Rich dynamics caused by diffusion

Abstract

It is an awfully difficult task to design an efficient numerical method for bifurcation diagrams, the graphs of Lyapunov exponents, or the topological entropy about discrete dynamical systems by linear/nonlinear diffusion with the Direchlet/Neumann- boundary conditions. Until now there are less works concerned with such a problem. In this paper, we propose a scheme about bifurcating analysis in a series of discrete-time dynamical systems with linear/nonlinear diffusion terms under the periodic boundary conditions. The complexity of dynamical behaviors caused by the diffusion term are to be determined. Bifurcation diagrams are shown by numerical simulation and chaotic behavior (chaotic Turing patterns) is demonstrated by computing the largest Lyapunov exponent. Our theoretical model can give an interesting case study about the phenomenon: the individuals exhibit a very simple dynamics but the groups with linear/nonlinear coupling can own a complex dynamics including fluctuation, periodicity and even chaotic behavior. We find that diffusion can trigger chaotic behavior in the present system and there exist multiple Turing patterns. It is interesting as regular or chaotic patterns can be reported in this study. Chaotic orbits emerge when exploring further in the diffusion coefficient space, and such a behavior is entirely absent in the corresponding continuous time-space system. The method proposed in the present paper is innovative and the conclusion is novel.

Appendices

Appendix I: $M\psi =-\gamma \lambda _{kl}\psi$.

Proof

It follows by direct computation that

$$\begin{aligned}{} & {} -4\cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} +\cos \frac{2(k-1)\pi }{s}\cos \frac{4(l-1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)\pi }{s}\cos \frac{2t(l-1)\pi }{t}\\{} & {} +\cos \frac{4(k-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} \quad +\cos \frac{2s(k-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)\pi }{t};\\{} & {} -4\cos \frac{2(k-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} + \cos \frac{2(k-1)\pi }{s}\cos \frac{(l-1)2(t-1)\pi }{t}\\{} & {} +\cos \frac{2s(k-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} +\cos \frac{2(k-1)\pi }{s}\cos \frac{(l-1)2\pi }{t}\\{} & {} \quad +\cos \frac{4(k-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2(k-1)\pi }{s}\cos \frac{l-1)2t\pi }{t};\\{} & {} -4\cos \frac{2(k-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} + \cos \frac{2s(k-1)\pi }{s}\cos \frac{2(l-1)(t-1)\pi }{t}\\{} & {} \quad +\cos \frac{2s(k-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} +\cos \frac{2(s-1)(k-1)\pi }{s}\cos \frac{2(l-1)t\pi }{2t}\\{} & {} \quad +\cos \frac{2(k-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2s(k-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} -4\cos \frac{(k-1)2s\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} +\cos \frac{(k-1)2s\pi }{s}\cos \frac{4(l-1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)(s-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} +\cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)s\pi }{s}\cos \frac{2t(l-1)\pi }{t}\\{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2(k-1)s\pi }{s}\cos \frac{2(l-1)\pi }{t} \end{aligned}$$

and for $j=2:t-1$,

$$\begin{aligned}{} & {} -4\cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} +\cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)(j-1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)(j+1)\pi }{t}\\{} & {} + \cos \frac{4(k-1)\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} \quad +\cos \frac{2s(k-1)\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2(k-1)\pi }{s}\cos \frac{2(l-1)j\pi }{t};\\{} & {} -4\cos \frac{(k-1)2s\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} +\cos \frac{(k-1)2s\pi }{s}\cos \frac{2(l-1)(j-1)\pi }{t}\\{} & {} \quad +\cos \frac{(k-1)2s\pi }{s}\cos \frac{2(l-1)(j+1)\pi }{t}\\{} & {} +\cos \frac{(k-1)2(s-1)\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} \quad +\cos \frac{(k-1)2\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} =-4\left( \sin ^2\frac{2(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{(k-1)2s\pi }{s}\cos \frac{2(l-1)j\pi }{t}; \end{aligned}$$

and for $i=2:s-1$,

$$\begin{aligned}{} & {} -4\cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)\pi }{t}\\ {}{} & {} +\cos \frac{2(k-1)i\pi }{s}\cos \frac{4(l-1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)i\pi }{s}\cos \frac{2t(l-1)\pi }{t} \\ {}{} & {} \cos \frac{2(k-1)(i+1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)(i-1)\pi }{s}\cos \frac{2(l-1)\pi }{t}\\ {}{} & {} =-4\left( \sin ^2\frac{2(k-1)\pi }{s}+\sin ^2\frac{2(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)\pi }{t};\\ {}{} & {} -4\cos \frac{(k-1)2i\pi }{s}\cos \frac{(l-1)2t\pi }{2t}\\ {}{} & {} +\cos \frac{(k-1)2i\pi }{s}\cos \frac{(l-1)2(t-1)\pi }{t}\\{} & {} \quad +\cos \frac{(k-1)2i\pi }{s}\cos \frac{(l-1)2\pi }{t} \\ {}{} & {} + \cos \frac{(k-1)2(i-1)\pi }{s}\cos \frac{(l-1)2t\pi }{t}\\{} & {} \quad +\cos \frac{(k-1)2(i+1)\pi }{s}\cos \frac{(l-1)2t\pi }{2t}\\ {}{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{(k-1)2i\pi }{s}\cos \frac{(l-1)2t\pi }{t}; \end{aligned}$$

and for $i=2:s-1$, $j=2:t-1$

$$\begin{aligned}{} & {} -4\cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\ {}{} & {} + \cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)(j+1)\pi }{t}\\{} & {} \quad +\cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)(j-1)\pi }{t}\\ {}{} & {} +\cos \frac{2(k-1)(i+1)\pi }{s}\cos \frac{2(l-1)j\pi }{t}\\{} & {} \quad + \cos \frac{2(k-1)(i-1)\pi }{s}\cos \frac{2(l-1)j\pi }{t} \\ {}{} & {} =-4\left( \sin ^2\frac{(k-1)\pi }{s}+\sin ^2\frac{(l-1)\pi }{t}\right) \\{} & {} \quad \cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)j\pi }{t}. \end{aligned}$$

$\square$

Appendix II: The existence of a standard orthogonal basis for A or discrete periodic boundary problems (2.2).

The orthogonality of $\{\psi ^{(k,l)}, k=1,\ldots , s, l=1,\ldots t\}$ comes from direct computation for t and t are both odd. Obviously $\psi ^{kl}\not =0$ because of $\psi ^{kl}_{st}=1$. Then, by Appendix I, $\{\psi ^{(k,l)}, k=1,\ldots , s, l=1,\ldots t\}$ is a orthogonal basis for A or discrete periodic boundary problems (2.2). The unitization is listed as follows

$$\begin{aligned}{} & {} e^{(1,1)}_{i,j}=\frac{1}{\sqrt{st}}\psi ^{(1,1)}=\frac{1}{\sqrt{st}}(1,1,\ldots ,1),\\ {}{} & {} e^{(k,1)}_{i,j}=\sqrt{\frac{2}{st}}\psi ^{k1},\ \text{ for } k\in {{\mathbb {N}}}(2,s),\\ {}{} & {} e^{(1,l)}_{i,j}=\sqrt{\frac{2}{st}}\psi ^{1l},\ \text{ for } l\in {{\mathbb {N}}}(2,t), \end{aligned}$$

and

$$\begin{aligned} e^{(k,l)}_{i,j}=\frac{2}{\sqrt{st}}\psi ^{kl},\ \text{ for } k\in {{\mathbb {N}}}(2,s), l\in {{\mathbb {N}}}(2,t). \end{aligned}$$

Now we prove that

$$\begin{aligned} \sum ^s_{i=1}\cos ^2\frac{2(k-1)i\pi }{s}={\left\{ \begin{array}{ll} s/2, &{}\cos \frac{4(k-1)\pi }{s}\not =1\\ s, &{}\cos \frac{4(k-1)\pi }{s}=1.\end{array}\right. }\end{aligned}$$

Since $\sum ^s_{i=1}\cos ^2\frac{2(k-1)i\pi }{s}=s/2+1/2\sum ^s_{i=1}\cos \frac{4(k-1)i\pi }{s}$, it holds true if

$$\begin{aligned} U^*\triangleq \sum ^s_{i=1}\cos ^2\frac{4(k-1)i\pi }{s}=0 \text{ or } 1. \end{aligned}$$

Set $\beta =\frac{4(k-1)\pi }{s}$. Then one can obtain

$$\begin{aligned}{} & {} 2\cos \beta \sum ^s_{i=1}\cos ^2\frac{2(k-1)i\pi }{s}\\ {}{} & {} \quad =\sum ^s_{i=1} (\cos \frac{4(k-1)(i+1)\pi }{s}+\cos \frac{4(k-1)(i-1)\pi }{s}. \end{aligned}$$

It follows directly from $\cos \frac{4(k-1)(s+1)\pi }{s}=\cos \frac{4(k-1)\pi }{s}$, $\cos \frac{4(k-1)(1-1)\pi }{s}=\cos \frac{4(k-1)s\pi }{s}$ that $2\cos \beta U^*=2U^*$. Eventually we find that $U^*=0$ for $\cos \beta \not =1$. If $\cos \beta =1$, it is natural that $\sum ^s_{i=1}\cos ^2\frac{2(k-1)i\pi }{s}=s$.

Next we are interested in the orthogonality of $\{\psi ^{kl}\}$. At first, it is interesting about the following result

$$\begin{aligned}{} & {} \sum ^s_{i=1}\cos \frac{2(k-1)i\pi }{s}\cos \frac{2(l-1)i\pi }{s}\\ {}{} & {} \quad =\left\{ \begin{array}{cl} 0, &{} \cos \frac{2(k+l-2)\pi }{s}\not =1;\\ s/2, &{}\cos \frac{2(k+l-2)\pi }{s} =1 \text{ and } \cos \frac{4(k-1)\pi }{s}\not =1;\\ s &{}\cos \frac{2(k+l-2)\pi }{s} =1 \text{ and } \cos \frac{4(k-1)\pi }{s} =1.\end{array}\right. \end{aligned}$$

Please note that $\cos \frac{2(k-l)\pi }{s}\not =1$. The proof follows from the same technique as above.

Therefore the vector set $\{\psi ^{kl}\}$ is a orthogonal basis of ${{\mathbb {R}}}^{s\times t}$ for s or t is an odd integer, otherwise it is invalid. It is linearly dependent because, among them, there are at least two same vectors related to the same eigenvalue with the multiplicity more than 1, for instance $k=s^*-1, l=t^*-1$, $k=s^*+3, l=t^*+3$ with $s=2\,s^*\ge 6$ and $t=2t^*\ge 6$.