Consider the stochastic differenctial equation: $dX_t=\frac34 X_t^2 dt-X_t^{3/2}dW_t$.

How to find a strong solution?

  • $\begingroup$ Consider $Y_t=F(X_t)=1/X_t$. Or more generally $F(X_t)=X_t^a$ and apply the Ito formula. $\endgroup$ – LutzL Jun 18 '14 at 18:03

If we want to solve an SDE of the form

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

i.e. an SDE which does not depend explicitly on the time $t$, the transformation

$$f(x) := \int^x \frac{1}{\sigma(y)} \, dy$$

is always worth a try. Here, $\sigma(y)=-y^{3/2}$; hence we consider $$f(x) =-\int_x^\infty \frac{1}{\sigma(y)} \, dy =\int_x^\infty y^{-3/2} \, dy = 2 \frac{1}{\sqrt{x}}.$$


$$f'(x) = -\frac{1}{x^{3/2}} \qquad f''(x) = \frac{3}{2} \frac{1}{x^{5/2}}.$$

By Itô's formula,

$$\begin{align*} f(X_t)-f(X_0) &= -\int_0^t \frac{1}{X_s^{3/2}} \, dX_s + \frac{3}{4} \int_0^t \frac{1}{X_s^{5/2}} \, d\langle X \rangle_s \\ &= \int_0^t dW_s - \frac{3}{4} \int_0^t X_s^{1/2} \, ds + \frac{3}{4} \int_0^t X_s^{\frac{1}{2}} \, ds \\ &= W_t \end{align*}$$

where we used that the quadratic variation $\langle X \rangle_s$ has infinitesimal variation $X_s^3 \, ds$. This shows that

$$X_t = f^{-1} \bigg( W_t+ f(X_0) \bigg) = \frac{4}{(W_t+f(X_0))^2} = \frac{X_0}{(1+\frac12\sqrt{X_0}\cdot W_t)^2}$$ for every $t\lt T$, where $$ T=\inf\{t\gt0\mid W_t=-2/\sqrt{X_0}\}. $$

  • $\begingroup$ +1. I took the liberty to remove a divergent integral (from $0$ to $x$), using the (convergent) integral from $x$ to $\infty$ instead. $\endgroup$ – Did Jun 19 '14 at 7:06
  • $\begingroup$ @Did Ah, right; thank you very much! $\endgroup$ – saz Jun 19 '14 at 7:52

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