0

I have two equations $\frac{dq}{dt}$= $Rq + K - 1$ and $\frac{dK}{dt}$=$ Nq + N$. Here, N and R are just constants so I was ignoring them and just assigning a certain arbitrary value. Solving nullclines for this gives me $q = (1-K)/r$ and $q=-1$. So both of them are going to intersect at the fourth quadrant.

My trouble right now is sketching the phase portrait for this. Are the two eigenvectors supposed to be the same as the two nullclines? I always thought eigenvectors are nullclines were more or less different ideas but is the point of intersection of the two nullcines the stable equilibrium path for this solution?

S.M.T
  • 742
GreenArrow
  • 1
  • You can write your system as $$\begin{pmatrix}q\ K\end{pmatrix}'=\begin{pmatrix}R&1\ N&0\end{pmatrix}\begin{pmatrix}q\ K\end{pmatrix}+\begin{pmatrix}-1\ N\end{pmatrix}$$ This system isn't homogeneous. Employing the eigenvector method would require that you enforce a substitution like $$\begin{pmatrix}x\ y\end{pmatrix}=\begin{pmatrix}q+1\ K-R-1\end{pmatrix}$$ This will transform your system into one thats homogeneous, then you can apply classic eigenvector method. – Matthew H. Jan 01 '22 at 01:44
  • so, will one of my eigenvectors be the nullcline itself? I think not but can that ever happen? – GreenArrow Jan 01 '22 at 23:05
  • One of your eigenvectors? Of what matrix? Generally speaking, if you have a homogeneous system $x'=Ax$ a nullcline need not be the span of some eigenvector of $A$, but it could be. Consider $x'=x,y'=y-x$. – Matthew H. Jan 01 '22 at 23:20

0 Answers0