Solving Difference Equations w/ Eigenvalues i & -i

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In summary, the solution to du/dt = Au = [ 0 -1; 1 0] u (eigenvalues i and -i) goes around in a circle: u = (cos t, sin t). The eigenvalues of I + A, (I-A)^-1, and (I-.5A)^-1(I+.5A) are 1,-.5, and 1.5, respectively.
  • #1
veronicak5678
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Homework Statement



The solution to du/dt = Au = [ 0 -1; 1 0] u (eigenvalues i and -i) goes around in a circle: u = (cos t, sin t). Suppose we approximate du/dt by forward, backward, and centered differences F, B, and C:
(F) U(n+1) - U(n) = AU(n)
(B) U(n+1) - U(n) = AU(n+1)
(C) U(n+1) - U(n) = .5A(U(n+1)+U(n))
Find the eigenvalues of I + A, (I-A)^-1, and (I-.5A)^-1(I+.5A). For which difference equation does the solution U(n) stay on a circle?

Homework Equations



Ax = lambda x

The Attempt at a Solution


I really don't know where to start...
 
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  • #2
Take (F) for example. You can rewrite it as

U(n+1) = (I+A)U(n)

Note the matrix multiplying U(n) is the first matrix you're asked to find the eigenvalues for. You can do the same thing for the other two relations, and you should find that the matrices that multiply U(n) are the other two matrices.

What the problem wants you to do is start by finding the eigenvalues of those three matrices.
 
  • #3
Thanks for answering. For the eigenvalues I got:
i) 1+i & 1-i
ii) .5 + .5i & .5 - .5i
iii) 3/5 + 4/5i & 3/5 - 4/5i

Do they look OK? How do I know which stays on a circle?
 
  • #4
You found the correct eigenvalues. Say you found the corresponding eigenvectors as well. How would you find the solution for each equation?
 
  • #5
I) x1=[.707 -.707i] x2=[.707 .707i]
II) x1=[-.707i -.707] x2=[.707i -.707]
III)x1=[.707 -.707i] x2=[.707 .707i]

Do you mean I need to use the diagonal matrix and find the inverse of each of these?
 
  • #6
No, I mean if you consider these systems in the basis where the matrix is diagonal, you should easily be able to deduce how the solutions will evolve. It's not necessary to actually solve the systems this way, but that might be the easiest way to see what the answer is if you can't already see what the solutions will look like.
 

Related to Solving Difference Equations w/ Eigenvalues i & -i

1. What are eigenvalues in difference equations?

An eigenvalue in a difference equation is a special value that, when multiplied by a corresponding eigenvector, produces a new vector that is parallel to the original vector. In other words, it represents a scalar factor by which the eigenvector is stretched or compressed.

2. How are eigenvalues used to solve difference equations?

Eigenvalues are used to solve difference equations by first finding the eigenvalues of the given difference equation. Then, these eigenvalues are used to construct a general solution to the difference equation, which can then be used to solve for specific values or initial conditions.

3. What do the complex values of i and -i represent in difference equations?

The complex values of i and -i represent the imaginary unit, which is used to solve difference equations that involve complex numbers. These values are typically seen in systems that involve oscillation or periodic behavior.

4. Can difference equations with eigenvalues i and -i have real solutions?

Yes, difference equations with eigenvalues i and -i can have real solutions. This is because the complex values of i and -i can be combined to form real solutions, as long as the initial conditions are also real numbers.

5. Are there any practical applications of solving difference equations with eigenvalues i and -i?

Yes, there are many practical applications of solving difference equations with eigenvalues i and -i. For example, they can be used to model and analyze systems that exhibit periodic behavior, such as population growth or oscillating electrical circuits. They are also commonly used in engineering and physics to study systems that involve complex numbers.

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