I derived the following equation in my research but I don't understand the equation very well. I hope to get some help here.
$\displaystyle \frac{\partial f(r,t)}{\partial t} = h_0(t) f(r,t) \Big( \bar{r}(t) - r \Big)$, where
- $r \ge 0$ is the physical quantity of interest
- $\displaystyle f(r,t)$ is the probability density function of $r$ at time $t$
- $h_0(t)$ is a given continuous nonnegative function: $h_0(t) \ge 0$ for $\forall t \in \mathbb{R}$
- $\displaystyle \bar{r}(t) \equiv \int_0^{\infty} r \, f(r,t) dr$ is the mean of $r$ at time t
My questions:
- Has anyone seen any equations like this one? What kind of equation is it? And can we say anything about its properties, in addition to the ones listed below?
- Any comments on the strategies to analytically solve the equation, given some initial condition $f(r, 0)$ (eg, log-normal distribution with parameters $\mu$ and $\sigma$)?
- To numerically solve the equation given $f(r, 0)$, currently I represent $f(r,t)$ as a vector over a fixed grid of $r$, and update it with Euler steps $f(r,t+dt) = f(r,t) + h_0(t) f(r,t) \big( \bar{r}(t) - r \big) dt$. It works, but seems a bit slow, and unstable: for some large $dt$ and $r \gg \bar{r}(t)$, $f(r, t+dt)$ can become negative, as opposed to staying very close to zero. Any comments on how to improve the numerical integration algorithm here?
- I am ultimately only interested in $\bar{r}(t)$, and am solving $f(r,t)$ just for $\bar{r}(t)$. Given the information presented here, is there a way to solve for $\bar{r}(t)$ directly?
Many thanks!
So far I have discovered the following properties of the equation:
- Conservation of probability mass: $\displaystyle \int_0^{\infty} f(r, t) dr = 1 \implies \int_0^{\infty} f(r, t+dt) dr = 1$
- Decay of mean: At time $t$, for $r < \bar{r}(t)$ its density increases because $\displaystyle\frac{\partial f(r,t)}{\partial t} \propto (\bar{r}(t) - r)>0$, while the opposite is true for $r > \bar{r}(t)$. As a result, the distribution of $r$ shifts to the left over time, towards a Dirac's delta function at $r=0$. In fact, one can derive that $\displaystyle \frac{d \bar{r}(t)}{dt} = - h_0(t) V(t)$, where $\displaystyle V(t) \equiv \int_0^{\infty} \big( r - \bar{r}(t) \big)^2 f(r,t) dr$ is the variance of $r$ at time $t$. That is, the mean of $r$ decays at a rate proportional to its variance.
- Invariance of Gamma distribution under the evolutionary operator: If $f(r,0)$ is Gamma distributed with shape parameter $k$ and scale parameter $\theta_0$, then $f(r,t)$ is also Gamma distributed with parameter $k$ and $\displaystyle \theta(t)=\frac{\theta_0}{\theta_0 \int_0^t h_0(\tau) d\tau +1}$.