1
$\begingroup$

I am having a lot of trouble understanding how to move from a system of ODE's to a dynamical systems point of view (that will allow me to make a phase-plane analysis).

Assume I want to write the following system (I invented it just as an example): $$ \begin{split} y''+2x+3y'+4x'=0\\ 2y''+2x-4y'+2x'=0 \end{split} $$ in a matrix form: $$ \frac{d}{dt} \begin{bmatrix} x\\x'\\y\\y' \end{bmatrix} = A\begin{bmatrix} x\\x'\\y\\y' \end{bmatrix} $$ where $A$ is a $4\times 4$ matrix.

I know I can write this system as: $$ \begin{bmatrix} 0&0&0&1\\0&0&0&2\end{bmatrix} \frac{d}{dt} \begin{bmatrix} x\\x'\\y\\y' \end{bmatrix} = \begin{bmatrix} -2&4&0&-3\\-2&-2&0&4\end{bmatrix} \begin{bmatrix} x\\x'\\y\\y' \end{bmatrix} $$ but it does not help me much (the inverse of the matrix in the LHS is not a well defined notion...).

In addition, I know that if I would only had the first equation, I would be able to write it as: $$ \frac{d}{dt} \begin{bmatrix} x\\x'\\y\\y' \end{bmatrix} = \begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&1\\-2&-4&0&-3 \end{bmatrix}\begin{bmatrix} x\\x'\\y\\y' \end{bmatrix} $$ but how can I combine the two equations into such a form (in order to make a phase-plane analysis) ??

Just to clarify- I know that one possible solution is to isolate $y''$ from the first equation, substitute in the second one, and then move to a matrix form, but I don't want to do it (too messy and not very helpful in case of a lot of variables and a lot of equations).

Will you please help me?

Thanks

$\endgroup$
5
  • $\begingroup$ What is the independent variable? You write $d/dx$, but $x$ also appears as un unknown function. $\endgroup$ Commented Mar 11, 2016 at 10:31
  • $\begingroup$ The system is $4$-dimensional. You cannot do a phase-plane analysis, since the plane is only $2$-dimensional. $\endgroup$ Commented Mar 11, 2016 at 11:04
  • $\begingroup$ You are right. Maybe I should rephrase my problem - I want to find the eigenvalues of the matrix $A$, when the system is viewed in a matrix notation (no need to make a further phase plane analysis). I really hope I made myself clear now. What do you think? Thanks you ! $\endgroup$ Commented Mar 11, 2016 at 11:08
  • $\begingroup$ Not a good example, maybe $x''$ instead of $y''$ in the second equation? $\endgroup$ Commented Mar 11, 2016 at 11:16
  • $\begingroup$ That's exactly the thing. I have $y''$ in both equations, and not one equation with $y''$ and the other with $x''$ ... Have you got an idea? Thanks! $\endgroup$ Commented Mar 11, 2016 at 15:16

2 Answers 2

Reset to default
1
$\begingroup$

Strictly speaking, if you decide just to write down two linear equations involving $x$, $y$ and their (first and second) time derivatives, there is no a priori reason to assume those two equations are consistent with each other. Therefore, the best thing to do is to write each equation separately as a four-dimensional linear system of the form $\mathbf{x}' = A \mathbf{x}$, and solve those two. If a solution obeys both original equations, it has to solve both dynamical systems simultaneously. By comparing their general solutions, you can then see if these overlap in some way.

Note that I didn't go into details regarding the example you gave, since I've got the impression your question is not about this example in particular, but about the general approach.

$\endgroup$
1
  • $\begingroup$ OK. I think I understand your answer . Thanks !!!! $\endgroup$ Commented Mar 11, 2016 at 15:17
0
$\begingroup$

Note that $\det \mathbf{A}=0$ and both $x'$ and $y'$ appear on both sides. How about in this way?

$$\left( \begin{array}{c} x' \\ y'' \end{array} \right)= \left( \begin{array}{cc} -\frac{1}{3} & -\frac{5}{3} \\ -\frac{2}{3} & \frac{11}{3} \end{array} \right) \left( \begin{array}{c} x \\ y' \end{array} \right)$$

$\endgroup$
3
  • $\begingroup$ What you actually did is to substitute some expressions from the first equation into the second, and vice versa, until you were able to write the system as you suggested, right? $\endgroup$ Commented Mar 11, 2016 at 15:19
  • $\begingroup$ @Bobby Yep sth like that. Find $(x',y'')$ in terms of $(x,y')$. $\endgroup$ Commented Mar 11, 2016 at 15:41
  • $\begingroup$ Thanks. I thought there is something easier, that does not involve any algebraic manipulations beforehand . Thanks ! $\endgroup$ Commented Mar 11, 2016 at 17:07

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.