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I want to draw phase plane diagram of the following differential equation

$$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10 y = 0.$$

Please check if my approach is correct. I have some doubts about it. They are in bold.

First of all I am changing it to a system of differential equation by putting $y_1 = y$, $y_1' = y_2$, $y_2' = 2y_2 - 10 y_1$. So the system of differential equation will be

$$\left[\begin{array}{c} y_1 \\ y_2\end{array}\right]' = \left[\begin{array}{c c} 0 & 1 \\ -10 & 2\end{array}\right]\left[\begin{array}{c} y_1 \\ y_2\end{array}\right]$$

Eigenvalues of the system will be $1 + 3i$ and $1 - 3i$. The eigenvector corresponding $1 + 3i$ is $(1, 1 + 3i)^t$. Computing a few steps we shall get the general solution

$$\left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right] = C_1 e^x \left(\begin{array}{c} \cos(3x) \\ \cos(3x) - 3\sin(3x) \end{array} \right) + C_2 e^x \left(\begin{array}{c} \sin(3x) \\ \sin(3x) + 3\cos(3x) \end{array} \right)$$

The system has only one critical point $(0,0)^t$. From the general solution we may say the critical point is a source, as $e^x \rightarrow \infty$ as $x \rightarrow 0$ so the trajectories will spiral out, I am predicting.

How to prove from the general solution that the trajectories will spiral out?

Let us transform the coordinate system. Transformation matrix will be $T = \left[ \begin{array}{c c} 1 & 0 \\ 1 & 3 \end{array} \right]$. New system of equation will be $$\left[\begin{array}{c} y_1 \\ y_2\end{array}\right]' = \left[\begin{array}{c c} 1 & 3 \\ -3 & 1\end{array}\right]\left[\begin{array}{c} y_1 \\ y_2\end{array}\right]$$

Corresponding general solution will be $$\left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right] = C_1 e^x \left(\begin{array}{c} \cos(3x) \\ \sin(3x) \end{array} \right) + C_2 e^x \left(\begin{array}{c} \sin(3x) \\ \cos(3t) \end{array} \right)$$

Here it is easy to prove that the trajectories will spiral out.

Will the trajectories will rotate clockwise or anticlockwise? How to determine?

I know that if $a + bi$ be the eigenvalue then the trajectories will rotate clockwise if $b > 0$ and anticlockwise if $b < 0$. But $a - bi$ will also be an eigenvalue which can be written as $a + (-b)i$. I am confused.

Why we are so much interested for the behavior of the system near the critical point?

Thank you for reading the long question and your help. Is there any short method?

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  • $\begingroup$ If possible please draw a phase plain diagram for me using a mathematics software like matlab, C etc. I do not know how to draw them. $\endgroup$
    – Supriyo
    Nov 27, 2013 at 5:11

2 Answers 2

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What you have looks good. Some comments to each of your questions:

Q1/Q2: Your eigenvalues/eigenvectors yield the general solution which you wrote and this tells the tale: you have a complex conjugate pair of eigenvalues (so trajectories spiral) and their real part is positive (so they spiral out).

Q3: Critical points correspond to the equilibrium solutions of the system. Typically one is interested in how a system responds when it is perturbed from equilibrium.

You might find these notes as well as this Wolfram Demonstration useful. Here is a rough sketch of the phase portrait for your problem using the latter:

enter image description here

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  • $\begingroup$ Thank you for your answer. For which solution the diagram is related to? The first solution or the second? $\endgroup$
    – Supriyo
    Nov 27, 2013 at 6:17
  • $\begingroup$ The entries of the matrix $A$ are displayed to the right of the slider bar: they are the entries of the original matrix you started with. $\endgroup$
    – JohnD
    Nov 27, 2013 at 6:24
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Long story short: suppose that $\pmatrix{a&b\\-b&a} = T^{-1}AT$ where $a,b$ are real numbers. If $\det(T)$ and $b$ have the same sign, then the trajectory will spiral clockwise. If they have opposite signs, then the trajectory will spiral counterclockwise.

The system will spiral out if $a^2 + b^2 > 1$, spiral in if $a^2 + b^2 < 1$, and will follow a stable orbit if $a^2 + b^2 = 1$


Full explanation:

The below is a work in progress. However, I thought the above response was helpful enough to merit posting this unfinished.

The key here is to understand matrices of the form $\pmatrix{a&b\\-b&a}$.

Let $re^{i\theta}$ be the polar form of $a+bi$. That is, let $\theta = \arctan(b/a)$ and let $r = \sqrt{a^2 + b^2}$. We have $$ \pmatrix{a&b\\-b&a} = \pmatrix{ r\cos \theta & r\sin\theta\\ -r\sin \theta & r\cos\theta }= \pmatrix{r&0\\0&r} \pmatrix{ \cos \theta & \sin\theta\\ -\sin \theta & \cos\theta } $$ That is, the matrix $\pmatrix{a&b\\-b&a}$ corresponds, in the standard basis, to a clockwise rotation by angle $\theta$, followed by a scaling by factor $r$.

However, it is important to note that with a different set of coordinates, it may be that this corresponds to a counterclockwise rotation. For example, $$ \pmatrix{0&1\\1&0} \pmatrix{ \cos \theta & \sin\theta\\ -\sin \theta & \cos\theta } \pmatrix{0&1\\1&0} = \pmatrix{ \cos \theta & -\sin\theta\\ \sin \theta & \cos\theta } $$ The trick is that as long as your chosen coordinate system is right-handed (that is, as long as $\det T > 0$), $\pmatrix{\cos \theta & \sin\theta\\-\sin \theta & \cos\theta}$ will correspond to a clockwise rotation by $\theta$, whereas if your coordinate system is left-handed (that is, if $\det T< 0 $), this will correspond to a counterclockwise rotation by $\theta$.

I hope that helps.


For this problem, our ODE was $y' = Ay$, with $$ A = \pmatrix{0&1\\-10&2} $$ And you found that $$ T = \pmatrix{1&0\\1&3}\implies \\ T^{-1}AT = \pmatrix{1 & 3\\ -3 & 1} $$ So, in this case, $1$ is our $a$ and $3$ would be our $b$. We find $\det T = 3 > 0$, which means that $b$ and $\det T$ have the same sign, which means that the trajectory will be clockwise.

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  • $\begingroup$ Thank you for answering and elaborating the main problem. In this case eigenvalues are $1-3i$ and $1+ 3i$ and the determinant is 10 > 0. So b = - 3 < 0$. So they have opposite signs. Shall we say the trajectory will rotate counterclockwise? $\endgroup$
    – Supriyo
    Nov 27, 2013 at 6:21
  • $\begingroup$ I've added a response to your question at the bottom of my answer. Does that make sense now? $\endgroup$ Nov 27, 2013 at 6:28
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    $\begingroup$ Thank you. It is clear now. $\endgroup$
    – Supriyo
    Nov 27, 2013 at 6:36

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