# 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 $$\hat u = \partial_x v$$?
• This is true since $u$ has 3 derivatives with respect to $x$. For example, if $b=0$, solve first (P2) and call $v$ the solution. Then $u(t,x)=u(t,0)+\int_0^x v(t,y)\, dy$, since both solve (P1). One can also use difference quotients with respect to $x$ to show that $u_{xx}$ is Lipschitz but then needs and argument for the existence of the third derivative at any point. For general $b$ it shoud be the same. Commented Jun 16, 2020 at 10:18
• Could you make your explanation into more detail? I could not understand. I tried similar way. Due to non-degeneracy, $\hat u\in C^{1,2}$. Set $u(t, x)= u(t, 0) + \int_0^x \hat u(t, y) dy$. Try to verify $u$ satisfies (P1), but the last step did not go through. Commented Jun 16, 2020 at 11:33
• Sure. Let v be the solution of (P2) with b=0 and set $w(t,x)=u(t,0)+\int_0^x v(t,y)dy$. Then $w(0,x)=0$ and $w_t(0,x)=u_t(t,0)+\int_0^x v_t(t,y) dy$. Using the equation for $v$ this last term equals $u_t(t,0)+w_{xx}+\ell-(v_x(t,0)+\ell (t,0))=w_{xx}+\ell$. This says that $w$ solves (P1) and then coincides with $u$. Commented Jun 16, 2020 at 11:50

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 $$$$\label{C1} (C1) \quad \begin{cases} g^{'}(t) = (b \hat u + \partial_{x} \hat u + l)(t, 0) \\ g(0) = 0 \end{cases}$$$$ where $$\hat u(t, x)$$ is the solution of equation (P2). We define $$$$g(t) = \int_{0}^{t} (b \hat u + \partial_{x} \hat u + l)(s, 0) \, d s, \quad \forall t \in \mathbb{R}^{+},$$$$ which satisfies the condition (C1). Thus $$\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).