How to solve the SDE $dX_t = aX_tdt + (b(t)-X_t^2)^{1/2}dW_t$?

Question

I need help on solve the following SDE: $\beta > 0$, $0<\gamma<1$, $X_0 = \frac{\sqrt{2}}{2}$ $$dX_t = -(\beta + \frac{1}{2}\gamma^2)X_tdt + \gamma\sqrt{e^{-2\beta t}-X_t^2}dW_t$$

I need help to find the transform function $f(t,x)$ and let $Y_t = f(t,X_t)$ so that $dY_t$ can be solved.

Thank you very much.

Answer 1

Hint:

$$f(t,x) := \frac{1}{\gamma} \arctan \left( \frac{x}{\sqrt{e^{-2\beta t}-x^2}} \right)$$

transforms the given SDE to a (very simple) linear SDE.

Remark: The transformation is a so-called variance transform. For an SDE of the form $$dX_t = b(t,X_t) \, dt + \sigma(t,X_t) \, dW_t \tag{1}$$ the transformation is given by

$$f(t,x) := \int_0^x \frac{dy}{\sigma(t,y)}.$$

In fact, there are necessary and sufficient conditions (in terms of derivatives of $b$ and $\sigma$) which state whether an SDE of the form $(1)$ can be transformed via $Y_t := f(t,X_t)$ into a linear SDE of the form $$dY_t = \bar{b}(t) \, dt + \bar{\sigma}(t) \, dW_t.$$ (See e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 19, for more details.)

Solution: $$X_t = \sqrt{\frac{e^{-2 \beta t} \tan^2(\gamma W_t+\pi/4)}{1+\tan^2(\gamma W_t+\pi/4)}} = e^{-\beta t} \sin(\gamma W_t+\pi/4).$$