Plotting a phase plane portrait of an oscillatory system?

Question

I am trying to plot a phase plane portrait of the following system:

dQ/dt = v*(z - Q)*(w^2 + Q^2)/(1 + Q^2) - Q

dz/dt = k - Q

where v = 1, k = 0.2, and w = 0.05.

So far, I have the following:

v = 1;k = 0.2;w = 0.05
QNullcline = Solve[{0 == v*(z - Q)*(w^2 + Q^2)/(1 + Q^2) - Q}, Q];
zNullcline = Solve[0 == k - Q, Q];



P1 = Plot[QNullcline[[1]][[1]][[2]], {z, 0, 10}, PlotStyle -> Blue];

P2 = Plot[QNullcline[[2]][[1]][[2]], {z, 0, 10}, PlotStyle -> Purple];

P3 = Plot[QNullcline[[3]][[1]][[2]], {z, 0, 10}, PlotStyle -> Green];

P4 = Plot[QNullcline[[1]][[1]][[2]], {z, 0, 10}, PlotStyle -> Red];

P5 = StreamPlot[{v*(z - Q)*(w^2 + Q^2)/(1 + Q^2) - Q, 
   k - Q}, {Q, 0, 10}, {z, 0, 10}];

Show[P1,P2,P3,P4,P5]

I have two questions:

1.) Did I plot the null clines properly or are they supposed to better match the stream plot?

2.) Is there a way to show that the model is oscillatory in the phase plane diagram?

Any help would be greatly appreciated!

Thank you!

Answer 1

update

answer #1:

(* examplary solution to verify streamplot*)
z0 = 10; Q0 = 8;
T = 50;
zQ = NDSolveValue[{Q'[t] == 
v*(z[t] - Q[t])*(w^2 + Q[t]^2)/(1 + Q[t]^2) - Q[t], 
z'[t] == k - Q[t], Q[0] == Q0, z[0] == z0}, {z, Q }, {t, 0, 50}]

Perhaps it's easier using ContourPlot to show the nullclines.

Try

Show[{StreamPlot[{k - Q, v*(z - Q)*(w^2 + Q^2)/(1 + Q^2) - Q}, {z, 0, 
    10}, {Q, -2, 5}, FrameLabel -> {z, Q }]
  , ParametricPlot[Through[zQ[t]], {t, 0, T}, 
   PlotRange -> {{0, 10}, {0, 10}}, PlotStyle -> Red]
   , ContourPlot[{0 == v*(z - Q)*(w^2 + Q^2)/(1 + Q^2) - Q, 
    0 == k - Q}, {z, 0, 10}, {Q, -2, 5}]}]

answer #2:

Phase plane should show closed curves around stable equilibrium points if the model is oscillatory.