Equilibrium points that are not the origin Problem set A Worked solutions

Find the equilibrium point of each system.

1. (a) \(\dfrac{dx}{dt}=2x − y − 5,\quad \dfrac{dy}{dt}=x + 3y − 6\)

   • Set both rates to zero: \(2x − y − 5=0\) and \(x + 3y − 6=0\)
   • From the calculator: \(x=3,\ y=1\)
   • Check: \(2(3)+(−1)(1)+(−5)=0\)

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

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

A system is \(\dfrac{dx}{dt}=−2x + y + 5,\quad \dfrac{dy}{dt}=x − 2y + 2\).

1. (a) Find the equilibrium point.

   • Both rates zero; from the calculator \(x=4,\ y=3\)

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

   • The matrix is \(\begin{pmatrix}-2&1\\1&-2\end{pmatrix}\); the constants do not change it
   • \(\lambda^2-(−4)\lambda+(3)=0\)
   • \(\lambda=−3\) and \(\lambda=−1\)

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

   • Real eigenvalues both negative: a stable node
   • stable node: the trajectories approach \((4, 3)\), so \(x\to4\) and \(y\to3\)

Two species of fish compete for food in a lake. Their numbers, in thousands, after \(t\) years satisfy \(\dfrac{dx}{dt}=x(10-x-y)\) and \(\dfrac{dy}{dt}=y(12-2x-y)\).

1. (a) Show that \((2, 8)\) is an equilibrium point.

   • In dx/dt, at \(x=2,\ y=8\): the bracket is \(10-2-8=0\)
   • In dy/dt the bracket is \(12-4-8=0\)
   • 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 \(10-x=0\): \((10, 0)\), only the first species
   • \(x=0\) and \(12-y=0\): \((0, 12)\), only the second species

The numbers of hares \(x\) and foxes \(y\) on an island, in hundreds, after \(t\) years satisfy \(\dfrac{dx}{dt}=0.5x-0.02xy\) and \(\dfrac{dy}{dt}=0.004xy-0.3y\).

1. (a) Find the equilibrium points.

   • \(x(0.5-0.02y)=0\) and \(y(0.004x-0.3)=0\)
   • \(y=25\) from the first, \(x=75\) from the second
   • Together with \(x=0,\ y=0\): (0, 0) and (75, 25)

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

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

3. (c) Start with \(x=80\) and \(y=25\). Use Euler's method with step length 0.1 to find \(x\) and \(y\) after one step. Is this start an equilibrium?

   • \(\dfrac{dx}{dt}=0.5(80)-0.02(80)(25)=0\), \(\dfrac{dy}{dt}=0.004(80)(25)-0.3(25)=0.5\)
   • \(x_1=80+0.1(0)=80\), \(y_1=25+0.1(0.5)=25.05\)
   • x does not move but y does, so the start is not an equilibrium

Two vats in a dye works are joined by a pipe. After \(t\) hours the vats hold \(x\) kg and \(y\) kg of dye, and satisfy \(\dfrac{dx}{dt}=−0.6x + 0.2y + 10,\quad \dfrac{dy}{dt}=0.3x − 0.5y + 11\).

1. (a) Find the equilibrium point.

   • Both rates zero; from the calculator \(x=30,\ y=40\)

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

   • \(\lambda^2-(−1.1)\lambda+(0.24)=0\)
   • \(\lambda=−0.8\) 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
   • \(x\to30\) and \(y\to40\)

A system is \(\dfrac{dx}{dt}=3x + y − 5,\quad \dfrac{dy}{dt}=2x + 2y − 6\).

1. (a) Find the equilibrium point.

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

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

   • \(\lambda^2-(5)\lambda+(4)=0\)
   • \(\lambda=1\): eigenvector \(\begin{pmatrix}1\\−2\end{pmatrix}\), gradient −2
   • \(\lambda=4\): 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 positive: an unstable node
   • Draw \(y=−2x+4\) and \(y=x+1\) through \((1, 2)\)
   • Arrows away from the point on both lines
   • Add a few curves that follow the arrows

A system is \(\dfrac{dx}{dt}=2x + y − 7,\quad \dfrac{dy}{dt}=kx + y − 5\), where \(k\) is a constant.

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

   • \(2x + y − 7=0\) and \(y − 5=0\)
   • \(y=5\), then \(x=1\)

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

   • The determinant of \(\begin{pmatrix}2&1\\k&1\end{pmatrix}\) is \(2-k\)
   • A saddle point has eigenvalues of opposite signs, so the determinant is negative
   • \(2-k\lt0\Rightarrow k\gt2\)

3. (c) Show that there is no equilibrium point when \(k=2\).

   • With \(k=2\) the equations are \(2x+y=7\) and \(2x+y=5\)
   • The left sides are the same, but the second equation says it equals \(5\)
   • \(7\ne5\), so no \((x, y)\) satisfies both