Coupled systems and phase portraits Problem set B One question at a time

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

• \(\det\begin{pmatrix}1-\lambda&3\\4&2-\lambda\end{pmatrix}=0\)
• \(\lambda^2-3\lambda-10=0\), so \(\lambda=-2\) or \(\lambda=5\)
• \(\lambda=-2\): \(3x + 3y=0\) gives \(\mathbf p=\begin{pmatrix}1\\-1\end{pmatrix}\)
• \(\lambda=5\): \(-4x + 3y=0\) gives \(\mathbf p=\begin{pmatrix}3\\4\end{pmatrix}\)

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

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

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

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

   • Both eigenvalues are positive: \(1\) and \(4\), so the origin is an unstable node
   • Draw the lines through the origin along \(\begin{pmatrix}2\\-1\end{pmatrix}\) and \(\begin{pmatrix}1\\1\end{pmatrix}\), with arrows pointing away from the origin
   • Curves leave the origin tangent to the \(\lambda=1\) line and become parallel to the \(\lambda=4\) line far from the origin

Two species of plant are introduced to an island. Their numbers, in hundreds, are modelled by \(\dot x=3x + y,\ \dot y=2x + 4y\), where t is in years. At \(t=0\) there are 5 hundred of species X and 1 hundred of species Y.

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

   • \(\det\begin{pmatrix}3-\lambda&1\\2&4-\lambda\end{pmatrix}=0\)
   • \(\lambda^2-7\lambda+10=0\), so \(\lambda=2\) or \(\lambda=5\)
   • \(\lambda=2\): \(x + y=0\) gives \(\mathbf p=\begin{pmatrix}1\\-1\end{pmatrix}\)
   • \(\lambda=5\): \(-2x + y=0\) gives \(\mathbf p=\begin{pmatrix}1\\2\end{pmatrix}\)

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

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

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. An island has limited space and resources, so the model is not suitable in the long term

The numbers of wolves (x) and deer (y) in a park, in hundreds, are modelled by \(\dot x=7x - 10y,\ \dot y=5x - 8y\), where t is in years. Initially there are 17 hundred wolves and 10 hundred deer.

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

   • \(\det\begin{pmatrix}7-\lambda&-10\\5&-8-\lambda\end{pmatrix}=0\)
   • \(\lambda^2+\lambda-6=0\), so \(\lambda=-3\) or \(\lambda=2\)
   • \(\lambda=-3\): \(10x - 10y=0\) gives \(\mathbf p=\begin{pmatrix}1\\1\end{pmatrix}\)
   • \(\lambda=2\): \(5x - 10y=0\) gives \(\mathbf p=\begin{pmatrix}2\\1\end{pmatrix}\)
   • \(t=0\): \(A+2B=17\) and \(A+B=10\), where \(A\) goes with \(\lambda=-3\)
   • \(x=3e^{-3t}+14e^{2t}\), \(y=3e^{-3t}+7e^{2t}\)

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 \(1\) and \(0.5\)
   • Start at \((17,10)\)
   • As t grows the \(e^{-3t}\) terms vanish, so the curve bends towards the \(\lambda=2\) line and leaves the window on the right

3. (c) What is the ratio of wolves to deer in the long term?

   • As \(t\to\infty\) the \(e^{2t}\) terms dominate, so \(x\approx14e^{2t}\) and \(y\approx7e^{2t}\)
   • The ratio of wolves to deer is \(2:1\), given by the eigenvector \(\begin{pmatrix}2\\1\end{pmatrix}\)

The deviations of the temperature (x, in degrees) and the pressure (y, in kPa) of a gas in a sealed tank from their set values are modelled by \(\dot x=-x + 4y,\ \dot y=-x - 3y\), where t is in minutes, with \(x(0)=-3\) and \(y(0)=4\).

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

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

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 \(t=0.18\)

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

   • \(\lambda=-2\pm1.73i\)
   • Both oscillate with smaller and smaller swings and tend to 0, so the temperature and the pressure return to their set values. 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 \((10,6)\).

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

   • Follow the curve from \((10,6)\): y rises first, then turns back
   • The highest y-value is about \(10.4\)

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

   • All the arrows point towards \((4,8)\)
   • The curve ends there: \(x\to4,\ y\to8\)

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 7 and y approaches 6.

   • All the arrows lead towards the point \((7,6)\), so every solution curve ends there: true

2. (b) y always moves towards 6.

   • Above \(y=6\) the arrows point down and below \(y=6\) they point up: true

3. (c) If x is greater than 7, then x decreases.

   • Just right of \(x=7\), high up in the diagram, the arrows point to the right: x increases there: false

4. (d) x always increases.

   • Right of \(x=7\) low in the diagram the arrows point to the left: x decreases there: false