Coupled systems and phase portraits Problem set A Worked solutions

Find the eigenvalues and eigenvectors of the matrix \(\begin{pmatrix}1&4\\2&3\end{pmatrix}\).

• \(\det\begin{pmatrix}1-\lambda&4\\2&3-\lambda\end{pmatrix}=0\)
• \(\lambda^2-4\lambda-5=0\), so \(\lambda=-1\) or \(\lambda=5\)
• \(\lambda=-1\): \(2x + 4y=0\) gives \(\mathbf p=\begin{pmatrix}2\\-1\end{pmatrix}\)
• \(\lambda=5\): \(-4x + 4y=0\) gives \(\mathbf p=\begin{pmatrix}1\\1\end{pmatrix}\)

A system is \(\dot x=4x + 3y,\ \dot y=x + 2y\).

1. (a) Find the eigenvalues and eigenvectors of the matrix of the system.

   • \(\det\begin{pmatrix}4-\lambda&3\\1&2-\lambda\end{pmatrix}=0\)
   • \(\lambda^2-6\lambda+5=0\), so \(\lambda=1\) or \(\lambda=5\)
   • \(\lambda=1\): \(3x + 3y=0\) gives \(\mathbf p=\begin{pmatrix}1\\-1\end{pmatrix}\)
   • \(\lambda=5\): \(-x + 3y=0\) gives \(\mathbf p=\begin{pmatrix}3\\1\end{pmatrix}\)

2. (b) Sketch the phase portrait, showing the directions of the eigenvectors, and name the type of equilibrium.

   • Both eigenvalues are positive: \(1\) and \(5\), so the origin is an unstable node
   • Draw the lines through the origin along \(\begin{pmatrix}1\\-1\end{pmatrix}\) and \(\begin{pmatrix}3\\1\end{pmatrix}\), with arrows pointing away from the origin
   • Curves leave the origin tangent to the \(\lambda=1\) line and become parallel to the \(\lambda=5\) line far from the origin

The numbers of two strains of bacteria in a culture, in thousands, are modelled by \(\dot x=2x + y,\ \dot y=3x + 4y\), where t is in hours. At \(t=0\) there are 6 thousand of strain X and 2 thousand of strain Y.

1. (a) Find the eigenvalues and eigenvectors of the matrix of the system.

   • \(\det\begin{pmatrix}2-\lambda&1\\3&4-\lambda\end{pmatrix}=0\)
   • \(\lambda^2-6\lambda+5=0\), so \(\lambda=1\) or \(\lambda=5\)
   • \(\lambda=1\): \(x + y=0\) gives \(\mathbf p=\begin{pmatrix}1\\-1\end{pmatrix}\)
   • \(\lambda=5\): \(-3x + y=0\) gives \(\mathbf p=\begin{pmatrix}1\\3\end{pmatrix}\)

2. (b) Write down the general solution, then find x(t) and y(t) for the starting values given.

   • \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{5t}\begin{pmatrix}1\\3\end{pmatrix}\)
   • \(t=0\): \(A+B=6\) and \(-A+3B=2\)
   • \(A=4,\ B=2\), so \(x=4e^{t}+2e^{5t}\) and \(y=-4e^{t}+6e^{5t}\)

3. (c) Comment on how suitable the model is in the long term.

   • Both eigenvalues are positive, so x and y grow without bound as t increases. A culture cannot grow for ever because food and space run out, so the model is not suitable in the long term

The numbers of foxes (x) and rabbits (y) on an island, in hundreds, are modelled by \(\dot x=8x - 5y,\ \dot y=10x - 7y\), where t is in years. Initially there are 14 hundred foxes and 18 hundred rabbits.

1. (a) Find x(t) and y(t).

   • \(\det\begin{pmatrix}8-\lambda&-5\\10&-7-\lambda\end{pmatrix}=0\)
   • \(\lambda^2-\lambda-6=0\), so \(\lambda=-2\) or \(\lambda=3\)
   • \(\lambda=-2\): \(10x - 5y=0\) gives \(\mathbf p=\begin{pmatrix}1\\2\end{pmatrix}\)
   • \(\lambda=3\): \(5x - 5y=0\) gives \(\mathbf p=\begin{pmatrix}1\\1\end{pmatrix}\)
   • \(t=0\): \(A+B=14\) and \(2A+B=18\), where \(A\) goes with \(\lambda=-2\)
   • \(x=4e^{-2t}+10e^{3t}\), \(y=8e^{-2t}+10e^{3t}\)

2. (b) Show the solution curve on a phase portrait for \(0\le x,y\le50\). Show the directions of the two eigenvectors.

   • Draw the lines through the origin with slopes \(2\) and \(1\)
   • Start at \((14,18)\)
   • As t grows the \(e^{-2t}\) terms vanish, so the curve bends towards the \(\lambda=3\) line and leaves the window at the top

3. (c) What is the ratio of foxes to rabbits in the long term?

   • As \(t\to\infty\) the \(e^{3t}\) terms dominate, so \(x\approx10e^{3t}\) and \(y\approx10e^{3t}\)
   • The ratio of foxes to rabbits is \(1:1\), given by the eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\)

The deviations of the numbers of two competing species, in hundreds, from their usual levels are modelled by \(\dot x=-3x - 4y,\ \dot y=2x + y\), where t is in years, with \(x(0)=5\) and \(y(0)=2\).

1. (a) Use technology to sketch the solution curve in the x-y plane.

   • \(x_{n+1}=x_n+0.1(-3x_n - 4y_n)\), \(y_{n+1}=y_n+0.1(2x_n + y_n)\)
   • From the calculator: scatter plot of \(y_n\) against \(x_n\)
   • The curve spirals in to the origin, anticlockwise

2. (b) Sketch the graphs of x and y against t for \(0\le t\le8\) on separate diagrams.

   • From the calculator: scatter plots of \(x_n\) and \(y_n\) against \(t_n\)
   • Both oscillate and the swings shrink; x first crosses the t-axis at about \(t=0.23\)

3. (c) Describe how x and y change with t.

   • \(\lambda=-1\pm2i\)
   • Both oscillate with smaller and smaller swings and tend to 0, so each population returns to its usual level. The eigenvalues have negative real part: a stable focus

The diagram shows the direction field for a system of differential equations in x and y, and the solution curve starting at \((10,12)\).

1. (a) Estimate the maximum value of x along the solution curve.

   • Follow the curve from \((10,12)\): x rises first, then turns back
   • The highest x-value is about \(13.1\)

2. (b) Estimate the long-term state of the system.

   • All the arrows point towards \((8,6)\)
   • The curve ends there: \(x\to8,\ y\to6\)

The diagram shows the direction field for a system of differential equations in x and y. Say whether each statement is true or false.

1. (a) Whatever the starting values, x approaches 9 and y approaches 7.

   • All the arrows lead towards the point \((9,7)\), so every solution curve ends there: true

2. (b) x always decreases.

   • Left of \(x=9\) the arrows point to the right, so x increases there: false

3. (c) If x is greater than 9, then x decreases.

   • Right of \(x=9\) every arrow points to the left: true

4. (d) If y is greater than 7, then y decreases.

   • Near the right-hand edge, just above \(y=7\), the arrows point upwards: y increases there: false