Phase Plane

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 5.4 – Introduction to the Phase Plane

Expected Educational Results

• Objective 17–1: I can solve systems of linear differential equations using a phase plane.

• Objective 17–2: I can determine the stability of linear systems.

• Objective 17–3: I can analyze almost linear systems.

• Objective 17–4: I can solve systems of linear differential equations using phase plane.

• Objective 17–5: I can describe nonlinear systems.

Phase Plane

Definition: Linear Systems of DEs

A $n\times n$$n\times n$ linear system of DEs has the form

$\begin{array}{c}{x}_1^{\prime }(t)=a_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n+f_1\\ x_2^{\prime }(t)=a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n+f_2\\ ⋮\\ x_n^{\prime }(t)=a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{nn}{x}_n+f_n\end{array}$$x_1^{\,\prime}(t)=a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n +f_1 \\ x_2^{\,\prime}(t)=a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n +f_2 \\ \vdots \\ x_n^{\,\prime}(t)=a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n +f_n$

Definition: Nonlinear Systems of DEs

A $n\times n$$n\times n$ nonlinear system of DEs is not in the above linear form.

Definition: 2D Phase Plane

1. The phase plane is the $x_1{x}_2$$x_1x_2$-plane, i.e., ${\mathbb{R}}^2$$\mathbb{R}^2$, for the system of DEs: ${\displaystyle x{x}^{\prime }(t)=AA(t)xx(t)+ff(t)}$$\displaystyle \pmb{x}^{\,\prime}(t)=\pmb{A}(t)\pmb{x}(t)+\pmb{f}(t)$.

Nullclines in 2D Phase Planes

Definition: Nullclines $x_1$$x_1$-nullclines are sets of points in a phase plane such that ${\displaystyle \frac{d{x}_1}{dt}=0}$$\displaystyle\frac{dx_1}{dt}=0$. $x_2$$x_2$-nullclines are sets of points in a phase plane such that ${\displaystyle \frac{d{x}_2}{dt}=0}$$\displaystyle\frac{dx_2}{dt}=0$.

Equilibrium Solutions in 2D Phase Planes

Definition: Equilibrium Solutions

Let $AA$$\pmb{A}$ be the [nonsingular, i.e., $\text{det}(AA)\ne 00$$\text{det}(\pmb{A})\ne\pmb{0}$] $2\times 2$$2\times 2$ coefficient matrix of a system of DEs.

Solutions $xx$$\pmb{x}$ for which $AxAx=00$$\pmb{Ax}=\pmb{0}$ are critical points and correspond to equilibrium solutions, a.k.a., fixed points, to the system of DEs.

Equilibrium points occur at the intersection(s) of the $x_1$$x_1$-nullclines with the $x_2$$x_2$-nullclines.

Example 01:

Find the nullclines and equilibrium solutions for

$\begin{array}{c}{x}_1^{\prime }(t)=\frac{x_2-x_1^2}{1+x_1^2+x_2^2}\\ x_2^{\prime }(t)=\frac{-x_1-x_2^2}{1+x_1^2+x_2^2}\end{array}$$x_1^{\,\prime}(t)=\frac{x_2-x_1^2}{1+x_1^2+x_2^2} \\ x_2^{\,\prime}(t)=\frac{-x_1-x_2^2}{1+x_1^2+x_2^2}$

Solution:

1. Find $x_1$$x_1$ nullclines: ${\displaystyle x_1^{\prime }(t)=\frac{x_2-x_1^2}{1+x_1^2+x_2^2}\stackrel{\text{set}}{=}0\Rightarrow x_2=x_1^2}$$\displaystyle x_1^{\,\prime}(t)=\frac{x_2-x_1^2}{1+x_1^2+x_2^2}\stackrel{\text{set}}{=}0\Rightarrow x_2=x_1^2$

2. Find $x_2$$x_2$ nullclines: ${\displaystyle x_2^{\prime }(t)=\frac{-x_1-x_2^2}{1+x_1^2+x_2^2}\stackrel{\text{set}}{=}0\Rightarrow x_1=-x_2^2}$$\displaystyle x_2^{\,\prime}(t)=\frac{-x_1-x_2^2}{1+x_1^2+x_2^2}\stackrel{\text{set}}{=}0\Rightarrow x_1=-x_2^2$

3. Find equilibrium solutions, solve for $x_1$$x_1$ and $x_2$$x_2$ using the above two equations, i.e., \newline substitute $x_2=x_1^2$$x_2=x_1^2$ into $x_1=-x_2^2$$x_1=-x_2^2$: ${\displaystyle x_1=-{\left(x_1^2\right)}^2\Rightarrow x_1^4+x_1=0\Rightarrow x_1(x_1^3+1)=0\Rightarrow x_1=0,-1}$$\displaystyle x_1=-\left(x_1^2\right)^2\Rightarrow x_1^4+x_1=0\Rightarrow x_1(x_1^3+1)=0 \Rightarrow x_1=0,-1$. Thus, the equilibrium points are $(0,0)$$(0,0)$ and $(-1,1)$$(-1,1)$.

Here is the graph of the nullclines in the phase plane for the above system of DEs.

Phase Portraits

Definition: Trajectory

A trajectory is an implicit solution to an IVP of a system of DEs.

Definition: Phase portrait

A phase portrait is a plot of several solutions, i.e., trajectories, to the system of DEs.

Here is the graph of the nullclines with several solution curves, i.e., trajectories, for the above system of DEs. Note the behavior of the solutions near the equilibrium solutions.

Here is the graph of the nullclines with several solution curves, i.e., trajectories, in the phase plane for the above system of DEs.

Classification of Equilibrium Points

Unstable Saddle

Asymptotically Stable Node

Unstable Node

Stable Center

Asymptotically Stable Spiral

Unstable Spiral

Equilibrium Solutions in $2\times 2$$2\times 2$ Linear Systems

Let $AA$$\pmb{A}$ be the $2\times 2$$2\times 2$ coefficient matrix of a linear system of DEs with eigenvalues ${\lambda }_1$$\lambda_1$ and ${\lambda }_2$$\lambda_2$.

Unstable Saddle

The eigenvalues of $AA$$\pmb{A}$ are of different signs, e.g., ${\lambda }_1<0<{\lambda }_2$$\lambda_1<0<\lambda_2$

Asymptotically Stable Node

The eigenvalues of $AA$$\pmb{A}$ are distinct and negative, e.g., ${\lambda }_1<{\lambda }_2<0$$\lambda_1<\lambda_2<0$

Unstable Node

The eigenvalues of $AA$$\pmb{A}$ are distinct and positive, e.g., ${\lambda }_2>{\lambda }_1>0$$\lambda_2>\lambda_1>0$

Stable Center

The eigenvalues of $AA$$\pmb{A}$ are complex with $\alpha =0$$\alpha=0$, i.e., ${\lambda }_{1,2}=\pm i\beta$$\lambda_{1,2}=\pm i\beta$ with $\beta >0$$\beta>0$

Asymptotically Stable Spiral

The eigenvalues of $AA$$\pmb{A}$ are complex, i.e., ${\lambda }_{1,2}=\alpha \pm i\beta$$\lambda_{1,2}=\alpha\pm i\beta$ with $\alpha <0$$\alpha<0$ and $\beta >0$$\beta>0$

Unstable Spiral

The eigenvalues of $AA$$\pmb{A}$ are complex, i.e., ${\lambda }_{1,2}=\alpha \pm i\beta$$\lambda_{1,2}=\alpha\pm i\beta$ with $\alpha >0$$\alpha>0$ and $\beta <0$$\beta<0$

Defective Node

The eigenvalues of $AA$$\pmb{A}$ are a non-zero, repeated real number with one linearly independent eigenvector. The defective node is unstable if $\lambda >0$$\lambda>0$ and asymptotically stable if $\lambda <0$$\lambda<0$

Star Node

The eigenvalues of $AA$$\pmb{A}$ are a non-zero, repeated real number with two linearly independent eigenvectors. The defective node is unstable if $\lambda >0$$\lambda>0$ and asymptotically stable if $\lambda >0$$\lambda>0$

Example 02:

Consider the system: $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =x_1(3x_1+x_2)\\ x_2^{\prime }(t)& =6x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= x_1\left(3x_1 + x_2\right) \\ x^{\,\prime}_2(t) &= 6x_2 \end{array}$

Solution:

NOTE: This system of DEs is non-linear since the first equation has $x_1^2$$x_1^2$.

NOTE: The phase portrait of the system is:

1
StreamPlot[{x1(3x1 + x2), 6x2}, {x1, -2, 2}, {x2, -2, 2}]

NOTE:

$x_1$$x_1$ nullclines: $x_1^{\prime }(t)=x_1(3x_1+x_2)\stackrel{\text{set}}{=}0\Rightarrow x_1=-\frac{x_2}{3}\text{ and }{x}_1=0$$x^{\,\prime}_1(t) = x_1\left(3x_1 + x_2\right) \stackrel{\text{set}}{=}0\Rightarrow x_1=-\frac{x_2}{3} \text{ and } x_1=0$. $x_2$$x_2$ nullclines: $x_2^{\prime }(t)=6x_2\stackrel{\text{set}}{=}0\Rightarrow x_2=0$$x^{\,\prime}_2(t) = 6x_2 \stackrel{\text{set}}{=}0\Rightarrow x_2=0$.

NOTE: By solving for all solutions to the three equations [nullclines]: $x_1=-\frac{x_2}{3}$$x_1=-\frac{x_2}{3}$, $x_1=0$$x_1=0$, and $x_2=0$$x_2=0$, the only equilibrium point is $(0,0)$$(0,0)$. The equilibrium point is an unstable saddle.

Investigation 01

For each system of DEs below, answer the following questions:

• Identify if the system of DEs is a linear system of DEs. Explain.

  • If the system is linear, then identify the eigenvalues and corresponding eigenvectors for the system of DEs.

• Use technology to sketch the phase portrait of the system in $x_1{x}_2$$x_1x_2$-plane, if valid.

• Identify the nullclines for the system.

• Identify and classify the equilibrium points of the system. Explain.

• Classify the equilibrium points of the system. Explain.

1. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =3x_1+x_2\\ x_2^{\prime }(t)& =4x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= 3x_1 + x_2 \\ x^{\,\prime}_2(t) &= 4x_2 \end{array}$

2. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =-5x_1+x_2\\ x_2^{\prime }(t)& =2x_1+5x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= -5x_1 + x_2 \\ x^{\,\prime}_2(t) &= 2x_1 + 5x_2 \end{array}$

3. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =6x_1+5x_2\\ x_2^{\prime }(t)& =7x_1-2x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= 6x_1 + 5x_2 \\ x^{\,\prime}_2(t) &= 7x_1 - 2x_2 \end{array}$

4. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =-x_1+2x_2\\ x_2^{\prime }(t)& =-2x_1-x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= -x_1 + 2x_2 \\ x^{\,\prime}_2(t) &= -2x_1 - x_2 \end{array}$

5. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =x_1-x_2\\ x_2^{\prime }(t)& =x_1+3x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= x_1 - x_2 \\ x^{\,\prime}_2(t) &= x_1 + 3x_2 \end{array}$

6. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =x_1-x_2\\ x_2^{\prime }(t)& =x_1+3x_2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= x_1 - x_2 \\ x^{\,\prime}_2(t) &= x_1 + 3x_2 \end{array}$

7. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =x_1{x}_2-6x_1\\ x_2^{\prime }(t)& =x_1{x}_2-2x_1+x_2-2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= x_1x_2 - 6x_1 \\ x^{\,\prime}_2(t) &= x_1x_2 - 2x_1 + x_2 - 2 \end{array}$

8. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =x_1^2-x_1{x}_2\\ x_2^{\prime }(t)& =x_1{x}_2-3x_1+2\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= x_1^2 - x_1x_2 \\ x^{\,\prime}_2(t) &= x_1x_2 - 3x_1 + 2 \end{array}$

9. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =-(x_1-x_2)(1-x_1-x_2)\\ x_2^{\prime }(t)& =x_1(2+x_2)\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= -(x_1 - x_2)(1 - x_1 - x_2) \\ x^{\,\prime}_2(t) &= x_1(2 + x_2) \end{array}$

10. $\begin{array}{c}{\displaystyle \begin{array}{lll}{x}_1^{\prime }(t)& =3x_1-2x_2,& x_1(0)=3\\ x_2^{\prime }(t)& =2x_1-2x_2,& x_2(0)=\frac{1}{2}\end{array}}\end{array}$$\displaystyle \begin{array}{lll} x^{\,\prime}_1(t) &= 3x_1 - 2x_2, & x_1(0)=3\\ x^{\,\prime}_2(t) &= 2x_1 - 2x_2, & x_2(0)=\frac{1}{2} \end{array}$

Investigation 02

For each system of DEs below, answer the following questions:

• Identify if the system of DEs is a linear system of DEs. Explain.

  • If the system is linear, then identify the eigenvalues and corresponding eigenvectors for the system of DEs.

• Identify the nullclines for the system.

• Identify and classify the equilibrium points of the system. Explain.

1. $\begin{array}{c}{\displaystyle \begin{array}{ll}{x}_1^{\prime }(t)& =3x_1+x_2-x_3\\ x_2^{\prime }(t)& =x_1+2x_2-x_3\\ x_3^{\prime }(t)& =3x_1+3x_2-x_3\end{array}}\end{array}$$\displaystyle \begin{array}{ll} x^{\,\prime}_1(t) &= 3x_1 + x_2 - x_3 \\ x^{\,\prime}_2(t) &= x_1 + 2x_2 - x_3 \\ x^{\,\prime}_3(t) &= 3x_1 + 3x_2 - x_3 \end{array}$

2. $\begin{array}{c}{\displaystyle \begin{array}{lll}{x}_1^{\prime }(t)& =-x_1+x_2,& x_1(0)=-2\\ x_2^{\prime }(t)& =x_1+2x_2+x_3,& x_2(0)=2\\ x_3^{\prime }(t)& =3x_2-x_3,& x_3(0)=1\end{array}}\end{array}$$\displaystyle \begin{array}{lll} x^{\,\prime}_1(t) &= -x_1 + x_2, & x_1(0)=-2 \\ x^{\,\prime}_2(t) &= x_1 + 2x_2 + x_3, & x_2(0)=2 \\ x^{\,\prime}_3(t) &= 3x_2 - x_3, & x_3(0)=1 \end{array}$

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Last Modified: Thursday, 15 October 2020 6:42 EDT