Recurrence relations with fractions

Question

I have the following two equations:

$$\alpha(t)=\frac{a}{b+\beta(t-1)}\\ \beta(t)=\frac{c}{d+\alpha(t-1)}$$ where $a,b,c,d$ are constants.

Question is, is there an analytical form for $\alpha$ as $t\to\infty$.

Furthermore, what are the conditions on $a,b,c,d$ for convergence?

Answer 1

The behaviour is easy: $\alpha(t)$ and $\beta(t)$ are both constant (more properly, this is true if you require them to be continuous; without this requirement, each $\alpha(t)$ and $\beta(t)$ can take two different values and convergence can fail).

If you substitute the second equation into the first one, you get $$ \alpha(t)=\frac{a}{b+\frac{c}{d+\alpha(t)}}=\frac{ad+a\alpha(t)}{bd+b\alpha(t)+c}. $$ This is a degree two equation on $\alpha(t)$, namely $$ b\alpha(t)^2+(bd+c-a)\alpha(t)-ad=0. $$

Answer 2

I don't see why this would be considered a recurrence relation.

You have $$ \alpha(t) = \frac{a}{b+\beta(t)} = \cfrac{a}{b+\cfrac{c}{d+\alpha(t)}}. $$ Multiplying both the numerator and the denominator of that last fraction by $d+\alpha(t)$, we get $$ \alpha(t) = \frac{a(d+\alpha(t))}{b(d+\alpha(t)) + c} = \frac{ad + a\alpha(t)}{(bd+c)+b\alpha(t)}. $$ Now multply both sides of the equation by the denominator of the fraction on the right, getting: $$ b(\alpha(t))^2 + (bd+c)\alpha(t) = ad+a\alpha(t). $$ Now it's just a quadratic equation.

And the solution does not indicate any dependence on $t$. Thus any function with values in the set of two solutions of the quadratic equation satifies the equation. It approaches a limit as $t\to\infty$ precisely if it's eventually constant. Otherwise its liminf and limsup are the two solutions of the quadratic equation.

Answer 3

This addresses the revised version of the question.

Using the expression of $\beta(t-1)$ as a function of $\alpha(t-2)$ in the expression of $\alpha(t)$ as a function of $\beta(t-1)$, one gets $\alpha(t)$ as a homographic function of $\alpha(t-2)$. Likewise for $\beta(t)$ as a homographic function of $\beta(t-2)$.

To study recursions of the form $x_{n+1}=\dfrac{ux_n+v}{x_n+w}$ with $v\ne wu$, one looks for a fixed point $z$ of the recursion. When $z$ exists, $z\ne u$ and $x_{n+1}-z=(u-z)\dfrac{x_n-z}{x_n+w}$ hence $$ \dfrac1{x_{n+1}-z}=\dfrac1{u-z}+\dfrac{w+z}{u-z}\dfrac1{x_n-z}. $$ This is an affine recursion of the form $y_{n+1}=Ay_n+B$. To iterate it, if $A=1$ one notes that $y_n=y_0+nB$ and if $A\ne1$ one looks once again for a fixed point, here $s=B/(1-A)$, hence $y_{n+1}-s=A(y_n-s)$ and $y_n=A^n(y_0-s)+s$.