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 Oct 11 comment Question about phase portrait and invariant subspaces @FemaleTank by definition all trajectories are invariant subspaces. Take any point in an orbit, take the action of the flow on that point in forward and backward time, it will give you the entire trajectory. Thus, eigenvectors are not the only invariant subspaces, however, they are important since you can obtain them explicitly without integrating the differential equation. Oct 11 comment Question about phase portrait and invariant subspaces First, I think that as you draw the directions, if $x_0\in L_1$ then $\lim_{t\to-\infty}e^{At}x_0=0$, right? (note the $-\infty$). Second, the same conclusions that you have for one eigenspace, say $L_1$ are applicable to $L_2$. So they you prove invariance in one, applies the same for the other just by changing time direction. Perhaps you can give more details if this is not enough clarification. Sep 12 comment A simple linear algebra question. It depends on the number of eigenvectors there are associated to the eigenvalue. It may happened that there are 1,2, or 3 linearly independent eigenvectors, which in turn will generate the eigenspace. For this, Jordan canonical form may be helpful. Aug 12 comment Conditions to preserve Laplacian matrix No, those are the principal properties that a Laplacian satisfies, in fact $\ker L=1_n$ is a consequence of $L1_n=0$ plus multiplicity assumptions. The definition of $L$ that I use is $L=D-A$ where $D$ is a diagonal "degree" matrix and $A$ is an adjacency matrix. Sorry, my syntax was misleading. Aug 12 comment Conditions to preserve Laplacian matrix Indeed, even more, I may assume (generically) that $\ker L=\left\{ 1_n\right\}$. However, concluding things from there is where I am stuck. Aug 12 comment Conditions to preserve Laplacian matrix Checking again your answer, I think it is not true since $W$ is or size $r\times n$. Then, the equation $Wx=0$ has non zero solutions. Aug 11 comment Diagonalization of A Dynamical System with Multiple Zero Eigenvalues exactly which part of Perko's book are you referring to? Under no extra assumption, what you claim may not be true (for example, depending on the multiplicity of the eigenvalues). It may interest you a result regarding the topological conjugacy (and decoupling) of systems having a center manifold, see e.g., staff.science.uu.nl/~kouzn101/NLDV/Lect12.pdf May 13 comment Poincaré lemma and conservative vector fields Of course! So silly of me, thank you. So should I thought as if $X$ satisfies $\frac{\partial X_i}{\partial x_j}=\frac{\partial X_j}{\partial x_i}$, then it is conservative? Mar 24 comment When does two curves do not intersect in the phase space Recall the existence and uniqueness of solutions of differential equations. If $F$ is at least $C^1$ in $x$ (one time differentiable with continuous derivative) then the equation $x'=F$ has a unique solution for a given initial condition $x(t_0)$. If two solutions were to intersect, that would mean that to a given initial condition (the point of intersection) correspond two orbits or solution curves. This contradicts unicity, and thus, it cannot happen. Feb 4 comment Why are equilibria so important? I guess Poincaré would be the best person to answer. I think he was the person who started to not care about explicit solution, but to look at the behaviour. Indeed, the main problem is that nonlinear systems are rarely solvable. Still, we do not quit in trying to understand them. So the next best thing to do is to study a simplified and local version of the nonlinear problem. For this, we may use the flow-box theorem away from equilibria, and so, it remains to study what happens near equilibria. Sep 24 comment How many solutions does the equation $2i+j+3k=l$ have in nonnegative integers? thanks a lot for this nice answer! Sep 18 comment critical points, differential equation To linearise you take derivative, or the so called Jacobian, and then evaluate the equilibrium point $(x,y)=(a,b/a)$. You can try it yourself and then ask if you have problems. Sep 18 comment critical points, differential equation there you go then. Sep 18 comment Proving that maximal interval of existence exists and that solution is unque Is Gronwall's inequality useful? encyclopediaofmath.org/index.php/Gronwall_lemma Sep 12 comment Stability analysis for a system of two differential equations You need to check the eigenvalues, and eigenvectors (at each equilibrium point), they will give you a local picture near each equilibrium point. Then you might be able to draw global conclusions. Of course everything will depend on the choice of the parameters, so you may have several distinct cases to consider, e.g. all parameters are positive, or all parameters are negative, and so on. Aug 26 comment Every solution of the system is attracted to the center manifold **tex text is the extension. Aug 24 comment Every solution of the system is attracted to the center manifold I made it on Inkscape together with an extension called text ext, which allows you to "write Latex". Aug 8 comment Central manifold theorem => Stable/unstable manifold? I don't think so, if the centre manifold is of dimension zero, then you have the common results of stable/unstable manifold theory. I must remark though that centre manifolds are very important as the dynamics there are given by the nonlinear terms of your system. Furthermore, centre manifolds are very useful, for example: 1) in singular perturbation problems, or the so called slow-fast systems; 2) In normal form theory; and related to it, 3) In reducing the dimension of the problem you are studying. Aug 4 comment Stable manifold for bidimensional nonlinear dynamic system with complex eigenvalues You are correct in your idea. Imagine a node. Every orbit is an invariant manifold. In fact every orbit is an invariant stable manifold. Their union form a 2 dimensional invariant stable manifold. The advantage of having real eigenvectors is that it is enough to study the generated spaces, in the sense that if you know what happens there, then you know what happens everywhere (in the linear case of course). For the complex case, recall that you can see a complex number in the real plane. So a 2d real system with complex eigs. can be seen in a 1d complex system (like passing to polar coords.) Jul 29 comment Doubt with smooth extensions I guess $\tilde\pi^{−1}(x)=(x,\pm\sqrt{-x})=(-y^2,\pm y)$ is the correct way of taking $\tilde\pi^{-1}$, but since I know which $y$ we are taking since the beginning, we could set $\tilde\pi^{-1}(y)=(-y^2,y)$.