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Steady-State Stability and the Second Derivative

  1. Nov 6, 2011 #1
    Hi. I have a question about steady-state stability and the second derivative test. I have been reading about it in a book on mathematical modeling, and the section concerns differential equations. I believe this forum is more appropriate than "General Math," but let me know if it is not.

    From what I was reading, my understanding is that if the second derivative is positive, then at a point where the first derivative is zero, then the system is unstable. Similarly, if the second derivative is negative, it is stable. Can someone please provide me with a reason why this is the case (either mathematically, graphically). Why would a function such as f(x)=x^2 (whose second derivative is positive) be unstable at its steady-state points?

    What confuses me is when I think about an example of energy in physics. Take the graph of f(x) = x^2. Moving slightly away from x = 0 (because f'(x)=0 at x=o), the system easily slides back down to the point of equilibrium. Whereas with f(x)= -x^2, a slight movement to either side causes the system to fall away from these points.

    Thanks!
     
  2. jcsd
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