User:EGM6341.s11.team1.Chiu/S10 Mtg41

Mtg 41: Tue, 13 Apr 10

[[media: Egm6341.s10.mtg41.djvu| page41-1]]


 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

HW:  [[media: Egm6341.s10.mtg40.djvu| page40-3]] continued


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

Initial condition $$ \mathbf{Z}(t_{0})= \mathbf{Z}_{0} \ $$  S+Z 2007 page 316


 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

Controls: $$ \mathbf{u}(t)= \left \lfloor T(t) \alpha(t) \right \rfloor \ $$


 * }


 * nm1.s11.Mtg41.pg1.fig1.svg


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

Solve for $$ \mathbf{Z}(t), \ t \in \left [ 0,40 \right ] \ $$  at   $$ \left \| \underset{ \color{blue} \underset{2 \ successive}{\uparrow}}{ \mathbf{Z}^{(k+1)}}- \underset{ \color{blue} \underset{numerical \ solutions}{\uparrow}}{ \mathbf{Z}^{(k)}} \right \|_{ \infty} \leq \theta(10^{-6}) \ $$


 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

with solution  $$ \color{blue} \mathbf{Z}^{(k+1)} \ $$   obtained with double number of nodes compose to   $$  \color{blue} \mathbf{Z}^{(k)}. \ $$

Run also matlab ode45.


 * }


 * }

[[media: Egm6341.s10.mtg41.djvu| page41-2]]

Question: Why did Simpson's rule pop out in HS algorithms? (Why not some other rules?)

Answer: Recall Simpson can integrate exactly polynomial of  $$ degree \leq 3. \ $$ (See [[media: Egm6341.s10.mtg41.djvu| page14-1]] for the theorem, and [[media: Egm6341.s10.mtg13.djvu| page13-3]] for the geometric interpretation of the area cancellation)

HS algorithm: collocation (enforcing compliance with differential equation at 2 end points and mid point); use cubic Hermitian interpretation.

Upon enforcing compliance with differential equation at mid point (closing gap     $$ \Delta \ $$), then Simpson's rule naturally pops out, since it can integrate exactly a cubic polynomial with 3 points. [[media: Egm6341.s10.mtg39.djvu| page34-3]]

End Question & Answer


 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

HW: Solve logistic equation using inconsistent trapezoid-Simpson algorithm: [[media: Egm6341.s10.mtg39.djvu| (3) page36-3]]


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ Z_{i+1}=Z_{i}+ \frac{ \frac{h}{2}}{3} \left [ f_{i}+4f_{i+ \frac{1}{2}}+f_{i+1} \right ] \ $$


 * }

[[media: Egm6341.s10.mtg41.djvu| page41-3]]


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

with     $$ Z_{i+ \frac{1}{2}}= \frac{1}{2}(Z_{i}+Z_{i+1}) \ $$      (linear interpretation) Trapezoid


 * }

Compare with HS results using same values of     $$ h \ $$


 * }