In section 5.1 Linear systems in A First Course in Differential Equations, Second Edition by J. David Logan , there is an example, Example 5.4. Here you consider

$ x'=2y\\ y'=x $

and want to find the shape of the orbits the solutions have in the phase plane. In the book, they have simply divided the two equations to get


The author also writes that he has used the chain rule: "Along an orbit $x=x(t), y=y(t)$ we also have $y$ as a function of $x$, or $y=y(x)$. Then the chain rule requires:

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ or $\frac{dy/dt}{dx/dt}=\frac{dy}{dx}$.

We can obtain the orbits by solving the differential equation above by separation of variables, as is done in the book. One then obtains the curves

$\frac{1}{2}x^2 -y^2=C$.

This corresponds to a phase plot with a saddle point. The plot is similiar to:

Phase plot illustatiton1


The curves that crosses the positive $x_1$ axis does not appear to be functions of $x_1$. How can then the above use of the chain rule be justified, where we have used that $x_2=f(x_1)$ or equivalently $y=f(x)$?

  • $\begingroup$ Is this connected with the Implicit function theorem? $\endgroup$ – FredikLAa Jul 13 '15 at 15:47

The "trick" here is that for curves that cross the positive $x_1$ axis it's more convenient to think of them as of functions of $x_2$. If you still insist on considering them as functions of $x_1$ you should take only one "branch" of the curve and you'll still have a some kind of problem at the point where branch crosses $x_1$-axis. If you are considering a point where either $x'_1$ or $x'_2$ doesn't vanish (it's a necessary and sufficient condition for existence of inverse for $\mathbb{R} \rightarrow \mathbb{R}$ functions) you have an inverse function $t = \tilde{f}(x_1)$ (or $t = \tilde{f}(x_2)$) and hence $x_2 = \tilde{g}(x_1)$ (or $x_1=\tilde{g}(x_2)$). And here you can again apply the chain rule.


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