# Nonlinear phase portrait and linearization

Consider the nonlinear system $x^{'} = y$, $y^{'}= -8 \sin x - 2y$

where $-2\pi$ < or = x < or = $2\pi$

Find the equilibrium points of the system.

$(-2\pi,0)$$(-\pi,0)$$(0,0)$ $(\pi,0)$$(2\pi,0) Linearize the system near every equilibrium point, and describe the behaviour of the linearized system. EDIT: i forgot to mention i used a taylor expansion to the first variable on \sin( x) thus y^{'}=-8x-2y J= \begin{bmatrix} 0 & 1 \\ -8 & -2 \\ \end{bmatrix} T= -2 \Delta=8 T^{2}-4\Delta <0 thus all fixed points are stable spirals this is clearly wrong see attached picture where i solve via nullclines Edit: as there was some confusion below THIS IS NOT CORRECT AT ALL there is no way the jacobian can be this for all 5 of those fixed points it doesn't make any sense. what i drew was a second attempt at this problem using nullclines. I want to know why this linearization is wrong? and if someone could please do it correctly! thx. ( we have not learned the method of nullclines in this course there mus tbe some way to get the correct answer via a linearization.) Sketch the phase portrait of the nonlinear system. see picture As you can see -2\pi,0,2\pi appear to be stable spirals and -\pi,\pi appear to be saddles where the stable and unstabel manifold flip with each iteration causing them to get sucked into the n\pi (where n is an odd iteger) nieghbour on each side please fix my math! ty • What exactly is the question? – apnorton Mar 26 '13 at 21:03 • the question is can u please tell me why my linearization of the system is baligerntly wrong? and if so fix it ? i think i should be lvieing it as -8\cos x in the y^{'} term of the jacobian if so why and if my phase portrait correct? – Faust Mar 26 '13 at 21:16 ## 1 Answer We are given the nonlinear system: \tag 1 x' = y, ~~ y'= -8 \sin x - 2y, where -2\pi \le x \le 2\pi. We start off by finding the fixed points of the system. To do this, we need to find those points where x' and y' are simultaneously equal to zero, over the given range in (1). So, x' = 0, when y = 0, and y' = -8 \sin x - 2y = 0 reduces to (since we must have y = 0 from the previous observation): y' = -8 \sin x -2(0) = 0 \rightarrow x = -2\pi, -\pi, 0, \pi, 2\pi. This gives us a total of five fixed (critical) points as:$$(-2\pi, 0), (-\pi, 0), (0,0), (\pi, 0), (2\pi, 0).$$In order to linearize the system, we have to investigate the behavior of the Jacobian of the system at those fixed points. So, the Jacobian matrix is:$$A = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ - 8 \cos x & -2 \end{bmatrix}$$Next, we need to evaluate the Jacobian matrix A at each of those fixed points. At (-2\pi, 0), we have:$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (-2\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$At (-\pi, 0), we have:$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (-\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$$At (0, 0), we have:$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (0) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$At (\pi, 0), we have:$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$$At (2\pi, 0), we have:$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (2\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$Notice, we have two identical sets of matrices given the periodicity of$\cos x$over our five fixed points. We are asked to linearize the system near every equilibrium point, and describe the behaviour of the linearized system. So, now we need to determine the behavior of these two matrices by looking at their eigenvalues. So, we get: •$A = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$, has eigenvalues$\lambda_1 = -4$and$\lambda_2 = 2$. These are real eigenvalues with opposite sign, so this is an unstable saddle node. •$A = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$, has eigenvalues$\lambda_1 = -1 + \sqrt{7}i$and$\lambda_2 = -1 - \sqrt{7}i$. These are a complex conjugate pair, with negative real part, so this is a stable spiral point. Note: The only thing that allows us to use this linearization is that we do not get borderline cases from the fixed points. You had better make sure you are clear on this last statement! Lastly, we can draw the phase portrait to show the direction field, the five fixed points and many solutions$x(t)$and$y(t)$in order to visualize this and to compare to our analyses. Of course, we should see two unstable saddle nodes and three stable spirals from the analyses above. Make sure to pick out the five fixed points$(x, y)\$ and that you see what we derived.

Regards

• ty it seems my sketch via nullclies was incorrect i assumed that the unstable manifold went into each saddle it appears it get sucked directly into the spiral ( not the stable manifold of the other fp) and the that the stable manifold merely tangents off to +- infinity very clever and well done sir! – Faust Mar 27 '13 at 4:09
• This is beautiful! What software to you use? +1 – Namaste Apr 13 '13 at 0:26
• @amWhy: yea, that one is a beautiful picture. I use the tools here (very easy, quick) and nice results. See DFIELD and PPLANE. Regards – Amzoti Apr 13 '13 at 2:18
• thanks for the links! It is a really slow Friday night! – Namaste Apr 13 '13 at 2:21
• @amWhy: You are very welcome and that is very useful code for lots of problems! Enjoy! – Amzoti Apr 13 '13 at 2:22