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I have problem and need your help. I must draw phase portraits of dynamical system which looks like this: $$\dot{x}_{1}(t) = -x_{1}(t) + x_{2}(t)$$ $$\dot{x}_{2}(t) = -x_{2}(t)$$ I know that the first I sould get eigenvector and eigenvalue of matrix but I'm not soure. Is this matrix will looks like this:

$$\begin{matrix} -1 & 1 \\ -1 & 0 \\ \end{matrix}$$ Am I right? What sould I do next?

Éric Guirbal
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Hadson
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  • A very helpful website and tool for drawing phase portraits can be found at http://math.rice.edu/~dfield/dfpp.html – Euler....IS_ALIVE Oct 18 '15 at 21:06
  • In your particular case the situation is a bit difficult because the eigenvalues are complex. They both have negative real part, which means the origin is ultimately stable, but you need to know how to interpret the complex eigenvalues/eigenvectors to properly sketch how the solutions spiral in towards the origin. An easier example to sketch would be $\begin{bmatrix} -1 & -1 \ -1 & 0 \end{bmatrix}$, which has a saddle point instead of a stable spiral at the origin. – Ian Oct 18 '15 at 21:22
  • Apparently I get down-voted for attempting to educate rather than hand over the answer on a plate, but the linearised system is wrong as stated in the question – Chris Kerridge Oct 18 '15 at 21:26
  • I didn't down-voted you but it was a bit hard to check is something is Hamiltonian when you hear this word first time in life ;). To find some information about dynamical systems in homeland language is really hard and my science english isn't so good, i read this like forth time and still don;t know how do to it ;/ – Hadson Oct 18 '15 at 21:40
  • Oh, there IS a typo: you actually have $\begin{bmatrix} -1 & 1 \ 0 & -1 \end{bmatrix}$. This is quite different from what you wrote, because this system is "defective", meaning that there is a single real eigenvalue of algebraic multiplicity $2$ and geometric multiplicity $1$. This comes with a certain characteristic behavior. – Ian Oct 18 '15 at 21:45

2 Answers2

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Not sure I understand what you are asking exactly but, if you want to draw a phase portrait, draw a phase portrait:

enter image description here

The general approach is to delineate the regions of the $(x_1,x_2)$-plane where $x'_1$ and $x'_2$ have a constant sign, and to deduce the variations of the solutions from this decomposition. These regions are limited by the so-called nullclines, which are the lines where $x'_1=0$ or $x'_2=0$.

In the present case, $x'_1=0$ corresponds to the first diagonal $x_2=x_1$ and $x'_2=0$ corresponds to the horizontal axis $x_2=0$. For example, at every point in the region $0<x_2<x_1$ (North-East to East angular sector), the dynamics points to the South-East ($x'_1<0$, $x'_2<0$). Likewise for the three other angular sectors and the four halflines which delimit them (whose union is the union of the nullclines), hence the diagram above.

Finally, in case the system you are interested in is actually $$x'_1=-x_1+x_2\qquad x'_2=-x_1$$ the same procedure applies, and yields the phase portrait below.

enter image description here

Did
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  • Did is describing the method of nullclines, which can be easily googled for more details. – Ian Oct 18 '15 at 21:23
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First thing should be to check if it's Hamiltonian which makes the whole process much easier as you just sketch level sets.

Find the stationary points, then classify them using the trace and determinant of the linearized system at those points. You can find the eigenvectors to find the orientation of sources/sinks/saddles/centres if it's not obvious from plugging in some values.

This should get you quite some way to a complete portrait.

Chris Kerridge
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  • That useful but could you say step by step or show on example how should i do that? This is my first time in this kind of math thats will be very grateful – Hadson Oct 18 '15 at 21:04