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Predicting and estimating probability density functions of chaotic systems

The research of Prof. Paweł Góra was supported by NSERC grant. Also, the research of Profs. Noah Rhee and Majid Bani-Yaghoub was supported by UMKC's funding for excellence program.
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  • In the present work, for the first time, we employ Ulam's method to estimate and to predict the existence of the probability density functions of single species populations with chaotic dynamics. In particular, given a chaotic map, we show that Ulam's method generates a sequence of density functions in L1-space that may converge weakly to a function in L1-space. In such a case we show that the limiting function generates an absolutely continuous (w.r.t. the Lebesgue measure) invariant measure (w.r.t. the given chaotic map) and therefore the limiting function is the probability density function of the chaotic map. This fact can be used to determine the existence and estimate the probability density functions of chaotic biological systems.

    Mathematics Subject Classification: Primary: 37A05; Secondary: 37A10, 37E05, 37M25, 92D25.

    Citation:

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  • Figure 1.  Graphs of the probability density function $f^*$ and its estimation $f_{2^5}$ for Example 2. Since $\{ f_{2^n} \}_{n \ge 1}$ is bounded by 2, by Theorem 4 $\{ f_{2^n} \}_{n \ge 1}$ is weakly pre-compact, and hence there exists a subsequence $\{ f_{2^{n_k}} \}_{n \geq 1}$ of $\{ f_{2^n} \}_{n \geq 1}$ which converges weakly to $f^* \in L^1[a,b]$. In fact, we may expect that $\{ f_{2^n} \}_{n \ge 1}$ converges strongly to $f^*$.

    Figure 2.  Graphs of the probability density function $f^*$ and its estimation $f_{2^9}$ for Example 3. Again we may expect that $\{ f_{2^n} \}_{n \ge 1}$ converges strongly to $f^*$.

    Figure 3.  Bifurcation diagram of Smith-Slatkin map for $\alpha = 1.5, \beta = 1, \gamma = 0.1$, as $p$ changes from 5 to 19.

    Figure 4.  Graph of Smith-Slatkin map for $\alpha = 1.5, \beta = 1, \gamma = 0.1$ and $p=18$ with $x^*=0.977725973073732$, $x_c=0.857817481013884$, $b=\tau(x_c)=1.295956066342802$, and $a=\tau^2(x_c)=0.147708628629491$. Note that $\tau: [a,b] \rightarrow [a,b]$.

    Figure 5.  Graph of $\frac{p^3-3p^2+5p-3}{6(p^2+1).}$ For $\alpha=1.5$ and $\gamma=0.1$, the condition $\frac\gamma\alpha < \frac{p^3-3p^2+5p-3}{6(p^2+1)}$ in Proposition 3 is fulfilled for all $p > 2.$

    Figure 6.  Graph of Smith-Slatkin map for $\alpha = 1.5, \beta = 1, \gamma = 0.1$ and $p=8$. Three first iterates of the critical point are shown.

    Figure 7.  Graph of Smith-Slatkin map for $\alpha = 1.5, \beta = 1, \gamma = 0.1$ and $p=8.5$. Three first iterates of the critical point are shown.

    Figure 8.  Graph of $f_{2^8}$ for Example 5. There are six peak values, where the first peak value occurs at $x=a$ and the last peak value at $x=b$. Although the closed form of the probability density function $f^*$ is unknown, the numerical simulations strongly show that $\{ f_{2^n} \}_{n \geq 1}$ is bounded by an integrable function. So, by Theorem 4, $\{ f_{2^n} \}_{n \ge 1}$ is weakly pre-compact, and hence there exists a sub-sequence $\{ f_{2^{n_k}} \}_{n \geq 1}$ of $\{ f_{2^n} \}_{n \geq 1}$ which converges weakly to $f^* \in L^1[a,b]$. By Proposition 2 the measure $\mu$ with density function $f^*$ is an absolutely continuous $\tau$-invariant measure. So $f^*$ is the PDF of the chaotic map $\tau$. In fact, we may expect that $\{ f_{2^n} \}_{n \geq 1}$ converges strongly to $f^*$.

    Table 1.  Maximum values of $f_{2^n}$

    $n$ $2^n$ maximum values of$f_{2^n}$
    $1$ $2$ $1.25$
    $2$ $4$ $1.5075$
    $3$ $8$ $1.7232$
    $4$ $16$ $1.8157$
    $5$ $32$ $1.9126$
    $6$ $64$ $1.9465$
    $7$ $128$ $1.9748$
    $8$ $256$ $1.9871$
    $9$ $512$ $1.9930$
    $10$ $1024$ $1.9964$
     | Show Table
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    Table 2.  Peak values of $f_{2^n}$ for $\tau$ in Example 3

    $n$ $2^n$ 1st peak value of $f_{2^n}$ 2nd peak value of $f_{2^n}$
    $5$ $32$ $1.2899$ $3.8291$
    $6$ $64$ $1.8337$ $5.4724$
    $7$ $128$ $2.4475$ $7.3235$
    $8$ $256$ $3.5149$ $10.5309$
    $9$ $512$ $4.8498$ $14.5401$
    $10$ $1024$ $6.9200$ $20.7534$
     | Show Table
    DownLoad: CSV

    Table 3.  The consecutive ratio of peak values of $f_{2^n}$ in Example 3

    $n$ $2^n$ $f_{2^n}/{f_{2^{n-1}}}$ at 1st peak value $f_{2^n}/{f_{2^{n-1}}}$ at 2nd peak value
    $6$ $64$ $1.4216$ $1.4292$
    $7$ $128$ $1.3347$ $1.3382$
    $8$ $256$ $1.4361$ $1.4380$
    $9$ $512$ $1.3798$ $1.3807$
    $10$ $1024$ $1.4269$ $1.4273$
     | Show Table
    DownLoad: CSV

    Table 4.  Peak values of $f_{2^n}$ for $\tau$ in Example 5

    $n$ $2^n$ The 1st peak values The 2nd peak values
    $8$ $256$ $9.9001$ $6.5190$
    $9$ $512$ $13.987$ $9.0862$
    $10$ $1024$ $20.609$ $12.770$
    $11$ $2048$ $28.397$ $16.860$
     | Show Table
    DownLoad: CSV

    Table 5.  Peak values of $f_{2^n}$ for $\tau$ in Example 5

    $n$ $2^n$ The 3rd peak values The 4th peak values
    $8$ $256$ $4.2530$ $2.7992$
    $9$ $512$ $5.8493$ $3.8154$
    $10$ $1024$ $8.1477$ $5.2147$
    $11$ $2048$ $10.711$ $6.8142$
     | Show Table
    DownLoad: CSV

    Table 6.  Peak values of $f_{2^n}$ for $\tau$ in Example 5

    $n$ $2^n$ The 5th peak values The 6th peak values
    $8$ $256$ $2.0871$ $4.1312$
    $9$ $512$ $2.8524$ $5.5220$
    $10$ $1024$ $3.5875$ $7.8931$
    $11$ $2048$ $4.5694$ $10.607$
     | Show Table
    DownLoad: CSV

    Table 7.  The consecutive ratio of the first two peak values of $f_{2^n}$ for $\tau$ in Example 5

    $n$ $2^n$ $f_{2^n}/{f_{2^{n-1}}}$ at 1st peak value $f_{2^n}/{f_{2^{n-1}}}$ at 2nd peak value
    $9$ $512$ $1.4128$ $1.3938$
    $10$ $1024$ $1.4735$ $1.4055$
    $11$ $2048$ $1.3779$ $1.3203$
     | Show Table
    DownLoad: CSV

    Table 8.  The consecutive ratio of the middle two peak values of $f_{2^n}$ for $\tau$ in Example 5

    $n$ $2^n$ $f_{2^n}/{f_{2^{n-1}}}$ at 3rd peak value $f_{2^n}/{f_{2^{n-1}}}$ at 4th peak value
    $9$ $512$ $1.3753$ $1.3630$
    $10$ $1024$ $1.3929$ $1.3668$
    $11$ $2048$ $1.3146$ $1.3067$
     | Show Table
    DownLoad: CSV

    Table 9.  The consecutive ratio of the last two peak values of $f_{2^n}$ for $\tau$ in Example 5

    $n$ $2^n$ $f_{2^n}/{f_{2^{n-1}}}$ at 5th peak value $f_{2^n}/{f_{2^{n-1}}}$ at 6th peak value
    $9$ $512$ $1.3667$ $1.3367$
    $10$ $1024$ $1.2577$ $1.4294$
    $11$ $2048$ $1.2737$ $1.3439$
     | Show Table
    DownLoad: CSV

    Table 10.  The estimates of ${\bar X}$ and $\sigma$ for Example 5

    $N$ ${\bar x}$ s
    $10^5$ $0.6658$ $0.3570$
    $10^6$ $0.6657$ $0.3574$
    $10^7$ $0.6657$ $0.3575$
     | Show Table
    DownLoad: CSV
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