### Author Topic: Q4--day section--problem 2  (Read 8550 times)

#### Alexander Jankowski

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• Karma: 19 ##### Q4--day section--problem 2
« on: March 23, 2013, 02:19:36 PM »
The second question was #9.2.17:

(a) Find an equation of the form $H(x,y) = c$ satisfied by the trajectories
$$\frac{dx}{dt} = 2y, \qquad \frac{dy}{dt} = 8x.$$
(b) Plot several level curves of the function $H$. These are trajectories of the given system. Indicate the direction of motion on each trajectory.

First, we determine the function $H(x,y)$:
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8x}{2y} \Longleftrightarrow ydy = 4xdx \Longrightarrow H(x,y) = \frac{1}{2} y^2 - 2x^2 = c,$$
where $c$ is a constant of integration. For $c = -2,-1,0,1,2$, we have: Therefore, for $c = 0$, the trajectories are two lines with slopes $2$ and $-2$ that intersect at the origin, and are separatrices. For $c \neq 0$, the trajectories are hyperbolÃ¦. In particular, for $c > 0$, the hyperbolÃ¦ lie along the ordinate; for $c < 0$, they lie along the abscissa. To determine the direction of the trajectories, we rewrite the system as the matrix equation
$$\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{cc} 0 & 2 \\ 8 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right),$$
and we plug in the vectors $(x,y)^T=(0,1)$ and $(x,y)^T=(0,-1)$. This yields, respectively,
$$\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 2 \\ 0 \end{array} \right), \qquad \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} -2 \\ 0 \end{array} \right).$$
We conclude that the hyperbolÃ¦ along the ordinate are directed counter-clockwise. For the hyperbolÃ¦ along the abscissa, we can use $(x,y)^T=(1,0)$ and $(x,y)^T=(-1,0)$ to get, respectively,
$$\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 0 \\ 8 \end{array} \right), \qquad \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 0 \\ -8 \end{array} \right).$$
Therefore, the hyperbolÃ¦ along the abscissa for $x < 0$ are directed downwards and those for $x > 0$ are directed upwards. This can be verified with a stream plot: « Last Edit: March 24, 2013, 06:20:06 PM by Alexander Jankowski »

#### Victor Ivrii ##### Re: Q4--day section--problem 2
« Reply #1 on: March 23, 2013, 02:34:58 PM »
For $C=1,2,4$, we have:

And what we get as $=0$?

#### Alexander Jankowski

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• Karma: 19 ##### Re: Q4--day section--problem 2
« Reply #2 on: March 23, 2013, 02:44:06 PM »
I updated my solution.

#### Victor Ivrii ##### Re: Q4--day section--problem 2
« Reply #3 on: March 23, 2013, 02:51:50 PM »
I updated my solution.

Correct--but what are these lines $H=0$? There are 4 of solutions: two escaping as $t\to+\infty$ but tending to the equilibrium point as $t\to -\infty$ and two escaping as $t\to-\infty$ but tending to the equilibrium point as $t\to +\infty$. They are separatrices

#### Alexander Jankowski

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• Karma: 19 ##### Re: Q4--day section--problem 2
« Reply #4 on: March 24, 2013, 06:10:31 PM »
Oh, okay. I was confused because the origin is a saddle point (as seen from the stream plot and the eigenvalues $\lambda = Â±4$) and there is no apparent basin of attraction. I should note that the textbook (9th ed.) discusses separatrices only in the context of basins of attraction. Anyway, I updated the solution again and plotted more level curves for completion.
« Last Edit: March 24, 2013, 06:22:07 PM by Alexander Jankowski »

#### Victor Ivrii ##### Re: Q4--day section--problem 2
« Reply #5 on: March 24, 2013, 06:22:36 PM »
Oh, okay. I was confused because the origin is a saddle point (the eigenvalues of the coefficient matrix are $\lambda = Â±4$) and there is no apparent basin of attraction. I should note that the textbook discusses separatrices only in that context. Anyway, I updated the solution again and plotted more level curves for completion.

Sure: unstable equilibrium points do not have basins of attractions as almost all trajectories go away. However if a linear system has a matrix with $n_+$ positive and $n_-$ negative eigenvalues (but not $0$ so $n_++n_-=n$) ) then the whole space breaks into direct sum of linear subspaces $V_+ \oplus V_-$ where all trajectories in $V_\pm$ tend to $0$ as $t\to \pm \infty$ and $V_+$, $V_-$ are unstable and stable subspaces respectively.

Similar picture  but with "curved" $W_\pm$ (called manifolds or (hyper)surfaces) instead of $V_pm$  happens for nonlinear systems. As $n=2$  we have $n_\pm=1$ and both of them are lines (separatrices) but for $n=3$ one of them is a surface and another a line and there are more possibilities as $n\ge 4$.