My question is, we give a meaning to the following expression: $$dX(t) = \mu(t,X(t))dt + \sigma(t,X(t))dW(t), \ \ X(0)=x.$$ where $W$ is a Wiener process.
This equation can be thought as $$\frac{dX(t)}{dt} = \mu(t,X(t)) + \sigma(t,X(t))\frac{dW(t)}{dt}, \ \ X(0)=x.$$
Now, if I wanted to differentiate again in an ODE I would write something like:
$$\frac{d^2X(t)}{dt^2} = \mu'(t,X(t)) + \sigma'(t,X(t))\frac{d^2W(t)}{dt^2}, \ \ X(0)=x.$$ But, my question is, can we do this in Stochastic ODE's? Does the term $\frac{d^2W(t)}{dt^2}$ have a meaning? What would one do in this case? Any idea, literature, reference? Thank you for your help!
When you write $$dX(t) = \mu(t,X(t))dt + \sigma(t,X(t))dW(t), \ \ X(0)=x.$$ this is an 'abuse of notation' and should be read as the equation
$$X(t)-x = \int_0^T\mu(t,X(t))dt + \int_0^T\sigma(t,X(t))dW(t)\qquad t\geq0.$$
This equation can't be solved pathwise since it would have no meaning. It has a more complex structure due to the stochastic integral $\int_0^T\sigma(t,X(t))dW(t)$ which is not differentiable in $T$.