How would you go about solving this? $$p(x,t)=C\exp\left[-x+\int_0^t\int_0^\infty y\,p(y,\tau)\,\mathrm{d}y\,\mathrm{d}\tau\right]$$

Here $p(x,t)$ is the time-dependent probability distribution of a variable $x$, so it should be normalized to 1, and positive everywhere. Also, we have the initial condition: $$p(x,0) = \exp\left[-x\right]$$

I hit across this equation in some work I'm doing, and I'm stuck. Don't know enough about integral equations to know where to begin.

Is there any hope at all to solve this? Where would you start?


1 Answer 1


Notice that the term $$\int_0^t \int_0^\infty y p(y,\tau) dy d\tau$$ only depends on $t$, so $$p(x,t) = C e^{-x} e^{f(t)}$$ with $$f(t) := \int_0^t \int_0^\infty y p(y,\tau) dy d\tau$$ replacing $p$ in the last equation $$f(t) = C \int_0^t e^{f(\tau)} d\tau \underbrace{\int_0^\infty y e^{-y} dy}_{1} = C\int_0^t e^{f(\tau)} d\tau$$.

$$f'(t) = C e^{f(t)}$$ and you can easily solve this DE.(notice that $f(0) = 0$)

  • $\begingroup$ Thanks themaker! Not so tricky for you after all. Going all the way through I find that $p(x,t)=e^{-x} \forall t$, by the way, so the original equation isn't time dependent at all! $\endgroup$ May 4, 2014 at 2:11
  • $\begingroup$ @AdrianoFerrari, Isn't that impossible unless the $\int_0^{\infty} (\cdots) dy=0$? For example, $\rho=C e^{-x} (1-C t)^{-1}$ satisfies the integral equation and setting $C=1$ satisfies your initial condition. (found using themaker's method) $\endgroup$
    – user18862
    May 4, 2014 at 4:24
  • 1
    $\begingroup$ @NeuroFuzzy, I obtained the same, and indeed it isnt normalizable for all $t$. $\endgroup$
    – themaker
    May 4, 2014 at 4:31

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