# Finding fixed points of a system of difference equations

Can anyone help me in finding the fixed points of the systems satisfying all the parameters are real, positive?

$$x_{n+1}=x_{n-d}e^{(r(1-\frac{x_{n-d}}{k})-\frac{\beta y_{n-d}}{(x_{n-d}+\gamma)}+x_{n-d}z_{n-d})}$$

$$y_{n+1}=y_{n-d}e^{(1-\frac{ay_{n-d}}{(bx_{n-d}+c)}-e+y_{n-d}z_{n-d})}$$

$$z_{n+1}=z_{n-d}e^{(-s+\frac{x_{n-d}}{y_{n-d}})}$$

Solve[x Exp[r (1 - x/k) - (\[Beta] y/(x + \[Gamma])) + x z] == x && y Exp[1 - (a y/(b x + c)) - e + y z] == y && z Exp[-s (x + y)] == z && r > 0 && k > 0 && \[Beta] > 0 && \[Gamma] > 0 && a > 0 && b > 0 && c > 0 && e > 0 && s > 0 && x > 0 && y > 0 && z > 0, {x, y, z}, Reals]


Output shows no solution exists. But I could see solution satisfying the conditions. Did I make any mistake?

• To solve a recurrence, you would use RSolve. However, I am not sure if this is possible with a symbolic delay d  . Solve treats e.g. x[n]as a single symbol. Mar 20 at 10:32
• The equations for $z$ in your Solve expression does not match your third recursion equation. Mar 20 at 12:39

It's easier to look for the zeros of the exponentials:

Solve[r (1 - x/k) - β y/(x + γ) + x z == 0 &&
1 - a y/(b x + c) - e + y z == 0 &&
-s + x/y == 0, {x, y, z}, Cubics -> False] // FullSimplify

(*    {{x -> Root[..., 1],
y -> 1/s Root[..., 1],
z -> ...},
{x -> Root[..., 2],
y ->  1/s Root[..., 2],
z -> ...},
{x -> Root[..., 3],
y -> 1/s Root[..., 3],
z -> ...}}             *)


If you don't want to work with Root objects, you can convert them to regular radicals with ToRadicals.

You'll still have to find out which one of these three solutions is the stable fixed-point of your recursion.