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My lecturer gave us the definition of a strong Lyapunov function. She then said that if $V$ is positive definite but $dV/dt$ is also positive definite (instead of negative definite) in a region containing an equilibrium point then the equilibrium point is unstable. If $dV/dt$ is positive definite can we conclude straightaway that the equilibrium point is unstable or do we also require $V$ being positive definite?

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No, function $V$ cannot be arbitrary. Only $dV/dt\geq 0$ is not sufficient to prove that it the equilibrium is unstable. For example let the stable simple linear system $$\dot{x}=-x$$ with $$V=-x^2$$ then $$\frac{dV}{dt}=-x\dot{x}=x^2$$ which is positive definite.

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  • $\begingroup$ Sorry, I think you've misunderstood the question. I was asking if we can find a function V such that $dV/dt$ is positive definite in a region containing an equilibrium point then is it true that the equilibrium point is unstable? $\endgroup$ – user152440 May 6 '15 at 18:04
  • $\begingroup$ @user152440 Well this is equivalent. I have edited my answer. $\endgroup$ – RTJ May 6 '15 at 19:58
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No you do not require $V$ to be positive definite. Read the instability theorem by Chetaev

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