# Transforming the solutions of $\dot x = f(\mathbf{x}, t)$ and $\dot x = B(\mathbf{x}, t)f(\mathbf{x}, t)$ into each other

How can I prove the following theorem?

If the function $B(\mathbf{x}, t)$ is strictly positive, then the solutions of the two differential equations $\dot x = f(\mathbf{x}, t)$ and $\dot x = B(\mathbf{x}, t)f(\mathbf{x}, t)$ can be transformed into each other by a strictly monotonic change in the time scale $\tau = \phi(t)$.

Note: This theorem is mentioned in Hofbauer and Sigmund's Evolutionary Games and Population Dynamics, p. 32.

Here is what I got so far, but I'm not sure how to proceed: \begin{align*} \frac{d \mathbf{x}}{d \tau} &= B(\mathbf{x}(\tau), \tau)f(\mathbf{x}(\tau), \tau) \\ \frac{d \mathbf{x}}{d t}\frac{dt}{d\tau} &= B(\mathbf{x}(\tau), \tau)f(\mathbf{x}(\tau), \tau) \\ f(\mathbf{x},t) \frac{dt}{d \tau} &= B(\mathbf{x}(\tau), \tau)f(\mathbf{x}(\tau), \tau) \\ \end{align*}

• Hm. It's strange, but I'm aware of only autonomous case of this theorem. Commented Sep 28, 2015 at 6:21
• Do you know where I can find the proof for the autonomous case? Commented Sep 28, 2015 at 13:16
• You almost did it by yourself :) In case of autonomous system $f(\mathbf{x}, t)$ and $f(\mathbf{x}(\tau), \tau)$ are the same (vector field doesn't depend on time; the point of phase space is the same for both expressions) and you can easily find $t(\tau)$ from its derivative. Commented Sep 28, 2015 at 13:37
• And I guess in the autonomous case B(x) doesn't have to be strictly positive, does it? Commented Sep 28, 2015 at 15:07
• No, even in autonomous case it has to have strict sign. There are two reasons for this: 1) technical: convergence of integral that defines $t(\tau)$; 2) methodical: you don't want to create additional equilibrium points, because this will clearly made two system 'not the same', right? Commented Sep 28, 2015 at 15:09