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AI HL · AHL 9

Differential Equations

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AHL 9.9

Phase portraits for real eigenvalues

AHL 9.9

Phase portraits for real eigenvalues

Target: I can find the eigenvalues of a 2 × 2 matrix from the characteristic equation. (+3 more)

By the end of this section

Key points

Eigenvalues

  • Solve \(\det(\mathbf M-\lambda\mathbf I)=0\): for a 2 × 2 matrix this is \(\lambda^2-(\text{trace})\lambda+\det\mathbf M=0\).
  • Check: the two eigenvalues add to the trace and multiply to the determinant.
  • If the determinant is negative, the eigenvalues are real and have opposite signs.

Phase portraits

  • A phase portrait shows how x and y change together over time: each curve is one trajectory, \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{\lambda_1t}\mathbf p_1+Be^{\lambda_2t}\mathbf p_2\) for one choice of A and B.
  • The arrows show the direction of travel as t increases, like a current carrying a small object.
  • The origin is the equilibrium point: if the start is at the origin, nothing moves.

Three types for real eigenvalues

  • Both eigenvalues positive: unstable node. Trajectories leave the origin.
  • Both eigenvalues negative: stable node. Trajectories approach the origin.
  • One positive, one negative (\(\lambda_1<0<\lambda_2\)): saddle point, which is unstable.

Drawing the sketch

  • Draw the eigenvector lines first, from the gradients of \(\mathbf p_1\) and \(\mathbf p_2\).
  • On the line for a positive eigenvalue the arrows point away from the origin. For a negative eigenvalue they point towards it.
  • For large t the curves approach the direction of the eigenvector with the larger eigenvalue. As t goes to minus infinity they approach the other.
  • You are not expected to give all the detail: two lines with arrows and a few curves is enough.

On the calculator

Eigenvalues and eigenvectors for the sketch

  • Enter the matrix of the system and find its eigenvalues. Check that they add to the trace and multiply to the determinant.
  • Find an eigenvector for each eigenvalue. The calculator may scale it: any non-zero multiple gives the same line.
  • Use the sign of each eigenvalue to decide the type of equilibrium and the direction of the arrows.

Screenshot and video to come

Explore: phase portraits

Choose a preset or type the four numbers of the system. Each curve shows how x and y change together as time passes, and the arrow shows the direction of travel. Drag the orange point to start a curve of your own, and say what it will do before you read the eigenvalues.

dx/dt =x +y
dy/dt =x +y



Eigenvalues
Eigenvectors
The origin is a

The straight teal lines are the eigenvector lines, when the eigenvalues are real.

Choose a preset to see its phase portrait.

  1. Choose the saddle. Which curves go towards the origin and which go away? How do they relate to the eigenvector lines?
  2. Choose the centre, then change d from −2 to −1.5, then to −2.5. What happens to the curves? Which number decides it: the trace, the determinant or the real part of the eigenvalue?
  3. Look at the two spirals. How does the sign of b decide whether the curves turn clockwise or anticlockwise? Check it with dx/dt at (0, 1).
  4. In the stable node, put the orange point on an eigenvector line. What does its curve do? Move it a little off the line. Why does the curve then bend?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Eigenvalues from the characteristic equation

Find the eigenvalues of the matrix of the system \(\dfrac{dx}{dt}=6x+2y\), \(\dfrac{dy}{dt}=-2x+y\).

Working
  1. The matrix

    \(\mathbf M=\begin{pmatrix}6&2\\-2&1\end{pmatrix}\)

    Read the coefficients of x and y in each equation.

  2. Characteristic equation

    \(\det(\mathbf M-\lambda\mathbf I)=(6-\lambda)(1-\lambda)-(2)(-2)=0\)

    Multiply the diagonal, subtract the product of the other two entries.

  3. Tidy

    \(\lambda^2-7\lambda+10=0\)

    The 7 is the trace and the 10 is the determinant.

  4. Solve

    \((\lambda-2)(\lambda-5)=0\Rightarrow\lambda=2\ \text{or}\ 5\)

    From the calculator: eigenvalues 2 and 5. Check that they add to 7 and multiply to 10.

Video walkthrough coming
An unstable node

For the system \(\dfrac{dx}{dt}=6x+2y\), \(\dfrac{dy}{dt}=-2x+y\), find the eigenvalues, say what kind of equilibrium point the origin is, and sketch the phase portrait.

−4−224−4−224xyλ = 2λ = 5
Working
  1. Eigenvalues

    \(\lambda=2\ \text{and}\ \lambda=5\)

    From the characteristic equation, as in the previous example. Both are positive.

  2. Classify

    Both eigenvalues positive: unstable node

    With positive exponents, every trajectory moves away from the origin as t increases.

  3. Eigenvectors

    \(\lambda=2:\ 4x+2y=0\Rightarrow\begin{pmatrix}1\\-2\end{pmatrix};\quad\lambda=5:\ x+2y=0\Rightarrow\begin{pmatrix}2\\-1\end{pmatrix}\)

    These give two straight-line trajectories, of gradient −2 and −1/2.

  4. Arrows

    On both lines the arrows point away from the origin.

    The direction of travel as t increases: like a current carrying a small object.

  5. Sketch

    Draw the two lines, then curves leaving the origin

    Near the origin the curves are tangent to the λ = 2 line. Far away they bend towards the λ = 5 direction. The exact shape is not needed.

Video walkthrough coming
A stable node

For the system \(\dfrac{dx}{dt}=-4x+y\), \(\dfrac{dy}{dt}=-3x\), find the eigenvalues, classify the origin and sketch the phase portrait.

−4−224−4−224xyλ = −3λ = −1
Working
  1. Eigenvalues

    \(\lambda^2+4\lambda+3=0\Rightarrow(\lambda+3)(\lambda+1)=0\Rightarrow\lambda=-3,\ -1\)

    The trace is −4 and the determinant is 3.

  2. Classify

    Both eigenvalues negative: stable node

    Negative exponents mean every trajectory moves towards the origin.

  3. Eigenvectors

    \(\lambda=-3:\ -x+y=0\Rightarrow\begin{pmatrix}1\\1\end{pmatrix};\quad\lambda=-1:\ -3x+y=0\Rightarrow\begin{pmatrix}1\\3\end{pmatrix}\)

    Gradients 1 and 3. Arrows point towards the origin on both.

  4. Large t

    \(e^{-t}\text{ decays more slowly than }e^{-3t}\)

    For large t the λ = −1 term dominates, so curves reach the origin tangent to the gradient 3 line.

  5. Sketch

    Two lines with inward arrows, curves running in towards the origin

    Far from the origin the curves run alongside the λ = −3 line (gradient 1).

Video walkthrough coming
A saddle point

For the system \(\dfrac{dx}{dt}=x+2y\), \(\dfrac{dy}{dt}=3x+2y\), find the eigenvalues, classify the origin and sketch the phase portrait.

−4−224−4−224xyλ = −1λ = 4
Working
  1. Eigenvalues

    \(\lambda^2-3\lambda-4=0\Rightarrow(\lambda-4)(\lambda+1)=0\Rightarrow\lambda=-1,\ 4\)

    The determinant is negative (2 − 6 = −4), so the eigenvalues have opposite signs.

  2. Classify

    Opposite signs: saddle point (unstable)

    A saddle is always unstable.

  3. Eigenvectors

    \(\lambda=-1:\ 2x+2y=0\Rightarrow\begin{pmatrix}1\\-1\end{pmatrix};\quad\lambda=4:\ -3x+2y=0\Rightarrow\begin{pmatrix}2\\3\end{pmatrix}\)

    Gradients −1 and 3/2.

  4. Arrows

    λ = −1 line: arrows towards the origin. λ = 4 line: arrows away.

    Trajectories along the positive-eigenvalue line move away from the origin.

  5. Sketch

    Curves come in near the λ = −1 line, then swerve away along the λ = 4 line

    They never cross the eigenvector lines.

Video walkthrough coming