Solve System with Repeated Eigenvalues

In summary: Should you not be dealing with the system \pmatrix{x_1'\\x_2'} = \pmatrix{7 & 1 \\ -4 & 3} \pmatrix{x_1\\x_2} + \pmatrix{t \\2t} ?
  • #1
faradayscat
57
8

Homework Statement


I want to solve this systemx' = [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?
 
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  • #2
faradayscat said:

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
stevendaryl said:
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
faradayscat said:

Homework Statement


I want to solve this systemx' = [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
Mark44 said:
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
faradayscat said:
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
faradayscat said:

Homework Statement


I want to solve this systemx' = [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
Mark44 said:
Just check that your solution satisfies the system of diff. equations.

I just checked, and its satisfied. Thanks!
 
  • #9
Ray Vickson said:
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.
 

What is a repeated eigenvalue?

A repeated eigenvalue occurs when a matrix has at least one eigenvalue that has a multiplicity greater than one, meaning it appears more than once. This means that there are multiple eigenvectors associated with the same eigenvalue.

Why is it important to solve systems with repeated eigenvalues?

Solving systems with repeated eigenvalues allows us to accurately analyze and predict the behavior of complex systems. It also helps us understand the underlying patterns and relationships within the data.

What methods can be used to solve systems with repeated eigenvalues?

Some common methods for solving systems with repeated eigenvalues include diagonalization, the Jordan canonical form, and the generalized eigenvector method. Each method has its own advantages and limitations, so the choice will depend on the specific problem at hand.

How do repeated eigenvalues affect the stability of a system?

Repeated eigenvalues can indicate a lack of distinct eigenvalues in the system, which can lead to instability and unpredictable behavior. In these cases, additional analysis and techniques may be necessary to accurately model and control the system.

Can repeated eigenvalues have complex values?

Yes, repeated eigenvalues can have complex values. In fact, complex eigenvalues often occur in systems with repeated eigenvalues, as they represent oscillatory behavior in the system. This can be seen in systems such as electrical circuits and mechanical systems with rotating components.

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