# Plot system difference equations

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,


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.