Coordinated Differential Equations

Subsection5.3.1The Case \(\lambda_1 \lt 0 \lt \lambda_2\)

The system

\begin{align*} x' \amp = x + 3y\\ y' \amp = x - y \end{align*}

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

\begin{equation*} A = \begin{pmatrix} 1 & 3 \\ 1 & -1 \end{pmatrix}. \end{equation*}

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 straight-line solutions must be lines containing \(\mathbf u\) and \(\mathbf v\) (Figure5.3.2).

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 \(c_2 e^{\lambda_2 t} (0, 1)\) lie on the \(y\)-axis and approach infinity as \(t \to \infty\text{.}\) The \(x\)-axis is a stable line of solutions, while the \(y\)-axis is an unstable line of solutions. 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 Figure5.3.3. The equilibrium point of such systems is called a saddle.

In general, a straight-line solution is called a stable line of solutions if all solutions approach \((0,0)\text{.}\) A straight-line solution is called an unstable line if all of the non-zero solutions approach infinity.

For the system in Example5.3.1, the unstable line of solutions is

\begin{equation*} {\mathbf x}_1(t) = c_1 e^{2t} \begin{pmatrix} 3 \\ 1 \end{pmatrix}. \end{equation*}

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

\begin{equation*} {\mathbf x}_2(t) = c_2 e^{-2t} \begin{pmatrix} 1 \\ - 1 \end{pmatrix}, \end{equation*}

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

\begin{equation*} {\mathbf x}(t) = c_1 e^{2t} \begin{pmatrix} 3 \\ 1 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} 1 \\ - 1 \end{pmatrix}. \end{equation*}

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 Figure5.3.5.

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 \(t \to - \infty\text{.}\)

Subsection5.3.2The 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 (Figure5.3.6). 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, \(\lambda_2\text{.}\)

Consider the system

\begin{equation*} \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} -5 \amp -2 \\ -1 \amp -4 \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}. \end{equation*}

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 \(\lambda_1 = -6\text{,}\) solutions tend towards the straight-line solution containing the vector \(\mathbf v_1 = (2,1 )\) more quickly (Figure5.3.8).

Subsection5.3.3The 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 (Figure5.3.9). In this case, we say that the equilibrium point is a source.

Consider the system

\begin{equation*} \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} 4 \amp -3 \\ -1 \amp 2 \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}. \end{equation*}

The eigenvalues of this system are \(\lambda_1 = 5\) and \(\lambda_2 = 1\) with eigenvectors \(\mathbf v_1 = (3, -1)\) and \(\mathbf v_2 = (1, 1)\text{,}\) respectively. Since the dominant eigenvalue is \(\lambda_1 = 5\text{,}\) solutions are closer to the straight-line solution containing the vector \(\mathbf v_2 = (3, -1)\) more as \(t \to \infty\) (Figure5.3.11).

Subsection5.3.4Important 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.