$

Writing out the derivative, we have inside the integral

$$\frac{F_af\partial_ag}{p^\text{eq}} - \frac{F_afg\partial_a p^\text{eq}}{(p^\text{eq})^2} - \frac{\sigma_a^2\partial_a f\partial_ag}{2p^\text{eq}} + \frac{\sigma_a^2\partial_a f g\partial_a p^\text{eq}}{2(p^\text{eq})^2} + \frac{F_mf\partial_mg}{p^\text{eq}} - \frac{F_mfg\partial_m p^\text{eq}}{(p^\text{eq})^2} - \frac{\sigma_m^2\partial_m f\partial_mg}{2p^\text{eq}} + \frac{\sigma_m^2\partial_m f g\partial_m p^\text{eq}}{2(p^\text{eq})^2} $$

Subtracting from this the once-integrated-by-parts $(f, \mathbb{W}g)$, all the symmetric terms drop and we are left with (after multiplying by $p^\text{eq}$):

$$\left(F_a - \frac{\sigma_a^2\partial_a p^\text{eq}}{2p^\text{eq}}\right)(f\partial_ag - g\partial_af) + \left(F_m - \frac{\sigma_m^2\partial_m p^\text{eq}}{2p^\text{eq}}\right) (f\partial_mg - g\partial_mf) = 0$$

Note that on the left we have stuff that just depends on $a$ and on the right things that depend on $m$. This is to say that each should be zero independent of the other. After straightforward manipulation, we arrive at the wanted result

$$\sigma_m^2\partial_m F_a = \sigma_a^2\partial_a F_m$$

I've never seen this relation before, nor did I find a reference for it. I'm a bit surprised it looks this simple.

Answer 2

What worries me, if I add a constant term $F_a = F_a + F$. I clearly drive the system out of equilibrium and detailed balance should be broken. However such a constant term does not influence the equation: $\sigma^2_m \partial_m F_a =\sigma^2_a \partial_a F_m $