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 August 1st, 2018, 02:54 PM #1 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 Determine nullclines and sketch them. So I am having a problem with this question. Can someone please help me? I have a system of two first-order ODEs: $\begin{cases} x^{\prime} = y \\ y^{\prime} = -kx - 2\beta y \end{cases}$ a) Find the eigenvalues of its matrix. For what values of $k$ and $\beta$ do you get two distinct real eigenvalues, and for what values do you get two non real complex eigenvalues? For this question, we have $M = \begin{bmatrix}0 & 1\\-k & -2\beta\end{bmatrix}$ then we set $det(M - \lambda I) = 0$, which leads to $\lambda ^2 + 2\beta \lambda + k = 0$, by using quadratic equation we will have: $\lambda _1 = -\beta + \sqrt{\beta ^2 - k}$ $\text{or } \lambda _2 = -\beta - \sqrt{\beta ^2 - k}$ For $\beta ^2 - k > 0$ or $k < \beta ^2$ there are two distinct real eigenvalues. For $\beta ^2 - k < 0$ or $k > \beta ^2$ there are two non real complex eigenvalues. For $k = \beta ^2$ there is only one eigenvalue which is $\lambda = -\beta$. b) Assume $k$ and $\beta$ are such that the two eigenvalues are non real complex numbers. Determine the nullclines for the system. Carefully sketch the nullclines and a few more trajectories. Is the equilibrium point $(0,0)$ stable, asymptotically stable, or nonstable? When $k > \beta ^2$ there are two non real complex eigenvalues. We set $x^{\prime} = 0$ which leads to a line $y = 0$. And $y^{\prime} = 0$ which leads to a line $y = -\dfrac{kx}{2\beta }$. Now, I think I have to draw two lines $y = 0$ and $y = -\dfrac{kx}{2\beta }$ but I do not know how to determine the direction of nullclines on a graph and how to determine stability of $(0,0)$. Can someone please help me? Thank you. Last edited by Shanonhaliwell; August 1st, 2018 at 03:01 PM. August 1st, 2018, 05:30 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 621 Thanks: 392 Math Focus: Dynamical systems, analytic function theory, numerics Nullclines don't have a "direction". They are simply the zeros of each coordinate of the derivative, which is what you have correctly computed. I think you are confusing this with the vector field direction along the nullclines which is easy to assess since in the plane these are always principal directions (which is one of the reasons nullclines are helpful). Anyway, you should sketch the vector field along the nullclines and in particular, notice that all trajectories are transverse. The stability is assessed from the eigenvalues directly. Namely, every hyperbolic linear continuous dynamical system is in a conjugacy class which is completely determined by its eigenvalues. Stable directions correspond to eigenvalues with negative real part and unstable directions correspond to eigenvalues with positive real part. Thanks from Shanonhaliwell Last edited by skipjack; August 2nd, 2018 at 05:37 PM. August 1st, 2018, 07:58 PM   #3
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Quote:
 Originally Posted by SDK Nullclines don't have a "direction". They are simply the zeros of each coordinate of the derivative, which is what you have correctly computed. I think you are confusing this with the vector field direction along the nullclines which is easy to assess since in the plane these are always principal directions (which is one of the reasons nullclines are helpful). Anyway, you should sketch the vector field along the nullclines and in particular, notice that all trajectories are transverse. The stability is assessed from the eigenvalues directly. Namely, every hyperbolic linear continuous dynamical system is in a conjugacy class which is completely determined by its eigenvalues. Stable directions correspond to eigenvalues with negative real part and unstable directions correspond to eigenvalues with positive real part.
You are right, nullclines don't have a direction. The "directions" here should be a direction of the velocity vectors along the nullclines. Thank you.

Last edited by skipjack; August 2nd, 2018 at 05:36 PM. Tags determine, nullclines, sketch Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post dthomas86 Algebra 4 November 21st, 2011 12:57 AM Scientific Calculus 3 October 31st, 2011 06:34 AM kolg Calculus 7 April 10th, 2010 08:21 AM rodjav305 Algebra 9 July 27th, 2009 06:30 PM Soha Algebra 1 December 26th, 2006 03:12 PM

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