Consider the following system of \(n\) linear ODEs: \[\overrightarrow{x}'=A\overrightarrow{x}, \;\;\;\;(53)\] where \(\overrightarrow{x}=[x_1,\; x_2,\cdots,\; x_n]^T\). \(\overrightarrow{x}\) is called an equilibrium solution of (53) if \(A\overrightarrow{x}=\overrightarrow{0}\). Obviously \(\overrightarrow{x}=\overrightarrow{0}\) is always an equilibrium solution of (53). Note that if \(A\) is invertible (i.e., \(\det A\neq 0\) or, all eigenvalues of \(A\) are nonzero), then \(A\overrightarrow{x}=\overrightarrow{0}\implies \overrightarrow{x}=A^{-1}\overrightarrow{0}=\overrightarrow{0}\) and \(\overrightarrow{x}=\overrightarrow{0}\) is the only equilibrium solution of (53).

Direction field/ tangent field: A solution curve or a trajectory of (53) can be visualized by plotting direction field of tangent vectors to solutions of (53). \(\overrightarrow{x}'=A\overrightarrow{x}\) is the tangent vector of a solution curve at \(\overrightarrow{x}\in \mathbb R^n\).

We mostly focus on the case \(n=2\) and plot trajectories on the \(x_1x_2\)-plane which is called the phase plane. A sketch of trajectories on the phase plane is called a phase portrait.

Example. \(\overrightarrow{x}'=\left[\begin{array}{rr} 1&2\\2&1\end{array} \right] \overrightarrow{x}\).

1. Draw direction field. Use the points: \((0,0),(\pm1,0),(0,\pm 1),(\pm 1,\pm 1)\).


2. Draw the phase portrait.

Solution. (a) At \(\overrightarrow{x}=\left[\begin{array}{r}0\\0\end{array} \right],\; \overrightarrow{x}'=\left[\begin{array}{rr} 1&2\\2&1\end{array} \right] \left[\begin{array}{r}0\\0\end{array} \right]= \left[\begin{array}{r}0\\0\end{array} \right].\)
At \(\overrightarrow{x}=\left[\begin{array}{r}\pm 1\\0\end{array} \right],\; \overrightarrow{x}'=\left[\begin{array}{rr} 1&2\\2&1\end{array} \right] \left[\begin{array}{r}\pm 1\\0\end{array} \right]= \pm \left[\begin{array}{r}1\\2\end{array} \right].\)
At \(\overrightarrow{x}=\left[\begin{array}{r}0\\ \pm 1\end{array} \right],\; \overrightarrow{x}'=\left[\begin{array}{rr} 1&2\\2&1\end{array} \right] \left[\begin{array}{r}0\\ \pm 1\end{array} \right]= \pm \left[\begin{array}{r}2\\1\end{array} \right].\)
At \(\overrightarrow{x}=\pm \left[\begin{array}{r}1\\1\end{array} \right],\; \overrightarrow{x}'=\pm \left[\begin{array}{rr} 1&2\\2&1\end{array} \right] \left[\begin{array}{r}1\\ 1\end{array} \right]= \pm \left[\begin{array}{r}3\\3\end{array} \right].\)
At \(\overrightarrow{x}=\pm \left[\begin{array}{r}1\\-1\end{array} \right],\; \overrightarrow{x}'=\pm \left[\begin{array}{rr} 1&2\\2&1\end{array} \right] \left[\begin{array}{r}1\\ -1\end{array} \right]= \pm \left[\begin{array}{r}-1\\1\end{array} \right].\) Now draw the tangent vector \(A\overrightarrow{x}\) at each \(\overrightarrow{x}\).

(b) The eigenvalues of the coefficient matrix are \(-1\) and \(3\) with corresponding eigenvectors \(\left[\begin{array}{r}1\\-1\end{array} \right]\) and \(\left[\begin{array}{r}1\\1\end{array} \right]\) respectively (show all the steps). So the general solution is \[\overrightarrow{x}(t)= c_1e^{-t}\left[\begin{array}{r}1\\-1\end{array} \right] +c_2e^{3t}\left[\begin{array}{r}1\\1\end{array} \right].\] When \(c_2=0\), \(\overrightarrow{x}=c_1\overrightarrow{x_1}=c_1e^{-t}\left[\begin{array}{r}1\\-1\end{array} \right]\). This gives the trajectory \(x_1=c_1e^{-t},\; x_2=-c_1e^{-t}\), i.e., the line \(x_2=-x_1\) (the line through the origin parallel to \(\overrightarrow{v_1}=\left[\begin{array}{r}1\\-1\end{array} \right]\)). This trajectory is in the quadrant 4 when \(c_1 > 0\) and in the quadrant 2 when \(c_1 < 0\). Note that \(\overrightarrow{x}=c_1\overrightarrow{x_1}=c_1e^{-t}\left[\begin{array}{r}1\\-1\end{array} \right]\) is far from the origin for large negative \(t\) and \(\overrightarrow{x}=c_1\overrightarrow{x_1}=c_1e^{-t}\left[\begin{array}{r}1\\-1\end{array} \right]\to \left[\begin{array}{r}0\\0\end{array} \right]\) as \(t\to \infty\).

When \(c_1=0\), \(\overrightarrow{x}=c_2\overrightarrow{x_2}=c_2e^{3t}\left[\begin{array}{r}1\\1\end{array} \right]\). This gives the trajectory \(x_1=c_2e^{3t},\; x_2=c_2e^{3t}\), i.e., the line \(x_2=x_1\) (the line through the origin parallel to \(\overrightarrow{v_2}=\left[\begin{array}{r}1\\1\end{array} \right]\)). This trajectory is in the quadrant 1 when \(c_2 > 0\) and in the quadrant 3 when \(c_2 < 0\). Note that \(\overrightarrow{x}=c_2\overrightarrow{x_2}=c_2e^{3t}\left[\begin{array}{r}1\\1\end{array} \right]\) is far from the origin for large positive \(t\) and \(\overrightarrow{x}=c_2\overrightarrow{x_2}=c_2e^{3t}\left[\begin{array}{r}1\\1\end{array} \right]\to \left[\begin{array}{r}0\\0\end{array} \right]\) as \(t\to -\infty\).

Note:

1. For these kind of trajectories (when eigenvalues are real of opposite signs) in the preceding example, the origin is called a saddle point which is unstable as trajectories move away from the origin as \(t\) increases.


2. When eigenvalues are distinct real of the same sign, the origin is called a node which is asymptotically stable when eigenvalues are negative (trajectories move toward the origin as \(t\) increases) and unstable when eigenvalues are positive (trajectories move away from the origin as \(t\) increases).

Example. \(\overrightarrow{x}'=\left[\begin{array}{rr} -5&4\\1&-2\end{array} \right] \overrightarrow{x}\).

Solution. The general solution is \[\overrightarrow{x}(t)= c_1e^{-t}\left[\begin{array}{r}1\\1\end{array} \right] +c_2e^{-6t}\left[\begin{array}{r}4\\-1\end{array} \right].\] \(c_2\overrightarrow{x_2}\) will dominate \(c_1\overrightarrow{x_1}\) for large negative \(t\) (i.e., when far from the origin). \(c_1\overrightarrow{x_1}\) will dominate \(c_2\overrightarrow{x_2}\) for large positive \(t\) (i.e., when near the origin). The origin is an asymptotically stable node.

Example. \(\overrightarrow{x}'=\left[\begin{array}{rr} 3&1\\2&4\end{array} \right] \overrightarrow{x}\).

Solution. The general solution is \[\overrightarrow{x}(t)= c_1e^{2t}\left[\begin{array}{r}-1\\1\end{array} \right] +c_2e^{5t}\left[\begin{array}{r}1\\2\end{array} \right].\] The origin is an unstable node.

Example. \(\overrightarrow{x}'=\left[\begin{array}{rr} -1&1\\1&-1\end{array} \right] \overrightarrow{x}\).

Solution. The general solution is \[\overrightarrow{x}(t)= c_1\left[\begin{array}{r}1\\1\end{array} \right] +c_2e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right].\] Each point on the line \(x_2=x_1\) is an equilibrium. The origin is a degenerate node.

Note. The origin is asymptotically stable when both eigenvalues are negative (trajectories move toward the origin as \(t\) increases) and unstable when at least one eigenvalue is positive (trajectories move away from the origin as \(t\) increases).

Last edited