User:Egm6341.s10.team3.sa/Mtg39

Mtg 39: Tue, 6 Apr 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 17:02, 12 August 2010 (UTC)

[[media: Egm6341.s10.mtg39.djvu | Page 39-1]]  Exact solution: $$\displaystyle x(t) = \frac{x_0 x_{max} e^{rt}}{x_{max}+x_0(e^{rt}-1)}$$


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HW:
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The exact expression of x(t) is :
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$$ i) Verify that:
 * $$\displaystyle \color{blue} x(t)= \frac{x_0x_{max}e^{rt}}{x_{max}+x_0(e^{rt}-1)}
 * $$\displaystyle \color{blue} x(t)= \frac{x_0x_{max}e^{rt}}{x_{max}+x_0(e^{rt}-1)}
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$$
 * $$\displaystyle \color{blue} \frac{dx}{dt}=rx \left( 1- \frac{x}{x_{max}}\right)
 * $$\displaystyle \color{blue} \frac{dx}{dt}=rx \left( 1- \frac{x}{x_{max}}\right)
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ii) Solve the above differential equation with hint on Lecture Notes [[media:Egm6341.s10.mtg40.djvu|p.40-1]]


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[[media: Egm6341.s10.mtg39.djvu | Page 39-2]]  DISCRETE POPULATION DYNAMICS: Logistic map $$\displaystyle \bar{x}_{i+1} = r\bar{x}_i (1-\bar{x}_i)$$ $$\displaystyle \bar{x} \in [0,1]$$ : Physical meaning range.

Scaling of continuous case: Eqn(1) on [[media: Egm6341.s10.mtg38.djvu | Page 38-3]] $$\displaystyle \bar{x}: = \frac{x}{x_{max}} \;\; \in [0,1]$$ $$\displaystyle x = {x_{max}}\bar{x} $$ $$\displaystyle dx = {x_{max}}d\bar{x} $$

Motivation: $$\displaystyle \dot{\bar{x}} = r\bar{x} $$

Forward Euler:(Explicit method - Unstable): $$\displaystyle \frac{\bar{x}_{i+1} - \bar{x}_i} {h} = \underbrace{r \bar{x}_i}_{at\; t=t_i} $$ $$\displaystyle \bar{x}_{i+1} = \bar{x}_i + hr \bar{x}_i = (1+hr) \bar{x}_i $$ 

Backward Euler:(Implicit method - Stable): $$\displaystyle \frac {\bar{x}_{i+1} - \bar{x}_i} {h} = \underbrace {r \bar{x}_{i+1}}_{at\; t=t_i}$$ $$\displaystyle \bar{x}_{i+1} = \frac{\bar{x}_i}{1-rh}$$ 

[[media: Egm6341.s10.mtg39.djvu | Page 39-3]] 

Fixed point of period 1: $$\displaystyle \bar{x}_{i+1} = \bar{x}_i $$ $$\displaystyle (Eq. 1) $$

$$\displaystyle \Rightarrow r\bar{x}_i (1-\bar{x}_i) = \bar{x}_i$$

2 fixed points: $$\displaystyle Trivial: \;\ \bar{x}=0 $$ $$\displaystyle (Eq. 2) $$

$$\displaystyle \bar{x}= \frac{r-1}{r} $$ $$\displaystyle (Eq. 3) $$

We want $$\displaystyle \bar{x} = \frac{r-1}{r} \in [0,1] \; \Rightarrow r>1 $$

Fixed point of period 2: $$\displaystyle \bar{x}_{i+1} = r \bar{x}_i(1 - \bar{x}_i) $$ $$\displaystyle (Eq. 4) $$

$$\displaystyle \bar{x}_{i+2} = r \bar{x}_{i+1}(1 - \bar{x}_{i+1})= \bar{x}_i$$ $$\displaystyle (Eq. 5) $$

Eliminate $$\displaystyle \bar{x}_{i+1} $$ from (Eq.4) and (Eq.5) to obtain, $$\displaystyle \bar{x}_i \bigg( \bar{x}_i - \frac{r-1}{r}\bigg) \bigg[ r^2\bar{x}_i^2 - r(1+r)\bar{x}_i + (1+r)\bigg]=0 $$ $$\displaystyle (Eq. 6) $$

Discriminant of (Eq.6) is a quadratic equation. $$\displaystyle r^2(r^2 - 2r -3) > 0 \; for\; r\; >\; 3$$. 

<p style="text-align:left;">[[media: Egm6341.s10.mtg39.djvu | Page 39-4]] 

For r>3, there exists fixed points with period 2. (Since solution of quadratic equation are real numbers).