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

Report 2

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Problem 1

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Problem Statement

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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

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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
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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
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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

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

Problem 2

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Problem Statement

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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

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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

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


Problem 3

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Problem Statement

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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

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Let  
 
 

Splitting into homogeneous and particular,
 
 

Inputting the initial conditions,
 
 


Honor Pledge

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

Problem 4

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Problem Statement

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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

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Therefore,  
 

Solve for constants   and  :
 
 

 
 
 
 

 
 
so,
 
 

Honor Pledge

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

Problem 5

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

Problem Statement

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

Solution

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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   so  

So y2(x) = x y1(x)


We get the general solution  

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


 

 

  ,  

 

       

Honor Pledge

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

Problem 6

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Problem Statement

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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.  

Solution

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Similarities
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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
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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

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