Phase portraits for real eigenvalues Problem set A One question at a time

Use the characteristic equation \(\det(\mathbf M-\lambda\mathbf I)=0\).

1. (a) Find the eigenvalues of \(\mathbf M=\begin{pmatrix}7&2\\-2&2\end{pmatrix}\).

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-9\lambda+18=(\lambda-3)(\lambda-6)=0\)
   • \(\lambda=3\ \text{or}\ 6\)
   • Check: the eigenvalues add to the trace 9 and multiply to the determinant 18.

2. (b) Find the eigenvalues of \(\mathbf M=\begin{pmatrix}-4&2\\3&-5\end{pmatrix}\).

   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2+9\lambda+14=(\lambda+7)(\lambda+2)=0\)
   • \(\lambda=-7\ \text{or}\ -2\)
   • Check: the eigenvalues add to the trace −9 and multiply to the determinant 14.

For the system \(\dfrac{dx}{dt}=4x+y\) and \(\dfrac{dy}{dt}=2x+3y\), find the eigenvalues and eigenvectors of the matrix of the system, say what kind of equilibrium point the origin is, and sketch the phase portrait. Show the eigenvector lines and the direction of the arrows.

• \(\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}\)
• \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-7\lambda+10=(\lambda-2)(\lambda-5)=0\)
• \(\lambda=2\ \text{or}\ 5\)
• \(\lambda=2:\ 2x+y=0\Rightarrow\begin{pmatrix}1\\-2\end{pmatrix}\)
• \(\lambda=5:\ -x+y=0\Rightarrow\begin{pmatrix}1\\1\end{pmatrix}\)
• Eigenvalues 2 and 5, both positive: the origin is an unstable node.
• Draw the line through the origin with direction \(\begin{pmatrix}1\\-2\end{pmatrix}\) (gradient −2) for \(\lambda=2\): arrows point away from the origin.
• Draw the line through the origin with direction \(\begin{pmatrix}1\\1\end{pmatrix}\) (gradient 1) for \(\lambda=5\): arrows point away from the origin.
• Other trajectories leave the origin tangent to the \(\lambda=2\) line and bend towards the direction of the \(\lambda=5\) line.

For the system \(\dfrac{dx}{dt}=-x+y\) and \(\dfrac{dy}{dt}=-3x-5y\), find the eigenvalues and eigenvectors of the matrix of the system, say what kind of equilibrium point the origin is, and sketch the phase portrait. Show the eigenvector lines and the direction of the arrows.

• \(\mathbf M=\begin{pmatrix}-1&1\\-3&-5\end{pmatrix}\)
• \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2+6\lambda+8=(\lambda+4)(\lambda+2)=0\)
• \(\lambda=-4\ \text{or}\ -2\)
• \(\lambda=-4:\ 3x+y=0\Rightarrow\begin{pmatrix}1\\-3\end{pmatrix}\)
• \(\lambda=-2:\ x+y=0\Rightarrow\begin{pmatrix}1\\-1\end{pmatrix}\)
• Eigenvalues −4 and −2, both negative: the origin is a stable node.
• Draw the line through the origin with direction \(\begin{pmatrix}1\\-3\end{pmatrix}\) (gradient −3) for \(\lambda=-4\): arrows point towards the origin.
• Draw the line through the origin with direction \(\begin{pmatrix}1\\-1\end{pmatrix}\) (gradient −1) for \(\lambda=-2\): arrows point towards the origin.
• Other trajectories come in from far away alongside the \(\lambda=-4\) line and reach the origin tangent to the \(\lambda=-2\) line.

For the system \(\dfrac{dx}{dt}=2x+y\) and \(\dfrac{dy}{dt}=4x-y\), find the eigenvalues and eigenvectors of the matrix of the system, say what kind of equilibrium point the origin is, and sketch the phase portrait. Show the eigenvector lines and the direction of the arrows.

• \(\mathbf M=\begin{pmatrix}2&1\\4&-1\end{pmatrix}\)
• \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2-\lambda-6=(\lambda+2)(\lambda-3)=0\)
• \(\lambda=-2\ \text{or}\ 3\)
• \(\lambda=-2:\ 4x+y=0\Rightarrow\begin{pmatrix}1\\-4\end{pmatrix}\)
• \(\lambda=3:\ -x+y=0\Rightarrow\begin{pmatrix}1\\1\end{pmatrix}\)
• Eigenvalues −2 and 3, opposite signs: the origin is a saddle point (unstable).
• Draw the line through the origin with direction \(\begin{pmatrix}1\\-4\end{pmatrix}\) (gradient −4) for \(\lambda=-2\): arrows point towards the origin.
• Draw the line through the origin with direction \(\begin{pmatrix}1\\1\end{pmatrix}\) (gradient 1) for \(\lambda=3\): arrows point away from the origin.
• Other trajectories come in near the \(\lambda=-2\) line, then swerve away and run along the \(\lambda=3\) line.

Two lakes are joined by a channel. After \(t\) years the amounts of a pollutant in the two lakes are \(x\) tonnes and \(y\) tonnes, modelled by \(\dfrac{dx}{dt}=-4x+y\) and \(\dfrac{dy}{dt}=2x-3y\).

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

   • \(\mathbf M=\begin{pmatrix}-4&1\\2&-3\end{pmatrix}\)
   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2+7\lambda+10=(\lambda+5)(\lambda+2)=0\)
   • \(\lambda=-5\ \text{or}\ -2\)
   • \(\lambda=-5:\ x+y=0\Rightarrow\begin{pmatrix}1\\-1\end{pmatrix}\)
   • \(\lambda=-2:\ -2x+y=0\Rightarrow\begin{pmatrix}1\\2\end{pmatrix}\)
   • Eigenvalues −5 and −2, both negative: the origin is a stable node.

2. (b) Describe the phase portrait: the eigenvector lines, the arrows and the shape of the curves.

   • Draw the line through the origin with direction \(\begin{pmatrix}1\\-1\end{pmatrix}\) (gradient −1) for \(\lambda=-5\): arrows point towards the origin.
   • Draw the line through the origin with direction \(\begin{pmatrix}1\\2\end{pmatrix}\) (gradient 2) for \(\lambda=-2\): arrows point towards the origin.
   • Other trajectories come in from far away alongside the \(\lambda=-5\) line and reach the origin tangent to the \(\lambda=-2\) line.

3. (c) At the start \(x=3\) and \(y=6\). Show that this point lies on an eigenvector line and describe how the two amounts change.

   • The start point \((3,6)\) is \(3\) times \(\begin{pmatrix}1\\2\end{pmatrix}\)
   • So it is on the \(\lambda=-2\) line: \(x=3e^{-2t},\ y=6e^{-2t}\)
   • As \(t\to\infty\) both tend to 0, along the line.

A toy boat floats on a stream. Its position relative to a post at the origin, \(t\) seconds after release, is \((x,y)\) metres, modelled by \(\dfrac{dx}{dt}=-x+2y\) and \(\dfrac{dy}{dt}=3x\).

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

   • \(\mathbf M=\begin{pmatrix}-1&2\\3&0\end{pmatrix}\)
   • \(\det(\mathbf M-\lambda\mathbf I)=\lambda^2+\lambda-6=(\lambda+3)(\lambda-2)=0\)
   • \(\lambda=-3\ \text{or}\ 2\)
   • \(\lambda=-3:\ 2x+2y=0\Rightarrow\begin{pmatrix}1\\-1\end{pmatrix}\)
   • \(\lambda=2:\ -3x+2y=0\Rightarrow\begin{pmatrix}2\\3\end{pmatrix}\)
   • Eigenvalues −3 and 2, opposite signs: the origin is a saddle point (unstable).

2. (b) The boat is released at \((2,-2)\). Describe its path.

   • The point \((2,-2)\) is \(2\) times \(\begin{pmatrix}1\\-1\end{pmatrix}\)
   • It is on the \(\lambda=-3\) line, which has arrows towards the origin.
   • It moves in towards the origin along the line, getting closer for ever but never reaching it (\(x=2e^{-3t},\ y=-2e^{-3t}\)).

3. (c) The boat is released at \((1,1)\) instead. Describe what happens in the long run.

   • General solution \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-3t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{2t}\begin{pmatrix}2\\3\end{pmatrix}\)
   • At t = 0: \(A=\tfrac{1}{5},\ B=\tfrac{2}{5}\)
   • The \(e^{2t}\) term grows and the other dies away, so the point leaves along \(\begin{pmatrix}2\\3\end{pmatrix}\).
   • It moves away from the origin: for large t it follows the line of gradient 3/2 (\(x\approx0.8e^{2t},\ y\approx1.2e^{2t}\)).

Consider the system \(\dfrac{dx}{dt}=3x+ky\) and \(\dfrac{dy}{dt}=2x+y\), where \(k\) is a constant.

1. (a) Find, in terms of \(k\), the product of the eigenvalues of the matrix of the system.

   • The product of the eigenvalues is the determinant.
   • \(\det\mathbf M=(3)(1)-k(2)=3-2k\)

2. (b) Hence find the values of \(k\) for which the origin is a saddle point.

   • A saddle point needs eigenvalues of opposite signs, so a negative product.
   • \(3-2k<0\Rightarrow k>\tfrac32\)

3. (c) For \(k=0\), find the eigenvalues and classify the origin.

   • \(k=0:\ \det(\mathbf M-\lambda\mathbf I)=(3-\lambda)(1-\lambda)=0\)
   • \(\lambda=1\text{ or }3\)
   • Both positive: unstable node.