I’m now faced with an SDE problem to prove the solution of the SDE: $$dX_t=(X_t-X_t^3)dt+\sigma dW_t, X_0=0$$ Is ergodic and estimate the bound of $EX_t^4$. I started to try to solve the SDE to get the explicit solution. Consider $Y_t=f(X_t)$, then we have:
$$dY_t=(f^{\prime}(X_t)(X_t-X_t^3)+\frac{\sigma}{2}f^{\prime\prime}(X_t))dt+\sigma f^\prime(X_t)dW_t$$
When $f^\prime(x)=exp(\frac{x^2}{\sigma}(\frac{x^2}{2}-1))$, we have:
$$dY_t=\sigma f^\prime (X_t)dW_t$$
And I don’t know how to go on now.
As suggested, to find the bound we can apply Ito to $f=X_{t}^{4}$, which, taking the exptectation and remembering that the stochastic part vanishes, yields:
$$ m(t) = \alpha(t)+\int_{0}^{t}4m(s)ds -4 \int_{0}^{t}\mathbb{E}[X_{s}^{6}] ds $$ which we can bound from above
$$ m(t) \le \alpha(t)+\int_{0}^{t}4m(s)ds $$
where $\alpha(t) = \int_{0}^{t}6 \sigma^2\mathbb{E}[X_{s}^{2}]ds$ and $m(t)= \mathbb{E}[X_{t}^{4}]$.
Since $4$ is a non-decreasing function, by the integral version of Gronwall Lemma we get:
$$ m(t) \le e^{4t} \alpha(t) $$