first order derivative of the parabolic equation Assume $b, \ell \in C_b^{1,2}(\mathbb R^2)$. We consider parabolic PDE
$$(P1)\quad \partial_t v = b \partial_x v + \partial_{xx} v + \ell, \ \forall (t, x) \in \mathbb R^+\times \mathbb R; \quad v(0, x) = 0, \forall x\in \mathbb R.$$
It is standard that there is a classical solution.
Assuming $\partial_{xxx} v$ exists, by taking $\partial_x$ to the equation, with $\hat v = \partial_x  v$, we obtain
$$\partial_t \hat v = b \partial_x \hat v + \partial_{xx} \hat v + \partial_x \ell + \hat v \partial_x b, \ \forall (t, x) \in \mathbb R^+\times \mathbb R; \quad \hat v(0, x) = 0, \forall x\in \mathbb R.$$
Therefore, by uniqueness of the solution, we conclude that

*

*The solution $\hat u$ of
$$(P2) \quad \partial_t \hat u = b \partial_x \hat u + \partial_{xx} \hat u + \partial_x \ell + \hat u \partial_x b, \ \hbox{ on } \mathbb R^+\times \mathbb R; \quad \hat u(0, x) = 0, \hbox{ on } \mathbb R,$$
satisfies $\hat u = \partial_x v$.

Now, I want to know if the above conclusion satisfies without assuming $v\in C^{1,3}$, that is

*

*Let $b, \ell \in C_b^{1,2}(\mathbb R^2)$, do $\hat u$ of (P2) and $v$ of (P1) satisfy $\hat u = \partial_x v$?

 A: We define
\begin{equation*}
    u(t, x) = g(t) + \int_{0}^{x} \hat u (t, y) \, d y, \quad \forall (t, x) \in (\mathbb{R}^{+}, \mathbb{R})
\end{equation*}
where $g(\cdot)$ is the function we want to find to make $u(t, x)$ is the solution of equation (P1). Suppose $u(t, x)$ is the solution of equation (P1), for the initial condition, we need
\begin{equation*}
    u(0, x) = g(0) + \int_{0}^{x} \hat u (0, y) \, d y = g(0) = 0.
\end{equation*}
And for $(t, x) \in (\mathbb{R}^{+}, \mathbb{R})$, we have
\begin{eqnarray*}
\partial_{t} u & = & g^{'} (t) + \int_{0}^{x} \partial_{t} \hat u(t, y) \, d y \\
& = & g^{'} (t) + \int_{0}^{x} (b \partial_{x} \hat u + \partial_{xx} \hat u + \partial_{x} l + \hat u \cdot \partial_{x} b ) (t, y) \, d y  \\
& = & g^{'} (t) + (b \hat u + \partial_{x} \hat u + l) |_{0}^{x}   \\
& = & g^{'} (t) + b \hat u + \partial_{x} \hat u + l - (b \hat u + \partial_{x} \hat u + l)(t, 0).
\end{eqnarray*}
By the definition of $u(t, x)$, we can get that
\begin{equation*}
    \partial_{x} u (t, x) = \hat u (t, x), \quad \forall (t, x) \in (\mathbb{R}^{+}, \mathbb{R}),
\end{equation*}
then we have
\begin{equation*}
    \partial_{t} u - (b \partial_{x} u + \partial_{xx} u + l) = g^{'}(t) - (b \hat u + \partial_{x} \hat u + l)(t, 0).
\end{equation*}
Thus the sufficient condition that $u(t, x)$ is the solution of equation (P1) is
\begin{equation} \label{C1}
(C1) \quad
    \begin{cases}
    g^{'}(t) = (b \hat u + \partial_{x} \hat u + l)(t, 0) \\
    g(0) = 0
    \end{cases}
\end{equation}
where $\hat u(t, x)$ is the solution of equation (P2). We define
\begin{equation}
    g(t) = \int_{0}^{t} (b \hat u + \partial_{x} \hat u + l)(s, 0) \, d s, \quad \forall t \in \mathbb{R}^{+},
\end{equation}
which satisfies the condition (C1). Thus
\begin{equation*}
    u(t, x) = \int_{0}^{t} (b \hat u + \partial_{x} \hat u + l)(s, 0) \, d s + \int_{0}^{x} \hat u (t, y) \, d y, \quad \forall (t, x) \in (\mathbb{R}^{+}, \mathbb{R})
\end{equation*}
is the solution of (P1) and it satisfies
\begin{equation*}
    \partial_{x} u (t, x) = \hat u (t, x), \quad \forall (t, x) \in (\mathbb{R}^{+}, \mathbb{R}).
\end{equation*}
What's more, we have $u(t, x) \in C_{b}^{1,3} (\mathbb{R}^{+}, \mathbb{R})$ and it is the unique solution of (P1).
