Saddle Point Stability How to Draw in Phase Plane

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Department v-six : Phase Plane

Before proceeding with actually solving systems of differential equations there's 1 topic that we demand to take a look at. This is a topic that's not e'er taught in a differential equations grade only in case you're in a form where it is taught we should cover it so that you are prepared for it.

Let'south kickoff with a general homogeneous organization,

\[\begin{equation}\vec x' = A\vec x\characterization{eq:eq1}\end{equation}\]

Find that

\[\vec x = \vec 0\]

is a solution to the organization of differential equations. What we'd like to ask is, practise the other solutions to the arrangement approach this solution every bit \(t\) increases or do they motion away from this solution? We did something similar to this when we classified equilibrium solutions in a previous section. In fact, what nosotros're doing here is simply an extension of this idea to systems of differential equations.

The solution \(\vec 10 = \vec 0\) is chosen an equilibrium solution for the system. As with the single differential equations case, equilibrium solutions are those solutions for which

\[A\vec x = \vec 0\]

We are going to assume that \(A\) is a nonsingular matrix and hence will take only 1 solution,

\[\vec ten = \vec 0\]

and and so nosotros will have only 1 equilibrium solution.

Dorsum in the unmarried differential equation case recall that we started by choosing values of \(y\) and plugging these into the role \(f(y)\) to determine values of \(y'\). Nosotros then used these values to sketch tangents to the solution at that detail value of \(y\). From this we could sketch in some solutions and use this data to classify the equilibrium solutions.

We are going to do something like here, but it volition exist slightly unlike also. Offset, we are going to restrict ourselves down to the \(2 \times 2\) case. So, we'll exist looking at systems of the form,

\[\begin{array}{*{20}{c}}\brainstorm{align*}{{10'}_1} & = a{x_1} + b{x_2}\\ {{ten'}_2} & = c{x_1} + d{x_2}\end{align*}&{\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\vec ten' = \left( {\begin{array}{*{20}{c}}a&b\\c&d\terminate{array}} \right)\vec 10}\terminate{array}\]

Solutions to this system will be of the grade,

\[\vec x = \left( {\begin{array}{*{20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\end{array}} \right)\]

and our unmarried equilibrium solution will exist,

\[\vec x = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\]

In the single differential equation example we were able to sketch the solution, \(y(t)\) in the y-t plane and encounter bodily solutions. However, this would somewhat difficult in this case since our solutions are actually vectors. What we're going to do here is think of the solutions to the system as points in the \({x_1}\,{x_2}\) plane and plot these points. Our equilibrium solution volition stand for to the origin of \({x_1}\,{x_2}\). airplane and the \({x_1}\,{x_2}\) plane is called the phase plane.

To sketch a solution in the phase plane we can pick values of \(t\) and plug these into the solution. This gives u.s. a betoken in the \({x_1}\,{x_2}\) or phase plane that we tin can plot. Doing this for many values of \(t\) volition and so give us a sketch of what the solution volition be doing in the phase plane. A sketch of a detail solution in the phase plane is chosen the trajectory of the solution. One time we take the trajectory of a solution sketched we can then ask whether or not the solution will arroyo the equilibrium solution as \(t\) increases.

We would like to exist able to sketch trajectories without actually having solutions in hand. At that place are a couple of ways to do this. Nosotros'll look at 1 of those here and we'll expect at the other in the next couple of sections.

I way to get a sketch of trajectories is to exercise something similar to what nosotros did the first time we looked at equilibrium solutions. Nosotros tin choose values of \(\vec x\) (annotation that these will be points in the phase aeroplane) and compute \(A\vec x\). This will give a vector that represents \(\vec x'\)at that item solution. As with the single differential equation case this vector will be tangent to the trajectory at that point. We tin sketch a bunch of the tangent vectors and and then sketch in the trajectories.

This is a adequately work intensive way of doing these and isn't the way to exercise them in general. All the same, it is a way to go trajectories without doing whatsoever solution work. All we need is the arrangement of differential equations. Let'south have a quick look at an case.

Example one Sketch some trajectories for the arrangement, \[\begin{assortment}{*{twenty}{c}}\begin{marshal*}{{10'}_1} & = {x_1} + ii{x_2}\\ {{x'}_2} & = 3{x_1} + ii{x_2}\finish{align*}&{\hspace{0.25in} \Rightarrow \hspace{0.25in}\vec x' = \left( {\begin{assortment}{*{20}{c}}ane&2\\3&2\end{array}} \right)\vec x}\end{array}\]

Show Solution

So, what we need to do is pick some points in the phase plane, plug them into the right side of the organisation. We'll do this for a couple of points.

\[\brainstorm{align*}\vec x & = \left( {\brainstorm{assortment}{*{20}{c}}{ - 1}\\one\finish{assortment}} \right) & \Rightarrow \hspace{0.25in}\vec x'& = \left( {\brainstorm{array}{*{20}{c}}1&ii\\3&2\end{array}} \correct)\left( {\begin{array}{*{20}{c}}{ - one}\\1\terminate{array}} \right) = \left( {\begin{assortment}{*{twenty}{c}}1\\{ - 1}\cease{array}} \right)\\ \vec x & = \left( {\begin{array}{*{twenty}{c}}two\\0\end{array}} \right) & \Rightarrow \hspace{0.25in}\vec x' & = \left( {\begin{assortment}{*{twenty}{c}}i&2\\3&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}ii\\0\terminate{assortment}} \right) = \left( {\begin{array}{*{20}{c}}two\\vi\end{array}} \correct)\hspace{0.25in}\\ \vec 10 & = \left( {\begin{array}{*{twenty}{c}}{ - 3}\\{ - two}\end{array}} \right) & \Rightarrow \hspace{0.25in}\vec x' & = \left( {\brainstorm{array}{*{xx}{c}}1&2\\three&2\stop{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\end{array}} \correct) = \left( {\brainstorm{array}{*{twenty}{c}}{ - 7}\\{ - 13}\end{array}} \correct)\hspace{0.25in}\end{align*}\]

So, what does this tell us? Well at the point \(\left( { - one,1} \right)\) in the phase plane at that place will exist a vector pointing in the direction \(\left\langle {1, - 1} \correct\rangle \). At the betoken \(\left( {2,0} \right)\) in that location will exist a vector pointing in the management \(\left\langle {ii,6} \right\rangle \). At the bespeak \(\left( { - 3, - 2} \right)\) there volition be a vector pointing in the direction \(\left\langle { - seven, - xiii} \right\rangle \).

Doing this for a big number of points in the phase plane will requite the following sketch of vectors.

Now all we need to do is sketch in some trajectories. To do this all we demand to do is remember that the vectors in the sketch above are tangent to the trajectories. Likewise, the management of the vectors give the management of the trajectory as \(t\) increases so we can evidence the fourth dimension dependence of the solution by adding in arrows to the trajectories.

Doing this gives the following sketch.

This sketch is chosen the phase portrait. Ordinarily phase portraits only include the trajectories of the solutions and non whatever vectors. All of our phase portraits form this indicate on will only include the trajectories.

In this case information technology looks like most of the solutions will kickoff away from the equilibrium solution and so as \(t\) starts to increase they motion in towards the equilibrium solution and and so somewhen beginning moving away from the equilibrium solution again.

There seem to be 4 solutions that have slightly different behaviors. It looks like two of the solutions will start at (or near at least) the equilibrium solution then motion straight abroad from it while ii other solutions start away from the equilibrium solution and then move straight in towards the equilibrium solution.

In these kinds of cases nosotros call the equilibrium point a saddle indicate and we call the equilibrium point in this case unstable since all but two of the solutions are moving away from it as \(t\) increases.

As nosotros noted before this is not by and large the style that we will sketch trajectories. All we really demand to get the trajectories are the eigenvalues and eigenvectors of the matrix \(A\). We volition meet how to practise this over the next couple of sections every bit we solve the systems.

Here are a few more phase portraits so you tin see some more possible examples. We'll actually be generating several of these throughout the grade of the next couple of sections.

Not all possible phase portraits take been shown hither. These are here to show you some of the possibilities. Make sure to notice that several kinds tin be either asymptotically stable or unstable depending upon the direction of the arrows.

Detect the difference between stable and asymptotically stable. In an asymptotically stable node or spiral all the trajectories will move in towards the equilibrium indicate equally t increases, whereas a center (which is e'er stable) trajectory volition only move around the equilibrium point just never really move in towards it.

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Source: https://tutorial.math.lamar.edu/classes/de/phaseplane.aspx

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