A student asked me:"I was looking at the past exam paper from 2010, and in question 03. (a) (iii) when drawing the phase plane, was wondering how we are supposed to conclude that it is a periodic orbit? I'm not completely sure how to proceed after finding the nullclines and the direction of the solution curve."

In the 2010 exam question 3(a)(iii), we have a Lyapunov function \(V\) with \(\frac{d}{dt}V=0\). This means that the level sets \(V_C:=\{x~|V(x)=C\}\) are flow-invariant and thus that every solution curve must lie inside one particular level set. As the level sets \(V_C\) with \(C>0\) in this example are closed curves that do not contain equilibria, necessarily these level sets must be periodic solutions. The level set \(V_0\) is the unique equilibrium \((x,y)=(1,1)\).

A similar situation was encountered in Test 2, question 2(ii).

In the 2010 exam question 3(a)(iii), we have a Lyapunov function \(V\) with \(\frac{d}{dt}V=0\). This means that the level sets \(V_C:=\{x~|V(x)=C\}\) are flow-invariant and thus that every solution curve must lie inside one particular level set. As the level sets \(V_C\) with \(C>0\) in this example are closed curves that do not contain equilibria, necessarily these level sets must be periodic solutions. The level set \(V_0\) is the unique equilibrium \((x,y)=(1,1)\).

A similar situation was encountered in Test 2, question 2(ii).

I would really thank you for the work that you have done and do on your blog, it has been a tremendous source of help to me and pretty sure many other people.

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