Is the System of ODEs Defined by Matrix B Decoupled?

  • Thread starter Thread starter jimmycricket
  • Start date Start date
  • Tags Tags
    Odes System
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
jimmycricket
Messages
115
Reaction score
2
Given the matrix [itex]b=\begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix}[/itex] decide if the system of ODEs, [itex]\frac{dx}{dt}=Bx[/itex] is decoupled. If yes find the general solution x=xh(t)

Homework Equations





The Attempt at a Solution


I would say the matrix is decoupled since the second equation involving [itex]2x[/itex]2(t) can be solved without the other two equations. Then the third equation can be solved without knowing [itex]x[/itex]1(t). We have:
[itex] x'_1 = -x_1 - x_3 \\ <br /> x'_2 = -4x_1 + 3x_2 - x_3 \\ <br /> x'_3 = -2x_3[/itex]
Im not sure where to go from here.
 
Physics news on Phys.org
You may want to look into matrix diagonalization [1]. If B can written as A D A-1, where D is a diagonal matrix, can you then use this to rewrite you ODE system to a new uncoupled variable basis?

[1] http://en.wikipedia.org/wiki/Diagonalizable_matrix
 
I have found the diagonal matrix,
[tex]D=<br /> \begin{pmatrix}<br /> -1 & 0 & 0\\<br /> 0 & 3 & 0\\<br /> 0 & 0 & -2<br /> \end{pmatrix}[/tex]
I thought the matrix B was already uncoupled though. Is this not the case?
 
jimmycricket said:
Given the matrix [itex]b=\begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix}[/itex] decide if the system of ODEs, [itex]\frac{dx}{dt}=Bx[/itex] is decoupled. If yes find the general solution x=xh(t)

Homework Equations





The Attempt at a Solution


I would say the matrix is decoupled since the second equation involving [itex]2x[/itex]2(t) can be solved without the other two equations.
?
Do you mean the third equation? It involves only x3' and x3.
jimmycricket said:
Then the third equation can be solved without knowing [itex]x[/itex]1(t). We have:
[itex] x'_1 = -x_1 - x_3 \\ <br /> x'_2 = -4x_1 + 3x_2 - x_3 \\ <br /> x'_3 = -2x_3[/itex]
Im not sure where to go from here.

jimmycricket said:
I have found the diagonal matrix,
[tex]D=<br /> \begin{pmatrix}<br /> -1 & 0 & 0\\<br /> 0 & 3 & 0\\<br /> 0 & 0 & -2<br /> \end{pmatrix}[/tex]
I thought the matrix B was already uncoupled though. Is this not the case?
The system of equations was not uncoupled. The purpose of finding a diagonal matrix that is similar to B gives you a system that is uncoupled. In an uncoupled system, each equation involves only a single variable and its derivative.
 
That was an error on my part, what I meant was the matrix is decoupled since the second equation involving [itex]-2x_3(t)[/itex] can be solved without the other two equations and then we can solve for [itex]x_1(t)[/itex] without knowing [itex]x_2(t)[/itex]