University of Florida/Egm4313/IEA-f13-team10/R2

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Report 2[edit]

Problem 1[edit]

Problem Statement[edit]

Show the equivalence between the two statements that define the linearity of differential operators.

Statement (1):
The definition of a linear operator, , is



for any two constants and , and for any two functions and

Statement (2):
The operator is linear iff


for any constant and for any function ,

and


for any two functions and .

Solution[edit]

We saw on piazza how to prove statement (1) from statement (2). To prove statement (2) from statement (1) we choose clever values for the constants and and for the functions and .

Proof for the Scaling Property[edit]

Since and can be any constant, and and can be any function, let us set



and



Then (1) becomes





Now let us say 2c is equal to some constant . The above equation becomes



which is the scaling property of a linear operator found in statement (2).

Proof for the Addition Property[edit]

Since and can be any constant, let us set



Then (1) becomes



which is the addition property of a linear operator found in statement (2).

Honor Pledge[edit]

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 2[edit]

Problem Statement[edit]

Build on my note in the same piazza post on differential operators on the notation
for general differential operators, and the notation
as defined in Eq.(1) in K 2011, sec.2.3, to explain the similarities and differences between these two notations.

Solution[edit]

The similarity with both and is that they look like they're both upper case D's. Both of these D's also symbolized the differential operator.
The difference between both of these D's is that
is a generalized differential operator, which can be seen as
and refers to the first differential operator:

Honor Pledge[edit]

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.


Problem 3[edit]

Problem Statement[edit]





Identify the polynomial as defined in K 2011, sec.2.3, for this problem.
Then rewrite the above problem in terms of the linear differential operator .
Use its linear property to demonstrate the above problem into a homogeneous part and a particular part.
How would the initial conditions be written in terms of the homogeneous solution and the particular solution?

Solution[edit]



Let



Splitting into homogeneous and particular,



Inputting the initial conditions,



Honor Pledge[edit]

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 4[edit]

Problem Statement[edit]

Given the two roots an the initial conditions:



Find the non-homogeneous L2-ODE-CC in standard form
and the solution in terms of the initial conditions
and the general excitation .
Consider no excitation

Plot the solution

Solution[edit]








Therefore,


Solve for constants and :










so,

Prob.4.PNG

Honor Pledge[edit]

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 5[edit]

R 2.5 K 2011, sec 2.2, p. 59, pb.5, with initial conditions

Problem Statement[edit]

Find the general solution to the ODE
With the initial values y(0) = 1 and y'(0) = .5

Solution[edit]


This is a linear, first order ODE with constant coefficients.

To find the general solution to this ODE set

so that and

Substituting in y to the ODE and factoring out we get:


Using the quadratic formula to solve for r we get

where and

Solving to get

Since we have a repeated root, we need to find v(x) so that y2(x) = v(x)y1(x)

Taking the first and second derivative of y2(x) we get:




solving for v(x) = 0 so v(x) = kx + c

So y2(x) = x y1(x)


We get the general solution

Now with the initial values y(0) = 1 and y'(0) = .5






,


0-1.JPG 0-5.JPG

Honor Pledge[edit]

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Problem 6[edit]

Problem Statement[edit]

Discuss the similarities and differences between Coulomb friction / damping (piazza post) with the damper used in the SDOF spring-mass-damper system with 2 ends fixed. 2.6.jpg

Solution[edit]

Similarities[edit]

With both ends fixed for SDOF system, we can say that (1)
In the theory of Coulomb friction/damping, the friction changes the direction of the oscillating object by "damping" the force of motion.

In the case of the SDOF system, the damper fixed to one of the ends is friction force shown in the theory.

Differences[edit]

The equation of motion for the SDOF system with two fixed ends is

The equation of motion for a typical coulomb damping system is

or

As seen above, the typical system involves a vague force to change the direction of an oscillating object while the SDOF system has a specific damper as function of .

Honor Pledge[edit]

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.