Question

Consider the autonomous DE$\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{/}\mathit{\pi }\mathbf{)}\mathit{y}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{y}$.Determine the critical points of the equation. Discuss a way ofobtaining a phase portrait of the equation. Classify the criticalpoints as asymptotically stable, unstable, or semi-stable.

Short answer

• The solutions converge on both sides of the critical point$y=0$ then it is asymptotically stable.
• The solutions diverge on both sides of the critical points $y=\pm \frac{\pi }{2}$then they are both unstable.

Step-by-step solution

Step 1:Critical points

The zeros of the function f in $dy/dx=f\left(y\right)$are of special importance. Wesay that a real number c is a critical point of the autonomous differential equation $dy/dx=f\left(y\right)$if it is a zero of f—that is, $f\left(c\right)=0$. A critical point is also called an equilibrium point or stationary point.

Find the value of critical points

To find the critical points, equate.$\frac{dy}{dx}=0$

$\begin{array}{l}\frac{dy}{dx}=\left(\frac{2}{\pi }\right)y-\sin y\\ \left(\frac{2}{\pi }\right)y-\sin y=0\\ y=0,\frac{\pi }{2},-\frac{\pi }{2}\end{array}$

As we can see by setting the differential side of the equation to zero the critical points are $y=0,\frac{\pi }{2},-\frac{\pi }{2}$.

Behaviour of critical points

We can determine the behaviour of the phase portrait by testing points around the critical points.

$\begin{array}{l}\frac{dy}{dx}=\left(\frac{2}{\pi }\right)\left(\pi \right)-\sin \left(\pi \right)\\ \frac{dy}{dx}=2>0\end{array}$

$\begin{array}{l}\frac{dy}{dx}=\left(\frac{2}{\pi }\right)(-\pi )-\sin \left(\pi \right)\\ \frac{dy}{dx}=-2<0\end{array}$

$\begin{array}{l}\frac{dy}{dx}=\left(\frac{2}{\pi }\right)(-\frac{\pi }{4})-\sin (-\frac{\pi }{4})\\ \frac{dy}{dx}=-\frac{1}{2}+\frac{\sqrt{2}}{2}>0\end{array}$

• If the result is negative then the plot is curving away from the critical point.

• If the result is positive then the plot is curving towards the critical point.
• Because the solutions converge on both sides of the critical point $y=0$then it is asymptotically stable.
• Because the solutions diverge on both sides of the critical points$y=\pm \frac{\pi }{2}$then they are both unstable.

Hence, the solutions converge on both sides of the critical point$y=0$then it is asymptotically stable.The solutions diverge on both sides of the critical points$y=\pm \frac{\pi }{2}$then they are both unstable.