I'm working through Nonlinear Dynamics and Chaos (Strogatz). In looking for fixed points the book prescribes $\dot{x}=f(x)=0$. So for $\dot{x}=x^{2}-1$, fixed points are $x^{*}=\pm1$.

In the Existence and Uniqueness section (2.5). It asks:

Discuss the existence and uniqueness of solutions to the initial value problem $\dot{x}=x^{2}+1$,$x(0)=x_{0}.$" Do solutions exist for all time?

I have two questions about the answer to the exercise (2.5.2).

  1. It says "Consider the case $x(0)=0$". How does it know that $x(0)=0$ since $f(x)=x^{2}+1$, but $x(t)$ is not defined?
  2. It goes on to say "Initial condition $x(0)=0$ implies $C=0$. Hence $x(t)=Tan(t)$ is the solution." Looking at the plot of $x^{2}+1$, at no point does $\dot{x}=0$. Look at the phase portrait, it seems that there are no fixed points. Is a "solution to the initial value problem" the same thing as fixed points?


$x^{2}+1$ Phase Portrait

EDIT: I think I understand it now. Solution to initial value problem solves the trajectory not for fixed points of a system (unless the initial value is at a fixed point). Like so (please correct me if I'm wrong though):

Phase portrait

  • 1
    $\begingroup$ The answer "No" to 2) makes 1) a non-question. There is a difference between solving initial-value problems and finding stationary points. The second can be a part of the first. $\endgroup$ Jul 4, 2017 at 22:57
  • $\begingroup$ Would you mind elaborating a bit more? Is the solution to the initial-value problem the trajectory of an initial point for $\dot{x}$? $\endgroup$ Jul 4, 2017 at 23:00
  • $\begingroup$ $\dot x(t)=f(t,x(t))$ is the differential equation. The data of that equation is the function $f$, the rest is symbolism (at the point of the problem statement). A solution is any differentiable function that together with its derivative satisfies this equation over its domain. An initial value $(t_0,x_0)$ is a selector among all possible solutions, i that $t_0$ has to be in the domain and $x(t_0)=x_0$. In that sense you can say that the IVP solution is the trajectory of $x_0$ under the dynamic defined by $f$. $\endgroup$ Jul 5, 2017 at 8:44

2 Answers 2


The existence and uniqueness theorem stated a page earlier in the book tells you that the initial value problem has a unique solution close to $t=0.$

When the book says "Consider," it is providing an example, being $x_0=0.$

Also note that the solution (in the case of $x_0=0$) is $x(t)=\tan(t)$ and not $x(t)=\tan(x)$.

  • $\begingroup$ I might add that in the case $x_0 = 0$, $x(t) = \tan(t)$ is a solution only for $t\in (-\pi/2, \pi/2)$. $\endgroup$
    – Michael L.
    Jul 5, 2017 at 0:05
  • $\begingroup$ @MichaelLee Yes, thank you. $\endgroup$ Jul 5, 2017 at 0:07

To find a solution to the initial value problem $\dot{x}=x^2+1$, consider $x(t)=\tan(t+C)$, where $C$ is some constant. Considering the case when $x(0)=0$, we see that $C=0$ since $\tan(0)=0$. So one could verify that $$ \begin{align*} \frac{dx}{dt}&=\dot{x} = \sec^2 t \\ &= \dfrac{\sin^2 t+ \cos^2 t}{\cos^2 t} \\ &= \tan^2 t + 1\\ &= x(t)^2 + 1, \end{align*} $$ where the solution $x(t)=\tan(t)$ exists only when $-\frac{\pi}{2}< t< \frac{\pi}{2}$.

To answer your second question, since $\dot{x}=x^2+1\geq 1>0$, i.e., $\dot{x}$ is always strictly greater than $0$, there are no fixed points.

Thus, a solution to the initial value problem is not the thing as fixed points of a system.


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