1
$\begingroup$

Let the non-linear differential equation be $\dot{x}= sin\left ( x \right )$

Plotting this on the graph, we have the standard sin curve with the exception that the horizontal axis is labelled $x$ and the vertical axis labelled as $\dot{x}$.

The text I'm using is Nonlinear dynamics and chaos by Steven H. Strogatz. He says

1) > a particle starting at $x_{0}=\pi/4$ moves to the right faster and faster until it crosses $x=\pi/4$(where sin reaches its maximum). Then the particle starts slowing down and eventually approaches the stable fixed point $x=\pi$ from the left.

Note that the curve is concave up at first then concave down; this corresponds to the initial acceleration for $x<\pi/2$, followed by the deceleration towards $x=\pi$

I've spent a good whole day returning back to HS precalculus but nothing about the italic made sense.

Let $f(x)=sin(x)$ Then, $f'(x)=Cos(x)$ The critical points are $x=0,\pi/2,3\pi/2$

Indeed, on the interval $I\in[0,\pi/2]$, we have $f'(x)>0$ so $f'(x)$ is concave down and $f(x)$ is increasing on this interval. But this contradicts(he claims concave up) with what the author has put forth. If the $f'(x)$ is concave up on the interval from $\pi/4$ to $\pi/2$, then, $f(x)$ cannot be increasing by theorem.

Author plots the solution to the nonlinear differential equation below: dash lines denotes $\pi/2$ Notice that on $I\in[\pi/4,\pi/2]$, the function $f(x)$ is increasing. Could someone help me out?

$\endgroup$
0

2 Answers 2

4
$\begingroup$

The claim is that $\frac{d^2}{dt^2} x(t) >0$ for $x$ near $\pi/4$ and $\frac{d^2}{dt^2} x(t) <0$ for $x$ near $\pi$. $\frac{d^2}{dt^2}x(t)=0$ at some point in between. (concave up/down is determined by looking at second derivatives)

We have $\frac{d^2}{dt^2} x = \cos(x)\sin(x)$ which is indeed positive at $x=\pi/4$ and negative between $x=\pi/2$ and $x=\pi$.

$\endgroup$
2
$\begingroup$

The question whether the curve of $x(t)$ is concave up/down is equivalent to the question whether the second derivative of $x(t)$ is positive (concave up) or negative (concave down). Since $\frac{\text{d} x}{\text{d} t} = \sin x(t)$, we have \begin{equation} \frac{\text{d}^2 x}{\text{d} t^2} = \frac{\text{d} x}{\text{d} t} \cos x(t) = \sin x(t) \cos x(t) = \frac{1}{2}\, \sin 2 x(t). \end{equation} Plotting the graph of $\sin 2 x$, it is clear that $\sin 2 x > 0$ for $\frac{\pi}{4} < x < \frac{\pi}{2}$ and $\sin 2 x < 0$ for $\frac{\pi}{2} < x < \pi$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .