User:Egm6322.s14.team2/report6

=Report 6 PEA 2 Spring 2014 Team 2=

Problem Statement
redo some problems in R5, particularly R5.4 (Navier’s equation in cylindrical coordinates), and R5.7 (R4.2; plotting true anomaly, together with eccentric anomaly and mean anomaly, versus time).

R5.4
The gradient of a scalar in cylindrical coordinates $$\phi$$ was derived in Report 4

It was also shown in sec.55 of the Lecture Notes that the divergence in general curvilinear coordinates is given by:

It was shown in Report 4 that for cylindrical coordinates:

To derive the vector Laplacian for cylindrical coordinates, the easiest way is to make use of the vector identity :

Where the cross product in general curvilinear coordinates is defined as :

In cylindrical coordinates:

Taking the curl of ($$):

Using ($$):

The r component of ($$) then becomes:

Similarly, the theta component is:

R5.7 True Anomaly Plots
In this section we revisit the problem R4.2 From Report 4 where we seek to accurately plot how the true anomaly (angle) changes a function of time.

Here we are asked to plot $$ \phi \text{ and } \mu $$ as functions of time. The results are presented below. Please note that the function $$ \mu $$ is linear with respect to time, so it is the line on the plots below.

The following relation was used to compute the values of the mean, eccentric, and true anomaly.

The analytic relation given by ($$) was found on a Wikipedia page about the True Anomaly. It can be derived using the relations and diagrams from the Wikipedia article on the Eccentric Anomaly.



Some observations about the figure above. Notice that the true anomaly crosses the mean anomaly at each half period. The true anomaly can be seen to be composed by shifting the alternating portions of the plots that go along the bottom of the plot. Those bottom plots do not follow the mean anomaly because the method required to compute the true anomaly uses the inverse cosine function. The range of this function is $$ 0 \text{ to } \pi $$, hence they can not go above $$ \pi $$. Thus an ad hoc method of adjusting the results from the inverse cosine function calculation was developed. The implementation is given in the source code below. Observe how the true anomaly has the shape of the $$ \theta_{original} $$ plot during odd half-periods and $$ \pi - \theta_{original} $$ during even half-periods.

Source:

Problem Statement
Problem taken from notes provided by Dr. Vu-Quoc. The problem is 70.11 in notes sec.70d.

Part A
From,

Show that

and thus the reason for the factor 2 as follows.

Part B
Consider a function $$ f(x) $$ such that

show that

and thus

Part C
Use ($$) and ($$) and the trigonometric identity,

to show that

Solution

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Part A
Let's start with ($$).

We can split the integral on the right-hand size of into the following two integrals.

We make the following substitution,

Now that we have expressed the integrand as a general function of $$ \theta $$ we make the following observation about the function. 1.) It is an even function. 2.) It has a period of $$ 2\pi $$. 3.) It is half-period even. We plot the function $$ F(\theta) $$ to verify our observations.



From the plot we can see that is is indeed an even function with a period of $$ 2\pi $$.

Thus ($$) takes the following form.

We make a change of variables of the following form for the second integrand in the equation above.

With this change of variables ($$) takes the following form.

Recall that the function F is half period even, thus $$ F(t) = F(t + \theta) $$. Therefore the two integrals are the same and can be written as follows.

Dividing both sides of the above result by 2 yields the following.

We see that this is in agreement with ($$)

Part B
We choose to prove ($$) by looking at the nature of the integrand. If we call the integrand $$ G(\theta) $$,

and recognize that the integration bounds are across a half period ( recall that the period of G is $$ 2 \pi $$. We also recall that the function G is half-period odd. Therefore integration over the half period yields 0.

Part C
For this part we begin by recognizing from ($$) that we wish to use the following form of the trigonometric identity in ($$),

where,

We therefore return to ($$) and the the substitution given by ($$) and ($$) to change the integrand to get the following.

The integral can be split into two parts now.

Note that the second integral was proven to be 0. Therefore ($$) can be written as follows.

Problem Statement
Problem taken from notes provided by Dr. Vu-Quoc. The problem is 70.12 in notes sec.70d.

King 2003 p. 66 follows a different, but related method to obtain the integral form of Bessel functions,

King 2003 p. 66 provides the following equations.

which are the same as equations (2) on p.70-29 and (4) p.70-28 in notes sec.70c.

Part 1
Show that the change of variable below

leads to the following expressions, and explain the reason for this change of variable:

What can you observe about the right hand side of ($$) and ($$)?

Part 2
Multiply ($$) by $$ \cos (m \eta ) $$ and integrate to obtain:

{{NumBlk|::|$$ \int_{\eta = 0}^\pi \cos( m \eta) \, \cos(x \cos \eta) \, d\eta = \left\{\begin{matrix} \pi J_{m} (x) \; \; \text{m even,} \\ 0 \; \; \; \; \; \; \; \; \text{m odd,} \end{matrix}\right. $$ |$$|}}

similarly,

{{NumBlk|::|$$ \int_{\eta = 0}^\pi \sin( m \eta) \, \sin(x \cos \eta) \, d\eta = \left\{\begin{matrix} 0 \; \; \; \; \; \; \; \; \text{m even,} \\ \pi J_{m} (x) \; \; \text{m odd,} \end{matrix}\right. $$ |$$|}}

Part 3
Use,

to arrive at,

Part 4
Compare step-by-step the methods on p.70-[30-31]l and the method on p.70-[31b-31c] in the notes sec.70d (i.e. discuss the similarities and the differences).

Part 5
Find the values of

Part 6
Show that $$ J_{0} (x) $$ is an even function.

Solution

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Part 1
Substituting $$\theta=\frac{\pi}{2}-\eta$$ into equations 3-2 and 3-3:

Part 2
Multiply ($$) by $$ \cos (m \eta ) $$ and integrate to obtain:

{{NumBlk|::|$$ \int_{\eta = 0}^\pi \cos( m \eta) \, \cos(x \cos \eta) \, d\eta = \left\{\begin{matrix} \pi J_{m} (x) \; \; \text{m even,} \\ 0 \; \; \; \; \; \; \; \; \text{m odd,} \end{matrix}\right. $$ |$$|}}

similarly,

{{NumBlk|::|$$ \int_{\eta = 0}^\pi \sin( m \eta) \, \sin(x \cos \eta) \, d\eta = \left\{\begin{matrix} 0 \; \; \; \; \; \; \; \; \text{m even,} \\ \pi J_{m} (x) \; \; \text{m odd,} \end{matrix}\right. $$ |$$|}}

Part 3
Summing the odd and even terms defined in Equations 3-15 and 3-16 and employing the trigonometric identity for the cosine of a sum of two angles a and b:

Part 4
The method on pages 31b and c are further explained in detail in King on page 66. The method on pages 70-30 and 70-31 uses the definitions of odd and even periodic functions.

Part 5
For n=0:

For n=1:

For n=2:

Part 6
The function is an even function if $$ J_{0} (x) =J_{0} (-x)$$.

Problem Statement
relation between Bessel functions with negative integer indices and Bessel functions with positive integer indices; rewrite the sine series of sin(z sin t) as infinite series with negative and positive odd integer indices; similarly rewrite the cosine series of cos(z sin t) as infinite series with negative and positive even integer indices; show the complement series---i.e., sine series of sin(z sin t) with even indices and cosine series of cos(z sin t) with odd indices---are zero; conclude that sin(z sin t) can be written as a sine series with all integer indices (negative / positive, odd / even); similarly for cos(z sin t). sec.70d, Pb-70.13.

Solution

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Part A
$$sin(z (sin(t))$$ can be rewritten as the following infinite series with positive odd indices:

Substituting in the bessel function this changes to:

Knowing that:

Then plugging ($$) into ($$) gives:

Simplifying the $$-1^{x}$$ terms gives:

$$cos(z (sin(t))$$ can be rewritten as the following infinite series with positive even indices:

Equation 4-4 is a generating function as shown below using the Taylor series expansion:

Using j=i+n where $$-\infty \le n \le \infty $$:

The generating function is invariant as $$t \to - \frac{1}{t}$$, therefore where n=-m:

Changing the dummy variable m back to n:

Therefore,

Writing out the summation for sin(z sinθ):

Therefore, the summation of the negative odds can be similarly expressed using Equation 4-8:

Therefore, after taking the summation of the positive and negative values of the odd n:

Writing out the summation for cos(z sinθ) in a similar fashion where $$J_{0} sin (0) = 0$$:

Therefore, the summation of the negative odds can be similarly expressed using Equation 4-8:

Therefore, after taking the summation of the positive and negative values of the even n:

The complement of the summation of the odd terms in the expansion of sin (z sin θ) is the summation of the evens which were found to be zero in problem R 6.2 and vice versa for cos (z sin θ):

Since the summations in Equations 4-15 and 4-16 are both zero, the expressions sin (z sin θ) and cos (z sin θ) can be simplified as follows with the dummy variable redefined as n (including both the odd and even terms):

Problem Statement
Knowing:

Evaluate $$h_0$$ by direct integration

Use

to deduce

by induction

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
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From ($$)

The first legendre polynomial is :

So using ($$):

It will now be proved by induction that:

Assume that the case holds when n=k:

When n=k+1, using ($$) but changing the index:

Substituting in $$:

($$) holds when k=0 (this can easily be seen from the results of ($$). Thus ($$) also holds for all n>0.

QED

Problem Statement
(a) An alternative definition of the weights of Gauss-Legendre quadrature is given by NIST Digital Library of Mathematical Functions

with

Show that

can be obtained from ($$) and ($$)

(b) Verify the formula for the error in

against the corresponding formula in the DLMF

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
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Part a
The Christoffel-Darboux identity specialized to the Legendre polynomials is given in (1) p.44b-16e as:

Where $$h_n$$ is defined as (see problem 5):

And $$\beta_n$$ is the ratio of the n+1 and n coefficients of $$P_n$$ given by :

Substituting the jth root for y in ($$):

Multiplying both sides by $$P_0(x)$$ and integrating from -1 to 1:

Using the orthogonality of the Legendre polynomials, all of the terms expect for k=0 on the right hand side are zero.

By orthogonality:

So, ($$) becomes:

Since

From ($$) and ($$):

Using ($$):

And using the definitions in (Eq 6-7})} and ($$):

Part b
The error given by DLFM is given by:

Where:

Thus,

Which matches exactly the error given in ($$)

=Contributing Team Members=


 * Cameron Stewart Solved problems 6.1, 6.5 and 6.6
 * Elizabeth Bartlett Solved problems 6.3 and 6.4
 * Kevin Frost Solved problem 6.5
 * Christopher Neal Solved problems 6.1, 6.2, and 5.7

=References=