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AI HL · AHL 9

Differential Equations

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AHL 9.11

Equilibrium points that are not the origin

AHL 9.11

Equilibrium points that are not the origin

Target: I can find the equilibrium point of a coupled system with constant terms by setting both rates to zero. (+4 more)

By the end of this section

Key points

Finding an equilibrium point

  • An equilibrium point is where both rates are zero: \(\dfrac{dx}{dt}=0\) and \(\dfrac{dy}{dt}=0\). If the system starts there, nothing changes.
  • For \(\dfrac{dx}{dt}=ax+by\) and \(\dfrac{dy}{dt}=cx+dy\) the equilibrium point is the origin. With constant terms, \(\dfrac{dx}{dt}=ax+by+e\) and \(\dfrac{dy}{dt}=cx+dy+f\), it moves to the point that solves \(ax+by+e=0\) and \(cx+dy+f=0\).
  • Solve the two equations together with the calculator, then check by putting the point back in.

Classifying it

  • Put \(u=x-x_0\) and \(v=y-y_0\), where \((x_0,y_0)\) is the equilibrium point. The constants cancel and \(\dfrac{du}{dt}=au+bv\), \(\dfrac{dv}{dt}=cu+dv\). The matrix is the same, so the eigenvalues and the type of point are the same as before.
  • The types are unchanged: stable or unstable node, saddle point, stable or unstable focus, centre. They are now about \((x_0,y_0)\), not the origin.
  • If the point is stable, \(x\) and \(y\) settle at \(x_0\) and \(y_0\) whatever the start. If it is unstable, they move away from it.

Nonlinear systems

  • A product such as \(xy\) gives several equilibrium points. Factorise each right side, then take every pair of factors, one from each equation.
  • In a predator-prey system \((0,0)\) means both die out. A point with both numbers positive is the pair of populations that can stay constant together.
  • To test a start, put it into both rates. If both are 0, Euler's method leaves it where it is. If not, one step shows which way it moves.

Sketching by hand

  • Mark the equilibrium point first. Find the eigenvalues and eigenvectors of the matrix.
  • Draw the two eigenvector lines through the equilibrium point, using their gradients.
  • Put arrows on each line: away from the point for a positive eigenvalue, towards it for a negative one.
  • Add a few curves that follow the arrows. Two lines with arrows and a few curves is enough.

On the calculator

Solving two equations

  • Write the two equations that make both rates zero, with the constants on the right side if the solver needs it.
  • Use the calculator's solver for a pair of linear equations in x and y.
  • Read x and y. Check by putting them back into both equations.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Finding where the system settles

Two ponds are joined by a stream. After \(t\) days the masses of phosphate in the ponds are \(x\) kg and \(y\) kg. They satisfy \(\dfrac{dx}{dt}=−0.4x + 0.1y + 6,\quad \dfrac{dy}{dt}=0.2x − 0.3y + 3\). Find the equilibrium point, say what kind it is, and say how much phosphate each pond holds in the long run.

Working
  1. Both rates zero

    \(−0.4x+0.1y+6=0,\quad 0.2x-0.3y+3=0\)

    At an equilibrium point nothing changes.

  2. Solve the pair

    From the calculator: \(x=21,\ y=24\)

    Check: −0.4(21) + 0.1(24) + 6 = 0.

  3. The matrix

    \(\begin{pmatrix}-0.4&0.1\\0.2&-0.3\end{pmatrix}\): \(\lambda^2+0.7\lambda+0.1=0\)

    The constants do not change the matrix.

  4. Eigenvalues

    \(\lambda=−0.2\) and \(\lambda=−0.5\)

    Both negative.

  5. Classify

    A stable node at (21, 24)

    Whatever the start, the ponds settle at 21 kg and 24 kg.

Video walkthrough coming
Shifting the origin

A system is \(\dfrac{dx}{dt}=x − 2y − 7,\quad \dfrac{dy}{dt}=2x + y − 4\). Find the equilibrium point. Write the system in terms of \(u=x-x_0\) and \(v=y-y_0\), where \((x_0,y_0)\) is the equilibrium point, and decide what kind of equilibrium point it is. Say which way the trajectories turn.

Working
  1. The equilibrium point

    \(x-2y-7=0,\ 2x+y-4=0\Rightarrow (x_0,y_0)=(3,-2)\)

    From the calculator.

  2. Shift

    \(u=x-3,\ v=y+2:\quad \dfrac{du}{dt}=u-2v,\ \dfrac{dv}{dt}=2u+v\)

    The constants cancel, so the matrix is the same as for x and y.

  3. Eigenvalues

    \(\lambda^2-2\lambda+5=0\Rightarrow\lambda=1\pm2i\)

    Complex, with real part 1 > 0.

  4. Classify

    An unstable focus at (3, −2)

    The trajectories spiral away from (3, −2), not from the origin.

  5. Direction

    At \((u,v)=(0,1)\): \(\dfrac{du}{dt}=−2\lt0\)

    Above the equilibrium point x decreases: the spiral turns anticlockwise.

Video walkthrough coming
A predator and prey model

The numbers of hares \(x\) and lynx \(y\) in a forest, in hundreds, after \(t\) years satisfy \(\dfrac{dx}{dt}=0.6x-0.03xy\) and \(\dfrac{dy}{dt}=0.01xy-0.4y\). Find the equilibrium points and say what each means.

Working
  1. Both rates zero

    \(x(0.6-0.03y)=0\) and \(y(0.01x-0.4)=0\)

    Factorise each right side.

  2. First equation

    \(x=0\) or \(y=20\)

    0.6 − 0.03y = 0 gives y = 20.

  3. Second equation

    \(y=0\) or \(x=40\)

    0.01x − 0.4 = 0 gives x = 40.

  4. Combine

    \(x=0\) and \(y=0\): (0, 0). \(y=20\) and \(x=40\): (40, 20)

    The other two combinations, x = 0 with x = 40 and y = 20 with y = 0, cannot both hold.

  5. Meaning

    (0, 0): both die out. (40, 20): 4000 hares and 2000 lynx, which both stay constant

    At (40, 20) the hares are eaten as fast as they breed, and the lynx die as fast as they are born.

Video walkthrough coming
Sketching a portrait about (2, 1)

A system is \(\dfrac{dx}{dt}=x + 2y − 4,\quad \dfrac{dy}{dt}=3x − 6\). Find the equilibrium point, say what kind it is, and sketch the phase portrait.

Working
  1. The equilibrium point

    \(x+2y-4=0,\ 3x-6=0\Rightarrow(2,1)\)

    Mark it first. Everything is drawn about this point.

  2. The matrix

    \(\dfrac{du}{dt}=u+2v,\ \dfrac{dv}{dt}=3u\): \(\lambda^2-\lambda-6=0\)

    u = x − 2 and v = y − 1.

  3. Eigenvalues

    \(\lambda=3\) and \(\lambda=−2\): a saddle point

    Opposite signs. A saddle is unstable.

  4. Eigenvector lines

    \(\lambda=3:\ \begin{pmatrix}1\\1\end{pmatrix}\), line \(y=x-1\). \(\lambda=−2:\ \begin{pmatrix}2\\-3\end{pmatrix}\), line \(y=−1.5x+4\)

    Both lines pass through (2, 1), with gradients 1 and −1.5.

  5. −112345−3−2−112345xyλ = 3λ = −2(2, 1)
    Arrows and curves

    Arrows leave along \(y=x-1\) (\(\lambda\gt0\)) and arrive along \(y=−1.5x+4\) (\(\lambda\lt0\)). Curves bend like hyperbolas between them.

    Two lines with arrows and a few curves is enough.

Video walkthrough coming