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Homework Help: Linearise the euqations near each equilibrium point?

  1. Jul 30, 2008 #1
    linearise the euqations near each equilibrium point?!?!?

    Write each of the following differential equations as two first order equations, find all the equlibrium points, and linearise the equations near each equlibrium point

    (a) X.. +x -1/4X3=0

    first part done getting X.=y and Y.=-x+1/4X3

    second part done getting (0,0) (2,0) and (-2,0)

    however i dont even know what the third part means let alone how to do it, assistance required please
  2. jcsd
  3. Jul 30, 2008 #2


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    Re: linearise the euqations near each equilibrium point?!?!?

    I don't understand your notation. What is X.. ? Which is the term representing the derivative? Maybe it's something I have not learnt before.
  4. Jul 30, 2008 #3


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    Re: linearise the euqations near each equilibrium point?!?!?

    I assume the two dots are supposed to be superscripts denoting derivatives. In ASCII it is better to use ' and " for derivatives: x"+ x- (1/4)x3. Letting y= x', y'= x" so that equation becomes y'+ x- (1/4)x3 or x'= y, y'= -x+ (1/4)x3 as you say.

    The equilibrium points are points where both derivatives are equal to 0: were x'= y= 0 and y'= -x+ (1/4)x3= 0. We can factor that as x((1/4)x2-1)= 0 and, yes, the equilibrium points are (0, 0), (2, 0), and (-2, 0).

    Now, "linearize" means simply to replace any nonlinear function with a linear approximation. The best linear approximation to x3 for x close to 0 is just 0 itself (because the slope of x3 at 0 is 0). At (0,0) the linear approximation is just
    x'= y, y'= -x.

    The derivative of x3 at x= 2 is 3(22)= 12 and 23= 8 so the tangent line to y= x3 at x= 2 is y= 12(x-2)+ 8. At (2, 0) the linearization is x'= y, y'= -x+ 12(x- 2)+ 8= 11x- 16.

    You could also do this by writing the "Jacobian" for the sytem and evaluating it at (0,0), (2,0) and (-2,0). Does your textbook say anything about the "Jacobian"?
  5. Jul 30, 2008 #4
    Re: linearise the euqations near each equilibrium point?!?!?

    .. no not X'' like X with 2 dots above it but i cant write that on here
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