Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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

share|cite|improve this question
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. – user13838 Oct 8 '11 at 19:14
@percusse, why not? If $0$ is an eigenvalue of the coefficient matrix, then every eigenvector for that eigenvalue is an equilibrium point. – Henning Makholm Oct 8 '11 at 19:22
@HenningMakholm That's correct. I think I'll call it a day and hit the bar. :) – user13838 Oct 8 '11 at 19:31
up vote 2 down vote accepted

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, $$ A = \pmatrix{\frac{-1}{5} &\frac{2}{5}\\\frac{2}{5} &\frac{-4}{5}} $$

share|cite|improve this answer
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? – Henning Makholm Oct 8 '11 at 20:33
@HenningMakholm I have added some more info. Let me know if it needs further clarification. – user13838 Oct 8 '11 at 20:46
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. – Henning Makholm Oct 8 '11 at 20:54
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 ... – Henning Makholm Oct 9 '11 at 3:12
@madmax s is the indeterminate, hence not a variable. I will edit my answer to show what I did. – 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!)

share|cite|improve this answer
I think your solution is the way without the Laplace transform. I like the way with laplace more ;) – madmax Oct 8 '11 at 22:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.