User:EML4500.f08.JAMAMA/NM/JA

= 7.1 =

Given: Verhulst Equation with initial conditions
Verhulst equation:

$$ \frac{d}{dt}x(t) = rx(1 - \frac{x}{x_{max}}) \ $$

$$ x_{max} = 15 \ $$ and $$ r = 1.4 \ $$

with initial condition of $$ x_0 \ $$

for time $$ t \in [0, 10] \ $$

Case 1:

$$ x_0 = 3 < \frac{1}{2} x_{max} \ $$

Case 2:

$$ x_0 = 9 > \frac{1}{2} x_{max} \ $$

Find: By using Hermite - Simpson rule, integrate given Verhulst equation
From given Prof. Vu Quoc's notes for

$$ z_{i+ 1} = z_i + \frac{h/2}{3} \left ( f_i + 4f_{i+\frac{1}{2}}[g(z_i,z_{i+1})],t_{i+\frac{1}{2}} + f_{i+1} \right ) \ $$

Solution
From given Prof. Vu Quoc's notes for

$$ z_{i+ 1} = z_i + \frac{h/2}{3} \left ( f_i + 4f_{i+\frac{1}{2}}[g(z_i,z_{i+1})],t_{i+\frac{1}{2}} + f_{i+1} \right ) \ $$

,where

$$ f_i = f(z_i, f_i) \ $$

$$ f_{i+1} = f(z_{i+1}, t_{i+1}) \ $$

$$ f_{i+\frac{1}{2}} = f[g(z_i,z_{i+1},t_{i+\frac{1}{2} })] = f(z_{i+\frac{1}{2}}, t_{i+\frac{1}{2}}) \ $$

,where $$ t_{i+\frac{1}{2}} = t_i + \frac{h}{2} \ $$ and as it was proven in HW 6.6 as well as given by Prof. Vu Quoc's

$$ z_{i+ \frac{1}{2}} = \frac{1}{2}(z_i + z_{i+1}) + \frac{h}{8}(f_i - f_{i+1}) \ $$

,note that $$ h \ $$ is the time step

Also specifying the equation to be of a form $$ F(z_{i+ 1})= 0 \ $$

$$ F(z_{i+ 1})= -z_{i+ 1} + z_i + \frac{h/2}{3} \left ( f_i + 4f([g(z_i,z_{i+1})],t_{i+\frac{1}{2}}) + f_{i+1} \right ) = 0\ $$

and, finally, as mentioned in

$$ z_{i+1}^{(k+1)} = z_{i+1}^{(k)} - \left [ \frac{d}{dz}F(z_{i+1}^{(k)}) \right ]^{-1} F(z_{i+1}^{(k)}) \ $$

in our case, for a given function, we have first order derivative. Therefore

$$ z_{i+1}^{(1)} = z_{i+1} - \left [ \frac{d}{dz}F(z_{i+1}) \right ]^{-1} F(z_{i+1}) \ $$

Then check the condition for convergence as it is shown in

$$ z_{i+1}^{(1)} - z_{i+1} ?< Tolerance \ $$

If condition is not met, choose $$ z_{i+1}^{(1)} = z_{i+2} \ $$ and iterate again to get $$ z_{i+2}^{(1)} \ $$ until the tolerance is met

Egm6341.s11.team3.rakesh 02:24, 21 April 2011 (UTC)