Is this way correct to plot system of difference equation. For example
StreamPlot[{(.2 (x^2) - x*y - .01 (y^2)), -.1 (y^2) + .9 x*
y + .02 (x^2)}, {x, -5, 5}, {y, -5, 5}, Axes -> True,
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1 Answer
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As far as I know, stream plot is for continuous time system (system of first order ode's) and not for difference equations (discrete system).
The system you posted is actually 2 ODE's and not difference equations. To show this, here is what you have
Clear["Global`*"];
StreamPlot[{(2 (x^2) - x*y - 1/100 (y^2)), -1/10 (y^2) + 9/10 x*y + 2/100 (x^2)},
{x, -.1, 2}, {y, -.1, 3},
Axes -> True,
StreamPoints -> {{{{1, 2}, Red}, Automatic}}]
The red trajectory above, is specific solution which passes through initial conditions $x(0)=1,y(0)=2$. To show this, here are the two ODE's, solved using NDSolve with initial conditions $x(0)=1,y(0)=2$ and the phase plot is given using ParametricPlot, which gives the same exact red solution curve above.
ode1 = x'[t] == (2 (x[t]^2) - x[t]*y[t] - 1/100 (y[t]^2));
ode2 = y'[t] == -1/10 (y[t]^2) + 9/10 x[t]*y[t] + 2/100 (x[t]^2);
{solX, solY} = NDSolveValue[{ode1, ode2, x[0] == 1, y[0] == 2}, {x, y}, {t, -10, 10}]
ParametricPlot[{solX[t], solY[t]}, {t, -10, 10},
PlotStyle -> Red, GridLines -> Automatic, GridLinesStyle -> LightGray]
Which gives the same solution from StreamPlot. This shows what you plotted is phase plot for 2 ODE's and not for difference equations.
It might be possible to do phase plot for difference equations ofcourse, but I do not think you can use StreamPlot for that. I do not know if Mathematica has special command for that, you might have to do it "manually". i.e. solve the two coupled difference equations, and do similar thing as ParametricPlot but for the discrete solutions.
