Invariant curves for a delay differential equation with a piecewise constant argument

(i) If \(-2< a<0\), then a continuous solutions ϕ of (4) with \(b=0\) is either of the piecewise linear form that \(f(x):= r_{i}x\) for \(x>0\), or \(:=0\) for \(x=0\), or \(:=r_{j}x\) for \(x<0\), where \(i,j=1, 2\), or given by where \(x_{n}=\frac{r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1}+r_{2}x_{0})\), \(n \in\mathbb{Z}\), with an arbitrarily chosen \(x_{0}\in(-\infty, +\infty)\) and \(x_{1} \in[r_{1}x_{0}, r_{2}x_{0}]\), and \(f_{n}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{n-1}^{-1}(x)\) for all \(x\in[x_{n}, x_{n +1})\), \(n=1, 2, \ldots \) , \(f_{-n-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}}f^{-1}_{-n}(x)\) for all \(x\in[x_{-n}, x_{-n+1})\), \(n=1, 2, \ldots\) , and \(f_{-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}}f_{0}(x)\), \(x\in[x_{0}, x_{1})\), with the arbitrarily chosen functions \(f_{0}\) such that \(f_{0}(x_{0})=x_{1}\), \(f_{0}(x_{1})=x_{2}\), and \(r_{1}\leq\frac{f_{0}(x)-f_{0}(y)}{x-y}\leq r_{2}\) for all \(x, y \in [x_{0}, x_{1})\). (ii) If \(a=0\), then (4) with \(b=0\) has a unique continuous solution f and \(f(x)=x+\beta\), where \(\beta\in\mathbb{R}\) is an arbitrary constant.

(i) If \(a<-2\), then (4) with \(b=0\) only has two continuous solutions f and \(f(x)=r_{1}x\) or \(r_{2}x\). (ii) If \(a=16\), (4) with \(b=0\) just has a continuous solution \(f(x)=-3x\). (iii) If \(a>16\), all continuous solutions f of (4) with \(b=0\) are given by where the sequence \(\{x_{n}\}\) is defined by \(x_{n}=\frac {r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1}+r_{2}x_{0})\), \(n\in\mathbb{Z}\), with an arbitrarily chosen \(x_{0}\in(0, +\infty)\) and \(x_{1} \in[r_{1}x_{0}, r_{2}x_{0}]\), and \(f_{2n-1}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n-2}^{-1}(x)\), \(x\in [x_{-2n+5},x_{-2n+3})\), \(n=1, 2, \ldots \) , \(f_{2n}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n-1}^{-1}(x)\), \(x\in [x_{-2n},x_{-2n+2})\), \(n=1, 2, \ldots \) , \(f_{-2n}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac{1}{r_{1}r_{2}} f^{-1}_{-2n+1}(x)\), \(x \in[x_{2n+3}, x_{2n+1})\), \(n=1, 2,\ldots\) , \(f_{-2n-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}} f^{-1}_{-2n}(x)\), \(x \in[x_{2n}, x_{2n+2})\), \(n=1, 2,\ldots\) , and \(f_{-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac{1}{r_{1}r_{2}} f_{0}(x)\), \(x\in[x_{0}, x_{2})\), with an arbitrarily chosen continuous function \(f_{0}\) on \([x_{0},x_{2})\) such that \(f_{0}(x_{0})=x_{1}\), \(f_{0}(x_{2})=x_{3}\), and \(r_{1}\leq\frac {f_{0}(x)-f_{0}(y)}{x-y}\leq r_{2}\), \(\forall x, y \in[x_{0}, x_{2})\).

Firstly, we consider (i). By Theorem 5 in [14], (4) with \(b=0\) only has two continuous solutions f and \(f(x)=r_{1}x\) or \(r_{2}x\), where the characteristic roots \(r_{1}\), \(r_{2}\) satisfy \(r_{1}<0<r_{2}\neq1\) and \(r_{1}\ne-r_{2}\) as shown in (C3). Next, we consider (ii). By Theorem 6 in [14], (4) with \(b=0\) just has a continuous solution \(f(x)=-3x\), where the characteristic roots \(r_{1}\), \(r_{2}\) satisfy \(r_{1}=r_{2}=-3\) as shown in (C4). Finally, we consider (iii). In order to piecewise construct all solutions of (4) with \(b=0\) we need a partition for the interval \((-\infty, \infty)\). For this purpose we consider a homogeneous linear difference equation which has the same coefficients as (4) with \(b=0\) correspondingly. Its characteristic equation is which has two characteristic roots \(r_{1}\) and \(r_{2}\) satisfying \(r_{1}< r_{2}<-1\) as shown in (C5). Thus, (4) and (6) can be, respectively, rewritten as If f is a solution of (4) with \(b=0\), we easily see that f is invertible. In fact, if \(f(x_{1})=f(x_{2})\), then \(f(f(x_{1}))=f(f(x_{2}))\). Thus, \(x_{1} = x_{2}\) by (4) because \(a\neq-2\), which implies that f is one to one. Next we only need to show that \(f(x)\rightarrow-\infty\) as \(x\rightarrow+\infty\) and \(f(x)\rightarrow+\infty\) as \(x \rightarrow-\infty\) because \(f(x)\rightarrow\pm\infty\) as \(x \rightarrow\pm\infty\), then the left-hand side of (4) with \(b=0\) tends to ±∞ by \(a>16\), but the right-hand side is equal to 0. Otherwise, \(f(x)\) has a finite limit as \(x \rightarrow\infty\), then \(f(f(x))-(2-\frac{a}{2})f(x)\) converges to a finite limit by the continuity of f on the whole of ℝ, but \((1+\frac{a}{2})x\) does not, which contradicts the requirement that \(f(f(x))-(2-\frac{a}{2})f(x)=-(1+\frac{a}{2})x\). Thus, we rewrite (4) in the following equivalent form: which is called the dual equation to (4) with \(b=0\). Solving the homogeneous linear difference (9) with arbitrarily chosen real initial values \(x_{0}\) and \(x_{1}\), we obtain Let \(x_{0}=x\) and \(x_{n+1}=f(x_{n})\) in (11), we have Furthermore, we can obtain where \(\Delta f^{n}(x, y)= \frac{f^{n}(x)-f^{n}(y)}{x - y}\) for any \(x\neq y\) and \(n\in\mathbb{Z}\). From (12) we can see that Since f is strictly monotonic, \(\Delta f^{n}(x, y)>0\) for even n, which implies \(\Delta f(x, y)-r_{1}\geq0\) and \(-\Delta f(x, y)+r_{2}\geq0\), that is, Moreover, we can see that \(f(0)=0\) from (13). In what follows, we arbitrarily choose \(x_{0}\in(0, +\infty)\) and \(x_{1}\in[r_{1}x_{0}, r_{2}x_{0}]\) and define a sequence \(\{x_{n}\}\), \(n\in\mathbb{Z}\), by (11). The sequences \(\{x_{2n}\}\), \(\{x_{2n+1}\}\), \(\{x_{-2n}\}\) and \(\{x_{-2n+1}\}\), where \(n=0, 1, 2,\ldots\) , are strictly monotone such that \(x_{2n}\rightarrow+\infty\), \(x_{2n+1}\rightarrow-\infty\), \(x_{-2n} \rightarrow0\), and \(x_{-2n+1} \rightarrow0\) as \(n \rightarrow\infty\). Thus, the sequence \(\{x_{n}\}\), \(n\in\mathbb{Z}\), is a partition of the interval \((-\infty, \infty)\). Next we arbitrarily choose a continuous function defined in the interval \([x_{0}, x_{2})\), satisfying \(f_{0}(x_{0})=x_{1}\), \(f_{0}(x_{2})=x_{3}\), and condition (14). We can recursively define the homeomorphisms \(f_{2n-1}: [x_{-2n+5}, x_{-2n+3})\rightarrow [x_{-2n}, x_{-2n+2})\), \(n=1, 2,\ldots\) , and \(f_{2n}: [x_{-2n}, x_{-2n+2})\rightarrow [x_{-2n+3}, x_{-2n+1})\), \(n=1, 2,\ldots\) , such that In fact, for \(f_{2n}\) defined satisfying (16) and (18), we let Obviously, \(f_{2n+1}(x_{-2n+3})=x_{-2n}\) and \(f_{2n+1}(x_{-2n+1})=x_{-2n-2}\). Making use of (18), we have \(\frac{1}{r_{2}} \leq\frac{f^{-1}_{2n}(x) - f^{-1}_{2n}(y)}{x-y} \leq\frac{1}{r_{1}}\) for \(x, y \in[x_{-2n}, x_{-2n+2})\). It is easy to deduce that Furthermore, we again let By the same argument we can see that By induction both \(f_{2n-1}\) and \(f_{2n}\) are well defined. Similarly, we can also recursively define the homeomorphisms \(f_{-2n+1}: [x_{2n-2}, x_{2n}) \rightarrow [x_{2n+3},x_{2n+1})\), \(n = 1, 2, \ldots\) , and \(f_{-2n}: [x_{2n+3}, x_{2n+1}) \rightarrow [x_{2n},x_{2n+2})\), \(n = 1, 2, \ldots\) . By the properties of the dual (10) we can obtain Therefore, Thus, we can define f is continuous on ℝ because \(f_{2n}(x_{-2n})=x_{-2n+1}=f_{2n+2}(x_{-2n})\), \(f_{2n+1}(x_{-2n+1})=x_{-2n-2}=f_{2n+3}(x_{-2n+1})\), where \(n=0, 1, 2, \ldots \) , \(f_{1}^{-1}(x_{0}) = f_{-1}(x_{0})\), \(f_{-2n}^{-1}(x_{2n+2})=x_{2n+3}=f_{-2n-2}^{-1}(x_{2n+2})\), and \(f_{-2n+1}^{-1}(x_{2n+3})=x_{2n+2}=f_{-2n-1}^{-1}(x_{2n+3})\), where \(n=1, 2, 3, \ldots\) . we can easily check that f defined in Theorem 2.2 satisfies (4) with \(b=0\) in ℝ. In fact, if \(x\in[x_{-2n},x_{-2n+2})\), \(n=0, 1, 2, \ldots \) , \(f^{2}(x)=f_{2n+1}(f_{2n}(x))=(r_{1}+r_{2})f_{2n}(x)-r_{1}r_{2}x=(r_{1}+r_{2})f(x)-r_{1}r_{2}x\), i.e., \(f^{2}(x)-(r_{1}+r_{2})f(x)-r_{1}r_{2}x=0\). Similarly, we can also check that f satisfies (4) with \(b=0\) for \(x\in[x_{-2n+3},x_{-2n+1})\), \(x\in[x_{2n+3}, x_{2n+1})\), \(x\in[x_{2n}, x_{2n+2})\) and \(x=0\), where \(n=1, 2, 3, \ldots\) . The proof is completed. □

In the case that \(b\ne0\), as indicated in [16] for (2) therein, (4) can be reduced equivalently to the equation the same type of equation as the one considered in Theorems 2.1 and 2.2 with vanishing b, by the replacement \(\tilde{f}(x)=f(x+\xi)-\xi\), where \(\xi=\frac{-b}{(1-r_{1})(1-r_{2})}\), if its characteristic roots \(r_{1}\), \(r_{2}\) are both real but neither of them is equal to 1. In this case solutions can be found from Theorems 2.1 and 2.2. So (4) with \(b\neq0\) can be reduced to (21) except for the case \(a=0\). For the case of \(a=0\) and \(b\neq0\), (4) has no real continuous solutions. In fact, by induction and (4) we can obtain \(f^{n}(x)=nf(x)-(n-1)x-\frac{n(n+1)}{2}b\), \(n\in\mathbb{Z}\). Furthermore, we have \(f^{n+1}(x)-f^{n}(x)=f(x)-x-(n+1)b\), \(n\in\mathbb{Z}\). For an arbitrary \(x\in\mathbb{R}\), \(f^{n+1}(x)-f^{n}(x)\) has the same sign when n takes the values N and −N, where N is a large positive integer, because f is strictly monotonic. But \(f(x)-x-(n+1)b\) has not, which contradicts the requirement \(f^{n+1}(x)-f^{n}(x)=f(x)-x-(n+1)b\), \(n\in\mathbb{Z}\).