first order derivative of the parabolic equation

Question

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 $\hat u = \partial_x v$?

Answer 1

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).