Coupled systems and phase portraits Problem set A Answers

\(\lambda=-1,\ \mathbf p=\begin{pmatrix}2\\-1\end{pmatrix}\); \(\lambda=5,\ \mathbf p=\begin{pmatrix}1\\1\end{pmatrix}\)

1. (a) \(\lambda=1,\ \mathbf p=\begin{pmatrix}1\\-1\end{pmatrix}\); \(\lambda=5,\ \mathbf p=\begin{pmatrix}3\\1\end{pmatrix}\)
2. (b) Unstable node: every curve leaves the origin. Near the origin the curves follow \(\begin{pmatrix}1\\-1\end{pmatrix}\) (the smaller eigenvalue); far away they follow \(\begin{pmatrix}3\\1\end{pmatrix}\)

1. (a) \(\lambda=1,\ \mathbf p=\begin{pmatrix}1\\-1\end{pmatrix}\); \(\lambda=5,\ \mathbf p=\begin{pmatrix}1\\3\end{pmatrix}\)
2. (b) \(x=4e^{t}+2e^{5t}\), \(y=-4e^{t}+6e^{5t}\)
3. (c) Both populations grow without limit (the eigenvalues are positive), which is impossible in a real culture: food and space are limited, so the model is only suitable for a short time

1. (a) \(x=4e^{-2t}+10e^{3t}\), \(y=8e^{-2t}+10e^{3t}\)
2. (b) The curve starts at \((14,18)\) and bends towards the line through the origin along \(\begin{pmatrix}1\\1\end{pmatrix}\), leaving the window at the top
3. (c) \(x:y=1:1\) (the eigenvector for \(\lambda=3\) is \(\begin{pmatrix}1\\1\end{pmatrix}\))

1. (a) A spiral going in to the origin, anticlockwise
2. (b) Both graphs oscillate with decreasing amplitude; x first crosses the t-axis at about \(t=0.23\)
3. (c) x and y oscillate about 0 with decreasing amplitude and both tend to 0: the populations return to their usual levels

1. (a) About \(13.1\) (any estimate within 0.5)
2. (b) \(x\to8\) and \(y\to6\)

1. (a) True
2. (b) False
3. (c) True
4. (d) False