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I need to find a Liapunov function to determine the stability of a point. Given the system $$\frac{du}{dt} = v, \ \frac{dv}{dt} = -u - u^4$$ Find a suitable Liapunov function to determine the stability of the point $(0,0)$.

Orginally, I was thinking of a Liapunov function $$V = au^2 + buv + cv^2.\ a,b,c \in \mathbb{R}$$ However, filling this in gives $$\dot{V} = V_u \frac{du}{dt} + V_v \frac{dv}{dt}$$ $$\dot{V} = (2au + bv)\cdot v + (bu + 2cv)\cdot(-u - u^4)$$ $$\dot{V} = a\cdot(2uv) + b \cdot (v^2-u^2-u^5)+c\cdot(-2uv-2uv^4)$$ And I cannot find any constant $a,b,c$ such that either $\dot{V} \leq 0$, $\dot{V} < 0$, $\dot{V} > 0$ holds, but I am unable to.

Can anyone help me out? Thanks in advance!

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Your equation is a particular case of an Hamiltonian system so for example the Hamiltonian: $$H(u,v)=v^2+\frac{1}{2} u^2+\frac{1}{5} u^5$$ is such that: $$ \frac{d }{dt}H(u(t),v(t)) =0$$.

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  • $\begingroup$ Hmm okay good to know. However, reading Wikipedia did not guide me to a Liapunov function. Can you help me a little further? Thanks in advance! $\endgroup$ – MathMan12345 Apr 9 '18 at 16:11
  • $\begingroup$ Why isn't $H$ a good Liapunov function ? $\endgroup$ – Delta-u Apr 9 '18 at 16:16
  • $\begingroup$ A I think $V = \frac{1}{2}v^2 + \frac{1}{2}u^2 + \frac{1}{5}u^5$ is a Lyapunov function, resulting in $\dot{V} = 0$, thus $(0,0)$ is asymptotically stable, right? Thanks! $\endgroup$ – MathMan12345 Apr 9 '18 at 16:21
  • $\begingroup$ No, $(0, 0)$ is NOT asymptotically stable (in Hamiltonian systems there are no asymptotically stable equilibria). $\endgroup$ – user539887 Apr 21 '18 at 21:29

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