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# Show that the equilibrium point x_0=0 for the differential equation x'=0 is stable but not asymptotically?

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### 1 Answer

- hfshawLv 79 years agoFavorite Answer
A stable equilibrium point is a solution to an autonomous differential equation (or system of equations) for which there exists some distance, ε, that if the initial conditions of the system are within ε of the equilibrium point, the system (values of the independent variable(s)) stays within a finite distance, δ, of the equilibrium point. For instance, the system could oscillate or circle around the equilibrium point, but the maximum distance from the equilibrium point is always bounded.

An asymptotically stable equilibrium point is a solution to an autonomous differential equation (or system of equations) for which there exists some distance, ε, that if the initial conditions of the system are within ε of the equilibrium point, the system (values of the independent variable(s)) will converge to the equilibrium point as the system evolves (as t -> ∞). Asymptotically stable equilibrium points "attract" the state of state of systems that start our sufficiently close to the equilibrium point.

For your differential equation, you should be able to see by inspection that all solutions are of the form, x(t) = constant. Furthermore, all the solutions are equilibrium points (because the definition of an equilibrium point is dx/dt = 0, and this is exactly the differential equation in question.)

None of the solutions, including x = 0, are asymptotically stable, though, because if the initial condition is x(0) = 0 + ε (i.e., the system starts off some small distance from x = 0), the system simply "sits" at x = ε for all time; it does not evolve toward x = 0 as t -> ∞. A similar argument can be made for every solution to this differential equation, so every solution is a stable equilibrium, but none of them are asymptotically stable.