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Stability of nonlinear ODE's

  1. Dec 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Find whether this system is stable or unstable at the steady state (x1,x2)=(0,0)
    dx1/dt = -x1+2sin(x1)+x2
    dx2/dt=2sin(x2)

    2. Relevant equations
    upload_2014-12-15_23-57-6.png
    3. The attempt at a solution
    z1=x1-0
    z2=x2-0
    dz1/dt=-z1+z2+2z1
    dz2/dt=2z2

    Jacobian =
    [ 1 1 ]
    [ 0 2 ]
    so the system is unstable.

    This problem is from my notes from class; I'm not 100% certain that it is written correctly. I am very confused about the part where z1 is set equal to x1 (I believe this is called linearization)? Could someone please clarify this step for me and how on earth
    dx2/dt=2sin(x2) becomes dz2/dt=2z2 ?

    The rest is pretty straightforward.
     

    Attached Files:

  2. jcsd
  3. Dec 16, 2014 #2

    Dick

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    If ##z_1=x_1-0## then ##x_1=z_1##. I don't see any problem with that. And the series expansion of ##sin(z)## around ##z=0## is ##z-z^3/3!+z^5/5!+...##. The linearization part is where you ignore the higher powers of ##z## and just replace ##sin(z)## with ##z##, since if ##z## is very close to 0 then the higher powers of ##z## are much smaller than ##z##.
     
  4. Dec 16, 2014 #3
    Thanks!
    If instead I had dx1/dt=-x13+x2, would it be valid to keep z1 as it is defined now (z1=x1-0) and would dz1/dt=-z13+x2 be true? Since the first term of the series expansion of x3 is x3
     
    Last edited: Dec 16, 2014
  5. Dec 16, 2014 #4

    Dick

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    No, 'linearization' mean you only keep first powers of variables around the steady state. You drop the higher powers.
     
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