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I would like to numerically solve a differential equation which contains a derivative of a random number (using a finite difference method with a time step $\Delta t$).

Let say I need to solve for $y(t)$ in $$ \frac{d^2}{dt^2}y(t) + A\frac{d}{dt}y(t) + By(t) = f(t) + C\frac{d}{dt}\xi_1(t) + D\xi_2(t)$$ where $\xi_1(t)$ and $\xi_2(t)$ is a random Gaussian variable (mean = 0, s.d. = 1). $A,B,C,D$ are constants and $f(t)$ is a known function of time.

Please could anyone tell what is the proper numerical treatment for the term $\frac{d}{dt}\xi_1(t)$?

user3886914
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2 Answers2

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The problem is that if $\xi_1(t)$ has $\mathcal N(0,1)$ distribution $\forall t$, than $\xi_1(t)$ is white noise, see wiki. But Gaussian white noise for continuous time does not even exist as a classical random variable! It is only weak random variable.

I think what you look for is theory of stochastic differential equation, many books could be found about this, example about SPDE, but you will have to learn the whole new theory about it and how to solve it.

iiivooo
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  • Hey, thanks for the answer. It seems that my random variables I was talking are in fact discrete. I may have to use other type of continuous randon variables inorder to properly solve my problem. – user3886914 Mar 16 '16 at 10:59
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If $\xi_1(t)$ is independent of $\xi_1(t')$ for $t\neq t'$ then $\xi_1(t)$ is not continuous so the integral over a given step size $\Delta t$ its derivative is meaningless. So for your equation to make sense $\xi_1(t)$ has to be a continuous function of $t$.
($\xi_1(t)$ need not be differentiable, just as $\xi_2(t)$ need not be continuous for the equation to have meaning. )

So if you let $\xi_1(t)$ be something like a Weiner distribution (perhaps modified to have some asymmetric drift depending on $t$), then the problem is meaningful. And now your contribution, in a step of size $\Delta t$, will be the usual $$ \frac{\xi_1(t+\Delta t)- \xi_1(t)}{\Delta t} $$ To work with a second-order method, it will be necessary for $\xi_1(t)$ to also be differentiable.

Mark Fischler
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  • Thank you for the answer. The Gaussian noise I was talking to is indeed discrete. I guess I need to study my problem more thoroughly and look for a more suitable random variable. – user3886914 Mar 16 '16 at 11:02