User:Egm6341.s10.Team4/HW7

= Problem 1: Linearization about Equilibrium Point $$\displaystyle \hat{x} = x_{max}$$ =

Given
Refer Lecture slide 38-4 for problem statement

The logistic equation for population dynamics is given by 38-3,


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$$\displaystyle \dot{x}(t) = f(x) = rx \left(1-\frac{x}{x_{max}}\right) $$
 * $$\displaystyle \longrightarrow (1) $$
 * }.
 * }.

Find
Linearize Equation (1) about the equilibrium point $$\displaystyle \hat{x} = x_{max}$$.

Solution
With a small perturbation $$\displaystyle y $$, the value of $$\displaystyle x $$ is given by,


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$$\displaystyle \begin{align} x &= \hat{x} + y \\ &= x_{max} + y \end{align} $$
 * }.
 * }.


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$$\displaystyle \begin{align} \Rightarrow \frac{dx}{dt} &= \frac{d(x_{max} + y)}{dt} \\ &= \frac{dy}{dt} \end{align} $$
 * }.
 * }.


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$$\displaystyle \begin{align} \Rightarrow \frac{dy}{dt} &= r(x_{max}+y) \left(1-\frac{(x_{max}+y)}{x_{max}}\right) \\ &= r(x_{max}+y) \left(-\frac{y}{x_{max}}\right) \\ &= -r(y+\frac{y^2}{x_{max}})

\end{align} $$
 * }.
 * }.

Since, $$\displaystyle y $$ is a small perturbation compared to $$\displaystyle x_{max} $$,


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$$\displaystyle \begin{align} \Rightarrow \frac{dy}{dt} &= -r(y+\cancelto{0}{\frac{y^2}{x_{max}}})\\ &= -ry \end{align} $$
 * }.
 * }.

And so, the solution


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$$\displaystyle \begin{align} \Rightarrow y = y_0 e^{-rt} \Rightarrow &y(t) \to 0 \ as\ t \to \infty \\ \Rightarrow &x(t) \to x_{max} \ as\ t \to \infty \end{align} $$
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.andy 06:16, 23 April 2010 (UTC)

Reviewed by - Egm6341.s10.Team4.roni 17:58, 23 April 2010 (UTC)

= Problem 2: Analytical equation for $$\displaystyle x(t) $$ in Newton-Raphson method =

Given
The relationship between population growth $$\displaystyle x $$ respect to time $$\displaystyle t $$ can be expressed as follows:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dot x=\frac{dx}{dt}=rx(1-\frac{x}{x_{max}})


 * }
 * }

Find
Find the analytical equation to show $$\displaystyle x $$ as a function of time $$\displaystyle t $$:

Solution

 * $$\displaystyle

\dot x=\frac{dx}{dt}=rx(1-\frac{x}{x_{max}}) $$


 * $$\displaystyle

\Rightarrow \frac{dx}{rx(1-\frac{x}{x_{max}}})=dt $$

For LHS we have:


 * $$\displaystyle

\frac{1}{x(1-\frac{1}{x_{max}}x)}=\frac{a}{x}+\frac{b}{1-\frac{1}{x_{max}}x}=\frac{a-\frac{a}{x_{max}}x+bx}{x(1-\frac{1}{x_{max}}x)} $$


 * $$\displaystyle

\Rightarrow a-\frac{a}{x_{max}}x+bx=1\Rightarrow x(b-\frac{a}{x_{max}})+a=1 $$


 * $$\displaystyle

\Rightarrow a=1 $$


 * $$\displaystyle

\Rightarrow b-\frac{a=1}{x_{max}}=0\Rightarrow b=\frac{1}{x_{max}} $$


 * $$\displaystyle

\Rightarrow \frac{1}{x(1-\frac{1}{x_{max}}x)}=\frac{1}{x}+\frac{\frac{1}{x_{max}}}{1-\frac{1}{x_{max}}x} $$


 * $$\displaystyle

\Rightarrow \frac{1}{r}\int\limits_{x_0}^{x}\frac{1}{x(1-\frac{1}{x_{max}}x)}\, dx=\frac{1}{r}\bigg[\int\limits_{x_0}^{x}\frac{1}{x}\,dx+\frac{1}{x_{max}}\int\limits_{x_0}^{x}\frac{1}{1-\frac{1}{x_{max}}x}\,dx\bigg]=\int\limits_{t_0}^{t}\,dt $$


 * $$\displaystyle

\Rightarrow \frac{1}{r}\bigg[ln(x)-ln(1-\frac{1}{x_{max}}x)]\Bigg ]^{x}_{x_0}=\frac{1}{r}\bigg[ln(\frac{x}{1-\frac{x}{x-{max}}})-ln(\frac{x_0}{1-\frac{x_0}{x_{max}}})\Bigg]=\int\limits_{t_0}^{t}\,dt $$


 * $$\displaystyle

\Rightarrow \frac{1}{r}ln(\frac{\frac{x}{1-\frac{x}{x_{max}}}}{\frac{x_0}{1-\frac{x_0}{x_{max}}}})=\frac{1}{r}ln(\frac{x(1-\frac{x_0}{x_{max}})}{x_0(1-\frac{x}{x_{max}})})=\int\limits_{t_0}^{t}\, dt=t-t_0 $$

Initial condition:


 * $$\displaystyle

t_0=0\Rightarrow x=x_0. $$


 * $$\displaystyle

\Rightarrow ln(\frac{x(1-\frac{x_0}{x_{max}})}{x_0(1-\frac{x}{x_{max}})})=r(t-t_0) $$


 * $$\displaystyle

\Rightarrow e^{rt}=\frac{x(1-\frac{x_0}{x_{max}})}{x_0(1-\frac{x}{x_{max}})}=\frac{x}{1-\frac{x}{x_{max}}}\times\frac{1-\frac{x_0}{x_{max}}}{x_0}=\frac{x_{max}.x}{x_{max}-x}\times\frac{x_{max}-x_0}{x_{max}.x_0} $$


 * $$\displaystyle

\Rightarrow \frac{x^2_{max}.x-x_{max}.x_0.x}{x^2_{max}.x_0-x_{max}.x_0.x}=e^{rt} $$


 * $$\displaystyle

\Rightarrow (x^2_{max}.x-x_{max}.x_0.x)=(x^2_{max}.x_0-x_{max}.x_0.x)e^{rt} $$


 * $$\displaystyle

\Rightarrow x(x^2_{max}-x_0.x_{max})=(x^2_{max}.x_0)e^{rt}-x_{max}.x_0.e^{rt}.x $$


 * $$\displaystyle

\Rightarrow x(x^2_{max}-x_0.x_{max}+x_{max}.x_0.e^{rt})=x^2_{max}.x_0.e^{rt} $$


 * $$\displaystyle

\Rightarrow x=\frac{x^2_{max}.x_0.e^{rt}}{x^2_{max}-x_0.x_{max}+x_{max}.x_0.e^{rt}}=\frac{x_{max}.x_0.e^{rt}}{x_{max}-x_0+x_0.e^{rt}} $$


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$$\displaystyle \Rightarrow x(t)=\frac{x_0.x_{max}.e^{rt}}{x_{max}+x_0(e^{rt}-1)} $$
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 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 15:50, 11 April 2010 (UTC) .

= Problem 3: Hermite-Simpson Rule to interpret the logistic Equation =

Given
Logistic Equation is given by $$\displaystyle \dfrac{dx}{dt}=rx(1-\dfrac{x}{x_{max}})$$

where $$\displaystyle x_{max}=10; r=1.2; x_0=2,7; t\in[0,10]$$

Find
Use Hermit Simpson rule to interpret the above equation.

Solution
This problem has been solved by writing a matlab code.

Hermit Simpson Code:

Case(1) : $$ \displaystyle x_0 = 2.0 $$ Plot: .

Case(2) : $$ \displaystyle x_0 = 7.0 $$ Plot: .

Author
Solved and typed by - Egm6341.s10.team4.anandankala 14:49, 23 April 2010 (UTC)

Proof read by -

= Problem 4: Discrete Population Dynamics - Logistic Map - Chaotic System - Period Doubling =

Given
Class notes and Pages 454-456 King et al 2003 - An Introduction to Chaotic Systems.

Given factor r -represents the growth from generation to the next [[media:Egm6341.s10.mtg39.djvu|39-3]]

Find
Run the given Code to reproduce plots 15.6 and 15.7 in King et al 2003.

Solution
original Code

Matlab Code:

Plot 1 shows the results from running the original code



Modification to the original Code

Matlab Code:

Plot 2 shows the results from running the modified code



Discussion:

From Plot 2 One can see that when r=4, a small change in the input of the initial x value would give apparently a random nature of solution and divergence after about 50 iterations. The 0 marker in blue are for x=0.1 starting point and the x markers in green are for x=0.1+1e-16. Initially (up to about 50 iterations) the x and o symbols fall one on to of the other (giving same solution) and diverging after about 50 iterations.

Author
Solved and typed by - Egm6341.s10.Team4.roni 18:49, 18 April 2010 (UTC)

Proof read by -

= Problem 5: Integration of logistic equation with $$h=2^kh $$=

Given
Refer Lecture slide 42-2 for problem statement

The logistic equation


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\frac{d\overline{x}}{dt} = rx\left(1- \overline{x}\right) $$ where,
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

$$\displaystyle \overline{x} = \frac{x}{x_{max}} $$

Find
Integrate logistic equation already found $$\hat{h}$$ such that Hermit-simpson algorithm yields error($$ 10^{-6}$$)

1) Run Hermit-Simpson with $$\displaystyle h= 2^k \hat{h}$$

2) Develop and Run Forward Euler with $$\displaystyle h= 2^k \hat{h}$$

3) Develop and Run Backward Euler with $$\displaystyle h= 2^k \hat{h}$$

Solution
1) Run Hermit-Simpson algo $$\displaystyle h= 2^k \hat{h}$$

See problem 3

2) Develop and Run Forward Euler with $$\displaystyle h= 2^k \hat{h}$$


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\begin{align} &\frac{d\overline{x}}{dt} = r\overline{x}\left(1- \overline{x}\right)\\ &\frac{\overline{x}_{i+1}-\overline{x}_i}{h}= r\overline{x_i}\left(1- \overline{x_i}\right)\\ &\overline{x}_{i+1}= hr\overline{x}_i\left(1- \overline{x}_i\right)+\overline{x}_i\\ &\overline{x}_{i+1}= (hr+1)\overline{x}_i-hr\overline{x}^2_i\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

MATLAB CODE:



3) Develop and Run Backward Euler with $$\displaystyle h= 2^k \hat{h}$$


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\begin{align} &\frac{d\overline{x}}{dt} = r\overline{x}\left(1- \overline{x}\right)\\ &\frac{\overline{x}_{i+1}-\overline{x}_i}{h}= r\overline{x}_{i+1}\left(1- \overline{x}_{i+1}\right)\\ &-\overline{x}_{i}= hr\overline{x}_{i+1}\left(1- \overline{x}_{i+1}\right)-\overline{x}_{i+1}\\ &=>hr\overline{x}^2_{i+1}+(1-hr)\overline{x}_{i+1}-\overline{x}_{i}= 0\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.


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\begin{align} &\overline{x}_{i+1}=\frac{hr-1 + \sqrt{(1-hr)^2+4hrx_i}}{2hr}\\ &\overline{x}_{i+1}=\frac{hr-1 - \sqrt{(1-hr)^2+4hrx_i}}{2hr}\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.



MATLAB CODE:

Conclusion

As the step size($$ 2^kh$$) increase, the Forwaord Euler method shows the unstable properties.

on the other hands, The Hermit simpson and Backworkd Euler shows the relatively stable properties.

Author
Solved and typed by - Egm6341.s10.Team4.yunseok 19:42, 23 April 2010 (UTC) .

= Problem 6 - Hermite-Simpson Integration of Coupled ODE's =

Given
This is a flight control problem. The dynamics of flight are well documented and will not be described in detail here.

General Method:

By using collocation to force our Hermite-Simpson interpolation to exactly solve the given ODE's at the mid point $$ \displaystyle z_{i+1/2}$$, we find that our time stepping method takes the form

$$ \displaystyle \mathbf{z_{i+1}}-\mathbf{z_{i}}=\frac{\tfrac{h}{2}}{3}(\mathbf{f_i}+4\mathbf{f_{i+1/2}+\mathbf{f_{i_1}}})$$

Additionally, using our interpolation $$ \displaystyle \mathbf{f_{i+1/2}}=\tfrac{1}{2}(\mathbf{z_i}+\mathbf{z_{i=1}}+\tfrac{h}{8}(\mathbf{f_i}-\mathbf{f_{i+1}}))$$.

However, no exact solution exists in order to determine $$ \displaystyle \mathbf{z_{i+1}}$$. Given $$ \displaystyle \mathbf{z_{i}}$$, the Newton-Rhapson method can be employed in order to find $$ \displaystyle \mathbf{z_{i+1}}$$. By converting $$ \displaystyle \mathbf{z_{i+1}}-\mathbf{z_{i}}=\frac{h}{6}(\mathbf{f_i}+4\mathbf{f_{i+1/2}+\mathbf{f_{i_1}}})$$ to $$ \displaystyle \mathbf{F(z_i,z_{i+1})}=\mathbf{0} $$, the Newton-Rhapson method takes the form

$$ \displaystyle \mathbf{z_{i+1}^{k+1}}=\mathbf{z_{i+1}^{k}}- \begin{bmatrix}\mathbf{\frac{dF(z_i,z_{i+1}^{k})}{dz_{i+1}}}\end{bmatrix}^{-1}\mathbf{F(z_i,z_{i+1}^{k})} $$

where

$$ \displaystyle \mathbf{F(z_i,z_{i+1})}= \mathbf{z_{i+1}}-\mathbf{z_{i}}- \frac{h}{6}(\mathbf{f_i}+4\mathbf{f_{i+1/2}+\mathbf{f_{i_1}}})=0 $$

Equations of Motion:

Let

$$ \displaystyle \begin{bmatrix} z \end{bmatrix} = \begin{bmatrix} x \\ y \\ v \\ \gamma\end{bmatrix}$$

The equations of motion that we are concerned with are

$$ \displaystyle \begin{bmatrix} \dot{z} \end{bmatrix} = \begin{bmatrix} \dot{x} \\ \dot{y} \\ \dot{v} \\ \dot{\gamma} \end{bmatrix} = \begin{bmatrix} v cos \gamma \\ v sin \gamma \\ \frac{1}{m}((T-D) cos \alpha - L sin \alpha - g sin \gamma)\\ \frac{T-D}{m}sin \alpha+\frac{L}{mv}cos \alpha -\frac{g}{v}cos \gamma \end{bmatrix} $$

The physical parameters for which we will solve using are

$$ \displaystyle D=\tfrac{1}{2}C_d\rho v^2 S_{ref} $$

$$ \displaystyle L=\tfrac{1}{2}C_l\rho v^2 S_{ref} $$

where

$$ \displaystyle C_d=a_1\alpha^2+a_2\alpha+a_3 $$

$$ \displaystyle C_l=b_1\alpha+b_2 $$

$$ \displaystyle \rho=c_1y^2+c_2y+c_3 $$

$$ \displaystyle a_1=-1.9431 \tfrac{1}{rad^2}$$

$$ \displaystyle a_2=-0.1499 \tfrac{1}{rad}$$

$$ \displaystyle a_3=0.2359 $$

$$ \displaystyle b_1=21.9 \tfrac{1}{rad}$$

$$ \displaystyle b_2=0.0 $$

$$ \displaystyle c_1=3.312E-9 \tfrac{kg}{m}$$

$$ \displaystyle c_2=-1.142E-4 \tfrac{kg}{m^2}$$

$$ \displaystyle c_3=1.224 \tfrac{kg}{m^3}$$

The initial conditions are

$$ \displaystyle x(t=0)=0.0 m$$

$$ \displaystyle y(t=0)=30.0 m $$

$$ \displaystyle v(t=0)=272.0 \tfrac{m}{s}$$

$$ \displaystyle \gamma(t=0)=0.0 rad $$

The flight is controlled using the angle of attack and the thrust, which are shown below



Find
We are asked to determine x,y,v, and \gamma using the Hermite-Simpson/Newton-Rhapson Method and then verify this with MATLAB's ode solver ode45.

Solution
In performing this analysis, we see that our Jacobian, $$ \displaystyle |\mathbf{\tfrac{dF}{dz_{i+1}}}|$$ can be described as

$$ \displaystyle \mathbf{\frac{dF(z_{i},z_{i+1})}{dz_{i+1}}}= \mathbf{\frac{dz_{i+1}}{dz_{i+1}}}-\mathbf{\frac{dz_{i}}{dz_{i+1}}}- \frac{h}{6}(\mathbf{\frac{df_{i}}{dz_{i+1}}}+4\mathbf{\frac{df_{i+1/2}}{dz_{i+1}}}+ \mathbf{\frac{df_{i+1}}{dz_{i+1}}})$$

Using the interpolation for z_{i+1/2} as a function of z_{i} and z_{i+1} described above, this becomes

$$ \displaystyle \mathbf{\frac{dF(z_{i},z_{i+1})}{dz_{i+1}}}= \mathbf{\frac{dz_{i+1}}{dz_{i+1}}}-\mathbf{\frac{dz_{i}}{dz_{i+1}}}- \frac{h}{6}(\mathbf{\frac{df_{i}}{dz_{i+1}}}+4\mathbf{\frac{df_{i+1/2}}{dz_{i+1/2}}\frac{dz_{i+1/2/2}}{dz_{i+1}}}+ \mathbf{\frac{df_{i+1}}{dz_{i+1}}})$$

In evaluating this equation, our Jacobian can be described as

$$ \displaystyle \mathbf{\frac{dF(z_{i},z_{i+1})}{dz_{i+1}}}= \mathbf{I}- \frac{h}{6}(3\mathbf{I}-\frac{h}{2}\mathbf{\frac{df_{i+1}}{dz_{i+1}}}) \mathbf{\frac{df_{i+1}}{dz_{i+1}}}$$

Thus, as long as $$ \displaystyle \tfrac{df}{dz} $$ can be calculated, the Jacobian can be found.

The obvious next step is to calculate $$ \displaystyle \tfrac{df}{dz} $$. The full expansion of $$ \displaystyle \tfrac{df}{dz} $$ can be seen to be

$$ \displaystyle

\frac{\mathbf{df}}{\mathbf{dz}}= \begin{bmatrix} 0 & 0 & cos(\gamma) & -v sin(\gamma) \\ 0 & 0 & sin(\gamma) & v cos(\gamma) \\ 0 & \frac{1}{m}(-\frac{dD}{dy}cos\alpha -\frac{dL}{dy}cos\alpha)& \frac{1}{m}(-\frac{dD}{dv}cos\alpha -\frac{dv}{dy}cos\alpha & \frac{1}{m}(-gcos\gamma)\\ 0 & \frac{1}{mv}(\frac{dD}{dy}sin\alpha+\frac{dL}{dy}cos\alpha)& -\frac{T}{mv^2}sin\alpha -\frac{1}{m}\frac{d}{dv}(\frac{D}{v})cos\alpha+\frac{1}{m}\frac{d}{dv}(\frac{L}{v})cos\alpha+\frac{g}{v^2}cos\gamma & \frac{g}{v}sin\gamma\\ \end{bmatrix}

$$

In performing the evaluations we can see that the two forms of numerical integration produce a very similar output.



Source Codes

Main Program

Calculate the Jacobian

Calculate f

Function for ode 45 to use

Author
Solved and typed by Egm6341.s10.Team4.riherd 15:24, 23 April 2010 (UTC)

= Problem 7: Logistic Equation: Inconsistent (Trapezoidal) Simpson's rule & Newton-Raphson =

Given
Refer Lecture slide 41-2 for problem statement

The logistic equation for population dynamics is given by 38-3,


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$$\displaystyle \dot{x}(t) = f(x) = rx \left(1-\frac{x}{x_{max}}\right) $$
 * $$\displaystyle \longrightarrow (1) $$
 * }.
 * }.

Here, $$\displaystyle x_{max}=10;\ r = 1.2,\ $$ and $$\displaystyle t \in [0,10] $$

The analytical solution is given by,


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$$\displaystyle x(t) = \frac{x_0x_{max}e^{rt}}{x_{max} + x_0(e^{rt}-1)} $$ $$
 * $$\displaystyle \longrightarrow (2)
 * }.
 * }.

Find
To solve logistic equation defined by Equation (1) for two initial conditions, $$\displaystyle x_0=2\ and\ x_0 = 7,\ $$, using Inconsistent(Trapezoidal) Simpson's rule given by,


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$$\displaystyle z_{i+1} = z_{i} + \frac{h/2}{3} \left[f_i + 4f_{(i+1/2)} +f_{i+1}\right] $$
 * $$\displaystyle \longrightarrow (3) $$
 * }.
 * }.

The Inconsistent rule (Trapezoidal) is given by,


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$$\displaystyle z_{i+1/2} = \frac{1}{2} [z_{i} + z_{i+1}] $$
 * }.
 * }.

Solution
Consider the Equation (1) in which $$\displaystyle f_{(i+1/2)}$$ is a function of $$\displaystyle z_{i}$$ and $$\displaystyle z_{i+1}$$, where $$\displaystyle z_i, z_{i+1} \rightarrow$$ values of $$\displaystyle x$$ at $$\displaystyle t = t_i$$ and $$\displaystyle t = t_{i+1}$$.

Since, this is an initial value problem, $$\displaystyle z_i$$ is known. So, Equation (3) as a whole becomes a function of $$\displaystyle z_{i+1}$$ given by,


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$$\displaystyle F(z_{i+1}) = 0 $$
 * $$\displaystyle \longrightarrow (4) $$
 * }.
 * }.

We find the root $$\displaystyle z_{i+1}$$ of Equation (4) using Newton-Raphson method 40-3,


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$$\displaystyle z_{i+1}^{(k+1)} = z_{i+1}^{(k)} - \left[\frac{dF(z_{i+1}^{(k)})}{dz}\right]^{-1} F(z_{i+1}^{(k)}) $$ $$
 * $$\displaystyle \longrightarrow (5)
 * }.
 * }.

The Newton-Raphson iteration, starting with an initial guess as $$\displaystyle z_{i+1}^{0} = z_{i}$$, is stopped once the absolute tolerance


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$$\displaystyle \begin{align} AbsTol &= ||z_{i+1}^{(k+1)} - z_{i+1}^{(k)}|| \\ &\le 10^{-6} \end{align} $$
 * }.
 * }.

is satisfied.

In the below Matlab code, for each of the h values $$\displaystyle x_{Actual} $$ is the vector calculated from Equation (3) in order to calculate the error max (for reference), $$\displaystyle e_{Max} = || x_{Actual} - x ||_{\infty} $$ where $$\displaystyle x $$ is calculated from Inconsistent Simpson's rule - Newton Raphson Method described above, for the given interval $$\displaystyle t \in [0,10] $$.

Matlab Code:

$$\displaystyle x_0 = 2 $$ :- A comparison on $$\displaystyle h $$ values:



The plot for $$\displaystyle h \approxeq 0.001 $$:



$$\displaystyle x_0 = 7 $$ :- A comparison on $$\displaystyle h $$ values:



The plot for $$\displaystyle h \approxeq 0.001 $$:



Author
Solved and typed by - Egm6341.s10.Team4.andy 05:49, 23 April 2010 (UTC) Reviewed by - Egm6341.s10.Team4.roni 18:00, 23 April 2010 (UTC)

= Problem 8: Rate of momentum change for optimal control problem =

Given
Envisage the below figure as free body diagram of an aircraft:



Find
Show that $$\displaystyle \bar{CD}=\bar{AB} $$ + higher order terms of $$\displaystyle d\gamma $$.

Solution
According to the above figure, we can survey these two cases at $$\displaystyle t $$ and $$\displaystyle t+dt $$, the velocity of the aircraft after $$\displaystyle dt $$ will reach to $$\displaystyle v+dv $$ and the angle between the aircraft and horizontal axis will reach to the $$\displaystyle \gamma+d\gamma $$. Thus, for two generated triangles $$\displaystyle \triangle OAB $$ and $$\displaystyle \triangle ODC $$ following relationships can be written:


 * $$\displaystyle

tan(d\gamma)=\frac{\bar {AB}}{\bar {OA}}=\frac{\bar {AB}}{v} $$

Likewise we have below relationship in $$\displaystyle \triangle ODC $$:


 * $$\displaystyle

sin(d\gamma)=\frac{\bar {CD}}{\bar {OD}}=\frac{\bar {CD}}{v+dv} $$

Hence, by applying Taylor series for these two cases:


 * $$\displaystyle

tan(d\gamma)=d\gamma+\frac{d\gamma^3}{3}+\frac{2d\gamma^5}{15}+...=\frac{\bar{AB}}{v} $$


 * $$\displaystyle

\Rightarrow \bar{AB}=v(d\gamma+\frac{d\gamma^3}{3}+...)=v.d\gamma+v\frac{d\gamma^3}{3}+v\frac{2d\gamma^5}{15}+... $$


 * $$\displaystyle

sin(d\gamma)=d\gamma-\frac{d\gamma^3}{3!}+\frac{d\gamma^5}{5!}+...=\frac{\bar{CD}}{v+dv} $$


 * $$\displaystyle

\Rightarrow \bar{CD}=(v+dv)(d\gamma-\frac{d\gamma^3}{3!}+\frac{d\gamma^5}{5!}...)=v.d\gamma-v\frac{d\gamma^3}{3!}+v\frac{d\gamma^5}{5!}+... $$


 * $$\displaystyle

\Rightarrow \bar{CD}-\bar{AB}=-\frac{1}{2}vd\gamma^3-\frac{1}{8}vd\gamma^5... $$


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

$$\displaystyle \Rightarrow \bar{CD}=\bar{AB}+v.f(d\gamma^3)+v.f(d\gamma^5)+... $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 22:19, 22 April 2010 (UTC) .

= Problem 9: Parameterization of ellipse=

Given
$$\displaystyle x=a cost; y=b sint.$$

Prove That
$$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$

Solution
We have $$\displaystyle x=acost; y=bsint.$$

$$\displaystyle x=a cost \Rightarrow x^2=a^2(cost)^2$$

$$\displaystyle y=b sint \Rightarrow y^2=b^2(sint)^2$$

$$\displaystyle \Rightarrow \frac{x^2}{a^2}=(cost)^2 and \frac{y^2}{b^2}=(sint)^2 $$

$$\displaystyle \Rightarrow \frac{x^2}{a^2}+ \frac{y^2}{b^2}=(cost)^2+(sint)^2 $$

$$\displaystyle \Rightarrow \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 $$

Author
Solved and typed by - Egm6341.s10.team4.anandankala 14:50, 23 April 2010 (UTC) .

= Problem 10: Parameterization of an Ellipse =

Given
Result from HW problem 7.9 [[media:Egm6341.s10.mtg42.djvu|42-1]], which is:

$$\displaystyle Eq. (1)  \qquad    \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 $$

Show
That the arc length dl comes to:

HW 7.10 on slide [[media:Egm6341.s10.mtg42.djvu|42-2]]

$$\displaystyle

Eq. (2) \qquad  dl= \sqrt{dx^2+dy^2} $$

Solution
Geometrically:


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


 * [[Image:RP_HW_7_10c.png]]
 * }.
 * }.
 * }.



Eq. (3) \qquad  dl^2= {dx^2+dy^2} $$



Eq. (4) \qquad  dl=  \sqrt{dx^2+dy^2} $$

and also:



Eq. (5) \qquad  dl =   \left( \sqrt{1+  \left( \frac{dy}{dx}\right)^2} \right) dx

$$


 * {| style="width:20%" border="0"

$$\displaystyle dl= \sqrt{dx^2+dy^2} $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |


 * style= |
 * }
 * }

Author
Solved and typed by - Egm6341.s10.Team4.roni 18:50, 18 April 2010 (UTC) Reviewed by -

= Problem 11: Circumference of Ellipse =

Given
Lecture notes [[media:Egm6341.s10.mtg42.djvu|p.42-2]]

Arc length
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dl = [dx+dy]^{1/2}\,dt$$
 * $$\displaystyle
 * $$\displaystyle


 * $$ \displaystyle x = a cost $$


 * $$ \displaystyle y = b sint $$


 * }
 * }

Find

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$$\displaystyle C = \int dl = a \int_{0}^{2\pi} [1-e^2cos^2t]^{1/2}\,dt$$ where, Eccentricity, $$ \displaystyle e = \Bigg[1-\frac{b^2}{a^2}\Bigg]^{1/2} $$
 * Show that,
 * Show that,
 * }
 * }

Solution

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$$\displaystyle \frac {dy}{dt} = b cos t $$  and
 * $$\displaystyle \frac {dx}{dt} = -a sin t $$  
 * $$\displaystyle \frac {dx}{dt} = -a sin t $$  
 * }
 * }


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\begin{align} dl &= [dx^2 + dy^2 ]^{1/2}\\ &= \Bigg[\bigg(\frac{dx}{dt}\bigg)^2 + \bigg(\frac{dy}{dt}\bigg)^2 \Bigg]^{1/2} dt\\ &=\Bigg[(-a sin t)^2 + (b cos t)^2\Bigg]^{1/2} dt\\ &= \Bigg[(a^2 sin^2 t) + (b^2 cos^2 t)\Bigg]^{1/2} dt\\ &= a \Bigg[sin^2 t + \left(\frac{b^2}{a^2} cos^2 t\right)\Bigg]^{1/2} dt\\ &= a \Bigg[sin^2 t + cos^2 t + \frac{b^2}{a^2} cos^2 t- cos^2 t \Bigg]^{1/2} dt\\ &= a \Bigg[1 -\left(1-\frac{b^2}{a^2}\right) cos^2 t \Bigg]^{1/2} dt\\ &= a \Bigg[1 - e^2 cos^2 t \Bigg]^{1/2} dt\\ \end{align} $$ So,
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle C = \int_{0}^{2\pi} dl = a \int_{0}^{2\pi} [1-e^2cos^2t]^{1/2}\,dt$$
 * }
 * }

Author
Solved and typed by - Egm6341.s10.Team4.yunseok 19:42, 23 April 2010 (UTC) .

= Problem 12 - Manipulation of Elliptical Integral =

Given
$$ \displaystyle C=a\int_{t=0}^{2\pi}\sqrt{1+e^2cos^2t} dt $$

Find
We are asked to show that $$ \displaystyle C=a\int_{t=0}^{2\pi}\sqrt{1+e^2cos^2t} dt =4 a\int_{t=0}^{\tfrac{\pi}{2}}\sqrt{1+e^2sin^2\alpha} d\alpha $$

Solution
Let $$ \displaystyle t=cos^{-1}(sin\alpha)$$ and $$ \displaystyle dt=cos\alpha\frac{1}{\sqrt{1-sin^2\alpha}}d\alpha=cos\alpha\frac{1}{cos\alpha}d\alpha=d\alpha $$.

Putting this into our equation,

$$ \displaystyle C = a \int_{t=0}^{2\pi} \sqrt{1+e^2 sin^2 \alpha} d\alpha $$

The function $$ \displaystyle \sqrt{1+e^2 sin^2 \alpha}$$ is periodic over a period of $$ \displaystyle \pi$$ and in this period the function is even around the point $$ \displaystyle \alpha=\tfrac{\pi}{2}$$. Thus, by symmetry, this integral can be broken up into 4 parts of equal magnitude such that

$$ \displaystyle C= 4 a\int_{t=0}^{\tfrac{\pi}{2}}\sqrt{1+e^2sin^2\alpha} d\alpha $$

Author
Solved and typed by Egm6341.s10.Team4.riherd 15:50, 23 April 2010 (UTC)

= Problem 13: Chebychev - Change of Variable =

Given
Refer Lecture slide 42-2 for problem statement

The integral for a function $$\displaystyle f(x)$$ is,


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$$\displaystyle I = \int_{-1}^{+1}f(x)dx $$
 * $$\displaystyle \longrightarrow (1)$$
 * }.
 * }.

between the interval $$\displaystyle x \in [-1,1]$$

Find
With a change of variable $$\displaystyle x = cos\theta$$, show that the integral in Equation (1) is,


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$$\displaystyle I = \int_{0}^{\pi} f(cos\theta)sin\theta d\theta $$
 * }.
 * }.

Solution
With a change of variable $$\displaystyle x = cos\theta \Rightarrow dx = -sin\theta d\theta ; cos(\pi) = x = -1 \Rightarrow \theta = \pi; cos(0) = x = 1 \Rightarrow \theta = 0$$ Equation (1) is,


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$$\displaystyle \begin{align}

I &= \int_{-1}^{1} f(x)dx \\ &= \int_{\pi}^{0} f(cos\theta) \left(-sin\theta d\theta \right) \\ &= - \int_{\pi}^{0}f(cos\theta) sin\theta d\theta \\ &= \int_{0}^{\pi}f(cos\theta) sin\theta d\theta \\

\end{align} $$
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.andy 06:32, 23 April 2010 (UTC)

= Problem 14: Obtaining $$\displaystyle a_k $$ in cosine fourier transform =

Given
The Cosine series as Fourier transform for function $$\displaystyle f(x) $$ is expressed as follows:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f(cos\theta)=\frac{a_0}{2}+\sum_{k=1} a_k.cos(k\theta)


 * }
 * }

Find
Demonstrate the following expression for $$\displaystyle a_k $$:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle a_k=\frac{2}{\pi}\int\limits_{0}^{\pi} f(cos\theta).cos(k\theta)\, d\theta


 * }
 * }

Solution
Considering $$\displaystyle x=cos(\theta) $$, cosine Fourier series for $$\displaystyle f(x) $$ is written as:


 * $$\displaystyle

f(x=cos(\theta))=\frac{a_0}{2}+\sum_{k=1} a_k.cos(k\theta) $$


 * $$\displaystyle

\xrightarrow{\times cos(k\theta)}f(x).cos(k\theta)=\frac{a_0}{2}.cos(k\theta)+\sum_{k=1} a_k.cos^2(k\theta) $$


 * $$\displaystyle

\xrightarrow{\int\limits_{0}^{\pi}}\int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta=\frac{a_0}{2}\int\limits_{0}^{\pi}cos(k\theta)\, d\theta+\sum_{k=1} a_k\int\limits_{0}^{\pi}cos^2(k\theta)\, d\theta $$


 * $$\displaystyle

\frac{a_0}{2}\times\underbrace{\frac{1}{k}[sin(k\theta)]_{0}^{\pi}}_{=0}+\sum_{k=1}a_k\int\limits_{0}^{\pi}cos^2(k\theta)\, d\theta=\sum_{k=1}a_k\int\limits_{0}^{\pi}cos^2(k\theta)\, d\theta $$


 * $$\displaystyle

cos(2\theta)=cos^2(\theta)-sin^2(\theta)=2cos^2(\theta)-1\Rightarrow cos^2(\theta)=\frac{cos(2\theta)+1}{2} $$


 * $$\displaystyle

\Rightarrow \int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta=\sum_{k=1} \lbrack a_k\times\frac{1}{2}\int\limits_{0}^{\pi}(cos(2\theta)+1)\, d\theta\rbrack=\frac{1}{2}\sum_{k=1}a_k\lbrack\int\limits_{0}^{\pi}cos(2\theta)\, d\theta+\int\limits_{0}^{\pi}\, d\theta]=\frac{1}{2}\sum_{k=1}a_k\lbrack[\underbrace{\frac{1}{2}sin(2\theta)]_{0}^{\pi}}_{=0}+\underbrace{[\theta]_{0}^{\pi}}_{\pi}\rbrack=\frac{\pi}{2}\sum_{k=1}a_k $$


 * $$\displaystyle

\Rightarrow a_k=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta $$


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$$\displaystyle a_k=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta $$
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 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 19:15, 18 April 2010 (UTC)

Reviewed by - Egm6341.s10.Team4.roni 18:02, 23 April 2010 (UTC)

= Problem 15: Integral Evaluation=

Given
$$\displaystyle f(cos\theta)= \frac{a_0}{2} + \sum_{k=1}^{\infty} a_kcos(k\theta)$$

Prove that
$$\displaystyle I= a_0 + \sum_{j=1}^{\infty} \frac{2a_{2j}}{1-(2j)^2}$$

Solution
$$\displaystyle I= \int_{-1}^{1}f(x)dx= \int_{0}^{\pi}f(cos\theta)sin\theta d\theta$$

$$\displaystyle \Rightarrow I= \int_{0}^{\pi}(\frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos(k\theta))sin\theta d\theta$$

$$\displaystyle \Rightarrow I= \int_{0}^{\pi} \frac{a_0}{2} sin\theta d\theta + \sum_{k=1}^{\infty}\int_{0}^{\pi} a_kcos(k\theta))sin\theta d\theta $$

$$\displaystyle \Rightarrow I= \dfrac{a_0}{2} [cos\theta]_{\pi}^{0}+ \sum_{k=1}^{\infty}\int_{0}^{\pi} \dfrac{a_k}{2} [sin((k+1)\theta) -sin((k-1)\theta)]d\theta $$

$$\displaystyle \Rightarrow I= a_0+ \sum_{k=1}^{\infty}\dfrac{a_k}{2} [\int_{0}^{\pi} sin((k+1)\theta)d\theta-\int_{0}^{\pi} sin((k-1)\theta)d\theta] $$

$$\displaystyle \Rightarrow I= a_0+ \sum_{k=1}^{\infty} \dfrac{a_k}{2} [[\dfrac{cos((k+1)\theta)}{(k+1)}]_{\pi}^{0} -[\dfrac{cos((k-1)\theta)}{(k-1)}]_{\pi}^{0}]$$

Since all the cosine terms for odd values of k become zero. The above expression is modified as below.

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} \dfrac{a_{2j}}{2} [[\dfrac{cos((2j+1)\theta)}{(2j+1)}]_{\pi}^{0} -[\dfrac{cos((2j-1)\theta)}{(2k-1)}]_{\pi}^{0}]$$

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} \dfrac{a_{2j}}{2} \dfrac{2}{(2j+1)}]-[\dfrac{2}{(2k-1)}$$

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} a_{2j}\dfrac{1}{(2j+1)}]-[\dfrac{1}{(2j-1)}$$

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} \dfrac{2a_{2j}}{1-(2j)^2}.$$

Hence the proof.

Author
Solved and typed by - Egm6341.s10.team4.anandankala 14:51, 23 April 2010 (UTC) .

= Contributing Team Members =

Egm6341.s10.Team4.andy 06:35, 23 April 2010 (UTC) - Author of Problems:1, 7, 13 ; Proof Read: 2, 3, 14, 15

Egm6341.s10.Team4.nimaa&amp;m 14:30, 23 April 2010 (UTC) - Author of Problems:2, 8, 14 ; Proof Read:15

Egm6341.s10.Team4.riherd 15:25, 23 April 2010 (UTC) - Author of Problem 6, 12. Proof read 10, 11.

Egm6341.s10.Team4.roni 18:04, 23 April 2010 (UTC) - Author of Problem 4, 10; Proof Read: 1,7,4.

Egm6341.s10.team4.anandankala 14:49, 23 April 2010 (UTC)- Author of Problem 3, 9, 15; Proof Read:2, 13, 14.

Egm6341.s10.Team4.yunseok 19:40, 23 April 2010 (UTC) - Author of Problem 5, 11, Proof Read:3