Differential Equations: Direction Fields and the Phase Line
Differential Equations: Direction Fields and the Phase Line

Section 3.3 Phase Plane Analysis of Linear Systems

In Section 3.2, we learned how to solve the system

provided the system has distinct real eigenvalues. If \(A\) has distinct real eigenvalues \(\lambda\) and \(\mu\) with eigenvectors \(\mathbf u\) and \(\mathbf v\text{,}\) respectively, then the general solution of the system is

Furthermore, we can use the general solution of such a system to find the straight-line solutions to the system. If \(c_2 = 0\text{,}\) then all solutions will lie along the line in the \(xy\)-plane that contains the vector \(\mathbf u\text{.}\) Similarly, if \(c_1 = 0\text{,}\) then all solutions will lie along the line in the \(xy\)-plane that contains the vector \(\mathbf v\text{.}\)

Subsection 3.3.1 The Case \(\lambda_1 \lt 0 \lt \lambda_2\)

Example 3.3.1.

The system

can be written in matrix form \(\mathbf x’ = A \mathbf x\text{,}\) where

The eigenvalues of \(A\) are \(\lambda = -2\) or \(\lambda = 2\) with eigenvectors \(\mathbf u = (1, -1)\) and \(\mathbf v = (3,1)\text{,}\) respectively. Therefore, the straightline solutions must be lines containing \(\mathbf u\) and \(\mathbf v\) (Figure 3.3.2).

Let us consider the special case of the system \({\mathbf x}’ = A {\mathbf x}\text{,}\) where \(\lambda_1 \lt 0 \lt \lambda_2\) and

Since this is a decoupled system,

we already know how to find the solutions. However, in keeping with the spirit of our investigation, we will find the eigenvalues of \(A\text{.}\) The characteristic equation of \(A\) is

and our eigenvalues are \(\lambda_1\) and \(\lambda_2\text{.}\) It is easy to see that we can associate eigenvectors \((1,0)\) and \((0, 1)\) to \(\lambda_1\) and \(\lambda_2\text{,}\) respectively. Thus, our general solution is

Since \(\lambda_1 \lt 0\text{,}\) the straight-line solutions of the form \(c_1 e^{\lambda_1 t} (1, 0)\) lie on the \(x\)-axis. These solutions approach zero as \(t \to \infty\text{.}\) On the other hand, the solutions \(c_2 e^{\lambda_2 t} (0, 1)\) line on the \(y\)-axis and approach infinity as \(t \to \infty\text{.}\) The \(x\)-axis is called the stable line, and the \(y\)-axis is called the unstable line. All other solutions

(with \(c_1, c_2 \neq 0\)) tend to infinity in the direction of the unstable line, since \({\mathbf x}(t)\) approaches \((0, c_2 e^{\lambda_2 t} )\) as \(t \to \infty\text{.}\) The phase portrait for the system

is given in Figure 3.3.3. The equilibrium point of such systems is called a saddle.

Example 3.3.4.

For the system in Example 3.3.1, the unstable line of solutions is

Each solution tends away from the origin as \(t \to \infty\text{.}\) The stable line of solutions is given by

and each solution on this line approaches the origin as \(t \to \infty\text{.}\) By the Principle of Superposition, the general solution to the system is

If \(c_1 \neq 0\text{,}\) we have \({\mathbf x}(t) \to {\mathbf x}_1(t)\) as \(t \to \infty\text{.}\) If \(c_2 \neq 0\text{,}\) we have \({\mathbf x}(t) \to {\mathbf x}_2(t)\) as \(t \to -\infty\text{.}\) Thus, we have the phase portrait in Figure 3.3.5.

For the general case, where \(A\) has eigenvalues \(\lambda_1 \lt 0 \lt \lambda_2\text{,}\) we always have a stable line of solutions and an unstable line of solutions. All other solutions approach the unstable line as \(t \to \infty\) and the stable line as \(t \to – \infty\text{.}\)

Subsection 3.3.2 The Case \(\lambda_1 \lt \lambda_2 \lt 0\)

Suppose \(\lambda_1 \lt \lambda_2 \lt 0\) and consider the diagonal system

The general solution of this system is

but unlike the case of the saddle, all solutions tend towards the origin as \(t \to \infty\text{.}\) To see how the solutions approach the origin, we will compute \(dy/dx\) for \(c_2 \neq 0\text{.}\) If

then

Since \(\lambda_2 – \lambda_1 \gt 0\text{,}\) the derivative, \(dy/dx\text{,}\) must approach \(\pm \infty\text{,}\) provided \(c_2 \neq 0\text{.}\) Therefore, the solutions tend towards the origin tangentially to the \(y\)-axis (Figure 3.3.6). We say that the equilibrium point for this system is a sink.

Since \(\lambda_1 \lt \lambda_2 \lt 0\text{,}\) we say that \(\lambda_1\) is the dominant eigenvalue. The \(x\)-coordinates of the solutions approach the origin much faster than the \(y\)-coordinates.

To see what happens in the general case, suppose that \(\lambda_1 \lt \lambda_2 \lt 0\text{,}\) the eigenvectors associated with \(\lambda_1\) and \(\lambda_2\) are \((u_1, u_2)\) and \((v_1, v_2)\text{,}\) respectively. The general solution of our system is

The slope of a solution curve at \((x, y)\) is given by

This last expression tends toward the slope \(v_2/v_1\) of the eigenvector of \(\lambda_2\) (unless \(c_2 = 0\)). If \(c_2 = 0\text{,}\) then we have the straight-line solution corresponding to the eigenvalue \(\lambda_1\text{.}\) Hence, all the solutions for this case (except those on the straight-line belonging to the dominant eigenvalue) tend toward the origin tangentially to the straight-line solution corresponding to the weaker eigenvalue, \(\lambda_2\text{.}\)

Example 3.3.7.

Consider the system

The eigenvalues of this system are \(\lambda_1 = -6\) and \(\lambda_2 = -3\) with eigenvectors \(\mathbf v_1 = (2,1 )\) and \(\mathbf v_2 = (1, -1)\text{,}\) respectively. Since the dominant eigenvalue is \(\lambda_2 = -6\text{,}\) solutions tend towards the straightline solution containing the vector \(\mathbf v_2 = (1, -1)\) more quickly (Figure 3.3.8).

Subsection 3.3.3 The Case \(\lambda_1 \gt \lambda_2 \gt 0\)

If \(\lambda_1 \gt \lambda_2 \gt 0\text{,}\) we can regard our direction field as the negative of the direction field of the previous case. The general solution and the direction field are the same, but the arrows are reversed (Figure 3.3.9). In this case, we say that the equilibrium point is a source.

Example 3.3.10.

Consider the system

The eigenvalues of this system are \(\lambda_1 = 5\) and \(\lambda_2 = 1\) with eigenvectors \(\mathbf v_1 = (3, -4)\) and \(\mathbf v_2 = (1, 1)\text{,}\) respectively. Since the dominant eigenvalue is \(\lambda_1 = 5\text{,}\) solutions are closer to the straightline solution containing the vector \(\mathbf v_2 = (3, -4)\) more as \(t \to \infty\) (Figure 3.3.11).

Subsection 3.3.4 Important Lessons

  • Given a system of linear differential equations\begin{equation*} \begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}, \end{equation*}we can use the eigenvalues of \(A\) to find and classify the solutions of the system.
  • If

    \begin{equation*} A = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}, \end{equation*}

    then \(A\) has two distinct real eigenvalues. The general solution to the system \({\mathbf x}’ = A {\mathbf x}\) is

    \begin{equation*} {\mathbf x}(t) = \alpha e^{\lambda_1 t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta e^{\lambda_2 t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}. \end{equation*}

    • For the case \(\lambda_1 \lt 0 \lt \lambda_2\text{,}\) the equilibrium point of the system \({\mathbf x}’ = A {\mathbf x}\) is a saddle.
    • For the case \(\lambda_1 \lt \lambda_2 \lt 0\text{,}\) the equilibrium point of the system \({\mathbf x}’ = A {\mathbf x}\) is a sink.
    • For the case \(0 \lt \lambda_1 \lt \lambda_2\text{,}\) the equilibrium point of the system \({\mathbf x}’ = A {\mathbf x}\) is a source.

Reading Questions 3.3.5 Reading Questions

1.

What is a stable line of solutions?

2.

For a \(2 \times 2\) linear system with distinct real eigenvalues, what are the three different possibilities for the phase plane of the system?

Exercises 3.3.6 Exercises

Phase Plane Analysis of Linear Systems with Distinct Real Eigenvalues

For each of the linear systems \(d\mathbf x/dt = A \mathbf x\) in Exercise Group 3.3.6.1–4

  1. Find the eigenvalues of \(A\text{.}\)
  2. What is the dominant eigenvalue?
  3. Find the eigenvectors for each eigenvalue of \(A\text{.}\)
  4. What are the straight-line solutions of \(d\mathbf x/dt = A \mathbf x\text{?}\)
  5. Describe the nature of the equilibrium solution at \(\mathbf 0\text{.}\)
  6. Sketch the phase plane and several solution curves.
1.
2.
3.
4.
5.

Solve each linear systems \(d\mathbf x/dt = A \mathbf x\) in Exercise Group 3.3.6.1–4 for the initial condition \(\mathbf x(0) = (2,2)\text{.}\)

6.

Consider the linear system \(d \mathbf x/dt = A \mathbf x\text{,}\) where

Suppose the initial conditions for the solution curve are \(x(0) = 1\) and \(y(0) = 1\text{.}\) We can use the following Sage code to plot the phase portrait of this system, including the straightline solutions and a solution curve.

Use Sage to graph the direction field for the system linear systems \(d\mathbf x/dt = A \mathbf x\) in Exercise Group 3.3.6.1–4. Plot a solution curve for the initial condition \(\mathbf x(0) = (2,2)\text{.}\) Be sure to show the corresponding straight-line solutions on your graph.

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