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Math Help - semistable equilibrium solutions

  1. #1
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    semistable equilibrium solutions

    Hi, I need help on this problem

    Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable

    Consider the equations
    dy/dt = k(1 - y)^2 where k is a positive constant. Show that y=1 is the only critical point, with the corresponding equilibrium solution theta(t) = 1.

    Thanks in advance
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Recklessid View Post
    Hi, I need help on this problem

    Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable

    Consider the equations
    dy/dt = k(1 - y)^2 where k is a positive constant. Show that y=1 is the only critical point, with the corresponding equilibrium solution theta(t) = 1.

    Thanks in advance
    i'm not sure about that "corresponding equilibrium solution theta(t) = 1" thing, but we can see that y = 1 is the only critical point since critical points occur when the derivative is 0 (or undefined, but this is a polynomial and hence defined everywhere, so we don't have to worry about that).

    so for critical points, dy/dt = 0
    => k(1 - y)^2 = 0
    => k = 0 or (1 - y)^2 = 0
    k cannot be zero, since it is a positive constant, therefore, (1 - y)^2 = 0
    => 1 - y = 0 ..............squared both sides
    => y = 1
    thus this is the only critical point since it is the only solution to dy/dt = 0
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Recklessid View Post
    Hi, I need help on this problem

    Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable

    Consider the equations
    dy/dt = k(1 - y)^2 where k is a positive constant. Show that y=1 is the only critical point, with the corresponding equilibrium solution theta(t) = 1.

    Thanks in advance
    Don't you want to show that y=1 is a semistable equilibrium?

    (you can do this by considering the sign of dz/dt for the perturbed equilibria z=1+/-epsilon)

    RonL
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