• Support PF! Buy your school textbooks, materials and every day products Here!

Repeated eigenvalues

  • #1

Homework Statement


I want to solve this system


x' = [itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex]x + [itex]\left( \begin{array}\\ t \\ 2t \end{array} \right)[/itex]

Homework Equations




The Attempt at a Solution



i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?
 
Last edited:

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,577

Homework Statement


I want to solve this system

x' = [(7,1),(-4,3)] + t[(1),(2)]
I apologize for being dense, but I don't understand this notation. Is the expression on the right a 2x2 matrix? If so, it would make it more readable if you used Tex notation:

[itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex]
 
  • #3
I apologize for being dense, but I don't understand this notation. Is the expression on the right a 2x2 matrix? If so, it would make it more readable if you used Tex notation:

[itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex]
Yes, that's right. I'm not really familiar with Latex, I'll have to read up on it. I've edited my post, thanks
 
  • #4
33,262
4,963

Homework Statement


I want to solve this system


x' = [itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex] + [itex]\left( \begin{array}\\ t \\ 2t \end{array} \right)[/itex]
As written, this makes no sense. You can't add a 2 x 2 array to a 2 x1 array (column matrix).

This would make more sense if it were something like this:
##\vec{x'} = \begin{bmatrix} 7 & 1 \\ -4 & 3 \end{bmatrix}\vec{x} + \begin{bmatrix} 1 \\ 2 \end{bmatrix}##

Here ##\vec{x}## means ##\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}## and similar for its derivative.
faradayscat said:

Homework Equations




The Attempt at a Solution



i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?
 
  • #5
As written, this makes no sense. You can't add a 2 x 2 array to a 2 x1 array (column matrix).

This would make more sense if it were something like this:
##\vec{x'} = \begin{bmatrix} 7 & 1 \\ -4 & 3 \end{bmatrix}\vec{x} + \begin{bmatrix} 1 \\ 2 \end{bmatrix}##

Here ##\vec{x}## means ##\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}## and similar for its derivative.
Yes that's that I meant, sorry for the confusion. So could i set up a fundamental matrix with the homogeneous solutions and solve the non-homogeneous system with, say, variation of parameters? I tried and this is what I get:

x(t) = c1e5t[itex]\left( \begin{array}\\ -1 \\ 2 \end{array} \right)[/itex] + c2e5t[itex]\left( \begin{array}\\ -t-1/2 \\ 2t \end{array} \right)[/itex] - (t/25)[itex]\left( \begin{array}\\ 1 \\ 18 \end{array} \right)[/itex] + (1/125)[itex]\left( \begin{array}\\ 3 \\ -26 \end{array} \right)[/itex]

I checked my solution on wolfram and it's slightly different, which annoys me.
 
  • #6
33,262
4,963
Yes that's that I meant, sorry for the confusion. So could i set up a fundamental matrix with the homogeneous solutions and solve the non-homogeneous system with, say, variation of parameters? I tried and this is what I get:

x(t) = c1e5t[itex]\left( \begin{array}\\ -1 \\ 2 \end{array} \right)[/itex] + c2e5t[itex]\left( \begin{array}\\ -t-1/2 \\ 2t \end{array} \right)[/itex] - (t/25)[itex]\left( \begin{array}\\ 1 \\ 18 \end{array} \right)[/itex] + (1/125)[itex]\left( \begin{array}\\ 3 \\ -26 \end{array} \right)[/itex]

I checked my solution on wolfram and it's slightly different, which annoys me.
Just check that your solution satisfies the system of diff. equations.
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


I want to solve this system


x' = [itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex] + [itex]\left( \begin{array}\\ t \\ 2t \end{array} \right)[/itex]

Homework Equations




The Attempt at a Solution



i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?
Isn't there a factor ##x## missing on the right? Should you not be dealing with the system
[tex] \pmatrix{x_1'\\x_2'} = \pmatrix{7 & 1 \\ -4 & 3} \pmatrix{x_1\\x_2} + \pmatrix{t \\2t} ? [/tex]
You can either plug in the matrix exponential in the solution
[tex] {\mathbf{x}} = e^{At} \int_0^t e^{-A \tau} {\mathbf{f}}(\tau) \, d \tau [/tex]
to your equation ##{\mathbf{x}}'(t) = A {\mathbf{x}}(t) + {\mathbf{f}}(t)##, or else use the Laplace-transform method.

As for the matrix exponential: you have ##A = P J P^{-1}##, where ##J## is the Jordan canonical form of ##A##:
[tex] J = \pmatrix{5 & 1 \\0 & 5} [/tex]
Furthermore, for any scalar ##x## we have ##e^{Ax} = P e^{Jx} P^{-1}##, and ##e^{Jx}## is easy to determine; see
webpages on matrix exponentials.
 
  • #8
Just check that your solution satisfies the system of diff. equations.
I just checked, and its satisfied. Thanks!
 
  • #9
Isn't there a factor ##x## missing on the right? Should you not be dealing with the system
[tex] \pmatrix{x_1'\\x_2'} = \pmatrix{7 & 1 \\ -4 & 3} \pmatrix{x_1\\x_2} + \pmatrix{t \\2t} ? [/tex]
You can either plug in the matrix exponential in the solution
[tex] {\mathbf{x}} = e^{At} \int_0^t e^{-A \tau} {\mathbf{f}}(\tau) \, d \tau [/tex]
to your equation ##{\mathbf{x}}'(t) = A {\mathbf{x}}(t) + {\mathbf{f}}(t)##, or else use the Laplace-transform method.

As for the matrix exponential: you have ##A = P J P^{-1}##, where ##J## is the Jordan canonical form of ##A##:
[tex] J = \pmatrix{5 & 1 \\0 & 5} [/tex]
Furthermore, for any scalar ##x## we have ##e^{Ax} = P e^{Jx} P^{-1}##, and ##e^{Jx}## is easy to determine; see
webpages on matrix exponentials.
Yes, I forgot the vector 'x' next to the coefficient matrix. Thanks for your tips, I actually did the variation of parameters method and everything worked out after I checked my solution as Mark44 suggested.
 
  • #10
Thanks everyone, I have no further questions.
 

Related Threads on Repeated eigenvalues

Replies
0
Views
781
Replies
4
Views
1K
Replies
3
Views
1K
Replies
4
Views
2K
  • Last Post
Replies
1
Views
832
  • Last Post
Replies
2
Views
3K
Replies
5
Views
2K
  • Last Post
Replies
4
Views
2K
Top