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AI HL · AHL 9

Differential Equations

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AHL 9.10

Complex eigenvalues: foci and centres

AHL 9.10

Complex eigenvalues: foci and centres

Target: I can find the complex eigenvalues \(\lambda=p\pm qi\) of a 2 × 2 system. (+4 more)

By the end of this section

Key points

Complex eigenvalues

  • Write the characteristic equation \(\lambda^2-(\text{trace})\lambda+\det=0\). If \((\text{trace})^2<4\det\) the roots are complex: \(\lambda=p\pm qi\).
  • Then \(p=\tfrac12(\text{trace})\) and \(q=\tfrac12\sqrt{4\det-(\text{trace})^2}\).
  • The factor \(e^{\pm iqt}\) gives \(\sin qt\) and \(\cos qt\), so x and y oscillate and the curves go round the origin. The origin is a focus.
  • You do not need the eigenvectors when the eigenvalues are complex.

Read the real part

  • \(p>0\): unstable focus. The curves spiral outwards.
  • \(p<0\): stable focus. The curves spiral inwards to the origin.
  • \(p=0\): neutral focus. The curves are closed ovals and keep the same size.
  • Only the real part decides stability. The imaginary part q decides how fast the curves go round.

The direction of the spiral

  • Find \(\dot x\) at the point \((0,1)\) on the positive y-axis.
  • If \(\dot x>0\) the curve moves to the right there: clockwise. If \(\dot x<0\) it moves to the left: anticlockwise.
  • For \(\dot x=ax+by\) the value at \((0,1)\) is just b.
  • The same idea answers "is x increasing here?": put the point into \(\dot x\) and read the sign.

The sketch

  • Mark the origin. Draw two or three curves with arrowheads.
  • Say the type in words (stable focus, unstable focus, neutral focus) and give the direction.

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.
A stable focus and its direction

Sketch the phase portrait for the system \(\dot x=-4x+6y,\ \dot y=-3x+2y\). Show the direction in which the curves go round the origin.

Working
  1. Characteristic equation

    \(\det\begin{pmatrix}-4-\lambda&6\\-3&2-\lambda\end{pmatrix}=0\) gives \(\lambda^2 + 2\lambda + 10 = 0\)

    Trace \(-2\) and determinant \(10\).

  2. Eigenvalues

    \(\lambda=-1\pm3i\)

    The roots are complex, so \(p=-1\) and \(q=3\). The origin is a focus.

  3. Stable or unstable

    \(\text{Re}(\lambda)=-1<0\), so the origin is a stable focus

    The curves spiral in towards the origin.

  4. Direction

    At \((0,1)\): \(\dot x=6>0\)

    On the positive y-axis x is increasing, so the curve moves to the right there: clockwise.

  5. −6−4−2246−6−4−2246xy
    Sketch

    Spirals going in, clockwise, with arrows

    You do not need the eigenvectors.

Video walkthrough coming
An unstable focus

Describe the phase portrait for the system \(\dot x=5x-5y,\ \dot y=5x-3y\): the type of equilibrium at the origin and the way the curves turn.

Working
  1. Eigenvalues

    \(\lambda^2 - 2\lambda + 10 = 0\), \(\lambda=1\pm3i\)

    Trace \(2\), determinant \(10\).

  2. Type

    \(\text{Re}(\lambda)=1>0\): unstable focus

    The curves spiral outwards, away from the origin.

  3. Direction

    At \((0,1)\): \(\dot x=5(0)-5(1)=-5<0\)

    On the positive y-axis x is decreasing, so the curve moves left: anticlockwise.

  4. −6−4−2246−6−4−2246xy
    Sketch

    Spirals going out, anticlockwise

    Start near the origin and draw outwards.

Video walkthrough coming
A neutral focus

A system has \(\dot x=2x+4y,\ \dot y=-2x-2y\). Find the eigenvalues, say what happens to the curves near the origin and give their direction.

Working
  1. Eigenvalues

    \(\det\begin{pmatrix}2-\lambda&4\\-2&-2-\lambda\end{pmatrix}=0\) gives \(\lambda^2+4=0\)

    The trace is \(0\) and the determinant is \(4\).

  2. Solve

    \(\lambda=\pm2i\)

    The real part is \(p=0\).

  3. Type

    \(p=0\): neutral focus

    The curves are closed ovals round the origin. They neither grow nor shrink.

  4. −6−4−2246−6−4−2246xy
    Direction

    At \((0,1)\): \(\dot x=4>0\): clockwise

    The same test works for a neutral focus.

Video walkthrough coming
Is x increasing here?

A system is \(\dot x=-3x+2y,\ \dot y=-2x+y\). At the point \((5,4)\), is x increasing or decreasing?

Working
  1. Use the first equation

    \(\dot x=-3(5)+2(4)\)

    x-dot is the rate of change of x. Substitute \(x=5,\ y=4\).

  2. Evaluate

    \(\dot x=-15+8=-7\)

  3. Conclude

    \(\dot x<0\): x is decreasing at \((5,4)\)

    A negative rate of change means x is falling there.

Video walkthrough coming