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Homework Help: Linearizing system of differential equations

  1. Oct 24, 2011 #1
    Linearize and classify the fixed points of
    [itex]\frac{d\theta1}{dt}[/itex] = [itex]\Omega[/itex] + sin [itex]\theta1[/itex] + [itex]\frac{1}{2}[/itex] (sin[itex]\theta[/itex]1 + sin[itex]\theta[/itex]2)

    [itex]\frac{d\theta2}{dt}[/itex] = [itex]\Omega[/itex] + sin [itex]\theta2[/itex] + [itex]\frac{1}{2}[/itex] (sin[itex]\theta[/itex]1 + sin[itex]\theta[/itex]2)

    I know that if the absolute value of omega is less than two, there will be 4 critical points, and if the absolute value of omega is greater than two, there will be no critical points. I have found the Jacobian,

    [itex]\frac{3}{2}[/itex] cos[itex]\theta1[/itex] [itex]\frac{1}{2}[/itex] cos[itex]\theta2[/itex]
    [itex]\frac{1}{2}[/itex] cos[itex]\theta1[/itex] [itex]\frac{3}{2}[/itex] cos[itex]\theta2[/itex]

    My problem is that I do not understand how to linearize this system of equations when I do not know the value of omega, because without knowing the value of omega, I cannot solve for [itex]\theta1[/itex] or [itex]\theta2[/itex]. Any advice regarding how to linearize this system, and from linearizing the system classify the fixed points, would be appreciated. Thanks in advance!
  2. jcsd
  3. Oct 24, 2011 #2


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    Well, here's my hack at this strange system. Setting [itex]\frac{d\theta_{1}}{dt} = 0[/itex] and [itex]\frac{d\theta_{2}}{dt} = 0[/itex] leads us to the simultaneous equations

    [itex] \frac{3}{2} sin \theta_{1} + \frac{1}{2} sin \theta_{2} = -\Omega[/itex] and [itex] \frac{1}{2} sin \theta_{1} + \frac{3}{2} sin \theta_{2} = -\Omega \Rightarrow sin \theta_{1} = sin \theta_{2} = -\frac{\Omega}{2} .[/itex]

    It is now clear why the character of the system changes at [itex]\Omega = 2[/itex] . We also see that in the "principal square" [itex][ 0 , 2\pi) \times [ 0 , 2\pi) [/itex] , there will be four critical points; for instance, with [itex]\Omega = 1[/itex] , the solution "angles" are [itex]\frac{7\pi}{6} [/itex] and [itex]\frac{11\pi}{6} [/itex] , so the critical points lie at [itex] ( \theta_{1} , \theta_{2} ) [/itex] , where the ordered pairs have every combination of the two solution angles.

    That indicates that someplace nice like the origin is not generally an equilibrium point, so we have to linearize around the (usually) four critical points. If we call the two "solution angles" [itex] \Theta_{1} , \Theta_{2} [/itex] , then the linearizations are going to look like

    [tex] \sin \theta_{1} = \sin \Theta_{1} + [ \cos \Theta_{1} \cdot ( \theta_{1} - \Theta_{1} ) ] , [/tex]

    and something similar for [itex]\sin \theta_{2} [/itex]. But we know that [itex]\sin \Theta_{1} = -\frac{\Omega}{2} [/itex] , so [itex]\cos \Theta_{1} = \pm \frac{1}{2} \sqrt{ 4 - \Omega^{2} } * [/itex] (which also gives us a better idea of what sort of change is going to take place at [itex]\Omega = 2[/itex] ). So the linearization in the neighborhood of a critical point is

    [tex] \sin \theta_{1} = ( -\frac{\Omega}{2} ) \pm [ \sqrt{ 4 - \Omega^{2} } \cdot ( \theta_{1} - \Theta_{1} ) ] , [/tex]

    and similarly for [itex] \sin \theta_{2} [/itex].

    * the plus-or-minus sign must be resolved according to the "appropriate quadrant" for the angles [itex] \Theta_{1} [/itex] or [itex] \Theta_{2} [/itex]

    Try it from here and see if that helps.

    EDIT: I pressed on somewhat further. Something very nice happens for the linearization in the [itex]\frac{1}{2} (\sin \theta_{1} + \sin \theta_{2}) [/itex] term , with the result that the entire expressions for [itex]\frac{d\theta_{1}}{dt} [/itex] and [itex]\frac{d\theta_{2}}{dt} [/itex] collapse into something fairly simple. That's a pleasant surprise! Just out of curiosity, what is the context or the application for this dynamical system?
    Last edited: Oct 24, 2011
  4. Oct 25, 2011 #3
    I had some questions regarding your solution.

    First of all, for the solution [tex] \sin \theta_{1} = ( -\frac{\Omega}{2} ) \pm [ \sqrt{ 4 - \Omega^{2} } \cdot ( \theta_{1} - \Theta_{1} ) ] , [/tex], should it not be [tex] \sin \theta_{1} = ( -\frac{\Omega}{2} ) \pm 1/2[ \sqrt{ 4 - \Omega^{2} } \cdot ( \theta_{1} - \Theta_{1} ) ] , [/tex]? Or if not, where did the 1/2 go?

    Additionally, in the same solution, I am a bit confused regarding the presence of both θ1 and Θ1. I expected to only have two angles (θ1, θ2), not four angles, and I am unsure how this effects my final solution.

    The application of this system is that it arises in the study of arrays of Josephson junctions. Beyond that, I am not exactly sure.
  5. Oct 25, 2011 #4


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    Sorry, yes, the 1/2 should be there (it got lost in all the other TeXing I was doing).

    The "lower-case" [itex] \theta_{1} [/itex] and [itex] \theta_{2} [/itex] are the variables. I am taking it from your problem statement that each is in the interval [itex] [ 0, 2\pi ) [/itex] , so what I am describing is a phase plot of [itex]\dot{\theta_{2}} [/itex] versus [itex]\dot{\theta_{1}} [/itex] over one period of each. Thus, there is a square [itex] [ 0, 2\pi ) \times [ 0, 2\pi ). [/itex] (This could, of course, be repeated endlessly over the phase plane.)

    The angles [itex] \Theta_{1} [/itex] and [itex] \Theta_{2} [/itex] are the two solutions to the equations [itex]\sin \theta = -\frac{\Omega}{2} [/itex] for the critical points . ( So, for [itex]\Omega = \sqrt{3} [/itex], we would have [itex] \Theta_{1} = \frac{4\pi}{3} [/itex] and [itex] \Theta_{2} = \frac{5\pi}{3} [/itex]. ) The four critical points in the phase plane are then [itex] ( \Theta_{1} , \Theta_{1} ) , ( \Theta_{1} , \Theta_{2} ) , ( \Theta_{2} , \Theta_{1} ) , [/itex] and [itex] ( \Theta_{2} , \Theta_{2} ) .[/itex] It is these points about which we are constructing the linearizations.

    Because [itex] \Theta_{1} [/itex] and [itex] \Theta_{2} [/itex] are in different "quadrants", their cosine values differ. For the example I gave here, with [itex]\Omega = \sqrt{3} [/itex],

    [tex] \sin \theta_{1,2} = ( -\frac{\sqrt{3}}{2} ) - \frac{1}{2}[ \sqrt{ 4 - (\sqrt{3})^{2} } \cdot ( \theta_{1,2} - \frac{4\pi}{3} ) ] = ( -\frac{\sqrt{3}}{2} ) - [ \frac{1}{2} \cdot ( \theta_{1,2} - \frac{4\pi}{3} ) ] [/tex] near [itex] \Theta_{1} = \frac{4\pi}{3} [/itex] and

    [tex] \sin \theta_{1,2} = ( -\frac{\sqrt{3}}{2} ) + \frac{1}{2}[ \sqrt{ 4 - (\sqrt{3})^{2} } \cdot ( \theta_{1,2} - \frac{5\pi}{3} ) ] = ( -\frac{\sqrt{3}}{2} ) + [ \frac{1}{2} \cdot ( \theta_{1,2} - \frac{5\pi}{3} ) ] [/tex] near [itex] \Theta_{2} = \frac{5\pi}{3} . [/itex]

    Just as an example of the linearization of one of the differential equations at a critical point, near [itex] ( \Theta_{1} , \Theta_{2} ) = ( \frac{4\pi}{3} , \frac{5\pi}{3} ) , [/itex] we have

    [tex]\dot{\theta_{1}}= \Omega + \frac{3}{2} \sin \theta_{1} + \frac{1}{2}\sin \theta_{2} = \sqrt{3} + \frac{3}{2} [ ( -\frac{\sqrt{3}}{2} ) - [ \frac{1}{2} \cdot ( \theta_{1} - \frac{4\pi}{3} ) ] ] + \frac{1}{2} [ ( -\frac{\sqrt{3}}{2} ) + [ \frac{1}{2} \cdot ( \theta_{2} - \frac{5\pi}{3} ) ] ] [/tex]

    [tex] = \sqrt{3} - \frac{3\sqrt{3}}{4} - [ \frac{3}{4} \cdot ( \theta_{1} - \frac{4\pi}{3} ) ] - \frac{\sqrt{3}}{4} + [ \frac{1}{4} \cdot ( \theta_{2} - \frac{5\pi}{3} )] = [ -\frac{3}{4} \cdot ( \theta_{1} - \frac{4\pi}{3} ) ] + [ \frac{1}{4} \cdot ( \theta_{2} - \frac{5\pi}{3} )] . [/tex]

    There would be a similar linearization for [itex]\dot{\theta_{2}} [/itex].

    You would have to do this for each of the four critical points, but happily, they are have the same form. To classify a critical point, you can substitute [itex]u = \theta_{1} - \frac{4\pi}{3} [/itex] and [itex]v = \theta_{2} - \frac{5\pi}{3} [/itex] (which give [itex]\dot{\theta_{1}} = \dot{u} [/itex] and [itex]\dot{\theta_{2}} = \dot{v} [/itex]) , so that the linearization near [itex] ( \Theta_{1} , \Theta_{2} ) = ( \frac{4\pi}{3} , \frac{5\pi}{3} ) [/itex] has [itex]\dot{u} = -\frac{3}{4} \cdot u + \frac{1}{4} \cdot v [/itex] and a similar result for [itex]\dot{v}[/itex].

    So it looks like a lot of work for the classification, but I think the cancellations all work the same way for the pair of differential equations near each of the four points.
  6. Oct 30, 2011 #5

    How do you get sinθ1=(−Ω2)±1/2[srt(4−Ω2)(θ1−Θ1)] as a solution. Is the solution for sinθ2=(−Ω1)±1/2[srt(4−Ω1)(θ2−Θ2)]?

    Also what happens when Ω=2?
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