Question about Euler equation and material derivative From the Euler equation
$$\partial_t\rho(x,t)+\partial_x(\rho(x,t) v(x,t))=0$$
we get the material derivative form;
\begin{equation}
    \frac{d}{dt}\rho=-\rho\partial_x v,
\end{equation}
where
$$\frac{d}{dt}=\partial_t+v(x,t)\partial_x$$
denotes the directional derivative along the direction
\begin{equation}
    \frac{dx}{dt}=v(x,t)
\end{equation}
for any point $(\bar{x},\bar{t})\in\mathbb{R}_+^2:={(x,t):x\in\mathbb{R},t\in\mathbb{R}_+},\mathbb{R}_+=(0,\infty)$, the integral curve of $\frac{dx}{dt}=v(x,t)$ through $(\bar{x},\bar{t})$ is denoted by $x=x(t;\bar{x},\bar{t})$.
At $t=0$, it passes through the point
$(x_0(\bar{x},\bar{t}),0):=(x(0;\bar{x},\bar{t}),0)$.
Along the curve $x=x(t;\bar{x},\bar{t})$, the solution of the ordinary differential equation with initial data:
$$\rho|_{t_0}=\rho_0(x_{0}(\bar{x},\bar{t}))$$
is
$$\rho(\bar{x},\bar{t})=\rho_0(x_0(\bar{x},\bar{t}))\exp\Big{(}-\int_0^{\bar{t}}\partial_xv(x(t;\bar{x},\bar{t}),t)dt\Big{)}>0$$
Here I have some questions;
first, clearly $\frac{dx}{dt}=\partial_tx+v(x.t)\partial_x x=v(x,t)$ hold, and the directional derivative along the direction $[1,v(x,t)]^T$ for scalar function $f=f(x,t)$ is
\begin{equation*}
    D_vf=Df(x,t) 
    \begin{pmatrix}
    1\\
    v(x,t)
    \end{pmatrix}
    =
    \begin{pmatrix}
    \partial_t f&\partial_x f
    \end{pmatrix}
     \begin{pmatrix}
    1\\
    v(x,t)
    \end{pmatrix}
    =\partial_t f+v(x,t)\partial_x f
\end{equation*}
In our paper, material derivative is interpreted as a directional derivative with respect to $v(x,t)$, but according to my calculations, it is considered to be exactly a directional derivative with respect to $[1\ \ v(x,t)]^T$. Is this thinking correct?
second,
for any point $(\bar{x},\bar{t})\in\mathbb{R}_+^2:={(x,t):x\in\mathbb{R},t\in\mathbb{R}_+},\mathbb{R}_+=(0,\infty)$, the integral curve of $\frac{dx}{dt}=v(x,t)$ through $(\bar{x},\bar{t})$ is denoted by $x=x(t;\bar{x},\bar{t})$.
Is there any equation which explicitly represent $x$ as function of$(t,\bar{x},\bar{t})$?
Third,
Here I'm not exactly sure what an integral curve is.
Along the curve $x=x(t;\bar{x},\bar{t})$, the solution of the ordinary differential equation $(\frac{d\rho}{dt}=-\rho\partial_x v)$ with initial data:
$$\rho|_{t_0}=\rho_0(x_{0}(\bar{x},\bar{t}))$$
is
$$\rho(\bar{x},\bar{t})=\rho_0(x_0(\bar{x},\bar{t}))\exp\Big{(}-\int_0^{\bar{t}}\partial_xv(x(t;\bar{x},\bar{t}),t)dt\Big{)}>0$$
Here I don't know why the solution for differential equation which is represented by material derivative is the function of $(\bar{x},\bar{t})$ rather than a function of $(x,t)$.
 A: To the last, the text seems to prefer to have $(\bar x,\bar t)$ as unchangeable points during the calculation, so that $t$ and $x$ are free as integration variable and flow function. You could shift that if, for example, you use $\tau$ for the integration variable and $\varphi$ for the flow function.

The rest seems ok, only I would order it or move the focus it differently. The primary data is $v$, from this vector field follows the flow $x$. Then the "material derivative" is just the time derivative along the flow lines/characteristic curves or integral (to the vector field) curves. This can indeed be seen as a directional derivative related to $v$.
Then along the characteristic curves the PDE reduces to an ODE, with the given solution formula.

Usually one would order the components of the Jacobian in the order of the arguments, that is, the tangent of the curve $[x(t),t]$ would be $[v(x(t),t),1]$.
A: This is not a direct answer to the question, but a different approach. Consider this as a comment but too long to be edited in the comments section.
$$\partial_t\rho(x,t)+\partial_x(\rho(x,t) v(x,t))=0\tag 1$$
$$\frac{\partial \rho(x,t)}{\partial t}+v(x,t)\frac{\partial \rho(x,t)}{\partial x}=-\frac{\partial v(x,t)}{\partial x}\rho(x,t)$$
I suppose that $v(x,t)$ is a known function (a given function). If not the equation $(1)$ would involve two unknown functions $v(x,t)$ and $\rho(x,t)$ which would make it not solvable because only one equation for two unknowns.
$v=v(x,t)$ being a known function implies that $v_x=v_x(x,t)$ is a known function.
On condensed scripf :
$$\frac{\partial \rho}{\partial t}+v\frac{\partial \rho}{\partial x}=-v_x\rho$$
This is a first order linear PDE wrt the unknown function $\rho(x,t)$.
Charpit-Lagrange characteristic ODEs :
$$\frac{dt}{1}=\frac{dx}{v}=\frac{d\rho}{-v_x}$$
A first characteristic equation comes from solving $\frac{dt}{1}=\frac{d\rho}{-v_x}$ :
$$\rho+\int v_x(x,t)dt=c_1$$
A second characteristic equation comes from solving $\frac{dx}{v}=\frac{d\rho}{-v_x}$ :
$$\rho+\ln|v|=c_2$$
The general solution of the PDE expressed on implicit form $F(c_1,c_2)=0$ is :
$$\boxed{F\left(\left(\rho+\int v_x(x,t)dt\right)\:,\:\big(\rho+\ln|v(x,t)|\big)\right)=0}$$
$F(X,Y)$ is an arbitrary function of two variables
where $\begin{cases}X=\rho+\int v_x(x,t)dt \\ Y=\rho+\ln|v(x,t)| \end{cases}$.
Without specifying some initial and/or boundary condition one cannot go further.
If an initial and/or boundary condition is given (or arbitrary chosen) one can expect to determine the function $F$ and find a solution which satisfies both the PDE and the condition.
