User:Egm6321.f12.team6/report7

=Report 7= ==Problem 7.1: Plotting F(5,-10;1;x) ==

Problem Statement

 * Use matlab to plot $$ \displaystyle F(5,-10;1;x) $$ near $$ X=0 $$ to display the local maximum (or maxima)in this region.
 * show that $$ \displaystyle F(5,-10;1;x)=(1-x)^{6}(1001x^{4}-1144x^{3}+396x^{2}-44x+1) $$

Given

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

$$ \displaystyle F(5,-10;1;x) $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle F(5,-10;1;x)=(1-x)^{6}(1001x^{4}-1144x^{3}+396x^{2}-44x+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2)
 * }

Nomenclature

 * $$ \displaystyle F(a,b;c;x):=\sum_{k=0}^\infty\frac{(a)_k\,(b)_k}{(c)_k}\,\frac{x^k}{k!} $$

Solution

 * This problem is also presented in Report 5 Problem 5.10. Many of groups presented the plotting of positive zone of the graph. As professor emphasized in the class lecture, the plotting of F(5,-10;1;x) is focused on the negative x value this time. In order to generate a graph, following MATLAB code were typed in.
 * {| style="width:100%" border="0" align="left"


 * style="width:30%; padding:10px; border:1px solid #888888" align="left" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="left" |
 * }
 * This will generate the following outcome. The first outcome is the function F that has been solved by MATLAB.
 * Result.jpeg
 * As you can see from above, we can conclude that F=(5,-10;1;x) can be expressed in the relationship as Equation 1.2.
 * Also, the second outcome will be the plot.
 * Hypergeometric Plot.jpg
 * The zoomed in plot is not as smooth. However, if the line is smooth, then we can anticipate that when x is 0, then it will go through 1. However, when x is negative values, you can see that the zoomed-out plot goes infinitively large.
 * {| style="width:100%" border="0"|-
 * The zoomed in plot is not as smooth. However, if the line is smooth, then we can anticipate that when x is 0, then it will go through 1. However, when x is negative values, you can see that the zoomed-out plot goes infinitively large.
 * {| style="width:100%" border="0"|-

Therefore, we can conclude that $$ \displaystyle F(5,-10;1;x)=(1-x)^{6}(1001x^{4}-1144x^{3}+396x^{2}-44x+1) $$.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

Also within the domain of -1 to 0, The local minimum will be at 1. However, there will not be any local maximum since the graph extends to infinity.
 * }

Part 2 n=3 use command "roots"

 * When n=3, according to (4.2), (4.1) becomes


 * Since $$t=0, z=50$$, so that $$\frac{1}{10}z F(1,\frac{1}{3};\frac{4}{3};\frac{-z^3}{5})=-k$$
 * I tried several times to solve the k:

1.Use Wolfram Alpha to solve for $$\frac{1}{10}z F(1,\frac{1}{3};\frac{4}{3};\frac{-z^3}{5})$$ when$$ z=50$$
 * It yields the following answer.


 * Matlab yields a similar result.

2.In order to solve the hypergeometric function with command "roots" in Matlab, expand the hypergeometric function into polynomials.
 * It yields the following result.


 * Solve for k when z=50.
 * It yields the following:
 * $$k=7.81205*10^{12}$$which differs a lot from the above.

3.Expand the hypergeometric funtion with more terms.
 * Solve for k when z=50.
 * It yields the following result:
 * $$k=-8.05433*10^{51}$$
 * According to the above facts, it is hard to find a suitable polynomial form for this hypergeomeric function. And I can't solve it with command "roots".

Part 3 n=3 use function "hypergeom"

 * Similar to Part2, solve k first.
 * Invoke Matlab to calculate k:


 * It yields $$k =-0.2067$$
 * So the relation between z and t becomes:


 * Get z-t numerical relation by change z gradually:


 * So the following graphic can be acquired.


 * Then we noticed that, when z is slightly under 0, some strange curve appears. Then we did a more detailed calculation around that part.


 * So the following graphic can be acquired.


 * After our discussion, we made the following explanation for the reversed curve.
 * When the object start to fall, the velocity increases due to the gravity. And meanwhile the resistance force increases with velocity. At a point, the gravity and resistance becomes equal. And that's the biggest falling velocity. As long as the value of z is lower than this max falling velocity, this z will never be reached. That's why the function gives unreasonable result. The correct graphic should be even after the reverse point. Refer to the following picture.




 * According to Matlab, the max falling speed is -0.5826. Adjust the code.

Problem Statement


Cylindrical coordinates ,which are shown in the above figure, are expressed as follows:

1) Find $$ \{dx_i \}=\{dx_1,dx_2,dx_3 \} $$
 * in terms of
 * $$ \{\xi_j \}=\{\xi_1,\xi_2,\xi_3 \} $$ and  $$ \{d\xi_k \}=\{d\xi_1,d\xi_2,d\xi_3 \} $$

2) Find $$ ds^2=\sum_{i}(dx_i)^2= \sum_{k}(h_k)^2(d\xi_k)^2 $$. Identify $$ \{h_i\} $$ in terms of $$ \{\xi_i\} $$

3) Find $$ \Delta u $$ in cylindrical coordinates

4) Use separation of variables to find the separated equations and compare to the Bessel equ.(1)P.27-1

Given
Laplace equation:


 * $$ \Delta u= 0 $$

Laplace operator in general curvilinear coordinates:

Bessel L2-ODE-VC:

Nomenclature

 * $$ (\xi_1,\xi_2,\xi_3) $$: General curvilinear coordinates
 * $$ f'(x)= \frac{df(x)}{dx} $$

Part 1) Find {dxi}
Expand $$ \{dx_i \} $$ by using product rule:

Part 2) Find (ds)^2 and Identify {hi}

 * $$ ds^2= (dx_1)^2+(dx_2)^2+(dx_3)^2 $$
 * $$ ds^2= (cos\xi_2\,d\xi_1-\xi_1\,sin\xi_2\,d\xi_2)^2+(sin\xi_2\,d\xi_1+\xi_1\,cos\xi_2\,d\xi_2)^2+(d\xi_3)^2 $$
 * $$ ds^2= cos^2\xi_2\,(d\xi_1)^2-2\,\xi_1\,sin\xi_2\,cos\xi_2\,d\xi_1\,d\xi_2+\xi_1^2\,sin^2\xi_2\,(d\xi_2)^2+sin^2\xi_2\,(d\xi_1)^2+2\,\xi_1\,sin\xi_2\,cos\xi_2\,d\xi_1\,d\xi_2+\xi_1^2\,cos^2\xi_2\,(d\xi_2)^2+(d\xi_3)^2 $$

Therefore, we have


 * (5.10)
 * }

Part 3) Find Laplace Operator in Cylindrical Coordinates
Eq.(5.9) gives

When $$ i=1 $$,

When $$ i=2 $$,

When $$ i=3 $$,

Pluging the results derived from eq.(5.11) to eq.(5.14) into eq.(5.4) yields


 * $$ \Delta u= \frac{1}{h_1h_2h_3}\sum_{i=1}^{3}\frac{\partial}{\partial\xi_i}[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}] $$
 * $$ \Delta u= \frac{1}{r} [\frac{\partial}{\partial r}(r\,\frac{\partial u}{\partial r})+\frac{1}{r}\,\frac{\partial^2 u}{\partial\theta^2}+r\,\frac{\partial^2 u}{\partial z^2}] $$


 * (5.15)
 * }

Part 4) Separation of Variables and Comparison with Bessel Equation
To solve Laplace Equation $$ \Delta u=0 $$, we need to employ the method of Separation of Variables introduced on lecture notes P.39-5. Assume that

Plug eq.(5.16) into eq.(5.15) and expand:

Divide eq.(5.17) by $$ R(r)\Theta(\theta)Z(z) $$:

Since the third term of eq.(5.18) is neither a function of $$ r $$ nor $$ \theta $$, we can denote it as a constant with respect to the first two terms.

Let $$ k_1= \frac{1}{Z}\frac{d^2Z}{d z^2} $$, and multiply eq.(5.18) with $$ r^2 $$, then we have

Now the second term of eq.(5.19) is not a function of $$ r $$, we can further denote it as a constant with respect to the first and third terms.

Let $$ k_2= \frac{1}{\Theta}\frac{d^2\Theta}{d\theta^2} $$, and expand eq.(5.19) using product rule:

Replace $$ r=x, R=y $$. Eq.(5.20) becomes

Which could be further simplified into:

Divide eq.(5.22) by $$ k_1 $$:

Recall Bessel's Equation:

Let $$ k_1=-\frac{1}{2}, k_2=\frac{v^2}{2} $$,eq(5.23) becomes


 * (5.24)
 * }

One can tell that the coeffcients of $$ y' $$ and $$ y $$ in eq.(5.24) correspond to those in Bessel's Equation.

However, the coefficient of $$ y' $$, which is $$ -2x^2 $$, is different from $$ 1-x^2 $$ in Bessel's Equation.

Problem Statement: The Laplacian in Spherical Coordinates Using the Math/Physics Convention

 * Express the Laplacian Operator in spherical coordinates using the Math Physics Convention.

Given

 * The Laplacian in General Curvilinear Coordinates;

Nomenclature

 * The Math Physics convention for spherical coordinates expresses three coordinate axes:
 * $$ (r, \bar \theta, \varphi) = (\xi_1, \xi_2, \xi_3)$$
 * In relation to the cartesian coordinate axes the Math Physics axes are measured as demonstrated in the figure below;


 * The vector, $$ \mathbf g_i $$, is a vector tangent to the curvilinear coordinate axis.

Solution

 * Step 1 -
 * The first objective is to relate the coordinates from the Cartesian coordinate system, to the Spherical system. This is done by projecting the vector formed from the Cartesian point, $$ (x, y, z) $$, in terms of the Spherical system, $$ (r, \bar \theta, \varphi) $$, onto each axis.  Inspection of Figure 1 yields the following results:


 * Step 2 -
 * The next step is to find the vectors tangent to the spherical coordinate axes. We can interpret equations, 6.3, 6.4 and 6.5 as functions such that, $$ x, \; y, \; z $$, are functions of $$ r, \; \bar \theta, \; \varphi $$. This can be expressed as $$ x(r, \bar \theta, \varphi), \; y(r, \bar \theta, \varphi), \; z(r, \bar \theta) $$. We can now infer that the derivative of equations 6.3, 6.4 and 6.5 define the tangent vector, $$ \mathbf g_i $$. Therefore the tangent vectors will have three components that define them. These components are in terms of the spherical coordinate axis, such that;


 * There are nine separate derivations that need to be performed. The results are found to be;


 * Step 3 -
 * Now that we have found the tangent vectors, we can find, $$ h_i $$, as defined from equation 6.2.
 * First we solve for $$ h_1 $$;


 * $$ h_1= \parallel \mathbf g_1 \parallel $$
 * $$ = \sqrt {\sin^2 \bar \theta \cos^2 \varphi + \sin^2 \bar \theta \sin^2 \varphi + \cos^2 \bar \theta} $$
 * $$ = \sqrt {\sin^2 \bar \theta ( \cos^2 \varphi + \sin^2 \varphi) + \cos^2 \bar \theta} $$
 * $$ = \sqrt {\sin^2 \bar \theta (1) + \cos^2 \bar \theta} $$
 * $$ = \sqrt {\sin^2 \bar \theta + \cos^2 \bar \theta} $$
 * $$ = \sqrt {1} $$


 * Next we solve for $$ h_2 $$;
 * $$ h_2= \parallel \mathbf g_2 \parallel $$
 * $$ = \sqrt {r^2 \cos^2 \bar \theta \cos^2 \varphi + r^2 \cos^2 \bar \theta \sin^2 \varphi + r^2 \sin^2 \bar \theta} $$
 * $$ = \sqrt {r^2 [\cos^2 \bar \theta ( \cos^2 \varphi + \sin^2 \varphi) + \sin^2 \bar \theta]} $$
 * $$ = \sqrt {r^2 [\cos^2 \bar \theta (1) + \sin^2 \bar \theta]} $$
 * $$ = \sqrt {r^2 (1)} $$


 * Last we solve for $$ h_3 $$;
 * $$ h_3= \parallel \mathbf g_3 \parallel $$
 * $$ = \sqrt {r^2 \sin^2 \bar \theta \sin^2 \varphi + r^2 \sin^2 \bar \theta \cos^2 \varphi} $$
 * $$ = \sqrt {r^2 \sin^2 \bar \theta ( \sin^2 \varphi + \cos^2 \varphi)} $$
 * $$ = \sqrt {r^2 \sin^2 \bar \theta (1)} $$


 * Step 4 -
 * The next operation is to solve the Laplacian operator in equation 6.1.
 * It will be convienient to solve for;


 * The summation in equation 6.1 will expand 3 times. We will solve for Laplacian for each expansion of the summation. Note that we take use of a nomenclature change in that;
 * $$ (r, \bar \theta, \varphi) = (\xi_1, \xi_2, \xi_3)$$


 * For the summation index, $$ i=1 $$, we have;


 * For the summation index, $$ i=2 $$, we have;
 * $$ \Delta u_2 = \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \bar \theta} \left ( \frac{r^2 \sin \bar \theta}{r^2} \frac{\partial u}{\partial \bar \theta} \right )$$


 * For the summation index, $$ i=3 $$, we have;
 * $$ \Delta u_3 = \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \varphi} \left ( \frac{r^2 \sin \bar \theta}{r^2 \sin^2 \bar \theta} \frac{\partial u}{\partial \varphi} \right )$$


 * We now can formulate our final equation;
 * $$ \Delta u = \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial r} \left ( r^2 \sin \bar \theta \frac{\partial u}{\partial r} \right ) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \bar \theta} \left ( \sin \bar \theta \frac{\partial u}{\partial \bar \theta} \right ) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \varphi} \left ( \frac{1}{ \sin \bar \theta} \frac{\partial u}{\partial \varphi} \right ) $$


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle \Delta u= \frac{1}{r^2 \sin \bar \theta}\left ( \frac{\partial }{\partial r} \left ( r^2 \sin \bar \theta \frac{\partial u}{\partial r} \right ) + \frac{\partial }{\partial \bar \theta} \left ( \sin \bar \theta \frac{\partial u}{\partial \bar \theta} \right ) + \frac{\partial }{\partial \varphi} \left ( \frac{1}{ \sin \bar \theta} \frac{\partial u}{\partial \varphi} \right ) \right )$$


 * (6.15)
 * }

Team Contribution

 * {| class="prettytable"


 * colspan="4" | Team Contribution Table
 * colspan="4" | Compiled by: Brandon Taylor
 * colspan="4" | Compiled by: Brandon Taylor


 * Problems
 * Solved and Typed By
 * Solved and Inspected By
 * Skimmed By
 * Problem 7.1
 * Kyung-Hoon Bang
 * Mutian Hu
 * All Team Members
 * Problem 7.2
 * Mutian Hu
 * Jiajun Han
 * All Team Members
 * Problem 7.3
 * CheMing Lee
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 7.4
 * Jiajun Han
 * Tianyu Gu
 * All Team Members
 * Problem 7.5
 * Tianyu Gu
 * Brandon Taylor
 * All Team Members
 * Problem 7.6
 * Brandon Taylor
 * CheMing Lee
 * All Team Members
 * }
 * Problem 7.6
 * Brandon Taylor
 * CheMing Lee
 * All Team Members
 * }
 * }