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What is the solution to this ODE (and SDE)?

  1. Apr 18, 2014 #1
    I'm trying to analyze the following Ito stochastic differential equation:

    $$dX_t = \|X_t\|dW_t$$

    where [itex]X_t, dX_t, W_t, dW_t \in \mathbb{R}^n[/itex]. Here, [itex]dW_t[/itex] is the standard Wiener process and [itex]\|\bullet\|[/itex] is the [itex]L^2[/itex] norm. I'm not sure if this has an analytical solution, but I am hoping to at least find an analytical expression for the expected value [itex]E[X_t][/itex].

    In order to gain intuition for this problem, I'm considering the following ordinary differential equation:

    $$\dot{z}(t) =\|z(t)\|b(t)$$

    where [itex]z(t), b(t) \in \mathbb{R}^n [/itex] and everything is completely deterministic. Does anyone know the analytical solution to this second equation, and under what conditions it exists?
     
  2. jcsd
  3. Apr 18, 2014 #2

    pasmith

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    It is most easily solved in spherical polar coordinates where [itex]\|z\| = r[/itex] so that [itex]z = r\mathbf{e}_r[/itex] and [itex]\mathbf{b} = b_r\mathbf{e}_r + \mathbf{b}_n[/itex] where [itex]\mathbf{b}_n \cdot \mathbf{e}_r = 0[/itex]. We then have [tex]
    \dot r \mathbf{e}_r + r \dot{\mathbf{e}_r} = r(b_r\mathbf{e}_r + \mathbf{b}_n).[/tex] Since [itex]\mathbf{e}_r[/itex] and [itex]\dot{\mathbf{e}_r}[/itex] are orthogonal we have [tex]
    \dot r = rb_r(t), \\
    \dot{\mathbf{e}_r} = \mathbf{b}_n.
    [/tex] The radial and angular components thus decouple and the radial component has solution [tex]
    r(t) = r(0) \exp\left( \int_0^t b_r(s)\,ds\right).
    [/tex] The angular component represents the motion of a point on the unit [itex](n-1)[/itex]-sphere. If [itex]\mathbf{b}[/itex] is continuous then a solution should exist, but finding a co-ordinate representation of it valid for all time may be impossible, as one can see from the angular components in [itex]\mathbb{R}^3[/itex], [tex]
    \dot \theta = b_\theta(t), \\
    \dot \phi = \frac{b_\phi(t)}{\sin \theta(t)}
    [/tex] with solution [tex]
    \theta(t) = \theta(0) + \int_0^t b_\theta(s)\,ds, \\
    \phi(t) = \phi(0) + \int_0^t \frac{b_\phi(s)}{\sin \theta(s)}\,ds,[/tex] which becomes invalid if ever [itex]\theta(t) < 0[/itex] or [itex]\theta(t) > \pi[/itex].
     
  4. Apr 18, 2014 #3
    Thanks a lot for the answer. Can you explain why [tex]
    \theta(t) = \theta(0) + \int_0^t b_\theta(s)\,ds, \\
    \phi(t) = \phi(0) + \int_0^t \frac{b_\phi(s)}{\sin \theta(s)}\,ds,[/tex] becomes invalid if
    ever [itex]\theta(t) < 0[/itex] or [itex]\theta(t) > \pi[/itex]?
     
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