4
$\begingroup$

If I have a system of difference equations

$$ \begin{cases} x(n + 1) = x(n) + x(n) y(n) + y(n)^2 \\ y(n + 1) = y(n) - x(n)^2 + y(n)^2 \end{cases}, \text{ with } \begin{cases} x(0) = 1\\ y(0) = 2 \end{cases}. $$

How can I solve it by Mathematica?

I tried this but I do not know if it is correct

RSolve[{x[n + 1] == x(n) + x(n) y(n) + y(n)^2,
        y[n + 1] == y(n) - x(n)^2 + y(n)^2},
       {x[n], y[n]}, n
      ]

Thank you for help!

$\endgroup$
2
  • $\begingroup$ I think you need to use [] and not (). But when I tried it, it could not solve it. eq1 = x[n + 1] == x[n] + x[n]*y[n] + y[n]^2; eq2 = y[n + 1] == y[n] - x[n]^2 + y[n]^2; ic = {x[0] == 1, y[0] == 2}; RSolve[{eq1, eq2, ic}, {x[n], y[n]}, n] because it is non-linear, it returned unevaluated. $\endgroup$
    – Nasser
    Commented Jun 27, 2020 at 4:37
  • $\begingroup$ Dear Nasser. you are right for linear.. but my broblem with not linear. Thank you for the reply ... Saud $\endgroup$
    – S A
    Commented Jun 27, 2020 at 5:13

1 Answer 1

5
$\begingroup$

Mathematica can not solve it since they are non-linear. The correct syntax is

eq1 = x[n + 1] == x[n] + x[n]*y[n] + y[n]^2; 
eq2 = y[n + 1] == y[n] - x[n]^2 + y[n]^2; 
ic  = {x[0] == 1, y[0] == 2}; 
sol = RSolve[{eq1, eq2, ic}, {x[n], y[n]}, n]

Another possibility is to try RecurrenceTable. Which gives you numerical value of x[n] and y[n] for as n increases. This shows your equations are not stable.

tbl = RecurrenceTable[{eq1, eq2, ic}, {x[n], y[n]}, {n, 0, 5}];

which gives

{{1., 2.}, {7., 5.}, {67., -19.}, {-845., -4147.}, 
 {2.0701*10^7, 1.64794*10^7}, {6.12712*10^14, -1.56959*10^14}}

You can see the solutions blow up very quickly. After only 4 iterations.

Graphics[Line[tbl], Axes -> True, AxesLabel -> {"x", "y"}, BaseStyle -> 12]

Mathematica graphics

So you might want to look at how you generated these equations. May be your model is wrong somewhere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.