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Problem 1: plot the local maximum and show equations
Report problem 5.10 from

Given
1.The hypergeometric series.

$$F(a,b;c;x)=\sum^\infty_{k=0}\frac{(a)_k(b)_k}{(c)_k}\frac{x^k}{k!}$$

2.Defination of local maximum.

A real-valued function f defined on a real line is said to have a local (or relative) maximum point at the point $$x_0$$, if there exists some$$\epsilon>0$$such that$$f(x_0)\geq f(x)$$ when$$|x-x_0|<\epsilon $$. The value of the function at this point is called maximum of the function. .

3.Determination of local maximum.

It is intuitively clear that the tangent line to the graph of a function at a local minimum or maximum must be horizontal, so the derivative at the point is 0, and the point is a critical point, which means $$f'(x_0)=0$$.

Find

 * 1) Use MATLAB to plot $$F(5,-10;1,x)$$ near $$ x=0 $$ to display the local maximum (or maxima) in this region.
 * 2) Show that :

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
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Part 1
Display the local maximum (or maxima) near $$x=0$$.

Choose $$x\in [-10,10] $$as the neighborhood of $$x $$near $$x_0=0$$. Get the data using function of hypergeom([a,b],c,d) in MATLAB.

In order to get the local maximum of $$F(a,b;c,x)$$, we can find the points of the data whose differential value is 0 and make further judgment through substituting the value back into $$F''(a,b;c;x)$$ to find the local maximum according to $$.Get the graph and codes as below.

So we can get local critical points and make judgment as below:

In short, $$F(5,-10;1;x)$$ near $$x=0$$ has local maximum points which are$$x_1=0.2288 ,x_2=0.6834$$. All of these maximum points are at the points when $$x>0$$. When $$x<0$$, $$F(5,-10;1;x)$$ are monotonic increasing to $$\infty $$ as $$x\rightarrow -\infty $$.

Part 2
Prove equality

If we expand the right hand side of $$. Then we get

According to $$, noticing that it contains ther term $$b+k-1 $$and $$b $$begins with -10, so it would reach 0 at the 12th term and makes each terms afterwards cancelled to 0. As a resualt,the left hand side of $$could be written as

So We get $$proved.

Problem 2: Application of method of variation of parameters
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Given: A L2-ODE-CC
The non-homogeneous linear 2nd order ordinary differential equation with constant coefficients(L2-ODE-CC)

Note: There is a typo in. The coefficient of $$y$$ given in the ODE should be $$a_0^2$$ instead of $$a_0$$.

Given the initial conditions being

Find: The final solution to the L2-ODE-CC
Find the final solution:

and compare ($$)~($$) to Eqns (2.4)~(2.5) in Dong 2012

Solution: Using method of variation of parameters

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

On our honors, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

First, we assume the homogeneous solution to the L2-ODE-CC is

with $$r$$ to be determined.

Instituting ($$) back to the left-hand-side of ($$) yields:

solve the characteristic equation

thus the homogenous solutions turn out to be

We take one of the homogeneous solution $$y_1=cosa_0t$$ in ($$) to form the following trial solution to the non-homogeneous L2-ODE-CC

where $$A(t)$$ is the unknown to be found.

Instituting ($$) into ($$) yeilds

Assume that

then ($$) turns to be

Note that ($$) is a L1-ODE-VC which could be solved directly or by IFM.

Noticing that

{{NumBlk|:|$$ \left\{\begin{matrix} \phi_{Bt}=-a_0sina_0t \\ \phi_{tB}=-2a_0sina_0t \end{matrix}\right.\Rightarrow \phi_{Bt}\neq \phi_{tB}$$|$$}}

i.e. ($$) is not exact and IFM should be used to solve it

Multiply ($$) with certain integration factor $$h(t)$$

Recall the 2nd exactness condition $$\bar{\phi}_{Bt}=\bar{\phi}_{tB}$$, we have:

i.e.

where $$k_1$$ is an integrating constant.

Plug ($$) back into ($$), we get the solution

where $$k_2$$ is an integrating constant.

Integrating ($$) yeilds

where $$k_3$$ is another integrating constant.

Expand ($$), we have:

Plug ($$) back into ($$), we get the final solution to ($$):

where $$C_1$$ and $$C_2$$ are two arbitrary constants.

Note the condition that $$t\geqslant t_0$$ and also integration range being $$[t_0, t]$$, the solution is modified to be

Consider two initial conditions as expressed in ($$), we have

{{NumBlk|:|$$ \left\{\begin{matrix} \Complex_1=y'_0 \\ \Complex_2=y_0 \end{matrix}\right. $$|$$}}

Therefore, the final solution to ($$) is

Compared with Eqns(2.5) in Dong 2012, we know they are in the same expression.

Problem 3: Derive infinitesimal length ds in spherical coord
Report problem 7.3 from.

Given
1. Infinitesimal length ds is given as

in Cartesian coord.

Cartesian Coord. in terms of spherical coord.:

$$h_i$$= magnituge of the tangent vector $$mathbf g_i$$ to the coordinate line $$\xi_i$$

2. Laplace operator in general curvilinear coord. is

in Spherical coord:

at i=1,

Find
1. Show that in spherical coord.

where

2. Laplace operator at i=2 and i=3 and the final form in spherical coordinates.

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

1. From ($$), ($$) and ($$)

From ($$),

where,

2. at i=2,

at i=3,

Thus, the Laplace operator in spherical coordinate is

Problem 4: plot and compare the equation of motion in air
Report problem 5.11 and 6.10 from

Given:
The schematic of the motion of a particle in the air :

Derive the equations of motion:

Consider the case $$k\neq 0$$ and $$v_{x0}=0$$: ($$) turns into

Which can be written as :

Separation of variables

For all values of $$n$$ :

Give particular values of some parameters :

Initial vertical velocity:$$z(t):=v_y(0)=50$$

Find:

 * For each value of $$n$$, find the vertical velocity $$z(t)$$ vs. time t, plot this function.
 * For each value of $$n$$, find the altitude $$y(t)$$ vs. time t, find the time when the projectile returns to the ground.

Use numerical methods to find the value of each given value of time t if the explicit expression cannot be obtained.


 * 1) $$n=2$$
 * 2) $$n=3$$. Use the matlab command "roots" to find the appropriate roots z for each given time t; verify with WA. Plot z versus t. Find $$y(t)$$by integrating $$z(t)$$ using the trapezoidal rule.
 * 3) $$n=3$$. Use the matlab function "hypergeom" to find the time t for each given value of z in the interval $$[-10,50]$$. Plot t versus z. Find $$y(t)$$by integrating $$z(t)$$ using the trapezoidal rule. Compare to Part 2.
 * 4) Verify the results in Parts 1 and 2 using the matlab in 2 steps:(a)use the command ode45 to integrate the L1-ODE-CC(1)p63-8 to obtain $$z(t)$$,(b)use the trapezoidal rule to integrate $$z(t)$$ to obtain $$y(t)$$.
 * 5) the equation of motion can be written as a system 1st order ODEs to be integrated using matlab ode45:$$\begin{align}z'&=-az^n-b\\y'&=z\end{align}$$.


 * 1) Verify Part 2.

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

According to $$, $$and$$, the uniform expression can be written as:

The inverse funciton of $$can be show as below in terms of $$t$$:

So we get:

When n=2
As what we have done in [Repot 6.10], it shows that:

the figure of z(t) vs t:

the figure of t vs z(t):

We can see that the expression of y(t) is a periodic function, the positive value which is nearest to 0 is $$0.6825$$, at which time the projectile reach the ground.

So can we solve this problem by numerical methods. The program and plot obtained by MATLAB are showed below

When n=3, use MATLAB command "roots"
The $$comes to be written as:

With the initial value of $$to obtain the valve of k:

The number of $$k\doteq0.20667$$ by WolframAlpha, so we just write the expression as below :

According to $$, we can get the expression written as a form of polynomial and calculate each value of the terms. Expand the hyper-geometric function into power series, and truncate to approximate:

Use MATLAB command "roots" to obtain the roots of z(t) according to each given t. So we get the roots and generate the figure as below.


 * {| class="prettytable"


 * Time t (s)
 * 0
 * 0.0715
 * 0.0720
 * 0.0725
 * 0.0730
 * 0.0735
 * 0.0740
 * 0.0745
 * 0.0750
 * $$\cdots$$
 * Velocity z(t) (m/s)
 * 50
 * 1.6608
 * 1.6342
 * 1.6139
 * 1.5966
 * 1.5811
 * 1.5669
 * 1.5536
 * 1.5411
 * $$\cdots$$
 * }
 * $$\cdots$$
 * }

Using to trapezoidal rule to integrate the z(t) to get y(t), showed below:

Because of truncating the polynomial series to finite powers, the data is not accurate or identical with the theoretical value.

When n=3, use MATLAB command "hypergeom"
The $$comes to be written as:

With the initial value of $$to obtain the valve of k:

The number of $$k\doteq0.20667022059582652141$$ by WolframAlpha, so we just write the expression as below :

With MATLAB command "hypergeom" directly solving the expression, so we can plot the figure of z(t) vs t as follow:

The inverse function of $$cannot be obtained by MATLAB, So numerical methods is needed.

Discard the part that contains imaginary number for these have no physical meaning, then exchange the axes of z(t) vs t to t vs z(t), get the graph like this:

Then integrate z(t) to get y(t). When the projectile reach ground, the value of y(t) should reach 0, which means that the accumulative sum of $$should be 0. The plot and MATLAB programs are showed below.

The answer, according to the $$, is $$0.5024$$, at which time the projectile reach the ground.

Use MATLAB command "ode45" to verify
According to $$, when n=3, the expression can be written as:

Input the expression in to MATLAB and solve it. We get the figure below shows that:

When n=3
The figures showed above are the same with that in Part 1 and Part 2.

Use MATLAB command "ode45" to verify Part 2 as a system 1st order ODE
According to $$, the expression can be written as:

They can be solved by MATLAB to obtain y(t).

The figures showed above is the same with that in Part 2.

Problem 5: Heat conduction on a cylinder
Problem R*7.5 from lecture notes section 40.

Given: A generalized coordinate system
Given the coordinate transformation:

Find: Various quantities

 * 1) Find $$\{dx_i \} = \{dx_1,dx_2,dx_3\}$$ in terms of $$\{ \xi _{j} \}= \{ \xi _{1},\xi _{2},\xi _{3}\}$$ and $$\{ d\xi _{j} \}= \{ d\xi _{1},d\xi _{2},d\xi _{3}\}$$.


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


 * 1) Find $$\Delta u$$ in cylindrical coordinates.


 * 1) Use separation of variables to find the separated equations and compare to the Bessel equation, given by:

Solution: Find the quantities

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

First, use the chain rule

Substituting in the coordinate transformations given by ($$) through ($$)

Now, plugging into the expression for $$ds^2$$ and simplifying with trigonometric identities and cancelations

Therefore,

The Laplace operator in general curvilinear coordinates is given by

Substituting yields

Now assume the solution is $$u = R(\xi _{1})\Theta(\xi _{2})Z(\xi _{3})$$, take the Laplacian and divide by $$R \Theta Z$$

Since the last term is the only term with $$\xi_3$$, and does not depend on the other curvilinear coordinates, it must independently be zero.

Multiplying the rest of the equation by $$\xi_{1}^{2}$$ again causes the first 2 terms to only depend on $$\xi_1$$ and the last term to only depend on $$\xi_2$$, therefore they are independently zero. Rearranging and simplifying yields the Bessel equation,

where

Problem 6: Finding the Laplacian in Spherical Coordinates using the Math/Physics Convention
Based on lecture notes [ http://upload.wikimedia.org/wikiversity/en/3/39/Pea1.f12.sec40.djvu Section 40]

Given: The Laplace Operator In General Curvilinear Coordinates
The Laplacian in general curvilinear coordinates is given as:

Also, the infinitesimal length ds is given as:

Lastly, the Math/Physics convention for spherical coordinates is defined as:

Where

$$\displaystyle \bar\theta = \frac{\pi}{2} - \theta $$

Find: The Laplacian in Spherical Coordinates
Find the Laplacian of u in spherical coordinates using the math/physics convention.

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

In order to obtain $$\displaystyle h_{1}, h_{2}, h_{3} $$ we must make the proper substitutions in ($$). We know from the cofunction identities in trigonometry that:

Furthermore,

($$) then becomes:

Here we see that

Employing this in ($$), we get:

This can be further simplified to arrive at the final solution:

Contribution
{| class=wikitable style="width:100%; text-align:center" ! colspan=3 | Problem Assignments ! Problem # !! Solved&Typed by!! Reviewed by
 * 1
 * Xiaolong Wu, Lin,Hao
 * All
 * 2
 * Goran Marjanovic, Zixiang Liu
 * All
 * 3
 * Zhou Zhe, Rui Kong, Anup Parikh
 * All
 * 4
 * Xiaolong Wu, Lin,Hao
 * All
 * 5
 * Zhou Zhe, Rui Kong, Anup Parikh
 * All
 * 6
 * Goran Marjanovic, Zixiang Liu
 * All
 * Zhou Zhe, Rui Kong, Anup Parikh
 * All
 * 6
 * Goran Marjanovic, Zixiang Liu
 * All
 * All