User:Egm6341.s10.team3.heejun/HW7

=  (4) Chaotic Systems (Chaos) =

Problem Statement
$$ \displaystyle Reproduce\, Fig.\, 15.6\, and\, 15.7\, in\, King\, et.\, al.\, 2003.\,\, (pp.\, 455-456)\, $$ Ref: Differential Equations: Amazon

$$ \displaystyle Logistic\,\, map\, $$

$$ \displaystyle x_{n+1}=r*x_n*(1-x_n) $$

$$ \displaystyle where\,\,\, r=constant,\, 0<r\leq 4\,\, and\,\, n=integer\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg40.djvu|p.40-1]]

Solution
$$ \displaystyle i)\, Reproduce\, the\, period\, doubling\, process\, as\, a\, bifurcation\, diagram\, using\, Matlab\, $$

$$ \displaystyle ii)\, Plot\, the\, bifurcation\, diagram\, $$



$$ \displaystyle iii)\, Reproduce\, a\, sequence\, of\, 100\, iterates\, of\, the\, logistic\, map\, with\, r=4,\, x_0=0.1\, and\, x_0=0.1+10^{-16}\, $$

$$ \displaystyle iiii)\, Plot\, the\, sequence\, $$



= ''' (7) Inconsistent Trap. Simpson algorithm '''=

Problem Statement
$$ \displaystyle i)\, Solve\, logistic\, equation\, using\, the\, inconsistent\, trap.\, simpson\, algorithm\, $$

$$ \displaystyle z_{i+1} = z_i + \frac{(\frac{h}{2})}{3}*[f_i + 4f_{i+\frac{1}{2}}+f_{i+1}]\, $$

$$ \displaystyle with\,\, z_{i+1} = \frac{1}{2}*(z_i + z_{i+1})\, $$

$$ \displaystyle ii)\, Compare\, with\, the\, results\, of\, the\, Hermite-Simpson\, Algorithm\, using\, same\, values\, of\, h\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg41.djvu|p.41-2]]

Solution
$$ \displaystyle i)\, Matlab\, Code-\,\,\, only\, difference\, between\, the\, Inconsistent\, Trap.\, Simpson\, algorithm\, and\, the\, Hermite-Simpson\, Algorithm\, for\, Matlab\, is\, a\, definition\, of\, H_{half}\, $$

$$ \displaystyle a)\, Inconsistent\, Trap.\, Simpson\, algorithm\, $$

$$ \displaystyle x_{half} = 0.5*(x(i)+x(i+1));\, $$

$$ \displaystyle b)\, Hermite-Simpson\, Algorithm\, $$

$$ \displaystyle x_{half} = 0.5*(x(i)+x(i+1)) + (h/8)*(f1-f2);\, $$

$$ \displaystyle ii)\, Case(1):\, Initial\, condition:\, x_0 = 2.0\, $$



$$ \displaystyle iii)\, Case(2):\, Initial\, condition:\, x_0 = 7.0\, $$



$$ \displaystyle iiii)\, Comparison\, with the\, Hermite-Simpson\, Algorithm\, $$

$$ \displaystyle a)\, Case(1):\, Initial\, condition:\, x_0 = 2.0\, $$



$$ \displaystyle b)\, Case(2):\, Initial\, condition:\, x_0 = 7.0\, $$



$$ \displaystyle Therefore,\, the\, Inconsistent\, Trap.\, Simpson\, algorithm\, and\, the\, Hermite-Simpson\, Algorithm\, have\, the\, same\, results\, $$

=  (8) Linear Momentum Term in the x' Direction with tan(dr) =

Problem Statement
$$ \displaystyle Show\, \overline{CD} = \overline{AB} + hot\,\, for\, small\, dr\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-1]]

Solution
$$ \displaystyle i)\, Let's\, consider\, x'\, and\, y'\, coordinate\, system.\, $$



$$ \displaystyle ii)\, Define\, length\, of\, \overline{CD}\, and\, \overline{AB}\, with\, tan(dr)\, in\, x'\, and\, y'\, coordinate\, system\, (projected\, to\, x-axis).\, $$

$$ \displaystyle \overline{AB} = tan(dr) * V\, $$ $$ \displaystyle \overline{CD} = tan(dr) * (V+dV)\, $$

$$ \displaystyle iii)\, Identify\, tan(dr)\, Using\, Taylor\, Series\, expansion\, to\, approximate\, the\, tangent\, of\, the\, small\, angle.\, $$

$$ \displaystyle tan(dr) \cong dr\, $$

$$ \displaystyle iiii) Rewrite\, \overline{CD}\, and\, \overline{AB}\, with\, the\, small\, angle\, approximation\, of\, tangent.\, $$

$$ \displaystyle \overline{AB} = dr * V\, $$ $$ \displaystyle \overline{CD} = dr * (V+dV)\, = dr*V + \cancelto{}{dr*dV}\, $$

$$ \displaystyle The\, higher\, order\, term (dr*dV)\, can\, be\, negligible.\, $$

$$ \displaystyle Therefore\, \overline{CD} = \overline{AB}\, $$

$$ \displaystyle iiiii)\, Verification\, of\, the\, small\, angle\, approximation\, of\, tangent.\, $$

$$ \displaystyle From\, the\, above\, results,\, the\, small\, angle\, approximation\, of\, tangent\, may\, work\, pretty\, well\, out\, up\, to\, about\, 10\, degrees.\, $$

$$ \displaystyle Therefore\, we\, can\, conclude\, that\, \overline{CD} = \overline{AB}\, with\, small\, dr.\, $$

=  (9) Parametrization of ellipse =

Problem Statement
$$ \displaystyle Show\, \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\, $$

$$ \displaystyle where\,\,\, x=a*cos(t)\,\, and\,\, y=b*sin(t)\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-1]]

Solution
$$ \displaystyle i)\, For L.H.S,\,\, \frac{x^2}{a^2}+\frac{y^2}{b^2} = (\frac{x}{a})^2+(\frac{y}{b})^2\,\,\,\,\,\,\,\, (1)\, $$

$$ \displaystyle ii)\, x=a*cos(t)\, \rightarrow\, (\frac{x}{a})=cos(t)\,\, and\,\, y=b*cos(t)\, \rightarrow\, (\frac{y}{b})=sin(t)\,\,\,\,\,\,\,\, (2)\, $$

$$ \displaystyle (1) + (2) is given by $$

$$ \displaystyle (cos(t))^2 + (sin(t))^2\,\,\,\,\,\,\,\,(3) $$

$$ \displaystyle iii) (3) is always 1 because of Pythagorean Identities.\, $$

$$ \displaystyle Thus,\, \frac{x^2}{a^2}+\frac{y^2}{b^2}\, is\, always\, 1.\, $$

=  (10) Arc Length =

Problem Statement
$$ \displaystyle Show\, dl=[dx^2+dy^2]^{\frac{1}{2}}\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-2]]

Solution
$$ \displaystyle i)\, Let's\, consider\, the\, Pythagorean\, theorem\, written\, by\, $$

$$ \displaystyle a^2 + b^2 = c^2\, $$

$$ \displaystyle where\, c\, represents\, the\, length\, of\, the\, hypotenuse,\, and\, a\, and\, b\, represent\, the\, lengths\, of\, the\, other\, two\, sides.\, $$

$$ \displaystyle ii)\, If\, the\, angle(dr)\, for\, dl\, is\, small,\, we\, can\, apply\, the\, Pythagorean\, theorem\, to\, the\, arc\, length\, formula\, like\, the\, below\, figure.\, $$



$$ \displaystyle iii) From\, the\, above\, figure,\, dl\, is\, given\, by\, $$

$$ \displaystyle dl^2 \cong dx^2 + dy^2\, $$

$$ \displaystyle Thus,\, dl = [dx^2 + dy^2]^{\frac{1}{2}}\, $$