User:Egm6321.f12.team6/report6 Updated

=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 can solve for a corresponding value of z for each value of t in Wolfram Alpha. We note that Wolfram Alpha does not produce results for t < 0.08. Also, for each value of t, Wolfram will produce 7 roots. Five of these roots are complex and are ignored right away. The remaining 2 real roots require us to develop a method to select the appropriate value for z. This is done by considering the function, $$ t(z) $$, over the interval, $$ -3 \ge z \ge 3 $$. From Calculus we recall that local maximum and minimum can be found by setting the first derivative to zero. We also recall that the second derivative can help us identify whether it is a maximum or a minimum. Doing this we have:
 * $$ \frac{dt(z)}{dz}=0.0008z^9-0.00399z^6+0.02z^3-0.1$$

Finding the minimum we have:
 * $$ 0=0.0008z^9-0.00399z^6+0.02z^3-0.1$$
 * $$ z=1.070926 $$

A second derivative test finds:
 * $$\frac{d^2t(z)}{dz^2}=.0072z^8-.02394z^5+.06z^2 $$

For the minimum point,
 * $$ z=1.070926 $$
 * $$ \frac{d^2t(z)}{dz^2}=.0475 > 0 $$

We can therefore deduce that the point at z=1.070926 is a local minimum. Now recall that we are actually interested in z(t) and not t(z). As we have already explained there are two real roots for each value of 't'. One root is on one side of the local minimum and the other root is on the other side. So with increasing 't', there are 2 trends that exist as we move away from the local minimum. The first is an increasing value of 'z', and the other is a decreasing value of 'z'. Because 'z' represents velocity, we expect that as 't' increases, 'z' will decrease. This deduction allows us to create a rule by which we can choose the proper root from Wolfram Alpha. Since we know the local minimum, $$ z=1.070926 $$, the proper root will be;
 * $$ z \le 1.070926 $$

With this rule in mind, and using Wolfram Alpha we can find the 'z' for a defined value of 't'. Our solutions are:
 * $$ t=0, z=50 $$
 * $$ t=0.08, z=1.439 $$
 * $$ t=0.09, z=1.2779 $$
 * $$ t=0.1, z=1.14 $$
 * $$ t=0.15, z=0.5722 $$
 * $$ t=0.2, z=0.067 $$
 * $$ t=0.3, z=-0.8976 $$
 * $$ t=0.4, z=-1.5147 $$
 * $$ t=0.5, z=-1.7948 $$
 * $$ t=0.6, z=-1.9516 $$
 * $$ t=0.7, z=-2.0577 $$
 * $$ \cdots $$

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