I am completely stuck on this:
The 2nd order system should be in this form: $\frac{dx}{dt}=Ax$ where A is the system matrix. $$x(t) = \begin{pmatrix} 2-e^{-t} \\ 1+2e^{-t} \end{pmatrix}$$ $$x(t=0) =: x_0 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$$

I have to obtain the system matrix A and the transition matrix $\Phi(t)$ But I have no clue what I can do here. I feel like I have too less information to solve this.

I was experimenting with $$\Phi(0) = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}$$ and $$x(t)=\Phi(t)x_0$$ but I am stuck here.

Any help appreciated.


$$ \pmatrix{\frac{s+2}{s+1}\\ \frac{3s+1}{s+1}}-A\pmatrix{\frac{s+2}{s(s+1)}\\ \frac{3s+1}{s(s+1)}} =\pmatrix{1\\3} \implies \pmatrix{\frac{1}{s+1}\\ \frac{-2}{s+1}}=A \pmatrix{\frac{s+2}{s(s+1)}\\ \frac{3s+1}{s(s+1)}} \implies \pmatrix{1\\ -2}=A \pmatrix{1 + \frac{2}{s}\\ 3 + \frac{1}{s}} $$

  • $\begingroup$ Are you sure that this question is asked as it is? Because it is a linear system and the equilibrium point is not the origin. In other words, $(2,1)$ point is also an equilibrium together with $(0,0)$ and this cannot happen in the case of a linear system. $\endgroup$ – user13838 Oct 8 '11 at 19:14
  • $\begingroup$ @percusse, why not? If $0$ is an eigenvalue of the coefficient matrix, then every eigenvector for that eigenvalue is an equilibrium point. $\endgroup$ – Henning Makholm Oct 8 '11 at 19:22
  • $\begingroup$ @HenningMakholm That's correct. I think I'll call it a day and hit the bar. :) $\endgroup$ – user13838 Oct 8 '11 at 19:31

The response of an autonomous system is indeed defined by the matrix exponential - transition matrix or the Laplace transformed version of the differential function which can be obtained through $\mathcal{L}(\dot x) = sX(s)-x(0)$ where $s$ being the indeterminate of the Laplace transform: $$ x(t) = e^{At} x(0) \text{ or}\quad X(s) = (sI-A)^{-1}x(0) $$ From this and after applying Laplace transform to the given time trajectories, we have, $$ \pmatrix{\frac{2}{s} - \frac{1}{s+1}\\\frac{1}{s}+\frac{2}{s+1}} = \pmatrix{\frac{s+2}{s(s+1)}\\ \frac{3s+1}{s(s+1)}}=(sI-A)^{-1} \pmatrix{1\\3}$$ Then, $$ (sI-A)\pmatrix{\frac{s+2}{s(s+1)}\\ \frac{3s+1}{s(s+1)}} =\pmatrix{1\\3} \implies \pmatrix{\frac{1}{s+1}\\\frac{-2}{s+1}}= A\pmatrix{\frac{s+2}{s(s+1)}\\ \frac{3s+1}{s(s+1)}}$$ Let $$A = \pmatrix{a &b\\c&d}$$ then $a(s+2)+b(3s+1) = s$ and $c(s+2) + d(3s+1) = -2s$. (Note that $s$ is cancelled out). These leads to $$ \pmatrix{1 &3\\2 &1}\pmatrix{a\\b} = \pmatrix{1\\0}\ , \ \pmatrix{1 &3\\2 &1}\pmatrix{c\\d} = \pmatrix{-2\\0} $$

Solving for $a,b,c,d$ gives, $$ A = \pmatrix{\frac{-1}{5} &\frac{2}{5}\\\frac{2}{5} &\frac{-4}{5}} $$

  • $\begingroup$ Wow, that is completely opaque to me. What is $s$? How is it related to $t$? Is $X(s)$ a typo for $x(s)$, or does a capital $X$ (of which there are exactly 1 in your answer) have some implied meaning here? Why are there no exponentials anywhere? $\endgroup$ – Henning Makholm Oct 8 '11 at 20:33
  • $\begingroup$ @HenningMakholm I have added some more info. Let me know if it needs further clarification. $\endgroup$ – user13838 Oct 8 '11 at 20:46
  • $\begingroup$ I still understand nothing at all. But the presence of script L's usually signifies that I'm out of my depth, so I will just be satisfied with my own version and not press the matter. $\endgroup$ – Henning Makholm Oct 8 '11 at 20:54
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    $\begingroup$ I think I can sort of see now what you're doing, after investigating the Laplace transform on Wikipedia. I didn't know about that before (somehow I managed to get a math B.S. without ever meeting any integral transform in a course setting). The $\mathcal L$'s that usually scare me away are Lagrangians in physics ... $\endgroup$ – Henning Makholm Oct 9 '11 at 3:12
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    $\begingroup$ @madmax s is the indeterminate, hence not a variable. I will edit my answer to show what I did. $\endgroup$ – user13838 Oct 9 '11 at 9:31

Here's how I would find $A$: Let $\beta(t)=\pmatrix{1\\e^{-t}}$. Then $$x(t) = \pmatrix{2&-1\\1&2}\beta(t) \text{ and } \frac{d}{dt}\beta(t)=\pmatrix{0&0\\0&-1}\beta(t)$$ Therefore, set $$X = \pmatrix{2&-1\\1&2}, \qquad D=\pmatrix{0&0\\0&-1}$$ Since differentiation is linear, it commutes with a linear transformation, so $$\frac{d}{dt}x(t) = \frac{d}{dt}X\beta(t) = X\frac{d}{dt}\beta(t) = XD\beta(t)$$ Thus the equation to be satisfied is $AX\beta(t)=XD\beta(t)$ for all $t$. We can achieve this by setting $AX=XD$ which gives $A=XDX^{-1}$.

I'm less sure what the canonical way to deal with the transition matrix is. One heuristic attempt would be to observe $$\beta(t)=\pmatrix{1&0\\0&e^{-t}}\beta(0)=e^{tD}\beta(0)$$ and therefore $$x(t)=X\beta(t)=Xe^{tD}\beta(0)=Xe^{tD}X^{-1}X\beta(0)=Xe^{tD}X^{-1}x(0)=e^{tXDX^{-1}}x(0)$$ suggesting $\Phi(t) = e^{tA}$. (Hmm.. this probably ought to be an explicit theorem of your text. Find and reference it instead of reproducing the above!)

  • $\begingroup$ I think your solution is the way without the Laplace transform. I like the way with laplace more ;) $\endgroup$ – madmax Oct 8 '11 at 22:14

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