Equilibrium points that are not the origin Problem set B One question at a time

Find the equilibrium point of each system.

1. (a) \(\dfrac{dx}{dt}=3x + 2y − 10,\quad \dfrac{dy}{dt}=−x + 2y + 6\)

   • Set both rates to zero: \(3x + 2y − 10=0\) and \(−x + 2y + 6=0\)
   • From the calculator: \(x=4,\ y=−1\)
   • Check: \(3(4)+(2)(−1)+(−10)=0\)

2. (b) \(\dfrac{dx}{dt}=−2x + 3y − 4,\quad \dfrac{dy}{dt}=x − 5y + 9\)

   • Set both rates to zero: \(−2x + 3y − 4=0\) and \(x − 5y + 9=0\)
   • From the calculator: \(x=1,\ y=2\)
   • Check: \(−2(1)+(3)(2)+(−4)=0\)

A system is \(\dfrac{dx}{dt}=−x − 3y + 5,\quad \dfrac{dy}{dt}=3x − y + 5\).

1. (a) Find the equilibrium point.

   • Both rates zero; from the calculator \(x=−1,\ y=2\)

2. (b) Find the eigenvalues of the matrix of the system.

   • The matrix is \(\begin{pmatrix}-1&-3\\3&-1\end{pmatrix}\); the constants do not change it
   • \(\lambda^2-(−2)\lambda+(10)=0\)
   • \(\lambda=−1\pm3i\)

3. (c) State the kind of equilibrium point. What happens to \(x\) and \(y\) in the long run?

   • Complex eigenvalues with negative real part: a stable focus
   • stable focus: the trajectories spiral towards \((−1, 2)\), so \(x\to−1\) and \(y\to2\)

Two species of beetle compete for leaves in a wood. Their numbers, in thousands, after \(t\) months satisfy \(\dfrac{dx}{dt}=x(8-x-y)\) and \(\dfrac{dy}{dt}=y(9-x-2y)\).

1. (a) Show that \((7, 1)\) is an equilibrium point.

   • In dx/dt, at \(x=7,\ y=1\): the bracket is \(8-7-1=0\)
   • In dy/dt the bracket is \(9-7-2=0\)
   • So both rates are 0 at this point

2. (b) Find the other equilibrium points and say what each one means.

   • Each rate is zero when a factor is zero
   • \(x=0\) and \(y=0\): (0, 0), both species absent
   • \(y=0\) and \(8-x=0\): \((8, 0)\), only the first species
   • \(x=0\) and \(9-2y=0\): \((0, 4.5)\), only the second species

The numbers of aphids \(x\) and ladybirds \(y\) on a rose bush, in tens, after \(t\) weeks satisfy \(\dfrac{dx}{dt}=0.8x-0.01xy\) and \(\dfrac{dy}{dt}=0.002xy-0.5y\).

1. (a) Find the equilibrium points.

   • \(x(0.8-0.01y)=0\) and \(y(0.002x-0.5)=0\)
   • \(y=80\) from the first, \(x=250\) from the second
   • Together with \(x=0,\ y=0\): (0, 0) and (250, 80)

2. (b) Say what each equilibrium point means for the two populations.

   • (0, 0): neither species is present
   • (250, 80): the prey are eaten as fast as they breed and the predators die as fast as they are born

3. (c) Start with \(x=260\) and \(y=80\). Use Euler's method with step length 0.1 to find \(x\) and \(y\) after one step. Is this start an equilibrium?

   • \(\dfrac{dx}{dt}=0.8(260)-0.01(260)(80)=0\), \(\dfrac{dy}{dt}=0.002(260)(80)-0.5(80)=1.6\)
   • \(x_1=260+0.1(0)=260\), \(y_1=80+0.1(1.6)=80.16\)
   • x does not move but y does, so the start is not an equilibrium

Two fields are joined by a drain. After \(t\) weeks they hold \(x\) kg and \(y\) kg of fertiliser, and satisfy \(\dfrac{dx}{dt}=−0.5x + 0.1y + 5,\quad \dfrac{dy}{dt}=0.2x − 0.4y + 16\).

1. (a) Find the equilibrium point.

   • Both rates zero; from the calculator \(x=20,\ y=50\)

2. (b) Find the eigenvalues of the matrix of the system and say what kind of equilibrium point it is.

   • \(\lambda^2-(−0.9)\lambda+(0.18)=0\)
   • \(\lambda=−0.6\) and \(\lambda=−0.3\)
   • Both negative: a stable node

3. (c) State the long-run amount in each, in kg.

   • A stable node attracts every trajectory
   • \(x\to20\) and \(y\to50\)

A system is \(\dfrac{dx}{dt}=−4x + y + 9,\quad \dfrac{dy}{dt}=2x − 3y − 7\).

1. (a) Find the equilibrium point.

   • Both rates zero; from the calculator \(x=2,\ y=−1\)

2. (b) Find the eigenvalues and eigenvectors of the matrix of the system.

   • \(\lambda^2-(−7)\lambda+(10)=0\)
   • \(\lambda=−2\): eigenvector \(\begin{pmatrix}1\\2\end{pmatrix}\), gradient 2
   • \(\lambda=−5\): eigenvector \(\begin{pmatrix}1\\−1\end{pmatrix}\), gradient −1

3. (c) Say what kind of equilibrium point it is and describe a sketch of the phase portrait: the lines to draw, the arrows and the curves.

   • Both eigenvalues negative: a stable node
   • Draw \(y=2x−5\) and \(y=−x+1\) through \((2, −1)\)
   • Arrows towards the point on both lines
   • Add a few curves that follow the arrows

A system is \(\dfrac{dx}{dt}=3x + y − 10,\quad \dfrac{dy}{dt}=kx + 2y − 8\), where \(k\) is a constant.

1. (a) For \(k=0\), find the equilibrium point.

   • \(3x + y − 10=0\) and \(2y − 8=0\)
   • \(y=4\), then \(x=2\)

2. (b) Find the values of \(k\) for which the equilibrium point is a saddle point.

   • The determinant of \(\begin{pmatrix}3&1\\k&2\end{pmatrix}\) is \(6-k\)
   • A saddle point has eigenvalues of opposite signs, so the determinant is negative
   • \(6-k\lt0\Rightarrow k\gt6\)

3. (c) Show that there is no equilibrium point when \(k=6\).

   • With \(k=6\) the equations are \(3x+y=10\) and \(6x+2y=8\)
   • 2 times the first equation is \(6x+2y=20\), but the second equation says it equals \(8\)
   • \(20\ne8\), so no \((x, y)\) satisfies both