Coordinated Differential Equations

Subsection5.4.1Complex Eigenvalues

Suppose that we have the system

where \(\beta \neq 0\text{.}\) The characteristic polynomial of this system is \(\det(A - \lambda I) = \lambda^2 + \beta^2\text{,}\) and so we have imaginary eigenvalues \(\pm i \beta\text{.}\) To find the eigenvector corresponding to \(\lambda = i\beta\text{,}\) we must solve the system

however, this reduces to solving the equation \(i \beta x = \beta y\text{.}\) Thus, we can find a complex eigenvector \((1, i)\text{.}\) Consequently,

is a complex solution to the system \({\mathbf x}' = A {\mathbf x}\text{.}\) The problem is that we have a real system of differential equations and would like real solutions. We can remedy the situation if we use Euler's formula, ¹If you are unfamiliar with Euler's formula, try expanding both sides as a power series to check that we do indeed have a correct identity.

Let us rewrite our solution as

and consider the real and imaginary parts of the solution:

Since

we know that \({\mathbf x}'_{\text{Re}}(t) = A{ \mathbf x}_{\text{Re}}(t)\) and \({\mathbf x}'_{\text{Im}}(t) = A {\mathbf x}_{\text{Im}}(t)\) by setting the real and imaginary parts equal. Thus, both \({\mathbf x}_{\text{Re}}(t)\) and \({\mathbf x}_{\text{Im}}(t)\) are solutions to our system. Moreover, since

we know that the appropriate linear combination of \({\mathbf x}_{\text{Re}}(t)\) and \({\mathbf x}_{\text{Im}}(t)\) will provide a solution to any initial value problem.

We claim that

is a general solution to our system. That is, we must be able to write every solution of our system as a linear combination of \({\mathbf x}_{\text{Re}}(t)\) and \({\mathbf x}_{\text{Im}}(t)\text{.}\) If

is another solution to the system \({\mathbf x}' = A {\mathbf x}\text{,}\) then

In other words, \(u'(t) = \beta v(t)\) and \(v'(t) = - \beta u(t)\text{.}\) Now, define \(f\) by

The derivative of \(f\) is

Therefore, \(f(t)\) is a complex constant and \(f(t) = (u(t) + i v(t)) e^{i\beta t} = a + bi\text{.}\) We can now write \(u(t) + iv(t) = (a + ib) e^{- i \beta t}\text{.}\) Thus,

Therefore,

Consequently,

Notice that the solutions

are periodic with period \(2 \pi / \beta\text{.}\) This type of system is called a center.

Consider the initial value problem

\begin{align*} \frac{dx}{dt} \amp = 2y\\ \frac{dy}{dt} \amp = -2x\\ x(0) \amp = 1\\ y(0) \amp = 2. \end{align*}

The eigenvalues of this system are \(\lambda = \pm 2i\text{.}\) Therefore, the general solution to the system is

\begin{align*} x(t) \amp = c_1 \cos 2t + c_2 \sin 2t\\ y(t) \amp = - c_1 \sin 2t + c_2 \cos 2t. \end{align*}

Using the initial conditions to solve for \(c_1\) and \(c_2\text{,}\) the solution to our initial value problem is

\begin{align*} x(t) \amp = \cos 2t + 2 \sin 2t\\ y(t) \amp = - \sin 2t + 2 \cos 2t. \end{align*}

The phase portrait is a circle of radius 2 about the origin (Figure5.4.3).

Subsection5.4.2Spiral Sinks and Sources

Now let us consider the system \({\mathbf x}' = A {\mathbf x}\text{,}\) where

and \(\alpha\) and \(\beta\) are nonzero real numbers. The characteristic equation of \(A\) is

so the eigenvalues are \(\lambda = \alpha \pm i \beta\text{.}\) We can use the equation

to show that \((1, i)\) is an eigenvector for \(\alpha + i \beta\text{.}\) Therefore, we have a complex solution of the form

As before, both

are real solutions to \({\mathbf x}' = A {\mathbf x}\text{.}\) Furthermore, these solutions are linearly independent. Indeed, \({\mathbf x}_{\text{Re}}\) cannot be a multiple of \({\mathbf x}_{\text{Im}}\) for all values of \(t\text{.}\) Thus, we have the general solution

The \(e^{\alpha t}\) factor tells us that the solutions either spiral into the origin if \(\alpha \lt 0\) or spiral out to infinity if \(\alpha \gt 0\text{.}\) In this case we say that the equilibrium points are spiral sinks and spiral sources, respectively.

Consider the initial value problem

\begin{align*} \frac{dx}{dt} \amp = -x/10 + y\\ \frac{dy}{dt} \amp = -x - y/10\\ x(0) \amp = 2\\ y(0) \amp = 2. \end{align*}

The matrix

\begin{equation*} \begin{pmatrix} -1/10 \amp 1 \\ -1 \amp -1/10 \end{pmatrix} \end{equation*}

has eigenvalues \(\lambda = -1/10 \pm i\text{.}\) The eigenvalue \(\lambda = -1/10 + i\) has an eigenvector \(\mathbf v = (1, i)\text{.}\) The complex solution of our system is

\begin{align*} \mathbf x(t) \amp = e^{(-1/10 + i)t} \begin{pmatrix} 1 \\ i\end{pmatrix}\\ \amp = e^{-t/10} e^{it} \begin{pmatrix} 1 \\ i \end{pmatrix}\\ \amp = e^{-t/10} (\cos t + i \sin t) \begin{pmatrix} 1 \\ i\end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} \cos t + i \sin t \\ - \sin t + i \cos t \end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} + i e^{-t/10} \begin{pmatrix} \sin t \\ \cos t \end{pmatrix} \end{align*}

The real and imaginary parts of this solution are

\begin{equation*} {\mathbf x}_{\text{Re}}(t) = e^{-t/10} \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} \quad \text{and} \quad {\mathbf x}_{\text{Im}}(t) = e^{-t/10} \begin{pmatrix} \sin t \\ \cos t \end{pmatrix}, \end{equation*}

respectively. Thus, we have the general solution

\begin{equation*} \mathbf x(t) = c_1 e^{-t/10} \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} + c_2 e^{-t/10} \begin{pmatrix} \sin t \\ \cos t \end{pmatrix}. \end{equation*}

Applying our initial conditions, we can determine that \(c_1 = 2\) and \(c_2 = 2\text{;}\) hence, the solution to our initial value problem is

\begin{equation*} \mathbf x(t) = 2 e^{-t/10} \begin{pmatrix} \cos t + \sin t \\ \cos t - \sin t \end{pmatrix}. \end{equation*}

The phase portrait of this solution indicates that we do indeed have a spiral sink (Figure5.4.5).

The initial value problem

\begin{align*} \frac{dx}{dt} \amp = x/10 + y\\ \frac{dy}{dt} \amp = -x + y/10\\ x(0) \amp = 0\\ y(0) \amp = 1/2. \end{align*}

The matrix

\begin{equation*} \begin{pmatrix} 1/10 \amp 1 \\ -1 \amp 1/10 \end{pmatrix} \end{equation*}

has an eigenvector \((1, -i)\) with eigenvalue \(\lambda = 1/10 - i\text{.}\) Thus, the complex solution is

\begin{equation*} \mathbf x(t) = e^{(1/10 - i)t} \begin{pmatrix} 1 \\ -i \end{pmatrix}. \end{equation*}

Following the procedure that we used in the previous example, the solution to our initial value problem is

\begin{equation*} \mathbf x(t) = \frac{1}{2} e^{t/10} \begin{pmatrix} \sin t \\ \cos t \end{pmatrix}, \end{equation*}

and he phase portrait is a spiral source (Figure5.4.7).

Subsection5.4.3Solving Systems with Complex Eigenvalues

Suppose that we have the linear system \(\mathbf x' = A \mathbf x\text{,}\) where

The characteristic polynomial of \(A\) is

If \((a + d)^2 - 4(ad - bc) \lt 0\text{,}\) then the eigenvalues of \(A\) are complex, and we cannot apply the strategy that we used to determine the general solution in the case of distinct real roots.

The nature of the equilibrium solution is determined by the factor \(e^{\alpha t}\) in the solution. If \(\alpha \lt 0\text{,}\) the equilibrium point is a spiral sink. If \(\alpha \gt 0\text{,}\) the equilibrium point is a spiral source. If \(\alpha = 0\text{,}\) the equilibrium point is a center.

Although we have outlined a procedure to find the general solution of \(\mathbf x' = A \mathbf x\) if \(A\) has complex eigenvalues, we have not shown that this method will work in all cases. We will do so in Section5.6.

Subsection5.4.4Important Lessons

• If

  \begin{equation*} A = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}, \end{equation*}

  then \(A\) has two complex eigenvalues, \(\lambda = \alpha \pm i \beta\text{.}\) The general solution to the system \({\mathbf x}' = A {\mathbf x}\) is

  \begin{equation*} {\mathbf x}(t) = c_1 e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + c_2 e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}. \end{equation*}

  If \(\alpha \lt 0\text{,}\) the equilibrium point is a spiral sink. If \(\alpha \gt 0\text{,}\) the equilibrium point is a spiral source.