Let $b: \mathbb R_+\times\mathbb R\to \mathbb R$ and $\sigma: \mathbb R_+\times\mathbb R\to (0,\infty)$ be functions as nice as possible (e.g. bounded and of bounded partial derivatives, and $\inf_{(t,x)}\sigma(t,x)>0$). Consider the Kolmogorov forward and backward PDEs:
\begin{eqnarray} \partial_tu(t,x) + \partial_x (bu)(t,x)-\frac{1}{2}\partial^2_{xx}(\sigma^2u)(t,x)=0,\quad \forall t>0, x>0,\\ \partial_tv(t,x) - b(t,x)\partial_x v(t,x)-\frac{\sigma(t,x)^2}{2}\partial^2_{xx}v(t,x)=0,\quad \forall t>0, x\in\mathbb R \end{eqnarray}
with $u(0,\cdot)=f$, $u(\cdot,0)=0$ and $v(0,\cdot)=g$.
Assuming both admit a classical solution. Then a straightforward computation yields
$$\partial_t(uv)(t,x) + \partial_x(buv)(t,x) - \partial_x\left(\frac{\partial_x(\sigma^2u)v}{2}-\frac{\sigma^2u\partial_xv}{2}\right)(t,x)=0,\quad \forall t>0, x>0,$$
and further by Fubini's theorem and integration by parts
\begin{eqnarray} 0&=&\int_0^T \int_0^\infty \left(\partial_t(uv)(t,x) + \partial_x(buv)(t,x) - \partial_x\left(\frac{\partial_x(\sigma^2u)v}{2}-\frac{\sigma^2u\partial_xv}{2}\right)(t,x)\right)dxdt \\ &=& \int_0^\infty uv(T,x)dx - \int_0^\infty fg(x)dx +\int_0^T \frac{\partial_x\big(\sigma^2u)v(t,0)}{2}dt\quad (\ast). \end{eqnarray}
My question is, if $u$ is a weak solution instead of the classical solution, does $(\ast)$ still hold? Here we may assume the initial conditions $f,g$ are also as nice as possible.
