lajos/Global attractivity

Global attractivity of the positive fixed point of a non-linear second order recursion

For any \(x>0\) and \(y>0\) let us define

\[ f(x,y):=\frac{x+y}{1+x+y} \]

and for any arbitrary \(\tau_1,\tau_2\in(0,+\infty)\) starting values we consider the second order non-linear difference equation

\[ \tau_n := f(\tau_{n-1},\tau_{n-2}),\quad n\ge 2. \]

Claim. \[\lim_{n\to +\infty} \tau_n = \frac{1}{2}.\]

Proof. For any \(x>0\) and \(y>0\) we have \(0<f(x,y)<1\), so \(\tau_n\in (0,1)\) if \(n\ge 3\). Moreover, for any \(x>0\) and \(y>0\), \(\partial_1 f(x,y)>0\) and \(\partial_2 f(x,y)>0\), so \(f\) is increasing in both variables. Thus, by omitting the first 2 terms of the \(\tau\)-sequence, we see that Theorem 1.6 of campus.mst.edu/ijde/contents/v3n1p1.pdf (with a 5-line, easy proof!) applies with \(I=(0,1)\). As a consequence, the even and odd subsequences of \(\tau_n\) are eventually monotonic, but since \(\tau_n\) is bounded, they both have a finite limit, clearly in \([0,1]\). Let \(\widetilde{\tau}_n\) denote any of these two subsequences. Since for any \(L\in[0,1]\), \(L=f(L,L)\) holds only if \(L=0\) or \(L=\frac{1}{2}\), the proof will be finished as soon as we have shown that \(\widetilde{\tau}_n\) cannot converge to \(0\).

Clearly, if \(\widetilde{\tau}_n\to 0\), then, by the above, the convergence can only be eventually monotone decreasing. So suppose that for some index \(N\) we have \[\frac{1}{2}>\widetilde{\tau}_N\ge \widetilde{\tau}_{N+1}\ge \widetilde{\tau}_{N+2}>0.\quad\quad (*)\] Then returning to the original sequence, there exists an index \(M\) such that \(\widetilde{\tau}_N=\tau_M\). To simplify the rest of the description, let us use the abbreviations \(x:=\tau_M\) and \(y:=\tau_{M+1}\).

Then (*) is equivalent to

\[ \frac{1}{2}>x\ge f(y,x)\ge f(f(f(y,x),y),f(y,x))>0 \]
with \(0<y<1\). But the above polynomial inequality does not have a solution, so the proof is complete.

For the sake of completeness, we give a proof of the only non-trivial, last statement. It is easily seen that \[\frac{1}{2}>f(y,x), \quad 0<y<1, \quad 0<x<\frac{1}{2} \] is equivalent to \((x,y)\in D\), where \[ D:=\{(\xi,\eta)\in\mathbb{R}^2\ :\ \left(0<\eta\le\frac{1}{2}\wedge 0<\xi<\frac{1}{2}\right) \vee \left(\frac{1}{2}<\eta<1\wedge 0<\xi<1-\eta\right)\}, \] further, \(f(y,x)\ge f(f(f(y,x),y),f(y,x))\) is just \(0\le P(x,y)\) with \[ P(x,y):= x^3 y+2 x^3+3 x^2 y^2+6 x^2 y+3 x y^3+6 x y^2-2 x y-x+y^4+2 y^3-2 y^2-2 y. \] We show that \(P\Big|_D<0\).

Since \(-5 y^3-30 y^2-6 y+95\), i.e., the resultant of \(\partial_1 P\) and \(\partial_2 P\) w.r.t. \(x\), does not have a root in \((0,1)\), \(P\) has no critical point in the interior of \(D\). So \(P\) attains its maximum on \(\partial D\), the boundary of \(D\). But \(\partial D\) is the union of 4 line segments, say, \(\ell_k\), \(k=1,2,3,4\). We check that the univariate polynomials satisfy \(P\Big|_{\ell_k}\le 0\), with equality only at certain (corner) points of \(\partial D\setminus D\).

We remark that this article also reduces global stability to deciding whether a multivariate polynomial is positive on a domain.