User:Egm6321.f12.team6/report6

=Report 6=

Problem Statement: Transformation of Variable Fifth Derivative

 * The principle of Transformation of Variables, causes a function, $$ y(x) $$, where, $$ x $$, is an independent variable, to become another function, $$ y(x(t)) $$. Find the fifth derivative,


 * in terms of the derivatives of the transformation variable, $$ t $$.

Given

 * The variable Transformation;


 * A function;

Nomenclature

 * Derivations follow the simplified notation convention below where, $$ q $$, is any variable:
 * $$ y_q:=\frac{dy}{dq} $$
 * $$ y_{qq}:=\frac{d^2y}{dq^2} $$
 * This convention is extended to include all higher order derivatives.

Solution
Given the transformation variable in $$, we recognize the derivation of, $$ y(x(t)) $$, must follow the chain rule as below.

Using $$, we can find that;
 * $$ \frac{dx}{dt}=e^t $$
 * We also have from our nomenclature;
 * $$ \frac{dy}{dt}=:y_t $$
 * $$ \frac{dy}{dx}=:y_x $$

Therefore applying $$, we have;
 * $$ y_t=y_xe^t $$
 * Rearranging terms gives us the first derivative;

To find the second derivative we can rearrange $$, to yield;

From $$, we must recognize that;


 * Now we are in a position to solve for, $$ y_{xx} $$.
 * $$ y_{xx}=\frac{dy_x}{dt}\frac{dt}{dx} $$

Apply $$, and $$, to give;
 * $$ y_{xx}=\left (\frac{d}{dt}(e^{-t}y_t)\right )e^{-t} $$
 * $$ y_{xx}=(e^{-t}y_{tt}-e^{-t}y_t)e^{-t} $$
 * $$ y_{xx}=e^{-t}(y_{tt}-y_t)e^{-t} $$


 * We will use the same method to solve for, $$ y_{xxx} $$
 * $$ y_{xxx}=\frac{dy_{xx}}{dt}\frac{dt}{dx} $$
 * $$ y_{xxx}=\left(\frac{d}{dt}(e^{-2t}(y_{tt}-y_t))\right)e^{-t} $$
 * $$ y_{xxx}=(-2e^{-2t}(y_{tt}-y_t)+e^{-2t}(y_{ttt}-y_{tt}))e^{-t} $$
 * $$ y_{xxx}=(e^{-2t}(-2y_{tt}+2y_t)+e^{-2t}(y_{ttt}-y_{tt}))e^{-t} $$


 * Using the same method to solve for, $$ y_{xxxx} $$, we have
 * $$ y_{xxxx}=\frac{dy_{xxx}}{dt}\frac{dt}{dx} $$
 * $$ y_{xxxx}=\left(\frac{d}{dt}(e^{-3t}(y_{ttt}-3y_{tt}+2y_t))\right)e^{-t} $$
 * $$ y_{xxxx}=(-3e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+e^{-3t}(y_{tttt}-3y_{ttt}+2y_{tt}))e^{-t} $$
 * $$ y_{xxxx}=(e^{-3t}(-3y_{ttt}+9y_{tt}-6y_t)+e^{-3t}(y_{tttt}-3y_{ttt}+2y_{tt}))e^{-t} $$


 * Finally deriving, $$ y_{xxxxx} $$, we have
 * $$ y_{xxxxx}=\frac{dy_{xxxx}}{dt}\frac{dt}{dx} $$
 * $$ y_{xxxxx}=\left(\frac{d}{dt}(e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t))\right)e^{-t} $$
 * $$ y_{xxxxx}=(-4e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}(y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt}))e^{-t} $$
 * $$ y_{xxxxx}=(e^{-4t}(-4y_{tttt}+24y_{ttt}-44y_{tt}+24y_t)+e^{-4t}(y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt}))e^{-t} $$


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

$$ \displaystyle y_{xxxxx}=e^{-5t}(y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_t) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * ($$)
 * }

Part 2 n=3
2.1 Find z(t)

When n=3, it is difficult to solve the integral in ($$) by hand, thus we need to use hypergeometric function to approximate the result

In ($$), we are given that


 * $$ \int\frac{dz}{az^n+b}= \frac{1}{b}\,z\,_2F_1 \left(1,\frac{1}{n};1+\frac{1}{n};-a\,\frac{z^n}{b}\right)+k $$

Substituting a,n,b with 2,3,10 yields

Now ($$)can be re-written as:

Next, we want to calculate the value for each term in ($$)

Use Pochhammer's symbol to expand hypergeometric function into power series, and truncate to approximate:
 * $$ \frac{1}{10}\,z\,_2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5} \right)= \frac{1}{10}\,z\,[1-\frac{1\cdot \frac{1}{3}}{\frac{4}{3}}\frac{\frac{z^3}{5}}{1!}+ \frac{1\cdot2\cdot\frac{1}{3} \cdot\frac{4}{3}}{\frac{4}{3}\cdot \frac{7}{3}}\frac{(\frac{z^3}{5})^2}{2!}-\frac{1\cdot2\cdot3\cdot\frac{1}{3}\cdot\frac{4}{3}\cdot\frac{7}{3}}{\frac{4}{3}\cdot\frac{7}{3}\cdot\frac{10}{3}}\frac{(\frac{z^3}{5})^2}{3!}+\cdots] $$

Initial value for the L.H.S. of ($$) can be calculated in Wolfram Alpha by typing "50/10 2F1(1,1/3;4/3;-1/5 50^3)":

($$) is re-written as

Since there is no explicit form of z in terms of t, we cannot solve for z(t) analytically.

Alternatively, we could solve for a corresponding value of z for each value of t in Wolfram Alpha:
 * $$ t=0, z=50 $$
 * $$ t=0.08, z=1.439 $$
 * $$ t=0.09, z=1.2779 $$
 * $$ t=0.1, z=1.14 $$
 * $$ t=0.15, z=0.5722 $$
 * $$ t=0.2, z=0.067 $$
 * $$ t=0.3, z=-0.8976 $$
 * $$ t=0.4, z=-1.5147 $$
 * $$ t=0.5, z=-1.7948 $$
 * $$ t=0.6, z=-1.9516 $$
 * $$ t=0.7, z=-2.0577 $$
 * $$ \cdots $$

Invoke Matlab to plot z against t:

Figure: Velocity vs. time



Let's use MotionGenesis and Matlab to verify the accuracy of previous steps

Numerical values for velocity z(t):
 * {| class="prettytable"


 * Time t (s)
 * Velocity z(t) (m/s)


 * 0
 * 50


 * 0.0010
 * 15.0712


 * 0.0020
 * 10.9025


 * 0.0030
 * 8.9679


 * 0.0040
 * 7.7923


 * }

Figure: Velocity vs. time



From the above figure, one can tell that hypergeoetric approximation is accurate after 0.08 sec, however there is an obvious error during the time interval 0-0.08 sec between hypergeoetric approximation and computer results. This is caused by failure in deriving any value of z when t is less than 0.08 sec from ($$) using Wolfram Alpha.

2.2 Find y(t)

Add the following codes to matlab

Numerical values for altitude y(t):
 * {| class="prettytable"


 * Time t (s)
 * Altitude y(t) (m)


 * 0.5080
 * 0.0030


 * 0.5090
 * 0.0013


 * 0.5100
 * -0.0003


 * 0.5110
 * -0.0020


 * 0.5120
 * -0.0037


 * }

Figure: Altitude vs. time




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

From the above figure and table of numerical values, we can tell that when $$ t=0.51 s, y=-0.00003496\approx 0 $$, which means that the particle returns to ground at the time 0.51s
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement: The Second Homogeneous Solution

 * Using the method of variation of parameters, show that the second homogeneous solution is:

Given

 * The second Homogeneous Solution;


 * The first Homogeneous Solution;


 * The Legendre Operator;


 * The Integrating Factor;


 * Homogeneous L2-ODE-VC;

Nomenclature

 * $$ \frac{dy}{dx}=y' $$
 * $$ \frac{d^2y}{dx^2}=y'' $$

Solution
First we will normalize $$, with respect to, $$ y'' $$, and identify the coefficients;

where;

Notice that by dividing the Legendre Equation through by, $$ 1-x^2 $$, we can put $$, in the form of $$, and thus identify the coefficients defined in $$. Therefore;

With the coefficients defined we can solve for the integrating factor found in $$;
 * $$ h(x)=u^2_1 \exp {\int a_1(x)dx} $$
 * $$ h(x)=\frac{1}{4}(2x^2-1)^2 \exp \left[{\int \frac {-2x}{1-x^2}dx}\right] $$
 * $$ h(x)=\frac{1}{4}(2x^2-1)^2 \exp [{\log (x^2-1)}] $$
 * $$ h(x)=\frac{1}{4}(2x^2-1)^2(x^2-1) $$

It is convenient to solve for;
 * $$ \frac{1}{h(x)}=\frac{1}{\frac{1}{4}(2x^2-1)^2(x^2-1)} $$

We can now solve for the second homogeneous solution by substitution of $$, and $$, into $$;
 * $$ u_2(x)=\frac{1}{2}(2x^2-1) \int \frac{4}{(2x^2-1)^2(x^2-1)}dx $$
 * $$ u_2(x)=\frac{1}{2}(2x^2-1)\left (\frac{4x}{2x^2-1}-\sqrt{2}\log(\sqrt{2}-2x)+2\log(1-x)-2\log(x+1)+\sqrt{2}\log(2x+\sqrt{2})\right) $$
 * $$ u_2(x)=\frac{1}{2}(2x^2-1)\left (\frac{4x}{2x^2-1}+2\log\left(\frac{1-x}{1+x}\right)+\sqrt{2}\log\left(\frac{\sqrt{2}+2x}{\sqrt{2}-2x}\right)\right) $$

Nomenclature
Nothing need to be nomenclatured.

Solution
For $$ When $$x=\pm 1$$, it is a 1st-ODE rather than a 2nd-ODE. Usually variation of parameters is not used to solve such problems.

When $$x\neq \pm 1$$, it can be written as the following form.

According to $$

According to $$, and Wolfram Alpha

According to $$, and Wolfram Alpha, another homogeneous solution can be achieved.

Since $$f(x)=1$$, according to $$, and Wolfram Alpha, the following particular solution can be acquired if assume the integral constants are 0.

The solution of the 2nd-ODE-VC is a linear combination of homogeneous solutions and particular solution. So according to $$, the solution would be