Phase Portraits (DP IB Applications & Interpretation (AI): HL): Revision Note

Phase portraits

What is a phase portrait for a system of coupled differential equations?

• Consider the system of coupled equations that can be represented in the matrix form $\dot{\mathit{x}}=\mathit{M}\mathit{x}$, where

  • $\dot{\mathit{x}}=\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)$

  • $\mathit{M}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

  • $\mathit{x}=\left(\begin{array}{c}x\\ y\end{array}\right)$

• A phase portrait is a diagram showing how the values of x and y change over time

  • On a phase portrait we will usually sketch several typical solution trajectories

  • The precise trajectory that the solution for a particular system will travel along is determined by the initial conditions for the system

• Suppose for the matrix $\mathit{M}$

  • ${\lambda }_1$ and ${\lambda }_2$ are the eigenvalues

  • ${\mathit{p}}_1$ and ${\mathit{p}}_2$ are the corresponding eigenvectors

• The overall nature of the phase portrait depends in large part on the values of ${\lambda }_1$ and ${\lambda }_2$

What does the phase portrait look like when the eigenvalues are real numbers?

• The phase portraits when the eigenvalues are real always include two lines representing the eigenvectors

  • These go through the origin

  • They are parallel to the eigenvectors

• The origin separates these two lines into four solution trajectories

• For example, if ${\mathit{p}}_1=\left(\begin{array}{c}1\\ 2\end{array}\right)$ and ${\mathit{p}}_2=\left(\begin{array}{c}-3\\ 4\end{array}\right)$ are eigenvectors

  • then these lines have the equations $y=2x$ and $y=-\frac{4}{3}x$ respectively

• The origin separates these two lines into four solution trajectories

  • Their directions depend on the signs of the corresponding eigenvalues

    • If an eigenvalue is positive then the corresponding lines are directed away from the origin

    • If an eigenvalue is negative then the corresponding lines are directed towards the origin

• The solution trajectories have the properties:

  • As $t\to -\infty$, the solution trajectory is parallel to the line corresponding to the smaller eigenvalue

  • As $t\to \infty$, the solution trajectory is parallel to the line corresponding to the larger eigenvalue

  • A solution trajectory never crosses a line corresponding to an eigenvector

Both eigenvalues are positive

• Suppose ${\lambda }_1>{\lambda }_2>0$

• All solution trajectories are directed away from the origin as t increases

• To draw a solution trajectory

  • Start at the origin

  • Draw a line roughly parallel to ${\mathit{p}}_2$

  • Move away from the origin and ${\mathit{p}}_2$

  • Draw a line roughly parallel to ${\mathit{p}}_1$

  • The line does not end

Both eigenvalues are negative

• Suppose $0>{\lambda }_1>{\lambda }_2$

• All solution trajectories are directed towards the origin as t increases

• To draw a solution trajectory

  • Start from somewhere at the edge of the graph

  • Draw a line roughly parallel to ${\mathit{p}}_2$

  • Move toward the origin and away from ${\mathit{p}}_2$

  • Draw a line roughly parallel to ${\mathit{p}}_1$

  • End at the origin

One positive and one negative eigenvalue

• Suppose ${\lambda }_1>0>{\lambda }_2$

• To draw a solution trajectory

  • Start from somewhere at the edge of the graph

  • Draw a line roughly parallel to ${\mathit{p}}_2$

  • Move toward the origin and away from ${\mathit{p}}_2$

  • Move away from the origin and toward ${\mathit{p}}_1$

  • Draw a line roughly parallel to ${\mathit{p}}_1$

  • The line does not end

What does the phase portrait look like when the eigenvalues are complex numbers?

• The phase portraits when the eigenvalues are complex always include curves orbiting the origin

  • They could be closed ellipses

  • They could be open spirals

• The real part of the complex eigenvalues affect the shape

  • Positive real parts cause the trajectories to spiral away from the origin

  • Negative real parts cause the trajectories to spiral towards the origin

  • Zero real parts cause the trajectories to form ellipses centred at the origin

• You can find the orientation of the trajectory by finding the value of $\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)$ at a specific point on one of the coordinate axes

  • For example, consider the system $\dot{\mathit{x}}=\left(\begin{array}{cc}1& -2\\ 1& -1\end{array}\right)\mathit{x}$

    • Pick the point $\left(1,0\right)$

    • $\dot{\mathit{x}}=\left(\begin{array}{cc}1& -2\\ 1& -1\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)$

    • From $\left(1,0\right)$, the trajectory is moving to the right and up

    • Therefore, it is counter-clockwise around the origin

Real part is equal to zero

• Suppose ${\lambda }_1=b\mathrm{i}$ and ${\lambda }_2=-b\mathrm{i}$

• To draw a solution trajectory

  • Find the direction at $\left(1,0\right)$ or $\left(0,1\right)$ to determine if it is clockwise or counter-counter

  • Draw ellipses that are centred at the origin

Real part is not equal to zero

• Suppose ${\lambda }_1=a+b\mathrm{i}$ and ${\lambda }_2=a-b\mathrm{i}$

• To draw a solution trajectory

  • Find the direction at $\left(1,0\right)$ or $\left(0,1\right)$ to determine if it is clockwise or counter-counter

  • Draw spirals from origin

  • Label the direction of the trajectories

    • If $a<0$ it spirals towards the origin

    • If $a>0$ it spirals away from the origin