-2
$\begingroup$

I am not able to integrate the following stochastic equation. The equation is $$\frac{dx}{dt}=g(1-x^2)x+\sqrt{g}(1-x^2)\xi(t)$$ $g$ is a constant and $x$ is defined between $-1$ and $+1$. $\xi(t)$ is a stochastic number drawn from a Gaussian distribution. Functional form of $\xi(t)$ is not very well known. Terms like $\int \xi(t)dt$ can appear in the solution. I guess some complicated variable substitution is needed but don't know exactly what ?

$\endgroup$
2
  • $\begingroup$ Is g a constant? And is $\xi(t)$ arbitrary? You should spend more time in formulating your question. $\endgroup$
    – MrYouMath
    May 3, 2016 at 9:12
  • $\begingroup$ thanks and I have edited my question. I think now its much clear to you. $\endgroup$ May 3, 2016 at 11:47

1 Answer 1

1
$\begingroup$

I tried the following (not sure it will work). Assuming an additive solution:

$$\frac{dx_1}{dt}=g(1-x_1^2)x_1$$ $$\frac{dx_2}{dt}=\sqrt{g}(1-x_2^2)\xi(t)$$

Both equations are separable.

Here is what i did with $x_1$: $$\frac{dx_1}{(1-x_1^2)x_1}=gdt$$ $$\left(\frac{1}{x_1}+\frac{1/2}{1-x_1}+\frac{-1/2}{1+x_1}\right)dx_1=gdt$$

Integrating and solving for $x_1$ resulsts in:

$$x_1=\frac{c_1\exp(gt)}{\sqrt{1+c_1^2\exp(2gt)}}$$

Here is what i did with $x_2$: $$\frac{dx_2}{(1-x_2^2)}=\sqrt{g}\xi(t)dt$$ $$1/2\left(\frac{1}{1-x_2}+\frac{1}{1+x_2}\right)dx_2=\sqrt{g}\xi(t)dt$$

Integrating both sides and solving for $x_2$ results in:

$$x_2=\frac{\exp(2\sqrt{2}\int\xi dt)-1}{\exp(2\sqrt{2}\int\xi dt)+1}$$

Now add both equations to get $x$

$$x=x_1+x_2=\frac{c_1\exp(gt)}{\sqrt{1+c_1^2\exp(2gt)}}+\frac{\exp(2\sqrt{2}\int\xi dt+c_2)-1}{\exp(2\sqrt{2}\int\xi dt+c_2)+1}$$

Now plug in $x$ into your ODE and see what relationship between $c_1$ and $c_2$ can be established (if it is possible at all) to eliminate one of the constants.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .