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August 1st, 2018, 02:54 PM   #1
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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.
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August 1st, 2018, 05:30 PM   #2
SDK
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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.
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August 1st, 2018, 07:58 PM   #3
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Quote:
Originally Posted by SDK View Post
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.
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