User:EGM6341.s11.team1.Chiu/S10 Mtg39

Mtg 39: Tue, 6 Apr 10

[[media: Egm6341.s10.mtg39.djvu| page39-1]]


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Exact:


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$$ x(t)= \frac{x_{0} x_{max} e^{rt}}{x_{max}+x_{0}(e^{rt}-1)} \ $$


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HW: Verify + solve differential equation [[media: Egm6341.s10.mtg40.djvu| Hints page40-1]]


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HW: Use HS algorithm [[media: Egm6341.s10.mtg35.djvu| page35-1]] to integrate [[media: Egm6341.s10.mtg40.djvu| (1) page38-3]]


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$$ x_{max}=10, r=1.2 \ $$ (time scale $$ \frac{1}{r}= \frac{1}{1.2} \cong 1 \ $$)


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2 cases:


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$$ x_{0}=2 < \frac{1}{2}x_{max} \ $$


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$$ x_{0}=7 > \frac{1}{2}x_{max} \ $$


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such that AbsTol $$ \color{red} = \theta(10^{-6}) \ $$ for $$ t \in \left [ 0,10 \right ] \ $$


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[[media: Egm6341.s10.mtg39.djvu| page39-2]]

Discrete population dynamics: Logistic map


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$$ \bar{x}_{i+1}=r \bar{x}_{i}(1- \bar{x}_{i}) \ $$


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$$ \bar{x} \in \left [ 0,1 \right ]: \ $$ Physical meaning range


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End Discrete population dynamics

Scaling of continuous case: [[media: Egm6341.s10.mtg38.djvu| (1)page38-3]]


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$$ \bar{x}:= \frac{x}{x_{max}} \in \left [ 0,1 \right ] \ $$


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$$ \left.\begin{matrix} x= x_{max} \cdot \bar{x} \\ dx= x_{max} \cdot d\bar{x} \end{matrix}\right\} \dot{\bar{x}}=r \bar{x}(1- \bar{x}) \ $$


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End Scaling of continuous case

Motiration: $$ \dot{\bar{x}}=r \bar{x} \ $$


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FE = Forward Euler algorithm (explicit, unstable): $$ \frac{ \bar{x}_{i+1}- \bar{x}_{i}}{h} = \underbrace{r \bar{x}_{i}}_{ \color{blue} at \ t=t_{i}} \ $$


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$$ \bar{x}_{i+1}= \bar{x}_{i}+hr \bar{x}_{i}=(1+hr) \bar{x}_{i} \ $$


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BE = Backward Euler algorithm (implicit, stable): $$ \frac{ \bar{x}_{i+1}- \bar{x}_{i}}{h} = \underbrace{r \bar{x}_{i+1}}_{ \color{blue} at \ t=t_{i+1}} \ $$


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$$ ( \frac{1}{h}-r) \bar{x}_{i+1}= \frac{ \bar{x}_{i}}{h} \ $$


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$$ \bar{x}_{i+1}= \frac{ \bar{x}_{i}}{1-hr} \ $$


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End Motiration

[[media: Egm6341.s10.mtg39.djvu| page39-3]]

Fixed points of period 1:


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$$ \underbrace{ \bar{x}_{i+1}}_{ \color{blue} r \bar{x}_{i}(1- \bar{x}_{i})}= \bar{x}_{i} \ $$

(1)
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2 fixed points: $$ \left\{\begin{matrix} \bar{x}=0 \ { \color{blue} Trivial} \ { \color{red} (2)} \\ \bar{x}= \frac{r-1}{r} \ { \color{red} (3)} \end{matrix}\right. \ $$


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Want  $$ \bar{x}= \frac{r-1}{r} \in \left [ 0,1 \right ] \ $$   Physical meaning ful range


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$$ \Rightarrow r>1 \ $$


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Fixed points of period 2:


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$$ \bar{x}_{i+1}=r \bar{x}_{i}(1- \bar{x}_{i}) \ $$

(4)
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$$ \underbrace{ \bar{x}_{i+2}}_{ \color{blue} r \bar{x}_{i+1}(1- \bar{x}_{i+1})}= \bar{x}_{i} \ $$

(5)
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Eliminate $$ \bar{x}_{i+1} \ $$ from (4)+(5) :


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$$ \bar{x}_{i}( \bar{x}_{i}- \frac{r-1}{r}) \underbrace{\left [ r^{2} \bar{x}_{i}^{2}-r(1+r) \bar{x}_{i}+(1+r) \right ]}_{ \color{blue} Discriminant \ of \ quadratic \ equation}=0 \ $$

(6)
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$$ r^{2}(r^{2}-2r-3)>0 \ $$  for   $$ r>3 \ $$


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[[media: Egm6341.s10.mtg39.djvu| page39-4]]

$$ \Rightarrow \ $$ for $$ r>3, \exists \ $$  fixed points with period 2.

(Since solution of quadratic equation are real number.)