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.

• What is the question? – Jacky Chong Oct 4 '16 at 19:36
• @JackyChong: I don't know how to show that the solution of the pde is really of the form written in the post. – user72829 Oct 4 '16 at 19:40

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}