Partial differential equation $[\partial_t + v(x)\partial_x - \rho(x)] D(t,x) = 0$ I'm stuck with the following problem I found (without a proof) in the Peskin and Schroeder textbook on quantum field theory (the differential equation mentioned below is equivalent to the Callan-Symanzik equation describing the evolution of the n-point correlation functions under variation of the energy scale at which the theory is defined):
Show that the solution of the equation
$$
[\partial_t + v(x)\partial_x - \rho(x)] D(t,x) = 0
$$
has the form
$$
D(t,x) = D(0,X_t(x)) \exp\left(\int_0^t d t' \rho(X_{t'}(x)) \right),
$$
where $X_t(x)$ is the solution of the equation
$$
\partial_t X_t(x) = - v(X_t(x))
$$
with the initial condition
$$
X_0(x) = x.
$$
Let's compute e.g.
$$
\partial_t \left[D(0,X_t(x)) \exp\left(\int_0^t d t' \rho(X_{t'}(x)) \right) \right]= \left[\partial_t D(0,X_t(x))\right] \exp\left(\int_0^t d t' \rho(X_{t'}(x)) \right)  + D(0,X_t(x)) \rho(X_t(x)) \exp\left(\int_0^t d t' \rho(X_{t'}(x)) \right) 
$$
I don't see any possible simplifications with other terms.
I wonder how to show that the solution of the mentioned differential equation is indeed of the above form.
 A: This is a simple transport type equation. We shall solve the PDE using the methods of characteristics. Suppose we could rewrite the PDE as follows
\begin{align}
\partial_t D + \nu(x) \partial_x D- \rho(x)D =&\ \partial_t D+ X'(t)\partial_x D -\rho(x)D \\
=&\ \frac{d}{dt}D(t, X(t)) -\rho(x)D = 0 
\end{align}
where $X'(t) = \nu(X)$. Solving for the characteristic $X$ leads to the solution
\begin{align}
\frac{X'}{\nu(X)} = 1 \ \ \Rightarrow \ \ \int \frac{dX}{\nu(X)} = t+ C
\end{align}
which really depends on $\nu(X)$ (nevertheless it exists). Hence let us assume that we found $X(t)$. Back to the PDE, we have
\begin{align}
\frac{d}{dt}D(t, X(t)) = \rho(X(t)) D(t, X(t)) \ \ \Rightarrow& \ \ \frac{d}{dt}\log|D(t, X(t))| =\rho(X(t))\\
\Rightarrow&\  \ \ \log|D(t, X(t))| -\log|D(0, X(0))| = \int^t_0 dt' \rho(X(t'))\\
\Rightarrow&\ \ D(t, X(t)) = D(0, X(0))\exp\left( \int^t_0 dt' \rho(X(t'))\right).
\end{align}
We are almost done, but not done. If we impose the initial condition that $X(0) = x$, then we have a unique curve that pasts through the point $(0, x)$ which will dictate the values of $D(t, x)$  when $(t, X)$ lies on the trace of $(t, X(t))$. Hence, we shall use the notation $X= X_t(x)$ to indicate the curve. Thus, your solution becomes
\begin{align}
D(t, X) = D(0,x)\exp\left( \int^t_0 dt' \rho(X_t(x)) \right). 
\end{align}
