In this section, we give some examples and their gures of iterations which agree with our main theorems mentioned in section 2. Moreover, these examples make our theorems applicable. From now on, we show their results by running 1500 iterations of F (see (6) and (7)) numerically and printing the points of the last 1000 times in the xy-plane.
Finally, they are arranged into three subsections. Notice that case 1 and 2 mentioned in the subsections mean the hypothesis of u of (1), i.e., u (t; x (t)) = f (x (t)) and u (t; x (t)) = f (x (t 1)) respectively.
9.1 Examples for theorems 2.2 and 2.3
First, consider the dynamic of an example for theorem 2.2. Let a = 1, b = 0:1, p = 10 and q = 0. De ne f (x) = sin x. It's clear that f is Lipschitz on R and the Lipschitz constant M = 1. We check whether a; b; p and q satisfy the hypothesis of theorem 2.2.
max ab; ap1 + max abq2 +bMa2 ;1a b apq + M
ja2pj = 0:3 < 1. So the hypothesis is satis ed. Set the initial point (x0; y0) = (2000; 1000) and let F iterate with it. Figure 1(a) exhibits that (0; 0) is the global attractor. One can see that the rst and second components of F both converge to 0 quickly as the iteration increases in gure 1(b) and 1(c) (notice that variable n represents the times of iteration).
Figure 1: the dynamic diagram and the iterate diagrams of two components of the example for theorem 2.2
Next, consider the dynamic of another example for theorem 2.3. De ne f (x) = jxj and let a = 2, b = 0:1, p = 10 and q = 0. Then M = 1 and max ab; ap1 + max abq2 +Ma;1a b apq = 0:55 < 1, which satis es the hypothesis of theorem 2.3. We also set the initial point (x0; y0) = (2000; 1000). Figure 2(a) exhibits (0; 0) is the global attractor and gure 2(b),(c) exhibit the situation that two components of F converge.
Figure 2: the dynamic diagram and the iterate diagrams of two components of the example for theorem 2.3
Therefore, the two examples verify the validity and the practicability of theorems 2.2 and 2.3.
9.2 Examples for theorems 2.6, 2.7, 2.8 and 2.9
First, we produce an example for theorem 2.6. Choose a set of special values of a; b; p; q and f as a = 1, b = 0:1, p = 0:01, q = 0 and f (x) = 0:95 sin x. Now we check whether such values and f can satisfy the hypotheses of the theorem. Clearly, 0:95 sin x has countably many simple zeros on R. We choose I = 2 ;32 and V =;. Then qx + f (x) = f (x) is C1 on 2 ;32 and has two simple zeros 0; in int 2 ;32 = 2 ;32 . The result p 2 (0; ) means that p approaches to 0 very closely. Since the system (6) is unde ned when p = 0, we just consider the dynamic for p = 0:01. Observe the dynamic in gure 3 for theorem 2.6 with an initial point (x0; y0) = (0:01; 0:02). We see that there is an irregular graph in the xplane. We can also see the graphs of iteration of x- and y-components of (6) in gures 4 and 5.
Figure 3: the dynamic diagram the example for theorem 2.6
Figure 4: the graph of iteration of x-component of (6) with iterating times from 1 to 500 for theorem 2.6
Figure 5: the graph of iteration of y-component of (6) with iterating times from 1 to 500 for theorem 2.6
To observe whether the perturbation of parameter does work, gure 6 shows the bifurcation about p around p = 0:01. We choose the interval of variation of p as [0:001; 1]
and also x q = 0. With a di erent value of p, the system begins with a randomly-selected initial point. Afterward, print the second component of the 1000th to 1500th iterations.
Figure 6: the bifurcation diagram of 2nd component of the map (6) about p of the example for theorem 2.6
Next, for theorem 2.7, we let a = 6; b = 1; p = 0:01; q = 0 and de ne f (x) = x(1 x).
Then f is C1 on [ 0:1; 1:1] and has two simple zeros 0; 1 in int ([ 0:1; 1:1]) = ( 0:1; 1:1).
Figure 7 informs us that the system (7) also exhibits a messy diagram with an initial point (0:01; 0:02). We also show the graphs of iteration of two components of (7) in gures 8 and 9.
Figure 7: the dynamic diagram the example for theorem 2.7
Figure 8: the graph of iteration of x-component of (7) with iterating times from 1 to 500 for theorem 2.7
Figure 9: the graph of iteration of y-component of (7) with iterating times from 1 to 500 for theorem 2.7
Additionally, we also show the bifurcation diagrams of the second component about p and q individually as follows (see gure 10 for p and gure 11 for q with random initial points). Choose the intervals of variation of p and q as [0:009; 0:012] and [ 0:01; 0:01]
respectively.
Figure 10: the bifurcation diagram of 2nd component of the map (7) about p of the example for theorem 2.7 and q = 0 xed
Figure 11: the bifurcation diagram of 2nd component of the map (7) about q of the example for theorem 2.7 and p = 0:01 xed
For theorem 2.8, since a cannot be zero, we consider the case that a = 0:0025, b = 0:001, p = 1, q = 0 and f (x) = 1:1 10 5sin x. It's clear that f is analytic on [0; ]n (r1; r2) for 0 < r1 < r2 < and 11 sin r1 = 11 sin r2 = . The parametric map gu(x) = u sin x has period-doubling property (see gure 12 with u from 0 to 35). We have
q
b2px + b12pf (x) = 11 sin x 34:56 sin x and 1:1 sin ([0; ]n (r1; r2)) = [0; ]. Since
q
b2px +b12pf (x) has a 3-periodic point, whose period is not a power of 2, by theorem A in [7], we get that htop b2qpx + b21pf (x) > 0. Thus, the hypothesis 2 of theorem 2.8 is satis ed.
Figure 12: the bifurcation diagram of gu with u from 0 to 35
Next, its dynamic diagram with the initial point (0:001; 0:002) is shown in gure 13.
We can see that its shape curls and the points on the graph are not uniformly dense.
Figure 13: the dynamic diagram the example for theorem 2.8
The graphs of the two components of this example are shown in gures 14 and 15. It is observable that the situations of iterating of them act not so regularly.
Figure 14: the graph of iteration of x-component of (6) with iterating times from 1 to 500 for theorem 2.8
Figure 15: the graph of iteration of y-component of (6) with iterating times from 1 to 500 for theorem 2.8
For theorem 2.9, consider a = 5, b = 1, p = 0:01, q = 1:11 and f (x) = sin x. Then f is analytic on [0; ]n (r1; r2) for 0 < r1 < r2 < and 1:1 sin r1 = 1:1 sin r2 = . Moreover, since 1qf (x) = 1:1 sin x, 1qf ([0; ]n (r1; r2)) = [0; ] and htop 1
qf > 0. Similarly, we also show its dynamic diagram and the graphs of iteration of components with the initial point (0:001; 0:002) (see gures 16, 17 and 18).
Figure 16: the dynamic diagram the example for theorem 2.9
Figure 17: the graph of iteration of x-component of (7) with iterating times from 1 to 500 for theorem 2.9
Figure 18: the graph of iteration of y-component of (7) with iterating times from 1 to 500 for theorem 2.9
We set its bifurcation diagram about the parameter p in the interval of variation [0:009; 0:02] in gure 19. The system (7) has the similar dynamic for a = 5, b = 1, q = 1:11 and f (x) = sin x when p is around 0:01.
Figure 19: the bifurcation diagram of 2nd component of the map (7) about p of the example for theorem 2.9
Therefore, theorems 2.8 and 2.9 are applicable.
9.3 Examples for theorems 2.10 and 2.11
We give two examples for theorems 2.10 and 2.11 respectively in this subsection.
For theorem 2.10, consider the case a = 3:91 ; b = 0; p = 3:9; q = 0 and f (x) = x (1 x).
Then we see that f is continuous on R and a2qpy + 1af ap1y = 3:9y (1 y). It's known that 3:9y (1 y) has positive topological entropy (it has 3-periodic points).
Now we see the dynamic diagram in gure 20 and the graphs of iteration of two components in gures 21 and 22. Choose the initial point (x0; y0) = (0:01; 0:02).
Figure 20: the dynamic diagram the example for theorem 2.10
Figure 21: the graph of iteration of x-component of (6) with iterating times from 1 to 500 for theorem 2.10
Figure 22: the graph of iteration of y-component of (6) with iterating times from 1 to 500 for theorem 2.10
The shape of gure 20 is like a part of a parabola and its distribution is not uniform.
So it satis es some characteristics of chaotic graphs. Figure 23 is the bifurcation diagram about b in [ 0:01; 0:01].
Figure 23: the bifurcation diagram of 2nd component of the map (6) about b of the example for theorem 2.10
For theorem 2.11, consider another case a = 1; b = 0; p = 1; q = 0 and f (x) = 1920 sin x.
Clearly, f is continuous on R. Since a21pf ( x) = f ( x) = f (x) and a1f ap1 x = f (x), we have that f 21;2021 = f 21 ;1920 21;2021 ( f 21 = f 2021 0:445, 21 0:15). The parametric map gu(x) = u sin (x) has also period-doubling property and htop(f ) > 0 (it has 3-periodic points). Thus, the hypothesis of theorem 2.11 is satis ed.
Next, see the dynamic diagram in gure 24. Moreover, gures 25 and 26 are the graphs of iteration of x- and y-components of (7).
Figure 24: the dynamic diagram the example for theorem 2.11
Figure 25: the graph of iteration of x-component of (7) with iterating times from 1 to 500 for theorem 2.11
Figure 26: the graph of iteration of y-component of (7) with iterating times from 1 to 500 for theorem 2.11
Due to the result p
b2+ q2 < 0 for some 0 > 0, we print two bifurcation diagrams about b and q respectively (see gures 27 and 28). So we can observe the chaotic property.
Figure 27: the bifurcation diagram of 2nd component of the map (7) about b of the example for theorem 2.11 and q = 0
Figure 28: the bifurcation diagram of 2nd component of the map (7) about q of the example for theorem 2.11 and b = 0
Conclusively, theorems 2.10 and 2.11 do work.
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