User:Egm6321.f11.team2/HW2

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R*2.1 - Homogenous Solutions of the Second-Order Legendre Equation[edit]

Given[edit]

The Legendre Differential Equation[1] is a second-order ordinary differential equation given as follows:

2.1.1


For this problem, we take the specific case n = 1, in which the Legendre Differential Equation takes the form

2.1.2


It is asserted that the following functions satisfy 2.1.2:

2.1.3

2.1.4


Find[edit]

Show that and satisfy 2.1.2.

Solution[edit]

This problem was solved without referring to past solutions.

is by far the simpler of the two functions, so we prove that it satisfies 2.1.2 first in order to make the process clear.
First, we find the first and second derivatives of :

2.1.5

2.1.6


Substituting 2.1.5 and 2.1.6 into 2.1.2 yields

which proves that satisfies equation 2.1.2.




The process for showing that satisfies 2.1.2 is exactly the same as above.
We begin by carefully taking the first two derivatives of :




Simplifying the last term yields

2.1.7


For the second derivative, we can easily differentiate the first term by using the results from taking the first derivative.


simplifying yields

2.1.8


Finally, substituting 2.1.7 and 2.1.8 into 2.1.2 yields

which proves that satisfies 2.1.2.

Since the Legendre Differential Equation is second-order, and we now possess two linearly independent solutions to the homogenous equation (see R2.6), every solution of the homogenous form of the equation can be expressed as a linear combination of and .

R*2.2- Reduced-Order Solution of an ODE with Missing Dependent Variable[edit]

Given[edit]





(2.2.1)



Solution is that

(2.2.2)



Find[edit]

Verify that eq (2.2.2) is indeed solution for eq (2.2.1).

Solution[edit]

Solved on my own


Equation 2.2.1 is that

(2.2.1)



And now Let





Plug in eq (2.2.1)

The equation 2.2.1 is following that













(2.2.2)



It shows that eq (2.2.2) is indeed solution for eq (2.2.1).

If you want double check,





Thus,


R*2.3- Linearity and Nonlinearity in General First-Order ODEs[edit]

Solved on my own

Given[edit]

An ODE of 1st order is read as


(2.3.1)


Find[edit]

show that the eqn (2.3.1) is linear in , and that it is in general an N1-ODE. But it is not the most general N1-ODE as represented by

(2.3.2)


Give an example of a more general N1-ODE.

Solution[edit]

Definition of Linear[edit]

If is a linear operator, then it must satisfy

Where are the independent variables, and







is linear in .



And it is easy to proof that



So, is in general an N1-ODE.

An example of the general N1-ODE[edit]


.

(2.3.3)


where is an arbitrary function of .

R*2.4- Verification of Nonlinearity in a Particular First-Order ODE[edit]

Given[edit]

(2.4.1)


Find[edit]

Verify that (2.4.1) is a N1-ODE.

Solution[edit]

Solved on my own


General nonlinear ODE of order 1 (N1-ODE) is

(2.4.2)


(2.4.1) is a statement of (2.4.2) showing that the highest order is one. Thus, (2.4.1) is a fist order differential equation.


Linearty has to satisfy the following condition:

(2.4.3)


Nonlinearty is the condition not satisfing the above statement as follows:

(2.4.4)


(2.4.1) can be rewritten as follows:

(2.4.5)

And then, the nonlinearity can be checked with the following two functions:

(2.4.6)

(2.4.7)


From (2.4.6) and (2.4.7), is found so that (2.4.1) is verified to be a nonlinear equation.


Finally it is concluded that (2.4.1) is a N1-ODE (Nonlinear First Oder Differential Equation).

R*2.5- Converting a General First-Order ODE into Particular Form[edit]

Given[edit]

2.5.1

Equation 2.5.1 satisfies the most general form , but can be converted into particular form .

Find[edit]

Using the hint:

2.5.2

and

2.5.3

Illustrate a way to convert equation 2.5.1 into particular form .

Solution[edit]

Solved on my own






2.5.4

Let

And


Equation 2.5.4 becomes which is a particular form of Equation 2.5.1 .

R*2.6 - Linear Independence of Solutions to the Second-Order Legendre ODE[edit]

Given[edit]

The Legendre Differential Operator/Equation takes the form:

(1)

(2)

Two homogeneous solutions for this equation are:

(3)

(4)

Find[edit]

Show that equations (3) and (4) are linearly independent, i.e., show that



i.e., for any given , show that



Plot and .

Solution[edit]

Solved on my own

Since we know there exists an and only one such that for



We also know that for any other point there exists an such that



The two homogeneous solutions are linearly independent if
For



Solving for we determine = -0.5677175.

For



Solving for we determine = -0.2571305.

Since the homogeneous solutions must be linearly independent. This can be seen by looking at their graph.

File:HomogeneousSolution.png
The two homogeneous solutions for the Legendre Differential Operator/Equation for n = 1



R*2.7 - Deriving Exact First Order Differential Equations from Level Sets[edit]

Given[edit]

The family of level sets

2.7.1


Find[edit]

Find

and show that it is an N1-ODE.

Solution[edit]

This solution was found without referring to past solutions.


We know from the chain-rule[2] that



Applying this directly yields



Thus

2.7.2

which, as can be seen, is already in the form .
Due to the way that it was derived, it is expected that 2.7.2 is exact, and indeed



It is obvious that G is an ordinary differential equation and that it is first-order.
To prove that it is nonlinear, suppose that and are functions of x. Then, slightly changing our notation to G(y) for clarity, we see that



The powers of y in G make it nonlinear.

R*2.8 - First Exactness Condition[edit]

Given[edit]

First exactness condition is that an ODE can be written in the form



Find[edit]

Does the following N1-ODE satisfy the first exactness condition?

(2.8.1)

Solution[edit]

Solved on my own

In order to satisfy the first exactness condition, it must be in the form of:



(2.8.2)



So rearrange eq (2.8.1) to fit eq (2.8.2)







is any function of y'. If the function has no explicit inverse

then it can not satisfy in the form of eq (2.8.2)

In this case, does not have exact inverse


Thus, This does not satisfy the first exactness condition.

R*2.9- Commutative Partial Derivatives[edit]

Solved on my own

Given[edit]

Review calculus, and differentiation

(2.9.1)


Find[edit]

Find the minimum degree of differentiability of the function such that the above equation is satisfied, State the full theorem and provide a proof.

Solution[edit]

Clairaut's theorem[edit]

According to the Schwarz's theorem,[3], if

has continuous second partial derivatives at any given point in , then for

And we then know that the minimum degree of differentiability of the function must be three when it satisfies eqn (2.9.1)

Proof of Clairaut's theorem[edit]

The result was first discovered by L.Euler around 1734 in connection with problems in hydrodynamics [4]. In order to prove this theorem, we shall first introduce the Mean Value theorem as follows:

Let be a continuous function on the closed interval , and derivative on the open interval , where , Then such that

(2.9.2)

Suppose Q is a closed rectangle with sides parallel to the coordinate axes, put

(2.9.3)





(2.9.4)


So, we put


,


Chose . if and are sufficiently small, we have



(2.9.5)

Fix , and let , then

(2.9.6)


Since was is arbitrary, and (2.9.6) holds for all sufficiently small


(2.9.7)


Since is an arbitrary point in , so




R*2.10-Verification of solution for N1-ODE[edit]

Given[edit]

(2.10.1)


Particular N1-ODE is

(2.10.2)

where and

Find[edit]

Verify that (2.10.1) is the solution for the (2.10.2).

Solution[edit]

Solved on my own


(2.10.1) would be rewritten as follows:

(2.10.3)


Differentiating the both sides of (2.10.3), (2.10.3) results in

(2.10.4)


Thus, (2.10.1) is verified to be the solution for (2.10.2.)

R 2.11-Solving PDE for Integrating Factor[edit]

Given[edit]

For

2.11.1

Find[edit]

Why solving equation 2.11.1 for the integrating factor is usually not easy ?

Solution[edit]

Solving equation 2.11.1 for the integrating factor is usually not easy because in general equation 2.11.1 is a non-linear partial differential equation for unknown function and it is having terms which makes it very difficult to solve for .

R*2.12 - Case 2: Finding as a Function of y[edit]

Given[edit]

A N1-ODE satisfies the first exactness condition if it takes the particular form:

(1)

The second exactness condition states that

(2)

If a N1-ODE satisfies the first exactness condition but not the second exactness condition the Euler Integrating factor method (IFM) can be used.
This requires an integrating factor be found such that the following N1-ODE is exact:

(3)

This can be written as:

(4)

To find we must apply the second exactness condition to obtain:

(5)

Find[edit]

Suppose , thus is a function of y only, then equation (5) becomes:

(6)

Find using equation (6).

Solution[edit]

Solved on my own

To find we can integrate equation (6) to obtain:

(7)





R*2.13 - Using an Integrating Factor to Render an Inexact L1-ODE-VC Exact[edit]

Given[edit]

2.13.1


Find[edit]

Show that h(x) = x is an integrating factor for 2.13.1, and that



is its solution.

Solution[edit]

This solution was found without reference to past solutions.


We first show that 2.13.1 is not already exact, or else its integrating factor would be !

Re-arranging 2.13.1, we arrive at the standard form


Now,

so 2.13.1 is not exact.

We thus proceed to multiply 2.13.2 by h(x) = x which yields

2.13.2

Now,

so 2.13.2 is exact and h(x) = x is the integrating factor that makes 2.13.1 exact.




Since every exact differential equation may be produced from a first integral such that

and


we may integrate and to find as follows:

2.13.3

2.13.4


Combining 2.13.3 and 2.13.4 yields



which upon solving for yields



While this already proves the assertion, we can also show that this is the correct expression for by plugging it back in to equation 2.13.1:



R*2.14 - General L1-ODE-VC[edit]

Given[edit]

(2.14.1)



(2.14.2)



(2.14.3)



Find[edit]

Solve the general L1-ODE-VC

1.





2. Find in term of

3.





Solution[edit]

Solved on my own

an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus, in this case often multiplying through by an integrating factor allows an inexact differential to be made into an exact differential.

From lecture note,



And applying second exactness condition, the equation is following that



If we assume that , then h is function of x only and the equation is following that



And we obtain some equations through rearranging equation above









And we can obtain h(x) through dividing by h



(2.14.1)



And in order to obtain y(x), we can apply the integrating factor to obtain the following equations



and



Thus,

(2.14.2)



part 1[edit]

1. In order to solve this problems, plug given information in eq (2.14.3)

the equation is following that



Also, eq (2.14.1) is following that





And eq (2.14.2) is following that







Let



Finally, the equation is that



In order to solve more, numerical method must be used.

part 2[edit]

From eq 2.14.3





and plug in eq 2.14.1





part 3[edit]





At this point,



Let





(2.14.4)



Then plug in eq 2.14.2



Let





Final solution is that



R*2.15- Integration constant[edit]

Solved on my own

Given[edit]

Since

(2.15.1)

(2.15.2)

is an L1-ODE_VC, there should be only one integration constant, not two.

Find[edit]

Show that the integration constant in

(2.15.3)


is not necessary, i.e., only in

(2.15.4)


is necessary.

Solution[edit]

Substituting the eqn (2.15.3) to eqn (2.15.4), we obtain the following expression





Then, we can define a new constant such that



hence, this new constant contains . In other words, is unnecessary.

R*2.16-Solution of L1-ODE-VC[edit]

Given[edit]

(2.16.1)


Integrating factor is

(2.16.2)


The derivated result in King 2003 p.512 is

(2.16.3)


Find[edit]

Show that (2.16.1) agrees with (2.16.3).

Solution[edit]

Solved on my own


General L1-ODE-VC is

(2.16.4)


In King 2003 p.512, the L1-ODE-VC is presented as follows:

(2.16.5)


From (2.16.3) and (2.164), the following relations can be derived.

(2.16.6)

(2.16.7)

(2.16.8)


From the relations, in King 2003 P512, the integrating factor is shwon as follows:

(2.16.9)


When (2.16.1) is substituted by (2.16.9), (2.16.1) becomes

(2.16.10)


Substituting the relation (2.16.8) for (2.16.10), (2.16.10) can be rewritten as follows:

(2.16.11)


(2.16.10) is the solution presented in King 2003 p.512 identifying A, and as follows:

(2.16.12)

(2.16.13)

(2.16.14)


Thus, it is verified that (2.16.1) agrees with (2.16.3).

R*2.17-Solving for homogenous counterpart[edit]

Given[edit]

Instead of identifying from .

Find[edit]

Solve the homogenous counterpart of .

Solution[edit]

Solved on my own













R*2.18 - Integrating Factor Method[edit]

Given[edit]

A L1-ODE-VC satisfies the first exactness condition if it takes the particular form:

(1)

The second exactness condition states that

(2)

If a L1-ODE-VC satisfies the first exactness condition but not the second exactness condition the Euler Integrating factor method (IFM) can be used.
This requires an integrating factor be found such that the following N1-ODE is exact:

(3)


Find[edit]

(4)





Find out whether equation (4) is exact. If it is not exact find the integrating factor to make it exact.

Solution[edit]

Solved on my own

Since equation (4) satisfies the first exactness condition we must test the second exactness condition.







Therefore, equation (4) is not exact. To make it exact an integration factor must be found such that:



Solving for h(x) produces:







References[edit]

Contributing Members[edit]

Team Contribution Table
Problem Number Assigned To Solved By Typed By Proofread By Signature
2.1 Jonathan Jonathan Jonathan all Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC)
2.2 kim kim kim all Egm6321.f11.team2.kim 19:51, 21 September 2011 (UTC)
2.3 Langshah Langshah Langshah all
2.4 Seounghyun Seounghyun Seounghyun all Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC)
2.5 Yuvraj Yuvraj Yuvraj all Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC)
2.6 Gary Gary Gary all
2.7 Jonathan Jonathan Jonathan all Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC)
2.8 Kim Kim Kim all Egm6321.f11.team2.kim 17:08, 21 September 2011 (UTC)
2.9 Langshah Langshah Langshah all
2.10 Seounghyun Seounghyun Seounghyun all Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC)
2.11 Yuvraj Yuvraj Yuvraj all Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC)
2.12 Gary Gary Gary all
2.13 Jonathan Jonathan Jonathan all Egm6321.f11.team2.epps 13:22, 21 September 2011 (UTC)
2.14 Kim Kim Kim all Egm6321.f11.team2.kim 17:08, 21 September 2011 (UTC)
2.15 Langshah Langshah Langshah all
2.16 Seounghyun Seounghyun Seounghyun all Egm6321.f11.team2.rho 19:09, 21 September 2011 (UTC)
2.17 Yuvraj Yuvraj Yuvraj all Egm6321.f11.team2.johri.y 17:36, 21 September 2011 (UTC)
2.18 Gary Gary Gary all

Egm6321.f11.team2.rho 20:41, 21 September 2011 (UTC)

  1. http://mathworld.wolfram.com/LegendreDifferentialEquation.html
  2. http://en.wikipedia.org/wiki/Chain_rule
  3. James, R.C. (1966) Advanced Calculus. Belmont, CA, Wadsworth.
  4. Marsden, J. and Weinstein, A. (1985) Calculus III. New York, Springer.pp769-772