$dX_t=-a^2\sin X_t\cos^3X_tdt+a\cos^2X_tdW_t$ with $X_0=x_0$

I think this is inverse type of SDE, refer to Solve the SDE $dX_t = \frac{1}{2}\sigma(X_t)\sigma'(X_t)dt+\sigma(X_t)dW_t$.

However, I can't find the inverse funcition.

My try is $(\cos^2X_t)'=-\sin X_t\cos^3X_t$ ,but I can't go forward anymore.

Could you tell me how to solve this?

  • $\begingroup$ Did you try out the approach I suggested in my answer (to the linked question)? $\endgroup$ – saz Jun 21 '14 at 14:26
  • $\begingroup$ I think $\sigma(y)=a\cos^2X_t$ by solution 2. $\sigma^{-1}(y)=\arccos \frac{y}{a}$? $\endgroup$ – user157209 Jun 21 '14 at 15:04
  • $\begingroup$ Sorry, but this doesn't make sense at all... Yes, $\sigma(y) = a \cos^2 y$, but we do not need the inverse of $\sigma$, but the inverse of $\int_0^{\cdot} \frac{1}{\sigma(y)} \, dy$, see my answer below. $\endgroup$ – saz Jun 22 '14 at 8:08

If we set

$$\sigma(x) = a \cdot \cos(x)^2$$

then $\sigma'(x) = -a \cdot \cos(x) \cdot \sin(x)$, i.e. we can rewrite the SDE in the following way:

$$dX_t = \sigma(X_t) \, dW_t + \sigma(X_t) \sigma'(X_t) \, dt.$$

This means that this SDE is in fact an inverse type SDE. Using the approach discussed in this answer (see Solution 2), we find that for

$$g(x) := \int_0^x \frac{1}{a \cos^2(y)} \, dy = \frac{1}{a} \tan(x)$$

we have

$$\frac{1}{a} \tan(X_t) - \frac{1}{a} \tan(x_0)= g(X_t) - g(x_0) = W_t.$$

Consequently, we obtain

$$X_t = \arctan \bigg( a W_t + \tan(x_0) \bigg).$$

It remains to check that $(X_t)_{t \geq 0}$ is in fact a solution to the SDE. This follows by applying Itô's formula.

  • $\begingroup$ Your solution is very easy to understand. Thank you. $\endgroup$ – user157209 Jun 22 '14 at 10:45

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