User:Egm6341.s10.team3.sa/Mtg35

Mtg 35: Tue, 30 Mar 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 15:23, 12 August 2010 (UTC)

[[media: Egm6341.s10.mtg35.djvu | Page 35-1]] 

OPTIMAL CONTROL : Hermite-Simpson Algorithm: Local coordinate for $$\displaystyle [t_i,t_{i+1}]: $$ s such that

$$\displaystyle t(s)= (1-s)t_i + st_{i+1} $$ $$\displaystyle (Eq. 1) $$

$$\displaystyle s=0\; \Rightarrow \; t(0)=t_i $$ $$\displaystyle s=1\; \Rightarrow \; t(1)=t_{i+1} $$

$$\displaystyle z(s)= c_0 + c_1s + c_2s^2 +c_3s^3 \;\; s\in [0,1] $$ $$\displaystyle (Eq. 2) $$

At the end points, $$\displaystyle t_i\; and\; t_{i+1} $$ enforce compliance with the differential equation.

Collocation:

$$\displaystyle \dot{(z_i)} = f_i:= f(z_i,t_i)$$ $$\displaystyle (Eq. 3) $$

$$\displaystyle \dot{(z_{i+1})} = f_{i+1}:= f(z_{i+1},t_{i+1})$$ $$\displaystyle (Eq. 4) $$

Question: What happened at other points in $$\displaystyle [t_i,t_{i+1}] $$

[[media: Egm6341.s10.mtg35.djvu | Page 35-2]] 

In general, $$\displaystyle \dot{z_t} \neq f_t \;\; t\in (t_i,t_{i+1}) $$ $$\displaystyle (Eq. 1) $$

Recall that z(t) is an approximation and not an exact solution. i.e, z(t) does not satisfy the differential equation at all points but only at certain discrete points.

Continuing from [[media: Egm6341.s10.mtg34.djvu | Page 34-4]], $$\displaystyle z(s) = \sum_{i=0}^{3} c_i s^i = \sum_{i=0}^{3}\underbrace{N_i(s)}_{Basis function} \underbrace{d_i}_{DOF}$$ $$\displaystyle (Eq. 2) $$

$$\displaystyle d_1=z_i\;\; d_2=\dot{z_i}\;\; d_3=z_{i+1}\;\; d_4=\dot{z_{i+1}}$$ $$\displaystyle (Eq. 3) $$

$$\displaystyle \dot{z} = \frac{dz}{dt} = \underbrace{\frac{dz}{ds}}_{z^'} \frac{ds}{dt}$$ $$\displaystyle (Eq. 4) $$

From (Eq.1) of [[media: Egm6341.s10.mtg35.djvu | Page 35-1]] , $$\displaystyle \frac{dt}{ds} = t_{i+1}-t_i = \underbrace{h}_{Step-size} $$ $$\displaystyle (Eq. 5) $$

$$\displaystyle \therefore \frac{ds}{dt} = \frac{1}{h} $$

[[media: Egm6341.s10.mtg35.djvu | Page 35-3]]  Goal: To find $$\displaystyle {[c_i]} $$ in terms of $$\displaystyle {[d_i]} $$

Using (Eq.2) and (Eq.3) in [[media: Egm6341.s10.mtg35.djvu | Page 35-2]], $$\displaystyle d_1 = z_i = z(s=0) = c_0$$ $$\displaystyle (Eq. 1) $$

$$\displaystyle d_2 = \dot{z_i} = \dot{z}(t=0) = {z_i}^' \frac{1}{h} = z^'(s=0)\frac{1}{h}$$ $$\displaystyle (Eq. 2) $$

$$\displaystyle z^'(s) = \sum_{i=1}^{3} ic_is^{i-1}$$ <p style="text-align:right;">$$\displaystyle (Eq. 3) $$

$$\displaystyle z^'(s=0) = c_1$$ <p style="text-align:right;">$$\displaystyle (Eq. 4) $$

Similarly,

$$\displaystyle d_3 = z_{i+1} = z(s=1) $$ <p style="text-align:right;">$$\displaystyle (Eq. 5) $$

$$\displaystyle d_4 = {z_{i+1}}^' \frac{1}{h} = z^'(s=1)\frac{1}{h} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6) $$

$$\displaystyle \begin{bmatrix} 1 &0 &0  &0 \\ 0 &1  &0  &0 \\ 1 &1  &1  &1 \\ 0 &1  &2  &3 \end{bmatrix}

\begin{Bmatrix} c_0\\ c_1\\ c_2\\ c_3\end{Bmatrix}=

\begin{Bmatrix} z_i\\ z'_i\\ z_{i+1}\\ z'_{i+1} \end{Bmatrix}$$ <br\>

<p style="text-align:right;">$$\displaystyle (Eq. 7) $$


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HW:
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Evaluate $$\displaystyle \color{blue} z_{i+1}$$ and $$\displaystyle \color{blue} z'_{i+1} in terms of $$ $$\displaystyle \color{blue} c_0,c_1,c_2,c_3 $$ and represent it in the coefficient matrix


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<p style="text-align:left;">[[media: Egm6341.s10.mtg35.djvu | Page 35-4]] 

$$\displaystyle \begin{Bmatrix} c_0\\ c_1\\ c_2\\ c_3\end{Bmatrix}=

\begin{bmatrix} 1 &0 &0  &0 \\ 0 &1  &0  &0 \\ -3 &-2  &3  &-1 \\ 2 &-1  &-2  &1 \end{bmatrix}

\begin{Bmatrix} z_i\\ z'_i\\ z_{i+1}\\ z'_{i+1} \end{Bmatrix}$$ <br\>

<p style="text-align:right;">$$\displaystyle (Eq. 1) $$ Note: $$\displaystyle z_i=d_1=\bar{d_1}\;\; z'_i = \bar{d_2} \;\; z_{i+1}=d_3=\bar{d_3} \;\; z'_{i+1}=\bar{d_4} $$


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HW:
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Find the inverse of given Matrix : A = $$\displaystyle \color{blue} \begin{bmatrix} 1 &0 &0  &0 \\ 0 &1  &0  &0 \\ 1 &1  &1  &1 \\ 0 &1  &2  &3 \end{bmatrix} $$


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HW:
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Identify the basis functions $$\displaystyle \color{blue} N_i(s) \;where\; i = 1,2,3,4 $$


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