I will phrase my question in terms of a concrete example, which involves the proof presented for Feynman-Kac formula, where one is interested in solving the following PDE:
\begin{gather}\frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) + f(x,t) = 0, \end{gather}
defined for all $x \in \mathbb{R}$ and $t \in [0, T]$, subject to the terminal condition
\begin{gather} u(x,T)=\psi(x), \end{gather}
where $\mu, \sigma, \psi, V, f$ are known functions, $T$ is a parameter and $ u:\mathbb{R}\times[0,T]\to\mathbb{R}$ is the unknown. The proof about the form of the solution starts as follows:
Let $u(x, t)$ be the solution to the above partial differential equation. Applying the product rule for Ito's processes to the process:
\begin{gather} Y(s) = e^{-\int_t^s V(X_\tau,\tau)\, d\tau} u(X_s,s)+ > \int_t^s e^{-\int_t^r V(X_\tau,\tau)\, d\tau}f(X_r,r) \, > dr\end{gather}
one gets
\begin{align} dY = {} & d\left(e^{- \int_t^s V(X_\tau,\tau)\, > d\tau}\right) u(X_s,s) + e^{- \int_t^s V(X_\tau,\tau)\, > d\tau}\,du(X_s,s) \\[6pt] & {} + d\left(e^{- \int_t^s > V(X_\tau,\tau)\, d\tau}\right)du(X_s,s) + d\left(\int_t^s e^{- > \int_t^r V(X_\tau,\tau)\, d\tau} f(X_r,r) \, dr\right) \end{align}
Since :\begin{gather}d\left(e^{- \int_t^s V(X_\tau,\tau)\, > d\tau}\right) =-V(X_s,s) e^{- \int_t^s V(X_\tau,\tau)\, d\tau} > \,ds,\end{gather} the third term is $O(dt \, du)$ and can be dropped.
My question is mathematically (aka rigorously) what does it mean to just drop $O(dt \, du)$?? I know the question has been raised before for this, but it seems OP there was convinced with yet another convergence argument, but my question is about the equality that is not based on asymptotic arguments. I assume in a rigorous sense $dt$ is a differential form. Can somebody in that language explain why such a cancellation takes place?
First, it's important to clarify what those differentials mean. If you are given the SDE $$dX_t = \mu(t,X_t) dt + \sigma(t,X_t) dB_t$$ where $B$ is a Brownian motion, then this is just shorthand for $$X_t - X_0 = \int_0^t \mu(s,X_s) ds + \int_0^t\sigma(s,X_s) dB_s$$ In this integral equation, the first integral is a pathwise Riemann integral and the second integral is an Itô integral. In particular, the differentials in this context have no intrinsic meaning.
Second: the reason why you can treat $du \, dv = (du)^2 = 0$ and $(dW_t)^2 = dt$ is because these formal products make Itô's formula work. In the proof of Itô's formula, these play the role of covariation, so that in particular:
$$\int_0^t (du)^2 = \lim_{n\to \infty} \sum_{i \in \pi^n} (u_i- u_{i-1})^2 \leq \lim_{n\to \infty} \sup_{i \in \pi^n} |u_i - u_{i-1}| \sum_{i \in \pi^n} |u_i- u_{i-1}| = 0$$ And, $$\int_0^t(dW_s)^2 = \lim_{n\to \infty} \sum_{i \in \pi^n} (W_{t_i} - W_{t_{i-1}})^2 = t = \int_0^t ds$$ Where $\pi = (\pi^n)$ is a refining partition whose mesh vanishes as $n\to\infty$.